Contoh Soal 12 (Segitiga dan Ketaksamaan)

$\begin{array}{ll}\\ 56.&(\textbf{OSK 2018})\\  &\textrm{Diketahui bilangan real}\: \: x\: \: \textrm{dan}\: \: y\\   &\textrm{yang memenuhi}\: \: \displaystyle \frac{1}{2}< \frac{x}{y}< 2\\  &\textrm{Nilai minimum}\: \: \displaystyle \frac{x}{2y-x}+\frac{2y}{2x-y}\: \: \textrm{adalah}\: ....\\\\    &\textbf{Jawab}\\    &\begin{aligned}&\color{red}\textrm{Alternatif 1}\\ &\textrm{Misal}\: \: t=\displaystyle \frac{x}{y}\\ &\textrm{Misalkan juga}\: \: f(t)=\displaystyle \frac{x}{2y-x}+\displaystyle \frac{2y}{2x-y}\\ &\textrm{maka}\: \: f(t)=\displaystyle \frac{2t^{2}-3t+4}{-2t^{2}+5t-2}\\ &\bullet \:\textrm{Agar minimum, maka}\: \: f'(t)=0\\ &\: \: \: \: \: \textrm{Sehingga}\\ &\: \: \: \: \:  f'(t)=4t^{2}+8t-14=0\\ &\: \: \: \: \Leftrightarrow t_{1,2}=-1\pm \displaystyle \frac{3}{2}\sqrt{2}\\ &\: \: \: \: \textrm{Pilih yang positif, yaitu}\: \: t=-1+ \displaystyle \frac{3}{2}\sqrt{2}\\ &\bullet \:  \textrm{Dengan proses substistusi harga}\: \: t\\ &\: \: \: \: \textrm{di atas, maka akan didapatkan }\\ &\: \: \: \: \textrm{nilai}\: \:  f(t)=1+\displaystyle \frac{4}{3}\sqrt{2} \end{aligned} \\ &\begin{aligned}&\color{red}\textrm{Alternatif 2}\\  &\textrm{Menurut bentuk}\: \: \displaystyle \frac{1}{2}< \frac{x}{y}< 2\\ &\textrm{jelas bahwa baik}\: \: 2y-x\: \: \textrm{dan}\: \: 2x-y \\ &\textrm{keduanya}\: \textbf{positif}\\ &\color{purple}\textrm{Lihat tabel berikut}\\ &\begin{array}{|c|c|c|}\hline \textrm{Bentuk}&\textrm{Pengecekan 1}&\textrm{Pengecekan 2}\\\hline \begin{aligned}&\displaystyle \frac{1}{2}< \frac{x}{y}< 2\\ &\textrm{Jelas bahwa}\\ &x,y\neq 0\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\textrm{Saat}\: \: (\times y)\\ &\textrm{Yaitu}:\\ &\displaystyle \frac{1}{2}(y)< \frac{x}{y}(y)< 2(y)\\ &\Leftrightarrow  \displaystyle \frac{1}{2}y< x< 2y\\ &\textrm{Jelas bahwa}\\ &2y-x>0\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\textrm{Saat dibali posisinya}\\ &\displaystyle \frac{1}{2}< \frac{y}{x}< 2\\ &\textrm{Saat}\: \:  (\times x)\\ &\textrm{Yaitu}:\\ &\displaystyle \frac{1}{2}(x)< \frac{y}{x}(x)< 2(x)\\ &\Leftrightarrow  \displaystyle \frac{1}{2}x< y< 2x\\ &\textrm{Jelas bahwa}\\ &2x-y>0 \end{aligned}\\\hline \end{array}\\ &\textrm{Saat masing-masing}\\ &\bullet \: \displaystyle \frac{x}{2y-x}=\displaystyle \frac{1}{3}+\frac{2}{3}\left ( \displaystyle \frac{2x-y}{2y-x} \right )\: \: \: \textrm{dan}\\ &\bullet \: \displaystyle \frac{2y}{2x-y}=\displaystyle \frac{2}{3}+\frac{4}{3}\left ( \displaystyle \frac{2y-x}{2x-y} \right )\\ &\color{blue}\textrm{Dengan ketaksamaan AM-GM diperoleh}\\ &\begin{aligned} \displaystyle \frac{x}{2y-x}+\displaystyle \frac{2y}{2x-y}&=1+\frac{2}{3}\left ( \displaystyle \frac{2x-y}{2y-x} \right )+\frac{4}{3}\left ( \displaystyle \frac{2y-x}{2x-y} \right )\\ &\geq 1+2\sqrt{\frac{2}{3}\left ( \displaystyle \frac{2x-y}{2y-x} \right )\frac{4}{3}\left ( \displaystyle \frac{2y-x}{2x-y} \right )}\\ &= 1+2\sqrt{\displaystyle \frac{8}{9}}\\ &=1+2\left ( \displaystyle \frac{2}{3} \right )\sqrt{2}\\ &=1+\displaystyle \frac{4}{3}\sqrt{2}     \end{aligned}   \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 57.&\textrm{Diketahui}\: \: a,b\: \: \textrm{bilangan real positif}\\ &\textrm{dan}\: \: a+b=1.\: \textrm{Tunjukkan bahwa}\\ &\qquad  \left ( a+\displaystyle \frac{1}{a} \right )^{2}+\left ( b+\displaystyle \frac{1}{b} \right )^{2}\geq \displaystyle \frac{25}{2}\\\\  &\textbf{Bukti}\\       \end{array}$.

$\: \: \: \quad\begin{aligned}&\color{purple}\textrm{Dengan AM-GM kita akan mendapatkan}\\ &1=a+b\geq 2\sqrt{ab}\Leftrightarrow \displaystyle \frac{1}{2}\geq \sqrt{ab}\Leftrightarrow 2\leq \displaystyle \frac{1}{\sqrt{ab}}\\ &\Leftrightarrow \displaystyle \frac{1}{\sqrt{ab}}\geq 2\Leftrightarrow \displaystyle \frac{1}{ab}\geq 4\\ &\color{red}\textrm{Perhatikan soal, dengan CS-Engel}\: \color{black}\textrm{akan}\\ &\textrm{didapatkan}\\ &\begin{aligned} &(1+1)\left (\displaystyle \frac{\left ( a+\frac{1}{a} \right )^{2}}{1} +\displaystyle \frac{\left ( b+\frac{1}{b} \right )^{2}}{1} \right )\geq \left (a+\frac{1}{a}+b+\frac{1}{b}  \right )^{2}\\ &\Leftrightarrow 2\left (\left (\displaystyle a+\frac{1}{a}  \right)^{2} +\left (b+\frac{1}{b}  \right )^{2} \right )\geq \left (a+b+\frac{1}{a}+\frac{1}{b}  \right )^{2}\\ &\Leftrightarrow \left (\displaystyle a+\frac{1}{a}  \right )^{2} +\left (b+\frac{1}{b}  \right )^{2}\geq \displaystyle \frac{1}{2}\left ( 1+\displaystyle \frac{a+b}{ab} \right )^{2}\\ &\Leftrightarrow \left (\displaystyle a+\frac{1}{a}  \right )^{2} +\left (b+\frac{1}{b}  \right )^{2}\geq\displaystyle \frac{1}{2}\left ( 1+\displaystyle \frac{1}{ab} \right )^{2}\\ &\Leftrightarrow \left (\displaystyle a+\frac{1}{a}  \right )^{2} +\left (b+\frac{1}{b}  \right )^{2}\geq\displaystyle \frac{1}{2}(1+4)^{2}\\ &\Leftrightarrow \left (\displaystyle a+\frac{1}{a}  \right )^{2} +\left (b+\frac{1}{b}  \right )^{2}\geq \displaystyle \frac{1}{2}\times 25\\ &\Leftrightarrow \left (\displaystyle a+\frac{1}{a}  \right )^{2} +\left (b+\frac{1}{b}  \right )^{2}\geq \displaystyle \frac{25}{2}\qquad \blacksquare    \end{aligned}      \end{aligned}$.

$\begin{array}{ll}\\ 58.&\textrm{Jika}\: \: a,b\: \: \textrm{bilangan positif, buktikan}\\ &\textrm{a}.\quad 2(a^{2}+b^{2})\geq (a+b)^{2}\\ &\textrm{b}.\quad 4(a^{3}+b^{3})\geq (a+b)^{3}\\ &\textrm{c}.\quad 8(a^{4}+b^{4})\geq (a+b)^{4}\\ &\textrm{d}.\quad 16(a^{5}+b^{5})\geq (a+b)^{5}\\ &\textrm{e}.\quad 32(a^{6}+b^{6})\geq (a+b)^{6}\\ &\textrm{f}.\quad 64(a^{7}+b^{7})\geq (a+b)^{7}\\ &\textrm{g}.\quad 128(a^{8}+b^{8})\geq (a+b)^{8}\\\\ &\textbf{Bukti}:\\  &\textrm{Akan ditunjukkan bukti poin 6.c saja}\\ &\textrm{untuk poin yang lain, silahkan pembaca}\\ &\textrm{sekalian untuk dibuktikan sendiri sebagai}\\ &\textrm{bahan latihan mandiri}.\\ &\textrm{Adapun bukti poin 6.c adalah sebagaimana}\\ &\textrm{berikut ini}\\ &\begin{aligned} &\color{red}\textrm{Dengan ketaksamaan CS-Engel}\: \color{black}\textrm{akan}\\ &\textrm{didapatkan}\\ &\bullet \: \: (1+1)(a^{4}+b^{4})\geq (a^{2}+b^{2})^{2}\\ &\: \quad\Leftrightarrow 2\left (a^{4}+b^{4}  \right )\geq (a^{2}+b^{2})^{2}\: \color{red}..........(1)\\ &\bullet \: \: (1+1)(a^{2}+b^{2})\geq (a+b)^{2}\\ &\: \quad\Leftrightarrow 2\left (a^{2}+b^{2}  \right )\geq \displaystyle (a+b)^{2},\quad (\textrm{kuadratkan})\\ &\: \quad \Leftrightarrow 4\displaystyle \left (a^{2}+b^{2}  \right )^{2}\geq (a+b)^{4}\\ &\: \quad \Leftrightarrow \left (a^{2}+b^{2}  \right )^{2}\geq \displaystyle \frac{(a+b)^{4}}{4}\: \color{red}...........(2)\\ &\textrm{Dari (1) dan (2) didapatkan hubungan}\\ &2\left (a^{4}+b^{4}  \right )\geq \left (a^{2}+b^{2}  \right )^{2}\geq \displaystyle \frac{(a+b)^{4}}{4}\\ &\Leftrightarrow 8\left ( a^{4}+b^{4} \right )\geq 4\left ( a^{2}+b^{2} \right )^{2}\geq (a+b)^{4}\\ &\Leftrightarrow 8\left ( a^{4}+b^{4} \right )\geq (a+b)^{4}\qquad \blacksquare   \end{aligned}    \end{array}$.

$\begin{array}{ll}\\ 59.&\textrm{Jika}\: \: x,y,z\: \: \textrm{adalah bilangan real positif}\\ &\textrm{dengan}\: \: x^{2}+y^{2}+z^{2}=27.\: \textrm{Tunjukkan}\\ & \textrm{bahwa}\: \: x^{3}+y^{3}+z^{3}\geq 81\\\\ &\textbf{Bukti}\\ &\color{red}\textrm{Pada contoh soal no.2 terdapat}\\ &(ax+by+cz)^{2}\leq (a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})\\ &\textrm{ganti mengganti}\: \:  a=b=c=1,\: \textrm{maka menjadi}\\ &(x+y+z)^{2}\leq (1^{2}+1^{2}+1^{2})(x^{2}+y^{2}+z^{2})\\ &\Leftrightarrow (x+y+z)^{2}\leq (3)(x^{2}+y^{2}+z^{2})\: \color{red}........(1)\\ &\textrm{Selanjutnya dengan mengganti dengan}\\ &x^{.^{\frac{3}{2}}},y^{.^{\frac{3}{2}}},z^{.^{\frac{3}{2}}}\: \: \textrm{dan}\: \: x^{.^{\frac{1}{2}}},y^{.^{\frac{1}{2}}},z^{.^{\frac{1}{2}}},\: \textrm{pada}\\ &(ax+by+cz)^{2}\leq (a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})\\ &\textrm{Kita akan dapatkan}\\ &(x^{2}+y^{2}+z^{2})\leq (x^{3}+y^{3}+z^{3})(x+y+z)\: \color{red}........(2)\\ &\textrm{Jika masing-masing ruas dikuadratkan, maka}\\ &(x^{2}+y^{2}+z^{2})^{4}\leq (x^{3}+y^{3}+z^{3})^{2}(x+y+z)^{2}\\ &\Leftrightarrow (x^{2}+y^{2}+z^{2})^{4} \leq 3(x^{3}+y^{3}+z^{3})^{2}\left ( x^{2}+y^{2}+z^{2} \right )\\ &\Leftrightarrow (x^{2}+y^{2}+z^{2})^{3} \leq 3(x^{3}+y^{3}+z^{3})^{2}\\ &\Leftrightarrow (27)^{3}\leq 3(x^{3}+y^{3}+z^{3})^{2}\\ &\Leftrightarrow 3^{9}\leq 3(x^{3}+y^{3}+z^{3})^{2}\\ &\Leftrightarrow 3^{8}\leq (x^{3}+y^{3}+z^{3})^{2}\\ &\Leftrightarrow (x^{3}+y^{3}+z^{3})^{2}\geq 3^{8}\\ &\Leftrightarrow (x^{3}+y^{3}+z^{3})\geq 3^{.^{\frac{8}{2}}}\\ &\Leftrightarrow (x^{3}+y^{3}+z^{3})\geq 3^{4}=81\qquad \blacksquare \end{array}$.

$\begin{array}{ll}\\ 60.&\textrm{Untuk}\: \: a,b,c,d\: \: \textrm{adalah bilangan real }\\ &\textrm{positif, tunjukkan bahwa}\\ &\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\geq \displaystyle \frac{64}{a+b+c+d}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\color{red}\textrm{Dengan ketaksamaan CS-Engel}\\ &(a+b+c+d)\left (\displaystyle \frac{1^{2}}{a}+\displaystyle \frac{1^{2}}{b}+\displaystyle \frac{2^{2}}{c}+\displaystyle \frac{4^{2}}{d}  \right )\geq (1+1+2+4)^{2}\\ &\Leftrightarrow \displaystyle \frac{1^{2}}{a}+\displaystyle \frac{1^{2}}{b}+\displaystyle \frac{2^{2}}{c}+\displaystyle \frac{4^{2}}{d}\geq \displaystyle \frac{(1+1+2+4)^{2}}{a+b+c+d}\\ &\Leftrightarrow \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\geq \displaystyle \frac{64}{a+b+c+d}\quad \blacksquare  \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Muslim, M.S. 2020. Kumpulan Soal dan Pembahasan Olimpiade Matematika SMA Tahun 2007-2019 Tingkat Kota/Kabupaten. Bandung: YRAMA WIDYA.
2. Widodo, T. 2018. Booklet OSN SMA 2018: Soal dan Solusi OSK, OSP, OSN SMA Bidang Matematika.
3. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.

Contoh Soal 11 (Segitiga dan Ketaksamaan)

$\begin{array}{ll}\\ 51.&\textrm{Diketahui rata-rata bilangan positif}\\ & m_{1},m_{2},...,m_{k}\: \: \textrm{adalah}\: \: A\\ &\textrm{Buktikan bahwa}\\  & \left (m_{1}+\displaystyle \frac{1}{m_{1}}  \right )^{2}+\left (m_{2}+\displaystyle \frac{1}{m_{2}}  \right )^{2}+...+\left (m_{k}+\displaystyle \frac{1}{m_{k}}  \right )^{2}\geq k\left ( A+\displaystyle \frac{1}{A} \right )^{2} \\\\  &\textbf{Bukti}\\  &\begin{aligned}&\textrm{Diketahui}\\ &\displaystyle \frac{m_{1}+m_{2}+m_{3}+...+m_{k}}{k}=A\\ &\color{blue}\textrm{Dengan ketaksamaan QM-AM akan diperoleh}\\ &\sqrt{\displaystyle \frac{(m_{1})^{2}+(m_{2})^{2}+...+(m_{k})^{2}}{k}}\geq \displaystyle \frac{m_{1}+m_{2}+...+m_{k}}{k}\\ &\Leftrightarrow \sqrt{\displaystyle \frac{(m_{1})^{2}+(m_{2})^{2}+...+(m_{k})^{2}}{k}}\geq A\\  &\Leftrightarrow \displaystyle \frac{(m_{1})^{2}+(m_{2})^{2}+...+(m_{k})^{2}}{k}\geq A^{2}\\ &\Leftrightarrow (m_{1})^{2}+(m_{2})^{2}+...+(m_{k})^{2}\geq kA^{2}\\ &\Leftrightarrow \displaystyle \sum_{i=1}^{k}(m_{i})^{2}\geq kA^{2}\: \color{red}........(1)  \end{aligned}\\ &\begin{aligned} &\color{blue}\textrm{Dengan ketaksamaan QM-HM diperoleh juga}\\ &\sqrt{\displaystyle \frac{\left ( \frac{1}{m_{1}} \right )^{2}+\left ( \frac{1}{m_{2}} \right )^{2}+...+\left ( \frac{1}{m_{k}} \right )^{2}}{k}}\geq \displaystyle \frac{k}{m_{1}+m_{2}+...+m_{k}}\\ &\Leftrightarrow \sqrt{\displaystyle \frac{\left ( \frac{1}{m_{1}} \right )^{2}+\left ( \frac{1}{m_{2}} \right )^{2}+...+\left ( \frac{1}{m_{k}} \right )^{2}}{k}}\geq \displaystyle \frac{1}{A}\\ &\Leftrightarrow \displaystyle \frac{\left ( \frac{1}{m_{1}} \right )^{2}+\left ( \frac{1}{m_{2}} \right )^{2}+...+\left ( \frac{1}{m_{k}} \right )^{2}}{k}\geq \displaystyle \frac{1}{A^{2}}\\ &\Leftrightarrow \left ( \frac{1}{m_{1}} \right )^{2}+\left ( \frac{1}{m_{2}} \right )^{2}+...+\left ( \frac{1}{m_{k}} \right )^{2}\geq \displaystyle \frac{k}{A^{2}}\\ &\Leftrightarrow \sum_{i=1}^{k}\left ( \displaystyle \frac{1}{m_{i}} \right )^{2}\geq \displaystyle \frac{k}{A^{2}}\: \color{red}........(2)  \end{aligned}\\ &\begin{aligned} &\color{blue}\textrm{Dengan (1) dan (2) akan diperoleh}\\ &\displaystyle \sum_{i=1}^{k}\left ( m_{i}+\displaystyle \frac{1}{m_{i}} \right )^{2}=\displaystyle \sum_{i=1}^{k}\left ( (m_{i})^{2}+2+\left ( \displaystyle \frac{1}{m_{i}} \right )^{2} \right )\\ &=\displaystyle \sum_{i=1}^{k}(m_{i})^{2}+\displaystyle \sum_{i=1}^{k}2+\displaystyle \sum_{i=1}^{k}\left ( \frac{1}{m_{i}} \right )^{2}\\ &\geq kA^{2}+2k+\displaystyle \frac{k}{A^{2}}\\ &= k\left ( A^{2}+2+\displaystyle \frac{1}{A^{2}} \right )\\ &=k\left ( A+\displaystyle \frac{1}{A} \right )^{2}\qquad \blacksquare   \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 52.&\textrm{Buktikan bahwa untuk bilangan asli}\: n>1\\  &\textrm{berlaku}\quad \sqrt[n]{1+\displaystyle \frac{\sqrt[n]{n}}{n}}+\sqrt[n]{1-\displaystyle \frac{\sqrt[n]{n}}{n}}<2\\\\   &\textbf{Bukti}\\   &\begin{aligned}&\textrm{Perhatikan bahwa untuk}\: \: \sqrt[n]{1-\color{red}\displaystyle \frac{\sqrt[n]{n}}{n}}\\  &\textrm{dengan}\: \: n>1,\: \: \textrm{maka}\: \: 0<\color{red}\displaystyle \frac{\sqrt[n]{n}}{n}\color{black}<1\\ &\color{blue}\textrm{Dengan ketaksamaan AM-GM dapat diperoleh}\\ &\begin{aligned} \displaystyle \frac{\left ( 1+\displaystyle \frac{\sqrt[n]{n}}{n} \right )+\underset{\textrm{sebanyak}\: \: (n-1)}{\underbrace{1+1+1+...+1}}}{n}&>\sqrt[n]{\left ( 1+\displaystyle \frac{\sqrt[n]{n}}{n} \right )111...1}\\ &=\sqrt[n]{\left ( 1+\displaystyle \frac{\sqrt[n]{n}}{n} \right )}\:\color{blue} ......(1)\\ \displaystyle \frac{\left ( 1-\displaystyle \frac{\sqrt[n]{n}}{n} \right )+\underset{\textrm{sebanyak}\: \: (n-1)}{\underbrace{1+1+1+...+1}}}{n}&>\sqrt[n]{\left ( 1-\displaystyle \frac{\sqrt[n]{n}}{n} \right )111...1}\\ &=\sqrt[n]{\left ( 1-\displaystyle \frac{\sqrt[n]{n}}{n} \right )}\: \color{blue} ......(2)\\    \end{aligned} \\ &\textrm{Jika ketaksamaan (1) dan (2) dijumlahkna, maka}\\ &\begin{aligned} \displaystyle \sqrt[n]{1+\displaystyle \frac{\sqrt[n]{n}}{n}}+\displaystyle \sqrt[n]{1-\displaystyle \frac{\sqrt[n]{n}}{n}}&<\displaystyle \frac{\underset{\textrm{sebanyak}\: \: (2n-2)}{\underbrace{1+1+1+...+1}}+\left ( 1+\displaystyle \frac{\sqrt[n]{n}}{n} \right )+\left ( 1-\displaystyle \frac{\sqrt[n]{n}}{n} \right )}{n}\\ &=\displaystyle \frac{\underset{\textrm{sebanyak}\: \: (2n-2)}{\underbrace{1+1+1+...+1}} +1+1}{n}\\ &=2\qquad \blacksquare     \end{aligned}   \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 53.&\textrm{Jika bilangan real positif}\: \: x,y,z\: \: \textrm{dengan}\\  &xyz=1\: .\: \textrm{tentukan nilai minimum dari}\\ &2x^{3}+12y^{2}+24z\\\\   &\textbf{Jawab}\\   &\textrm{Misal}\\ & A=2x^{3}+12y^{2}+24z\\ &\Leftrightarrow A=x^{3}+x^{3}+4y^{2}+4y^{2}+4y^{2}+\underset{\textrm{sebanyak}\: \: 6\: \: \textrm{kali}}{\underbrace{4z+4z+...+4z}}\\ &\begin{aligned}&\color{blue}\textrm{Dengan ketaksamaan AM-GM dapat diperoleh}\\ &\displaystyle \frac{A}{11}\geq \sqrt[11]{(x^{3})(x^{3})(4y^{2})(4y^{2})(4y^{2})\underset{\textrm{sebanyak}\: \: 6}{\underbrace{(4z)...(4z)}} }\\ &\Leftrightarrow \displaystyle \frac{A}{11}\geq \sqrt[11]{(xyz)^{6}(4)^{9}}\\ &\Leftrightarrow A\geq 11\sqrt[11]{4^{9}}  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 54.&(\textbf{OSK 2019})\\ &\textrm{Tentukan bilangan real terbesar}\: \: M,\\ & \textrm{sehingga untuk setiap}\: \: x\: \: \textrm{positif berlaku}\\   &(x+1)(x+3)(x+5)(x+11)\geq Mx\\\\   &\textbf{Jawab}\\   &\textrm{Diketahui}\\ &(x+1)(x+3)(x+5)(x+11)\geq Mx\\ &\Leftrightarrow x^{4}+20x^{3}+122x^{2}+268x+165\geq Mx\\ &\color{blue}\textrm{Dengan AM-GM kita akan mendapatkan}\\ &\displaystyle \frac{x^{4}+\overset{\textrm{sebanyak}\: \: 20}{\overbrace{x^{3}+x^{3}+...+x^{3}}}+\overset{\textrm{sebanyak}\: \: 122}{\overbrace{x^{2}+x^{2}+...+x^{2}}}+\overset{\textrm{sebanyak}\: \: 268}{\overbrace{x+x+...+x}}+\overset{\textrm{sebanyak}\: \: 165}{\overbrace{1+1+...+1}}}{1+20+122+268+165}\\ &\geq \sqrt[576]{x^{4}\left (\underset{\textrm{sebanyak}\: \: 20}{\underbrace{x^{3}x^{3}...x^{3}}}  \right )\left (\underset{\textrm{sebanyak}\: \: 268}{\underbrace{x^{2}x^{2}...x^{2}}}  \right )\left (\underset{\textrm{sebanyak}\: \: 165}{\underbrace{11...1}}  \right )}\\ &\Leftrightarrow \displaystyle \frac{x^{4}+20x^{3}+122x^{2}+268x+165}{576}\geq \sqrt[576]{x^{576}}\\ &\Leftrightarrow x^{4}+20x^{3}+122x^{2}+268x+165\geq 576x\\ &\textrm{Jadi, nilai}\: \: M=\color{red}576 \end{array}$.

$\begin{array}{ll}\\ 55.&(\textbf{OMITS 2012})\\ &\textrm{Jika diketahui}\: \: x_{1},x_{2},x_{3},...,x_{2012}\in \left ( \displaystyle \frac{1}{4}\: ,1 \right ),\\ & \textrm{maka nilai minimum dari bentuk}\\  &^{.^{x_{1}}}\log \left ( x_{2}-\displaystyle \frac{1}{4} \right )+^{.^{x_{2}}}\log \left ( x_{3}-\displaystyle \frac{1}{4} \right )+...+^{.^{x_{2012}}}\log \left ( x_{1}-\displaystyle \frac{1}{4} \right ) \\\\  &\textbf{Jawab}:\\   &\begin{aligned}&\textrm{Perhatikan bahwa}\\ &\bullet \: \textrm{untuk}:\: x\geq 1,\: \textrm{berlaku}\: \left ( x-\displaystyle \frac{1}{2} \right )^{2}\geq 0\\ &\: \quad \textrm{sehingga berlaku juga}\: \left (x-\displaystyle \frac{1}{4}  \right )\leq x^{2}\\ &\bullet \: \textrm{untuk}:\: x_{1},x_{2},x_{3},...,x_{2012}\in \left ( \displaystyle \frac{1}{4}\: ,1 \right )\\ &\: \quad \textrm{kita akan memperoleh fakta bahwa}\\ &\: \quad ^{.^{x_{i}}}\log \left ( x_{i+1}-\displaystyle \frac{1}{4} \right )\geq ^{.^{x_{1}}}\log x_{x_{i+1}}^{2}=2.^{.^{x_{1}}}\log x_{x_{i+1}}\\ &\textrm{Selanjutnya}\\ &^{.^{x_{1}}}\log \left ( x_{2}-\displaystyle \frac{1}{4} \right )+^{.^{x_{2}}}\log \left ( x_{3}-\displaystyle \frac{1}{4} \right )+...+^{.^{x_{n}}}\log \left ( x_{1}-\displaystyle \frac{1}{4} \right )\\ &=\displaystyle \sum_{i=1}^{n} .^{.^{x_{i}}}\log \left ( x_{i+1}-\displaystyle \frac{1}{4} \right )\geq 2\displaystyle \sum_{i=1}^{n}.^{.^{x_{i}}}\log (x_{i+1})=2\displaystyle \sum_{i=1}^{n}\displaystyle \frac{\log x_{i+1}}{\log x_{i}}\\ &\color{blue}\textrm{Dengan AM-GM kita mendapatkan}\\ &\displaystyle \sum_{i=1}^{n} .^{.^{x_{i}}}\log \left ( x_{i+1}-\displaystyle \frac{1}{4} \right )\\ &\geq 2\displaystyle \sum_{i=1}^{n}\displaystyle \frac{\log x_{i+1}}{\log x_{i}}\geq 2n.\sqrt[n]{\prod_{i=1}^{n}\displaystyle \frac{\log x_{i+1}}{\log x_{i}}}\\ &= 2n.\sqrt[n]{\displaystyle \frac{\log x_{2}}{\log x_{1}}.\frac{\log x_{3}}{\log x_{2}}.\frac{\log x_{4}}{\log x_{3}}...\frac{\log x_{n}}{\log x_{n-1}}.\frac{\log x_{1}}{\log x_{n}}}\\ &=2n.1=2n\\ &\textrm{Sehingga nilai minimum dari}\\ &^{.^{x_{1}}}\log \left ( x_{2}-\displaystyle \frac{1}{4} \right )+^{.^{x_{2}}}\log \left ( x_{3}-\displaystyle \frac{1}{4} \right )+...+^{.^{x_{2012}}}\log \left ( x_{1}-\displaystyle \frac{1}{4} \right )\\ &=2n=2(2012)=4024 \end{aligned} \end{array}$. 

DAFTAR PUSTAKA

1. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
2. Muslim, M.S. 2020. Kumpulan Soal dan Pembahasan Olimpiade Matematika SMA Tahun 2007-2019 Tingkat Kota/Kabupaten. Bandung: YRAMA WIDYA.
3. Sidi, A.B. 2010. Aljabar: Alhaqibiyyah Attadribiyyah lita'hil Attullab litasfiyat Oulimbiyat Arriyadliyyat bi Hazakhistan. Saudi Arabia.

Contoh Soal 10 (Segitiga dan Ketaksamaan)

 $\begin{array}{ll}\\ 46.&\textrm{Misalkan}\: \: a,b,c\: \: \textrm{bilangan real non negatif}\\ &\textrm{dengan}\: \: a+b+c=1,\: \: \textrm{Tunjukkan bahwa}\\ &\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq \displaystyle \frac{9}{2}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Diketahui}\: \: a+b+c=1\\ &\color{blue}\textrm{Dengan AM-GM kita memiliki}\\ &2(a+b+c)\left (\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}  \right )\\ &= ((a+b)+(b+c)+(c+a))\left (\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}  \right )\\ &\geq 3\sqrt[3]{(a+b)(b+c)(c+a)}\times 3\sqrt[3]{\displaystyle \frac{1}{(a+b)(b+c)(c+a)}}\\ &=9\sqrt[3]{\displaystyle \frac{(a+b)(b+c)(c+a)}{(a+b)(b+c)(c+a)}}=9\sqrt[3]{1}=9\\ &\textrm{Sehingga}\\ &2\left (\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}  \right )\geq 9\\ &\Leftrightarrow \:  \left (\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}  \right )\geq \displaystyle \frac{9}{2}\qquad \blacksquare   \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 47.&(\textbf{OSN 2013})\\ &\textrm{Tentukan semua bilangan real}\: \: M\\  &\textrm{sedemikian sehingga untuk sebarang}\\  &\textrm{bilangan real}\: \: a,b,c\: \: \textrm{paling sedikit}\\ &\textrm{satu di antara tiga bilangan berikut}\\ &\qquad a+\displaystyle \frac{M}{ab},\: b+\displaystyle \frac{M}{bc},\: c+\displaystyle \frac{M}{ca}\\ &\textrm{bernilai lebih dari atau sama dengan}\\ & 1+M \\\\   &\textbf{Jawab}:\\    &\begin{aligned} &\textrm{Diketahui bahwa}\\ &\textrm{min}\left \{ a+\displaystyle \frac{M}{ab},\: b+\displaystyle \frac{M}{bc},\: c+\displaystyle \frac{M}{ca} \right \}\geq 1+M\\ &\textrm{Perhatikan bahwa}\\ &a+\displaystyle \frac{M}{ab}+ b+\displaystyle \frac{M}{bc}+ c+\displaystyle \frac{M}{ca}\\ &=a+b+c+\displaystyle \frac{M}{ab}+\displaystyle \frac{M}{bc}+\displaystyle \frac{M}{ca}\\ &=\displaystyle \frac{1}{2}\left ( 2(a+b+c)+\displaystyle \frac{2M}{ab}+\displaystyle \frac{2M}{bc}+\displaystyle \frac{2M}{ca} \right )\\ &=\displaystyle \frac{1}{2}\left ( a+b+\displaystyle \frac{2M}{ab}+b+c+\displaystyle \frac{2M}{bc}+c+a+\displaystyle \frac{2M}{ca} \right )\\ &\color{blue}\textrm{Dengan AM-GM akan diperoleh bentuk}\\ &\geq \displaystyle \frac{1}{2}\left ( 3\sqrt[3]{2M}+3\sqrt[3]{2M}+3\sqrt[3]{2M} \right )\\ &=\displaystyle \frac{9}{2} \sqrt[3]{2M}\\ &\color{red}\textrm{Pilih nilai minimum}\\ &\left \{ a+\displaystyle \frac{M}{ab},\: b+\displaystyle \frac{M}{bc},\: c+\displaystyle \frac{M}{ca} \right \}=1+M=\displaystyle \frac{3\sqrt[3]{2M}}{2}\\ &\textrm{Selanjutnya}\\ &1+M=\displaystyle \frac{3\sqrt[3]{2M}}{2}\\ &\Leftrightarrow \: M=\displaystyle \frac{1}{2} \end{aligned}    \end{array}$.

$\begin{array}{ll}\\ 48.&(\textbf{OSK 2017})\\ &\textrm{Diketahui bilangan real positif}\: \: a,b,c\\  & \textrm{yang memenuhi}\: \: a+b+c=1.\: \textrm{Nilai}\\  &\textrm{minimum dari}\: \: \displaystyle \frac{a+b}{abc}\: \: \textrm{adalah}\: .\: ...\\\\  &\textbf{Jawab}\\    &\begin{aligned}&\textrm{Diketahui bahwa}\: \: a,b,c\in R^{+}\\ &\textrm{dengan}\: a+b+c=1.\: \textrm{kita dapat peroleh}\\ &a+b=1-c\\ &\textrm{Selanjutnya}\\ &\displaystyle \frac{a+b}{abc}=\displaystyle \frac{a}{abc}+\frac{b}{abc}=\displaystyle \frac{1}{bc}+\frac{1}{ac}\\ &\color{blue}\textrm{Sebelumnya ingat ketaksamaan AM-HM}\\ &\displaystyle \frac{m+n}{2}\geq \displaystyle \frac{2}{\displaystyle \frac{1}{m}+\frac{1}{n}}\Leftrightarrow \displaystyle \frac{1}{m}+\frac{1}{n}\geq \displaystyle \frac{4}{m+n}\\ &\textrm{Sehingga}\\ &\displaystyle \frac{a+b}{abc}=\displaystyle \frac{1}{bc}+\frac{1}{ac}\geq \displaystyle \frac{4}{ac+bc}= \displaystyle \frac{4}{c(a+b)}\\ &\qquad \geq \displaystyle \frac{4}{c(1-c)}=\displaystyle \frac{4}{c-c^{2}}=\displaystyle \frac{4}{\displaystyle \frac{1}{4}-\displaystyle \frac{1}{4}+c-c^{2}}\\ &\qquad =\displaystyle \frac{4}{\displaystyle \frac{1}{4}-\left ( c-\displaystyle \frac{1}{2} \right )^{2}}\\ &\textrm{Saat}\: \: c-\displaystyle \frac{1}{2}=0,\: \textrm{maka akan diperoleh}\\ &\textrm{nilai minimum yaitu}:\\ &\left ( \displaystyle \frac{a+b}{abc} \right )_{\textrm{minimum}}=\displaystyle \frac{4}{\displaystyle \frac{1}{4}-0}=16   \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 49.&\textrm{Jika pada soal 48 diubah dengan}\\ &a,b,c\: \: \textrm{adalah sisi segitiga}\: \: ABC\\  & \textrm{yang memenuhi}\: \: a+b+c=1.\\ & \textrm{Tentukan nilai dari}\: \: \displaystyle \frac{a+b}{abc}\\\\  &\textbf{Jawab}\\    &\begin{aligned}&\textrm{Diketahui bahwa}\: \: a,b,c\: \: \textrm{sisi}\: \: \bigtriangleup ABC\\ &\textrm{dengan}\: a+b+c=1.\: \textrm{kita dapat peroleh}\\ &a+b=a+b\\ &\textrm{Ingat bahwa dalam}\: \: \bigtriangleup ABC,\: \textrm{berlaku}\\ &\begin{cases} \bullet  & a+b>c \\  \bullet  & a+c>b \\  \bullet  & b+c>a  \end{cases}\\ &\textrm{maka}\: \:  a+b+a+b> a+b+c\\ &\Leftrightarrow \: 2(a+b)> 1\Leftrightarrow a+b> \displaystyle \frac{1}{2}\\ &\color{blue}\textrm{Dan dengan AM-GM diperoleh}\\ &\displaystyle \frac{a+b+c}{3}\geq \sqrt[3]{abc}\Leftrightarrow \displaystyle \frac{1}{3}\geq \sqrt[3]{abc}\\ &\Leftrightarrow \displaystyle \frac{1}{27}\geq abc\\ &\textrm{Selanjutnya}\\ &\displaystyle \frac{a+b}{abc}> \displaystyle \frac{\frac{1}{2}}{\frac{1}{27}}=\displaystyle \frac{27}{2}=13,5\\ &\textrm{Jadi, nilai}\: \: \displaystyle \frac{a+b}{abc}>13,5  \end{aligned}    \end{array}$.

$\begin{array}{ll}\\ 50.&\textrm{Diketahui bilangan real positif}\: \: x_{1},x_{2},...,x_{k}\\ & \textrm{memenuhi persamaan}\: \: x_{1}+x_{2}+...+x_{k}=1\\  &\textrm{Buktikan bahwa}\\ &\displaystyle \frac{1}{(x_{1})^{2013}}+\frac{1}{(x_{2})^{2013}}+...+\frac{1}{(x_{k})^{2013}}\geq k^{2014}  \\\\  &\textbf{Bukti}\\   &\begin{aligned}&\color{blue}\textrm{Dengan ketaksamaan AM-GM kita}\\ &\textrm{memiliki}\\ &\displaystyle \frac{x_{1}+x_{2}+...+x_{k}}{k}\geq \sqrt[k]{x_{1}.x_{2}...x_{k}}\\ &\Leftrightarrow \: \displaystyle \frac{1}{k} \geq \sqrt[k]{x_{1}.x_{2}...x_{k}}\\ &\Leftrightarrow \: \sqrt[k]{x_{1}.x_{2}...x_{k}}\leq \displaystyle \frac{1}{k}\\ & \Leftrightarrow \: x_{1}.x_{2}...x_{k}\leq \left ( \displaystyle \frac{1}{k} \right )^{k}\\ &\textrm{Selanjutnya}\\ &\displaystyle \frac{\displaystyle \frac{1}{(x_{1})^{2013}}+\frac{1}{(x_{2})^{2013}}+...+\frac{1}{(x_{k})^{2013}}}{k}\\ &\geq \sqrt[k]{\displaystyle \frac{1}{(x_{1})^{2013}.(x_{2})^{2013}...(x_{k})^{2013}}}\\ &\displaystyle \frac{1}{(x_{1})^{2013}}+\frac{1}{(x_{2})^{2013}}+...+\frac{1}{(x_{k})^{2013}}\\ &\geq k\sqrt[k]{\displaystyle \frac{1}{(x_{1})^{2013}.(x_{2})^{2013}...(x_{k})^{2013}}}\\ &\geq k\sqrt[k]{\displaystyle \frac{1}{(x_{1}.x_{2}...x_{k})^{2013}}}\geq k.\: \sqrt[k]{\displaystyle \frac{1}{\left ( \left ( \displaystyle \frac{1}{k} \right )^{k} \right )^{2013}}}\\ &\geq k.\sqrt[k]{k^{2013k}}\\ &\geq k.k^{2013}\\ &\geq k^{2014}\qquad \blacksquare \end{aligned}  \end{array}$.

DAFTAR PUSTAKA

1. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
2. Muslim, M.S. 2020. Kumpulan Soal dan Pembahasan Olimpiade Matematika SMA Tahun 2007-2019 Tingkat Kota/Kabupaten. Bandung: YRAMA WIDYA.
3. Widodo, T. 2013. Pembahasan OSN Matematika SMA Tahun 2013 Seleksi Tingkat Nasional. 

Lanjutan 4 Materi Ketaksamaan : Ketaksamaan Chebyshev

2. Ketaksamaan Chebyshev

Jika $a_{1}\leq a_{2}\leq a_{3}\leq \cdots \leq a_{n}$ dan $b_{1}\leq b_{2}\leq b_{3}\leq \cdots \leq b_{n}$ adalah merupakan kumpulan bilangan yang monoton naik atau $a_{1}\geq a_{2}\geq a_{3}\geq \cdots \geq a_{n}$ dan $b_{1}\geq b_{2}\geq b_{3}\geq \cdots \geq b_{n}$ adalah merupakan kumpulan bilangan yang monoton turun, maka

$\begin{aligned}&\left (\displaystyle \frac{a_{1}b_{1}+ a_{2}b_{2}+a_{3}b_{3}+ \cdots + a_{n}b_{n}}{n}  \right )\\ &\geq \left (\displaystyle \frac{a_{1}+a_{2}+a_{3}+...+a_{n}}{n}  \right )\\ &\times \left (\displaystyle \frac{b_{1}+b_{2}+b_{3}+...+b_{n}}{n}  \right ) \end{aligned}$.

$\begin{aligned}&\color{red}\textbf{atau}\\\\ &n\left (a_{1}b_{1}+ a_{2}b_{2}+a_{3}b_{3}+ \cdots + a_{n}b_{n}  \right )\\ &\geq \left (a_{1}+a_{2}+a_{3}+...+a_{n}  \right )\times \left (b_{1}+b_{2}+b_{3}+...+b_{n} \right ) \end{aligned}$.

Tetapi jika kumpulan bilangan di atas memiliki kemonotonan yang berbeda, yaitu $a_{1}\leq a_{2}\leq a_{3}\leq \cdots \leq a_{n}$  dan  $b_{1}\geq b_{2}\geq b_{3}\geq \cdots \geq b_{n}$ atau  $a_{1}\geq a_{2}\geq a_{3}\geq \cdots \geq a_{n}$  dan $b_{1}\leq b_{2}\leq b_{3}\leq \cdots \leq b_{n}$, maka ketaksamaan akan menjadi

$\begin{aligned}&\left (\displaystyle \frac{a_{1}b_{1}+ a_{2}b_{2}+a_{3}b_{3}+ \cdots + a_{n}b_{n}}{n}  \right )\\ &\leq \left (\displaystyle \frac{a_{1}+a_{2}+a_{3}+...+a_{n}}{n}  \right )\\ &\times \left (\displaystyle \frac{b_{1}+b_{2}+b_{3}+...+b_{n}}{n}  \right ) \end{aligned}$.

$\begin{aligned}&\color{red}\textbf{atau}\\\\ &n\left (a_{1}b_{1}+ a_{2}b_{2}+a_{3}b_{3}+ \cdots + a_{n}b_{n}  \right )\\ &\leq \left (a_{1}+a_{2}+a_{3}+...+a_{n}  \right )\times \left (b_{1}+b_{2}+b_{3}+...+b_{n} \right ) \end{aligned}$.

Bukti

Pada kumpulan bilangan yang memiliki kemonotonan yang sama yaitu:

$a_{1}\leq a_{2}\leq a_{3}\leq \cdots \leq a_{n}$ dan $b_{1}\leq b_{2}\leq b_{3}\leq \cdots \leq b_{n}$, dengan mengaplikasikan ketaksamaan Renata akan diperoleh

$\begin{aligned}a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}&=a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}\\ a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}&\geq a_{1}b_{2}+a_{2}b_{3}+\cdots +a_{n}b_{1}\\ a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}&\geq a_{1}b_{3}+a_{2}b_{4}+\cdots +a_{n}b_{2}\\ &\vdots \\ a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}&\geq a_{1}b_{n}+a_{2}b_{1}+\cdots +a_{n}b_{n-1}\\ \textrm{Jika ketaksamaan di at}&\textrm{as ditmabahkan, maka}\\ n(a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n})&\geq (a_{1}+a_{2}+\cdots +a_{n})(b_{1}+b_{2}+\cdots +b_{n})\\ \textrm{Bentuk terakhir adalah}&\: \textrm{bukti dari ketaksamaan ini} \end{aligned}$.

$\begin{aligned}\color{blue}\textrm{Sebagai}&\: \color{blue}\textrm{misal, andaikan}\\ \bullet \: \: n=2&\: \Leftrightarrow \: 2(a_{1}b_{1}+a_{2}b_{2})\geq (a_{1}+a_{2})(b_{1}+b_{2})\\ \bullet \: \: n=3&\: \Leftrightarrow \: 3(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})\geq (a_{1}+a_{2}+a_{3})(b_{1}+b_{2}+b_{3})\\ &\vdots \\ \bullet \: \: n=k&\: \Leftrightarrow \: k(a_{1}b_{1}+a_{2}b_{2}+\cdots  +a_{k}b_{k})\geq (a_{1}+a_{2}+\cdots +a_{k})(b_{1}+b_{2}+\cdots +b_{k})\\ \end{aligned}$ .

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Jika}\: \: a,b,x,y\: \textrm{bilangan real positif}\\ &\textrm{sehingga}\: \: a\geq b\: \: \textrm{dan}\: \: x\geq y.\\ &\textrm{Tunjukkan bahwa}\\ &\qquad  (ax+by)\geq \displaystyle \frac{1}{2}(a+b)(x+y)\\\\ &\textbf{Bukti}:\\ &\textrm{Dengan}\: \textbf{ketaksamaan Chebyshev}\\ &\textrm{dapat diperoleh bentuk}\\ &2(ax+by)\geq (a+b)(x+y)\\ &\Leftrightarrow (ax+by)\geq \displaystyle \frac{1}{2}(a+b)(x+y)\quad \blacksquare  \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: a,b>0\: ,\: \textrm{tunjukkan bahwa}\\ & 2(a^{2}+b^{2})\geq (a+b)^{2}\\\\ &\textbf{Bukti}:\\ &\color{red}\textrm{Alternatif 1}\\ &\textrm{Dengan}\: \textbf{ketaksamaan Chebyshev}\\ &\textrm{untuk}\: \: a\geq b,\: \textrm{dapat diperoleh bentuk}\\ &2(a.a+b.b)\geq (a+b)(a+b)\\ &\Leftrightarrow 2(a^{2}+b^{2})\geq (a+b)^{2}\quad \blacksquare\\ &\color{red}\textrm{Alternatif 2}\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan CS-Engel}\\ &(1+1)(a^{2}+b^{2})\geq (1.a+1.b)^{2}\\ &\Leftrightarrow 2(a^{2}+b^{2})\geq (a+b)^{2}\quad \blacksquare\\ &\color{red}\textrm{Alternatif 3}\\ &\textrm{Perhatikan bahwa}\: \: (a-b)^{2}\geq 0\\ &\Leftrightarrow \: a^{2}+b^{2}-2ab\geq 0\\ &\Leftrightarrow \: a^{2}+b^{2}\geq 2ab\\ &\Leftrightarrow \: 2(a^{2}+b^{2})\geq a^{2}+b^{2}+2ab\\ &\Leftrightarrow \: 2(a^{2}+b^{2})\geq (a+b)^{2}\quad \blacksquare   \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Jika}\: \: a,b,c>0\: ,\: \textrm{tunjukkan bahwa}\\ &3(a^{2}+b^{2}+c^{2})\geq (a+b+c)^{2}\\\\ &\textbf{Bukti}:\\ &\textrm{Dengan}\: \textbf{ketaksamaan Chebyshev}\\ &\textrm{untuk}\: \: a\geq b\geq c,\: \textrm{dapat diperoleh bentuk}\\ &3(a.a+b.b+c.c)\geq (a+b+c)(a+b+c)\\ &\Leftrightarrow 3(a^{2}+b^{2}+c^{2})\geq (a+b+c)^{2}\qquad \blacksquare  \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: a,b,c>0\: ,\: \textrm{dengan}\: \: a^{2}+b^{2}+c^{2}=1\\&\textrm{tunjukkan bahwa}\: \: \: a+b+c\leq \sqrt{3}\\\\ &\textbf{Bukti}:\\ &\textrm{Dengan}\: \textbf{ketaksamaan Chebyshev}\\ &\textrm{Kurang lebih caranya sama dengan no.3 di atas}\\ &\textrm{untuk}\: \: a\geq b\geq c,\: \textrm{dapat diperoleh bentuk}\\ &3(a.a+b.b+c.c)\geq (a+b+c)(a+b+c)\\ &\Leftrightarrow 3(a^{2}+b^{2}+c^{2})\geq (a+b+c)^{2}\\ &\Leftrightarrow 3(1)\geq (a+b+c)^{2}\\ &\Leftrightarrow \sqrt{3}\geq (a+b+c)\\ &\Leftrightarrow (a+b+c)\leq \sqrt{3} \qquad \blacksquare  \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Diberikan}\: \: a,b,c>0,\: \: \textrm{tunjukkan kebenaran}\\ &\textbf{ketaksamaan Nesbitt}\: \textrm{berikut}\\ &\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}\\\\ &\textbf{Bukti}:\\  &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}&\textrm{Asumsikan},\\ &\color{red}\begin{cases} & a\geq b\geq c \\ & \displaystyle \frac{1}{b+c}\geq \frac{1}{a+c}\geq \frac{1}{a+b} \end{cases} \\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan Chebyshev}\\ &\displaystyle \frac{\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}{3}\geq \displaystyle \frac{(a+b+c)}{3}\left ( \displaystyle \frac{\left (\displaystyle \frac{1}{b+c}+ \frac{1}{a+c}+ \frac{1}{a+b}  \right )}{3} \right )\\ &\textrm{Dengan AM-HM dan}\: K=\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\\ &\Leftrightarrow \displaystyle \frac{K}{3}\geq \displaystyle \frac{(a+b+c)}{3}\left ( \displaystyle \frac{3}{(b+c)+(a+c)+(a+b)} \right )\\ &\Leftrightarrow K\geq \displaystyle \frac{3(a+b+c)}{2(a+b+c)}\\ &\Leftrightarrow K\geq \displaystyle \frac{3}{2}\\ &\Leftrightarrow \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \displaystyle \frac{3}{2}\qquad \blacksquare  \end{aligned}\\ &\color{blue}\textrm{Alternatif 2}\\ &\begin{aligned}&\color{magenta}\textbf{Pertama},\: \color{black}\textrm{asumsikan}\\ &\color{red}\begin{cases} & a\geq b\geq c \\ & \displaystyle \frac{1}{b+c}\geq \frac{1}{a+c}\geq \frac{1}{a+b} \end{cases} \\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan Chebyshev}\\ &3\left (\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}  \right )\geq (a+b+c)\left (\displaystyle \frac{1}{b+c}+ \frac{1}{a+c}+ \frac{1}{a+b}  \right )\\ &\color{magenta}\textbf{Kedua},\: \color{black}\textrm{asumsikan}\\ &\color{red}\begin{cases} & a+b\geq a+c\geq b+c \\ & \displaystyle \frac{1}{a+b}\leq \frac{1}{a+c}\leq \frac{1}{b+c} \end{cases} \\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan Chebyshev}\\ &3\left (\displaystyle \frac{a+b}{a+b}+\frac{a+c}{a+c}+\frac{b+c}{b+c}  \right )\leq (a+b+a+c+b+c)\left (\displaystyle \frac{1}{a+b}+ \frac{1}{a+c}+ \frac{1}{b+c}  \right )\\ &\Leftrightarrow 3(1+1+1)\leq 2(a+b+c)\left (\displaystyle \frac{1}{a+b}+ \frac{1}{a+c}+ \frac{1}{b+c}  \right )\\ &\Leftrightarrow \displaystyle \frac{9}{2}\leq (a+b+c)\left (\displaystyle \frac{1}{a+b}+ \frac{1}{a+c}+ \frac{1}{b+c}  \right )\\ &\textrm{Dari dua ketaksamaan di atas didapatkan}\\ &3\left (\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}  \right )\geq \displaystyle \frac{9}{2}\\ &\Leftrightarrow \left (\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}  \right )\geq \displaystyle \frac{3}{2}\qquad \blacksquare     \end{aligned}   \end{array}$ 

$\begin{array}{ll}\\ 6.&\textrm{Diberikan}\: \: a,b,c,d>0,\: \: \textrm{tunjukkan bahwa}\\  &(a+b+c+d)\left (\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}  \right )\geq 16\\\\ &\textbf{Bukti}:\\ &\begin{aligned}&\textrm{Asumsikan}\\ &\begin{cases} & a\geq b\geq c\geq d \\ & \displaystyle \frac{1}{d}\geq \frac{1}{c}\geq \frac{1}{b}\geq \frac{1}{a} \end{cases} \\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan Chebyshev}\\ &\displaystyle \frac{\displaystyle \frac{a}{a}+\frac{b}{b}+\frac{c}{c}+\frac{d}{d}}{4}\leq \displaystyle \left ( \displaystyle \frac{(a+b+c+d)}{4} \right )\left ( \displaystyle \frac{\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}{4} \right )\\ &\Leftrightarrow 1\leq \displaystyle \left ( \displaystyle \frac{(a+b+c+d)}{4} \right )\left ( \displaystyle \frac{\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}{4} \right )\\ &\Leftrightarrow 16\leq (a+b+c+d)\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \right )\\ &\Leftrightarrow (a+b+c+d)\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \right )\geq 16\quad \blacksquare  \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Jika}\: \: a,b,c>0\: ,\: \textrm{dengan}\: \: a\neq b\neq c\\ &\textrm{tunjukkan bahwa}\\ &\left ( a^{3}+b^{3}+c^{3} \right )> \displaystyle \frac{(a+b+c)^{3}}{9}> 3abc\\\\ &\textbf{Bukti}:\\ &\textrm{Dengan}\: \textbf{ketaksamaan Chebyshev}\\ &\textrm{untuk}\: \: a\geq b\geq c,\: \textrm{dapat diperoleh bentuk}\\ &\displaystyle \frac{\left ( a^{3}+b^{3}+c^{3} \right )}{3}> \left ( \displaystyle \frac{a+b+c}{3} \right )\left ( \displaystyle \frac{a+b+c}{3} \right )\left ( \displaystyle \frac{a+b+c}{3} \right )\\ &\Leftrightarrow \displaystyle \frac{\left ( a^{3}+b^{3}+c^{3} \right )}{3}> \left ( \displaystyle \frac{a+b+c}{3} \right )^{3}> \left ( \displaystyle \frac{3\sqrt[3]{abc}}{3} \right )^{3}\\ &\Leftrightarrow \displaystyle \frac{\left ( a^{3}+b^{3}+c^{3} \right )}{3}> \displaystyle \frac{(a+b+c)^{3}}{27}> abc\\ &\Leftrightarrow \left ( a^{3}+b^{3}+c^{3} \right )> \displaystyle \frac{(a+b+c)^{3}}{9}> 3abc\quad \blacksquare   \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{tunjukkan bahwa untuk}\: \: n\: \: \textrm{bilangan asli}\\ &\textrm{berlaku}\\ &1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n}\leq n\sqrt{\displaystyle \frac{n+1}{2}}\\\\ &\textbf{Bukti}:\\ &\textrm{Dengan}\: \textbf{ketaksamaan Chebyshev}\\  &\textrm{dapat diperoleh bentuk berikut}\\ &\displaystyle \frac{1+2+3+\cdots +n}{n}\geq \displaystyle \frac{\left (1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n} \right )}{n}\times \displaystyle \frac{\left (1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n} \right )}{n}\\ &\Leftrightarrow \displaystyle \frac{1+2+3+\cdots +n}{n}\geq \displaystyle \frac{\left ( 1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n} \right )^{2}}{n^{2}}\\ &\Leftrightarrow  \displaystyle \frac{\left ( 1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n} \right )^{2}}{n^{2}}\leq  \displaystyle \frac{1+2+3+\cdots +n}{n}\\ &\Leftrightarrow \left (1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n}  \right )^{2}\leq n(1+2+3+\cdots +n)\\ &\Leftrightarrow \left (1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n}  \right )^{2}\leq n\left ( \displaystyle \frac{n(n+1)}{2} \right )\\ &\Leftrightarrow \left (1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n}  \right )\leq n\sqrt{\displaystyle \frac{n+1}{2}}\quad \blacksquare  \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{tunjukkan bahwa untuk}\: \: n\: \: \textrm{bilangan asli}\\&\textrm{berlaku}\\ &\displaystyle \frac{1}{\sqrt{n}}\left (1+\sqrt{\displaystyle \frac{1}{2}}+\sqrt{\displaystyle \frac{1}{3}}+\cdots +\sqrt{\displaystyle \frac{1}{n}}  \right )\leq (2n-1)^{.^{\frac{1}{4}}}\\\\ &\textbf{Bukti}:\\ &\textrm{Dengan}\: \textbf{ketaksamaan Chebyshev}\\&\textrm{untuk}:\: \left ( 1\geq \displaystyle \frac{1}{2}\geq \frac{1}{3}\geq \cdots \geq \frac{1}{n} \right )\\  &\textrm{dapat diperoleh bentuk berikut}\\ &\left (1+ \displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right )^{2}\leq n\left ( 1+\displaystyle \frac{1}{2.2}+\frac{1}{3.3}+\cdots +\frac{1}{n.n} \right )\\ &\Leftrightarrow \left (1+ \displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right )^{2}\leq n\left ( 1+\displaystyle \frac{1}{1.2}+\frac{1}{2.3}+\cdots +\frac{1}{(n-1).n} \right )\\ &\Leftrightarrow \left (1+ \displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right )^{2}\leq n\left ( 1+\left (1-\displaystyle \frac{1}{2}  \right )+\left (\displaystyle \frac{1}{2}-\frac{1}{3}  \right )+\cdots +\left (\displaystyle \frac{1}{(n-1)}-\frac{1}{n}  \right ) \right )\\ &\Leftrightarrow \left (1+ \displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right )^{2}\leq n\left ( 1+1-\displaystyle \frac{1}{n} \right )=n\left ( 2-\displaystyle \frac{1}{n} \right )\\&\Leftrightarrow \left (1+ \displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right )\leq \sqrt{2n-1}\quad \color{red}..........(1)\\ &\textrm{Gunakan lagi}\: \textbf{ketaksamaan Chebyshev}\\ &\left (1+\sqrt{\displaystyle \frac{1}{2}}+\sqrt{\displaystyle \frac{1}{3}}+\cdots +\sqrt{\displaystyle \frac{1}{n}}  \right )^{2}\leq n\left (1+ \displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right )\\ &\Leftrightarrow \displaystyle \frac{1}{n}\left (1+\sqrt{\displaystyle \frac{1}{2}}+\sqrt{\displaystyle \frac{1}{3}}+\cdots +\sqrt{\displaystyle \frac{1}{n}}  \right )^{2}\leq \left (1+ \displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right )\: \: \color{red}...(2)\\&\textrm{Dari ketaksamaan (1) dan (2), dapat diperoleh}\\ & \displaystyle \frac{1}{n}\left (1+\sqrt{\displaystyle \frac{1}{2}}+\sqrt{\displaystyle \frac{1}{3}}+\cdots +\sqrt{\displaystyle \frac{1}{n}}  \right )^{2}\leq \sqrt{2n-1}=(2n-1)^{.^{\frac{1}{2}}}\\ &\Leftrightarrow \displaystyle \frac{1}{\sqrt{n}}\left (1+\sqrt{\displaystyle \frac{1}{2}}+\sqrt{\displaystyle \frac{1}{3}}+\cdots +\sqrt{\displaystyle \frac{1}{n}}  \right )\leq (2n-1)^{.^{\frac{1}{4}}}\qquad \blacksquare    \end{array}$.

$\begin{array}{ll}\\ 10.&(\textbf{OSN 2011})\\ &\textrm{Jika}\: \: a,b,c>0\: ,\: \textrm{dengan}\: \: abc=1\\ &\textrm{Jika diketahui}\\ &a^{2011}+b^{2011}+c^{2011}< \displaystyle \frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}\\ &\textrm{tunjukkan bahwa}\\ &(a+b+c)> \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\\\\ &\textbf{Bukti}:\\ &\textrm{Asumsikan}\\&\color{red}\begin{cases} &a\geq b\geq c \\ &\displaystyle \frac{1}{c}\geq \frac{1}{b}\geq \frac{1}{a} \end{cases}\\ &\textrm{Dengan}\: \textbf{ketaksamaan Chebyshev}\\ &\textrm{Perhatikan}\\ &\begin{aligned}&\displaystyle \frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}\geq  \displaystyle \frac{1}{3}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\left ( \displaystyle \frac{1}{a^{2010}}+\frac{1}{b^{2010}}+\frac{1}{c^{2010}} \right )\\ &\Leftrightarrow \displaystyle \frac{1}{a^{2010}}+\frac{1}{b^{2010}}+\frac{1}{c^{2010}}\geq  \displaystyle \frac{1}{3}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\left ( \displaystyle \frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}} \right )\\  &\Leftrightarrow \displaystyle \frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}\geq  \displaystyle \frac{1}{3}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\left ( \displaystyle \frac{1}{a^{2008}}+\frac{1}{b^{2008}}+\frac{1}{c^{2008}} \right )\\  &\qquad\qquad \vdots \\ &\Leftrightarrow \displaystyle \frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}\geq  \displaystyle \frac{1}{3}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\left ( \displaystyle \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \right )\\ &\Leftrightarrow \displaystyle \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\geq  \displaystyle \frac{1}{3}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\\ &\textrm{Sehingga}\\ &\displaystyle \frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}\geq  \displaystyle \frac{1}{3^{2010}}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^{2011}\: \: \color{red}.....(1) \end{aligned} \\&\color{red}\textrm{dan}\\ &\begin{aligned}&a^{2011}+b^{2011}+c^{2011}\geq  \displaystyle \frac{1}{3}(a+b+c)(a^{2010}+b^{2010}+c^{2010})\\ &\Leftrightarrow a^{2010}+b^{2010}+c^{2010}\geq \displaystyle \frac{1}{3}(a+b+c)(a^{2009}+b^{2009}+c^{2009})\\  &\Leftrightarrow a^{2009}+b^{2009}+c^{2009}\geq \displaystyle \frac{1}{3}(a+b+c)(a^{2008}+b^{2008}+c^{2008})\\&\qquad\qquad \vdots \\&\Leftrightarrow a^{3}+b^{3}+c^{3}\geq \displaystyle \frac{1}{3}(a+b+c)(a^{2}+b^{2}+c^{2})\\&\Leftrightarrow a^{2}+b^{2}+c^{2}\geq \displaystyle \frac{1}{3}(a+b+c)(a+b+c)\\ &\textrm{Sehingga}\\ &\Leftrightarrow a^{2011}+b^{2011}+c^{2011}\geq \displaystyle \frac{1}{3^{2010}}(a+b+c)^{2011}\: \: \color{red}........(2)  \end{aligned}\\ &\begin{aligned}&\color{purple}\textrm{Dari ketaksamaan (1) dan (2) didapatkan}\\ &\frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}\geq  \displaystyle \frac{1}{3^{2010}}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^{2011}\\ &>a^{2011}+b^{2011}+c^{2011}\geq \displaystyle \frac{1}{3^{2010}}(a+b+c)^{2011}\\ &\color{blue}\textrm{atau}\\ &\displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{1}{a^{2011}}\geq \displaystyle \frac{1}{3^{2010}}\left ( \displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{1}{a}\right )^{2011}> \displaystyle \sum_{\textrm{siklik}}^{.}a^{2011}\geq \displaystyle \frac{1}{3^{2010}}\left ( \displaystyle \sum_{\textrm{siklik}}^{.}a \right )^{2011}\\ &\Leftrightarrow \displaystyle \frac{1}{3^{2010}}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^{2011}>\displaystyle \frac{1}{3^{2010}}(a+b+c)^{2011}\\ &\Leftrightarrow \left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )> (a+b+c)\\&\Leftrightarrow \: a+b+c<  \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\qquad \blacksquare    \end{aligned}  \end{array}$.

DAFTAR PUSTAKA

1. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.

Contoh Soal 9 (Segitiga dan Ketaksamaan)

$\begin{array}{ll}\\ 41.&\textrm{Jika}\: \: a,b,c> 0\: ,\: \textrm{tunjukkan bahwa}\\ &\qquad  \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \displaystyle \frac{3}{2}\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\left ( (a+b)+(b+c)+(c+a) \right )\left ( \displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right )\geq 9\\ &\Leftrightarrow \: 2(a+b+c)\left ( \displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right )\geq 9\\ &\Leftrightarrow \: (a+b+c)\left ( \displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right )\geq \displaystyle \frac{9}{2}\\ &\Leftrightarrow \: \displaystyle \frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\geq \displaystyle \frac{9}{2}\\ &\Leftrightarrow \: 1+\displaystyle \frac{c}{a+b}+1+\displaystyle \frac{a}{b+c}+1+\displaystyle \frac{b}{c+a}\geq \displaystyle \frac{9}{2}\\ &\Leftrightarrow \: 3+\displaystyle \frac{c}{a+b}+\displaystyle \frac{a}{b+c}+\displaystyle \frac{b}{c+a}\geq \displaystyle \frac{9}{2}\\ &\Leftrightarrow \: \displaystyle \frac{c}{a+b}+\displaystyle \frac{a}{b+c}+\displaystyle \frac{b}{c+a}\geq \displaystyle \frac{9}{2}-3\\ &\Leftrightarrow \: \displaystyle \frac{c}{a+b}+\displaystyle \frac{a}{b+c}+\displaystyle \frac{b}{c+a}\geq \displaystyle \frac{3}{2}\quad \blacksquare  \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 42.&\textrm{Jika}\: \: a,b,c\: \: \textrm{adalah sisi-sisi segitiga}\\  &\textrm{ABC, tunjukkan bahwa}\\ & \quad  \displaystyle \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\geq 3\\\\  &\textbf{Bukti}\\   &\begin{aligned}&\textrm{Perhatikan bahwa dengan AM-HM diperoleh}\\ &(a+b+c)\left (\displaystyle \frac{1}{b+c-a}+\frac{1}{c+a-b}+\frac{1}{a+b-c} \right )\geq 9\\ &\Leftrightarrow \: \left ( (b+c-a)+(c+a-b)+(a+b-c) \right )\\ &\quad \times \left (\displaystyle \frac{1}{b+c-a}+\frac{1}{c+a-b}+\frac{1}{a+b-c} \right )\geq 9\\ &\Leftrightarrow \:  \displaystyle \frac{a+b+c}{b+c-a}+\frac{a+b+c}{c+a-b}+\frac{a+b+c}{a+b-c}\geq 9\\ &\Leftrightarrow \: 1+\displaystyle \frac{2a}{b+c-a}+1+\displaystyle \frac{2b}{c+a-b}+1+\displaystyle \frac{2c}{a+b-c}\geq 9\\ &\Leftrightarrow \: 3+\displaystyle \frac{2a}{b+c-a}+\displaystyle \frac{2b}{c+a-b}+\displaystyle \frac{2c}{a+b-c}\geq 9\\ &\Leftrightarrow \: \displaystyle \frac{2a}{b+c-a}+\displaystyle \frac{2b}{c+a-b}+\displaystyle \frac{2c}{a+b-c}\geq 6\\ &\Leftrightarrow \: \displaystyle \frac{a}{b+c-a}+\displaystyle \frac{b}{c+a-b}+\displaystyle \frac{c}{a+b-c}\geq 3\quad \blacksquare  \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 43.&\textrm{Diketahui bilangan real positif}\: \: a,b,c\\ &\textrm{dengan}\: \:  abc=1\: .\: \textrm{Buktikan bahwa}\\  &\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3  \\\\   &\textbf{Bukti}\\    &\begin{aligned}&\textrm{Diketahui bahwa}\: \: abc=1\\ &\textrm{Perhatikan bahwa}\\ &\begin{cases} \bullet  & \displaystyle \frac{1+ab}{1+a}+\displaystyle \frac{abc+ab}{1+a}=ab\left ( \displaystyle \frac{1+c}{1+a} \right ) \\  \bullet  & \displaystyle \frac{1+bc}{1+b}+\displaystyle \frac{abc+bc}{1+b}=bc\left ( \displaystyle \frac{1+a}{1+b} \right ) \\  \bullet  & \displaystyle \frac{1+ca}{1+c}+\displaystyle \frac{abc+ca}{1+c}=ac\left ( \displaystyle \frac{1+b}{1+c} \right )  \end{cases}\\ &\color{red}\textrm{Jika hasilnya dijumlahkan}\color{black},\: \textrm{maka}\\ &\textrm{kita akan mendapatkan hasil}\\ &ab\left ( \displaystyle \frac{1+c}{1+a} \right )+bc\left ( \displaystyle \frac{1+a}{1+b} \right )+ac\left ( \displaystyle \frac{1+b}{1+c} \right )\\ &\color{blue}\textrm{Selanjutnya dengan AM-GM akan diperoleh}\\ &\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\\ &=ab\left ( \displaystyle \frac{1+c}{1+a} \right )+bc\left ( \displaystyle \frac{1+a}{1+b} \right )+ac\left ( \displaystyle \frac{1+b}{1+c} \right )\geq 3\sqrt[3]{(abc)^{2}}\\ &\Leftrightarrow \: \displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3\sqrt[3]{(abc)^{2}}\\ &\Leftrightarrow \: \displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3\sqrt[3]{1}\\ &\Leftrightarrow \: \displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3\qquad \blacksquare  \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 44.&(\textbf{SEAMO III-Malaysia})\\ &\textrm{Misalkan}\: \: a,b,c\: \: \textrm{bilangan real positif}\\ &\textrm{Tunjukkan bahwa}\\ &\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\geq \displaystyle \frac{3}{1+abc}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\\ &\displaystyle \frac{1}{a(1+b)}+\frac{1}{1+abc}=\frac{1}{1+abc}\left ( \displaystyle \frac{1+a}{a(1+b)}+\frac{b(1+c)}{(1+b)} \right )\\ &\textrm{Sehingga}\\ &\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc}\\ &=\displaystyle \frac{1}{1+abc}\left ( \displaystyle \frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a} \right )\\ &\color{blue}\textrm{Secara AM-GM kita mendapatkan}\\ &\left ( \displaystyle \frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a} \right )\geq 6\\ &\textrm{Kembali ke soal}\\ &\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc}\geq \displaystyle \frac{1}{1+abc}(6)\\ &\Leftrightarrow \: \displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\geq \displaystyle \frac{3}{1+abc}\qquad \blacksquare  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 45.&\textrm{Misalkan}\: \: a,b,c\: \: \textrm{bilangan real non negatif}\\ &\textrm{dengan}\: \: a+b+c=1,\: \: \textrm{Tunjukkan bahwa}\\ &\displaystyle \frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}\geq \displaystyle \frac{9}{10}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Diketahui}\: \: a+b+c=1\\ &\color{blue}\textrm{Dengan AM-GM kita memiliki}\\ &\displaystyle \frac{a+b+c}{3}\geq \sqrt[3]{abc}\Leftrightarrow \displaystyle \frac{1}{3}\geq \sqrt[3]{abc}\Leftrightarrow \displaystyle \frac{1}{27}\geq abc\\ &\textrm{Sehingga}\\ &\displaystyle \frac{1}{1+3abc}\geq \displaystyle \frac{1}{1+3\left ( \displaystyle \frac{1}{27} \right )}\geq \displaystyle \frac{1}{1+\displaystyle \frac{1}{9}}\geq \displaystyle \frac{9}{10}\\ &\color{red}\textrm{Selanjutnya dengan perluasan AM-HM}\\ &\textrm{Sebagaimana bentuk berikut}\\ &\left ( \displaystyle \frac{p_{1}a_{1}+p_{2}a_{2}+p_{3}a_{3}}{p_{1}+p_{2}+p_{3}} \right )\geq \left ( \displaystyle \frac{p_{1}+p_{2}+p_{3}}{p_{1}a_{1}^{-1}+p_{2}a_{2}^{-1}+p_{3}a_{3}^{-1}} \right )\\ &\textrm{maka}\\ &\displaystyle \frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}\geq \displaystyle \frac{1}{a(1+bc)+b(1+ac)+c(1+ab)}\\ &\qquad\qquad\qquad\qquad\qquad\quad  \geq \displaystyle \frac{1}{a+b+c+3abc}\\ &\qquad\qquad\qquad\qquad\qquad\quad  \geq \displaystyle \frac{1}{1+3abc}\\ &\qquad\qquad\qquad\qquad\qquad\quad  \geq \displaystyle \frac{9}{10}\qquad \blacksquare  \end{aligned}   \end{array}$.

DAFTAR PUSTAKA

1. Bambang, S. 2012. Materi, Soal dan Penyelesaian Olimpiade Matematika Tingkat SMA/MA. Jakarta: BINA PRESTASI INSANI.
2. Bintari, N., Gunarto, D. 2007. Panduan Menguasai Soal-Soal Olimpiade Matematika Nasional dan Internasional. Yogyakarta: INDONESIA CERDAS.
3. Tung, K.Y. 2013. Ayo Raih Medali Emas Olimpiade Matematika SMA. Yogyakarta: ANDI.

Contoh Soal 8 (Segitiga dan Ketaksamaan)

 $\begin{array}{ll}\\ 36.&\textrm{Jika}\: \: a,b,c> 0\: ,\: \textrm{dengan}\: \: a+b+c=1\\ &\textrm{tunjukkan bahwa}\\ &\qquad  (1-a)(1-b)(1-c)\geq 8abc\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\textrm{Perhatikan bahwa}\\ &\begin{cases} 1-a &=b+c \\  1-b & =a+c \\  1-c & =a+b  \end{cases}\\ &\textrm{Kurang lebih seperti pembuktian}\\ &\textrm{pada no.35 di atas dengan tetap}\\ &\textrm{menggunakan AM-GM akan}\\ &\textrm{didapatkan}\\ &\bullet \: \: a+b\geq 2\sqrt{ab}\\ &\bullet \: \: b+c\geq 2\sqrt{bc},\: \: \textrm{dan}\\ &\bullet \: \: c+a\geq 2\sqrt{ca}\\ &\textrm{Selanjutnya}\\ &(1-a)(1-b)(1-c)= (b+c)(a+c)(a+b)\\ &\: \: \: \qquad\qquad\qquad\quad\quad \geq 2\sqrt{bc}\times 2\sqrt{ac}\times 2\sqrt{ab}\\ & \: \: \: \qquad\qquad\qquad\quad\quad \geq 8\sqrt{a^{2}b^{2}c^{2}}\\ &\: \: \:  \qquad\qquad\qquad\quad\quad \geq 8abc\qquad \blacksquare  \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 37.&\textrm{Misalkan}\: \: a,b,c\: \: \textrm{bilangan real positif}\\ &\textrm{dengan nilai}\: \: abc=1.\: \: \textrm{Nilai terkecil}\\  &\textrm{dari}\: \: (a+2b)(b+2c)(ac+1)\: \: \textrm{tercapai}\\ &\textrm{ketika}\: \: a+b+c\: \: \textrm{bernilai}\: ...\: .\\\\  &\textbf{Jawab}:\\   &\begin{aligned}&\textrm{Dengan AM-GM kita dapatkan}\\ &\bullet \: a+2b\geq 2\sqrt{2ab}\\ &\bullet \: b+2c\geq 2\sqrt{2bc}\\ &\bullet \: ac+1\geq 2\sqrt{ac}\\ &\textrm{Selanjutnya}\\ &(a+2b)(b+2c)(ac+1)\geq 2\sqrt{2ab}\times 2\sqrt{2bc}\times 2\sqrt{ac}\\ &\: \qquad\qquad\qquad\qquad\qquad \geq 8\sqrt{4a^{2}b^{2}c^{2}}\\ &\: \qquad\qquad\qquad\qquad\qquad \geq 8\times 2abc\\ &\: \qquad\qquad\qquad\qquad\qquad \geq 16\times 1\\ &\: \qquad\qquad\qquad\qquad\qquad \geq 16\\ &\textrm{Karena}\: \: abc=1,\: \textrm{dengan cara coba-coba}\\ &\textrm{dapat kita peroleh nilai}\\ &a=2,\: b=1,\: \: \textrm{dan}\: \: c=\displaystyle \frac{1}{2}\\ &\textrm{Kita cek ke}\: \: (a+2b)(b+2c)(ac+1)\geq 16\\ &(2+2)(1+1)(1+1)\geq 16\: \: \textrm{adalah}\: \textbf{benar}\\ &\textrm{Sehingga nilai}\: \: a+b+c=2+1+\displaystyle \frac{1}{2}=3\displaystyle \frac{1}{2}=\displaystyle \frac{7}{2}  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 38.&\textrm{Diketahui}\: \: a,b,c\: \: \textrm{bilangan real positif}\\ &\textrm{dan}\: \: (a+1)(b+1)(c+1)=8.\\ &\textrm{tunjukkan bahwa}\quad  abc< 1\\\\  &\textbf{Bukti}\\   &\begin{aligned}&\textrm{Diketahui bahwa}\\ &(a+1)(b+1)(c+1)=8\\ &abc+(ab+bc+ca)+(a+b+c)+1=8\\ &\color{blue}\textrm{Dengan AM-GM kita akan mendapatkan}\\ &abc+3(abc)^{.^{\frac{2}{3}}}+3(abc)^{.^{\frac{1}{3}}}+1< 8\\ &\Leftrightarrow \: \left ( (abc)^{.^{\frac{1}{3}}}+1 \right )^{3}< 2^{3}\\ &\Leftrightarrow \: (abc)^{.^{\frac{1}{3}}}+1< 2\\ &\Leftrightarrow \: (abc)^{.^{\frac{1}{3}}}< 1\qquad \blacksquare   \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 39.&\textrm{Jika}\: \: a,b,c> 0\\ &\textrm{Tunjukkan bahwa}\\ &\qquad  \displaystyle \frac{a}{a^{2}+1}+\displaystyle \frac{b}{b^{2}+1}+\displaystyle \frac{c}{c^{2}+1}\leq  \displaystyle \frac{3}{2}\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\textrm{Perhatikan bahwa}\\ &(a-1)^{2}\geq 0\\ &\Leftrightarrow a^{2}-2a+1\geq 0\\ &\Leftrightarrow  a^{2}+1\geq 2a\\ &\displaystyle \frac{1}{2}\geq \displaystyle \frac{a}{a^{2}+1}\\ &\textrm{atau}\\ &\displaystyle \frac{a}{a^{2}+1}\leq \displaystyle \frac{1}{2}\: \: ..................(1)\\ &\textrm{dengan cara yang sama akan} \\ &\textrm{pula jiika}\: \: b\: \: \textrm{dan}\: \: c\: \: \textrm{dikondisikan}\\ &\textrm{akan didapatkan ketaksamaan}\\ &\displaystyle \frac{b}{b^{2}+1}\leq \displaystyle \frac{1}{2}\: \: ..................(2)\\ &\displaystyle \frac{c}{c^{2}+1}\leq \displaystyle \frac{1}{2}\: \: ..................(3)\\ &\textrm{Jika ketaksamaan}\: \: (1),\: (2),\: \&\: \: (3)\\ &\textrm{dijumlahkan akan menghasilkan}\\ &\displaystyle \frac{a}{a^{2}+1}+\displaystyle \frac{b}{b^{2}+1}+\displaystyle \frac{c}{c^{2}+1}\leq \displaystyle \frac{1}{2}+\displaystyle \frac{1}{2}+\displaystyle \frac{1}{2}\\ &\qquad\qquad\qquad\qquad\qquad\quad \leq 3\left ( \displaystyle \frac{1}{2} \right )\\ &\qquad\qquad\qquad\qquad\qquad\quad \leq \displaystyle \frac{3}{2}\: \quad \blacksquare \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 40.&\textrm{Jika}\: \: a,b,c\: \: \textrm{adalah sisi-sisi segitiga}\\ &\textrm{ABC, tunjukkan bahwa}\\ & \qquad  \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< 2\\\\  &\textbf{Bukti}\\  &\begin{aligned}&\textrm{Perhatikan bahwa pada segitiga}\\ &\textrm{ABC berlaku}\\ &\begin{cases} \color{red}a+b & >\color{red}c \\  a+c & >b \\  b+c & >a  \end{cases}\\ &\textrm{Misalkan}\: \: 2s=a+b+c,\: \: \textrm{maka}\\ &a+b=a+b\\ &\Leftrightarrow \: a+b+\color{red}a+b\color{black}>a+b+\color{red}c\\ &\Leftrightarrow \: 2(a+b)> 2s\\ &\Leftrightarrow \: (a+b)>s\\ &\textrm{Demikian juga akan berlaku}\\ &\begin{cases} b+c &>s \\  c+a &>s  \end{cases}\\ &\textrm{Sehingga}\\ &\begin{cases} \displaystyle \frac{a}{b+c} & <\displaystyle \frac{a}{s} \\  \displaystyle \frac{b}{c+a} & <\displaystyle \frac{b}{s} \\  \displaystyle \frac{c}{a+b} & <\displaystyle \frac{c}{s}  \end{cases}\\ &\textrm{Selanjutnya kita kembali ke soal}\\ &\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< \displaystyle \frac{a+b+c}{s}\\ &\: \: \: \qquad\qquad\qquad\qquad\quad <\displaystyle \frac{2s}{s}\\ &\: \: \: \qquad\qquad\qquad\qquad\quad <2\qquad \blacksquare  \end{aligned}  \end{array}$.

Contoh Soal 7 (Segitiga dan Ketaksamaan)

$\begin{array}{ll}\\ 31.&(\textbf{Russia 1992})\\ &\textrm{Jika}\: \: a,b> 1\: ,\: \textrm{buktikan bahwa}\\ &\qquad  \displaystyle \frac{x^{2}}{y-1}+\displaystyle \frac{y^{2}}{x-1}\geq 8\\\\  &\textbf{Bukti}\\   &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}&\textrm{Gunakan AM-GM untuk mendapatkan}\\ &\displaystyle \frac{x^{2}}{y-1}+\displaystyle \frac{y^{2}}{x-1}\geq 2\sqrt{\displaystyle \frac{x^{2}y^{2}}{\sqrt{(x-1)(y-1)}}}\\ &\: \qquad\qquad\qquad \geq \displaystyle \frac{2xy}{\sqrt{(x-1)(y-1)}}\\ &\textrm{Sebelum kita lanjutkan, ingat bahwa}\\ &(x-2)^{2}\geq 0\Leftrightarrow x^{2}-4x+4\geq 0\\ &\Leftrightarrow x^{2}\geq 4x-4\Leftrightarrow x^{2}\geq 4(x-1)\\ &\Leftrightarrow \displaystyle \frac{x^{2}}{x-1}\geq 4\Leftrightarrow \displaystyle \frac{x}{\sqrt{x-1}}\geq 2\\ &\textrm{Demikian juga}\: \: \displaystyle \frac{y}{\sqrt{y-1}}\geq 2\\ &\textrm{Selanjutnya kembali ke semula yaitu}\\ &\: \qquad\qquad\qquad \geq \displaystyle \frac{2xy}{\sqrt{(x-1)(y-1)}}\\ &\: \qquad\qquad\qquad \geq \displaystyle 2.2.2\\ &\displaystyle \frac{x^{2}}{y-1}+\displaystyle \frac{y^{2}}{x-1}\geq 8\qquad \blacksquare  \end{aligned}\\ &\color{blue}\textrm{Alternatif 2}\\ &\begin{aligned}&\textrm{Misalkan saja}\\ &\begin{cases} a & =x-1 \Rightarrow x=a+1\\  b & =y-1 \Rightarrow y=b+1 \end{cases}\\ &\textrm{Maka}\\ &\displaystyle \frac{x^{2}}{y-1}+\displaystyle \frac{y^{2}}{x-1}=\displaystyle \frac{(a+1)^{2}}{b}+\displaystyle \frac{(b+1)^{2}}{a}\\ &\textrm{Dengan AM-GM akan diperoleh}\\ &\displaystyle \frac{(a+1)^{2}}{b}+\displaystyle \frac{(b+1)^{2}}{a}\geq 4\left ( \displaystyle \frac{a}{b}+\displaystyle \frac{b}{a} \right )\\ &\textrm{Bentuk di atas didapatkan dari}\\ &(a-1)^{2}\geq 0\Leftrightarrow a^{2}-2a+1\geq 0\\ &\Leftrightarrow a^{2}+2a-4a+1\geq 0\\ &\Leftrightarrow a^{2}+2a+1\geq 4a\Leftrightarrow (a+1)^{2}\geq 4a\\ &\textrm{Demikian juga yabf satunya, yaitu}\\ &(b+1)^{2}\geq 4b.\: \textrm{Serta bentuk}\\ &\left ( \displaystyle \frac{a}{b}+\displaystyle \frac{b}{a} \right )\geq 2\sqrt{\displaystyle \frac{a}{b}.\frac{b}{s}}=2\\ &\textrm{Selanjutnya kembali ke soal, yaitu}:\\ &\displaystyle \frac{(a+1)^{2}}{b}+\displaystyle \frac{(b+1)^{2}}{a}\geq 4\left ( \displaystyle \frac{a}{b}+\displaystyle \frac{b}{a} \right )\\ &\: \,  \qquad\qquad\qquad\qquad \geq 4(2)\\ &\: \,  \qquad\qquad\qquad\qquad \geq 8\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 32.&\textrm{Untuk}\: \: a>0 \: ,\: \textrm{tentukan nilai }\\ &\textrm{minimum dari bentuk}\: \: \displaystyle \frac{4a^{2}+8a+13}{6+6a}\\\\ &\textbf{Jawab}:\\   &\begin{aligned}&\textrm{Diketahui}\: \: a>0, \: \textrm{maka bentuk}\\ &\displaystyle \frac{4a^{2}+8a+13}{6+6a}=\displaystyle \frac{4(a+1)^{2}+9}{6(a+1)}\\ &=\displaystyle \frac{2(a+1)}{3}+\frac{3}{2(a+1)}\\ &\textrm{Dengan AM-GM akan diperoleh}\\ &\displaystyle \frac{2(a+1)}{3}+\frac{3}{2(a+1)}\\ &\geq 2\sqrt{\displaystyle \frac{2}{3}\times \frac{3}{2}\times \frac{(a+1)}{(a+1)}}=2\\ &\textrm{Jadi},\: \: \displaystyle \frac{2(a+1)}{3}+\frac{3}{2(a+1)}\geq 2  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 33.&\textrm{Untuk}\: \: 0\leq a< 6 \: ,\: \textrm{tentukan nilai }\\ &\textrm{minimum dari bentuk}\: :\: a\left ( 6-a \right )^{2}\\\\  &\textbf{Jawab}:\\   &\begin{aligned}&\textrm{Diketahui}\: \: 0\leq a< 6, \: \textrm{maka bentuk}\\ &a\left ( 6-a \right )^{2}=\displaystyle \frac{1}{2}(2a)(6-a)(6-a)\\ &\textrm{Dengan GM-AM akan kita peroleh}\\ &\displaystyle \frac{1}{2}(2a)(6-a)(6-a)\leq \displaystyle \frac{1}{2}\left ( \displaystyle \frac{2a+6-a+6-a}{3} \right )^{3}\\ &\: \: \qquad\qquad\qquad\quad\quad \leq \displaystyle \frac{1}{2}\left (\displaystyle \frac{12}{3}  \right )^{3}\\ &\: \: \qquad\qquad\qquad\quad\quad \leq \displaystyle \frac{1}{2}\left (4  \right )^{3}\\ &\: \: \qquad\qquad\qquad\quad\quad \leq \displaystyle \frac{1}{2}\left (64  \right )\\ &\: \: \qquad\qquad\qquad\quad\quad \leq 32\\  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 34.&\textrm{Jika}\: \: a,b> 0\: ,\: \textrm{tunjukkan bahwa}\\ &\qquad  (a+1)(b+1)(ab+1)\geq 8ab\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\textrm{Perhatikan bahwa dengan AM-GM}\\ &\bullet \: \: a+1\geq 2\sqrt{a}\\ &\bullet \: \: b+1\geq 2\sqrt{b},\: \: \textrm{dan}\\ &\bullet \: \: ab+1\geq 2\sqrt{ab}\\ &\textrm{Selanjutnya}\\ &(a+1)(b+1)(ab+1)\geq 2\sqrt{a}\times 2\sqrt{b}\times 2\sqrt{ab}\\ &\: \: \qquad\qquad\qquad\qquad\quad \geq 8\sqrt{a^{2}b^{2}}\\ &\: \: \qquad\qquad\qquad\qquad\quad \geq 8ab\qquad \blacksquare  \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 35.&\textrm{Jika}\: \: a,b,c> 0\: ,\: \textrm{tunjukkan bahwa}\\ &\qquad  (a+b)(b+c)(c+a)\geq 8abc\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Kurang lebih seperti pembuktian}\\ &\textrm{pada no.34 di atas dengan tetap}\\ &\textrm{menggunakan AM-GM akan}\\ &\textrm{didapatkan}\\ &\bullet \: \: a+b\geq 2\sqrt{ab}\\ &\bullet \: \: b+c\geq 2\sqrt{bc},\: \: \textrm{dan}\\ &\bullet \: \: c+a\geq 2\sqrt{ca}\\ &\textrm{Selanjutnya}\\ &(a+b)(b+c)(c+a)\geq 2\sqrt{ab}\times 2\sqrt{bc}\times 2\sqrt{ca}\\ & \: \: \: \qquad\qquad\qquad\quad\quad \geq 8\sqrt{a^{2}b^{2}c^{2}}\\ &\: \: \:  \qquad\qquad\qquad\quad\quad \geq 8abc\qquad \blacksquare  \end{aligned}   \end{array}$.

DAFTAR PUSTAKA

1. Bambang, S. 2012. Materi, Soal dan Penyelesaian Olimpiade Matematika Tingkat SMA/MA. Jakarta: BINA PRESTASI INSANI.
2. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
3. Manfrino, R.B., dkk. 2009. Inequalities A Mathematical Olympiad Approach. Basel: Birkhauser Verlag AG.

Contoh Soal 6 (Segitiga dan Ketaksamaan)

$\begin{array}{ll}\\ 26.&\textrm{Diketahui}\: \: a,b\: \: \textrm{bilangan real positif}\\ &\textrm{dan}\: \: a+b=1.\: \textrm{Tunjukkan bahwa}\\ &\qquad  \left ( a+\displaystyle \frac{1}{a} \right )^{2}+\left ( b+\displaystyle \frac{1}{b} \right )^{2}\geq \displaystyle \frac{25}{2}\\\\  &\textbf{Bukti}\\   &\begin{aligned}&\textrm{Diketahui bahwa}\: \: a+b=1\\ &\color{red}\textrm{Dengan QM-AM kita akan mendapatkan}\\ &\sqrt{\displaystyle \frac{x^{2}+y^{2}}{2}}\geq \displaystyle \frac{x+y}{2}\Leftrightarrow \displaystyle \frac{x^{2}+y^{2}}{2}\geq \left ( \displaystyle \frac{x+y}{2} \right )^{2}\\ &\textrm{Kita misalkan}\: \: \color{purple}\begin{cases} x & =a+\displaystyle \frac{1}{a} \\\\  y & =b+\displaystyle \frac{1}{b}  \end{cases}\\ &\textrm{maka}\\ &\begin{aligned}\displaystyle \frac{\left ( a+\displaystyle \frac{1}{a} \right )^{2}+\left (b+\displaystyle \frac{1}{b}  \right )^{2}}{2}&\geq \left ( \displaystyle \frac{a+\displaystyle \frac{1}{a}+b+\displaystyle \frac{1}{b}}{2} \right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( a+b+\displaystyle \frac{1}{a}+\frac{1}{b} \right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( 1+\displaystyle \frac{1}{a}+\frac{1}{b} \right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( 1+ \displaystyle \frac{a+b}{ab}\right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( 1+\displaystyle \frac{1}{ab} \right )^{2} \end{aligned}\\ &\color{blue}\textrm{Dari GM-AM kita akan mendapatkan}\\ &\sqrt{ab}\leq \displaystyle \frac{a+b}{2}\Leftrightarrow ab\leq \left ( \displaystyle \frac{a+b}{2} \right )^{2}\\ &\Leftrightarrow ab\leq \left ( \displaystyle \frac{1}{2} \right )^{2}\Leftrightarrow ab\leq \displaystyle \frac{1}{4} \\ &\textrm{Sehingga}\\ &\begin{aligned}\left ( a+\displaystyle \frac{1}{a} \right )^{2}+\left (b+\displaystyle \frac{1}{b}  \right )^{2}&\geq \displaystyle \frac{1}{2}\left ( 1+\displaystyle \frac{1}{ab} \right )^{2}\\ &\geq \displaystyle \frac{1}{2}\left ( 1+ \displaystyle \frac{1}{\displaystyle \frac{1}{4}}\right )^{2}\\ &\geq \displaystyle \frac{1}{2}\left ( 1+4 \right )^{2}\\ &\geq \displaystyle \frac{1}{2}\times 25\\ &\geq \displaystyle \frac{25}{2}\qquad \blacksquare  \end{aligned}  \end{aligned}  \end{array}$.

$.\: \: \: \quad \textbf{versi cara yang lain}\: $ silahkan klik di sini

$\begin{array}{ll}\\ 27.&\textrm{Jika}\: \: a,b> 0\: ,\: \textrm{tunjukkan bahwa}\\ &\qquad  (a^{2}+1)(b^{2}+1)\geq (a+b)^{2}\\\\  &\textbf{Bukti}\\   &\begin{aligned}&\textrm{Dipilih}\: \: (ab-1)^{2}\geq 0\\ &\textrm{Selanjutnya}\\ &(ab-1)^{2}=a^{2}b^{2}-2ab+1\geq 0\\ &\Leftrightarrow \: a^{2}b^{2}+1\geq 2ab\\ &\Leftrightarrow \: a^{2}b^{2}+a^{2}+b^{2}+1\geq a^{2}+b^{2}+2ab\\ &\Leftrightarrow \: (a^{2}+1)(b^{2}+1)\geq (a+b)^{2}\qquad \blacksquare   \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 28.&\textrm{Jika}\: \: a,b> 0\: ,\: \textrm{tunjukkan bahwa}\\ &\qquad  \displaystyle \frac{a^{3}+b^{3}}{2}\geq \left ( \displaystyle \frac{a+b}{2} \right )^{3}\\\\  &\textbf{Bukti}\\   &\begin{aligned}&\textrm{Dipilih}\: \: (a-b)^{2}\geq 0\\ &\textrm{Selanjutnya}\\ &(a+b)(a-b)^{2}\geq 0\\ &\Leftrightarrow \: (a+b)(a^{2}-2ab+b^{2})\geq 0\\ &\Leftrightarrow \: a^{3}+b^{3}-a^{2}b-ab^{2}\geq 0\\ &\Leftrightarrow \: a^{3}+b^{3}\geq a^{2}b+ab^{2}\\ &\Leftrightarrow \: 3a^{3}+3b^{3}\geq 3a^{2}b+3ab^{2}\\ &\Leftrightarrow \: 4a^{3}+4b^{3}\geq a^{3}+b^{3}+3a^{2}b+3ab^{2}\\ &\Leftrightarrow \: 4a^{3}+4b^{3}\geq (a+b)^{3}\\ &\Leftrightarrow \: \displaystyle \frac{1}{8}\left (4a^{3}+4b^{3}  \right )\geq \displaystyle \frac{1}{8}(a+b)^{3}\\ &\Leftrightarrow \: \displaystyle \frac{a^{3}+b^{3}}{2}\geq \left ( \displaystyle \frac{a+b}{2} \right )^{3}\qquad \blacksquare   \end{aligned}    \end{array}$ .

$.\: \: \: \quad \textbf{versi cara yang lain}\: $ silahkan klik di sini

$\begin{array}{ll}\\ 29.&\textrm{Jika}\: \: a\geq b> 0\: ,\: \textrm{tunjukkan bahwa}\\  &\qquad  a+\displaystyle \frac{1}{b(a-b)}\geq 3\\\\  &\textbf{Bukti}\\   &\begin{aligned}a+\displaystyle \frac{1}{b(a-b)}&=a-b+b+\displaystyle \frac{1}{b(a-b)}\\ \color{blue}\textrm{Dengan AM}&\color{blue}-\textrm{GM diperoleh}\\ &=a-b+b+\displaystyle \frac{1}{b(a-b)}\\ &\geq 3\sqrt[3]{(a-b)\times b\times \displaystyle \frac{1}{b(a-b)}}\\ &\geq 3\sqrt[3]{1}\\ &\geq 3\qquad \blacksquare   \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 30.&\textbf{(OSN 2008)}\\ &\textrm{Jika}\: \: a,b> 0\: ,\: \textrm{tunjukkan bahwa}\\  &\quad  \displaystyle \frac{1}{\left ( 1+\sqrt{a} \right )^{2}}+\displaystyle \frac{1}{\left ( 1+\sqrt{b} \right )^{2}}\geq \displaystyle \frac{2}{a+b+2}\\\\  &\textbf{Bukti}\\   &\begin{aligned}&\textrm{Dipilih}\\ &\begin{aligned}\left ( 1+\sqrt{a} \right )^{2}&=1+a+2\sqrt{a}\\ &=2+2a-1-a+2\sqrt{a}\\ &=2+2a-(1+a-2\sqrt{a})\\ &=2+2a-(1-\sqrt{a})^{2}\\ \textrm{sehingga}\: \: &\\ \left ( 1+\sqrt{a} \right )^{2}&=2+2a-(1-\sqrt{a})^{2}\\ \left ( 1+\sqrt{a} \right )^{2}&\leq 2+2a\\ \textrm{atau}\qquad\: &\\ \displaystyle \frac{1}{\left (1+\sqrt{a}  \right )^{2}}&\geq \displaystyle \frac{1}{2+2a} \end{aligned}\\ &\textrm{Selanjutnya kembali ke bentuk soal}\\ &\textrm{yaitu}:\\ &\begin{aligned}\displaystyle \frac{1}{\left ( 1+\sqrt{a} \right )^{2}}+\displaystyle \frac{1}{\left ( 1+\sqrt{b} \right )^{2}}&\geq \displaystyle \frac{1}{2+2a}+\frac{1}{2+2b}\\ &\geq \displaystyle \frac{1}{2}\left ( \displaystyle \frac{1}{1+a}+\frac{1}{1+b} \right )\\ \textrm{Dengan AM-HM dipero}&\textrm{leh}\\ &\geq \displaystyle \frac{1}{2}\left (\displaystyle \frac{2.2}{\displaystyle \frac{1}{\frac{1}{1+a}}+\displaystyle \frac{1}{\frac{1}{1+b}}} \right )\\ &\geq \displaystyle \frac{2}{(1+a)+(1+b)}\\ &\geq \displaystyle \frac{2}{a+b+2}\qquad \blacksquare  \end{aligned}   \end{aligned}  \end{array}$.

DAFTAR PUSTAKA

1. Bambang, S. 2012. Materi, Soal dan Penyelesaian Olimpiade Matematika Tingkat SMA/MA. Jakarta: BINA PRESTASI INSANI.
2. Yohanes, S., Panji, R. 2008. Mahir Olimpiade Matematika SMA. Yogyakarta: KENDI MAS MEDIA.
3. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.

Contoh Soal 5 (Segitiga dan Ketaksamaan)

$\begin{array}{ll}\\ 21.&\textrm{Diketahui}\: \: a,\: b,\: c\: \: \textrm{sisi segitiga}\\ &\textrm{dengan}\: \: abc=1,\: \textrm{tunjukkan bahwa}\\  &a+b+c\geq 1\\\\ &\textbf{Bukti}\\ &\textrm{perhatikan bukti no.18 berikut}\\   &\begin{aligned}&a+b+c\geq 3\sqrt[3]{abc}\\ &\textrm{dengan mengganti nilai}\: \: abc=1,\\ &\textrm{maka akan didapatkan}\\ &a+b+c\geq 3\sqrt[3]{1} \\ &\Leftrightarrow \: a+b+c\geq 3.1\\ &\Leftrightarrow \: a+b+c\geq 3\qquad \blacksquare   \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textrm{Jika}\: \: a\: \: \textrm{bilangan real positif}\\  &\textrm{tunjukkan bahwa}\: \: a^{2}+\displaystyle \frac{2}{a}\geq 3\\\\ &\textbf{Bukti}\\   &\begin{aligned}&\textrm{Diketahui}\: \: a^{2}+\displaystyle \frac{2}{a}\geq 3,\: \textrm{dengan}\: a> 0\\ &\textrm{maka dengan mengalikan dengan}\: \: a\\ &\textrm{bentuknya menjadi}\: \: a^{3}+2\geq 3a\\ &\textrm{Dengan AM-GM}\\ &\displaystyle \frac{a^{3}+1+1}{3}\geq \sqrt[3]{a^{3}\times 1\times 1}\\ &\Leftrightarrow \: \displaystyle \frac{a^{3}+2}{3}\geq \sqrt[3]{a^{3}}\\ &\Leftrightarrow \: \displaystyle \frac{a^{3}+2}{3}\geq a\\ &\Leftrightarrow \: \: a^{3}+2\geq 3a\\ &\Leftrightarrow \: \qquad a^{3}\geq 3a-2 \qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 23.&\textrm{Buktikan bahwa untuk bilangan}\: \: a\: \: \textrm{tidak}\\ &\textrm{negatif}\: \:  \textrm{berlaku}\: \: a^{3}\geq 3a-2\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\color{blue}\textrm{Alternatif 1}\\ &a^{3}\geq 3a-2\\ &\Leftrightarrow \: a^{3}\geq a+2a-2\\ &\Leftrightarrow \: a^{3}-a\geq 2a-2\\ &\Leftrightarrow \: a(a^{2}-1)\geq 2(a-1)\\ &\Leftrightarrow \: a(a+1)(a-1)\geq 2(a-1)\\ &\Leftrightarrow \: (a^{2}+a)(a-1)\geq 2(a-1)\\ &\Leftrightarrow \: (a^{2}+a)(a-1)- 2(a-1)\geq 0\\ &\Leftrightarrow \: (a-1)(a^{2}+a-2)\geq 0\\ &\Leftrightarrow \: (a-1)(a-1)(a+2)\geq 0\\ &\Leftrightarrow \: (a-1)^{2}(a+2)\geq 0\qquad \blacksquare \\\\  &\color{blue}\textrm{Alternatif 2}\\ &\textrm{Lihat bukti no.22 di atas} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 24.&\textrm{Jika}\: \: a,b,m,n\: \: \textrm{bilangan real yang}\\  &\textrm{memenuhi}\: \: a^{2}+b^{2}=1\: \: \textrm{dan}\: \: m^{2}+n^{2}=1\\ &\textrm{Tunjukkan bahwa}\quad  \left | am+bn \right |\leq 1\\\\ &\textbf{Bukti}\\   &\begin{aligned}&\color{blue}\textrm{Perhatikan bahwa}\\ &(a^{2}+b^{2})(m^{2}+n^{2})-(am+bn)^{2}\\ &=a^{2}m^{2}+a^{2}n^{2}+b^{2}m^{2} +b^{2}n^{2}-a^{2}m^{2}-b^{2}n^{2}-2abmn\\ &=a^{2}n^{2}+b^{2}m^{2}-2abmn\\ &=(an-bm)^{2}\\ &\color{red}\textrm{Sekarang kita punya}\\ &(an-bm)^{2}\geq 0\\ &\Leftrightarrow \: (a^{2}+b^{2})(m^{2}+n^{2})-(am+bn)^{2}\geq 0\\ &\Leftrightarrow \: \quad 1\times 1-(am+bn)^{2}\geq 0\\ &\Leftrightarrow \: \quad\qquad 1-(am+bn)^{2}\geq 0\\ &\Leftrightarrow \: \quad\qquad (am+bn)^{2}-1\leq  0\\ &\Leftrightarrow \: \: \: \: \: \qquad\qquad (am+bn)^{2}\leq  1\qquad \blacksquare  \end{aligned}  \end{array}$

$\begin{array}{ll}\\ 25.&\textrm{Jika}\: \: a+b=1\: \: \textrm{dengan}\: \: a,b> 0\\ &\textrm{Tunjukkan bahwa}\\ &\qquad\qquad  \displaystyle \frac{a^{2}}{a+1}+\displaystyle \frac{b^{2}}{b+1}\geq \displaystyle \frac{1}{3}\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\textrm{Diketahi}\: \: a+b=1,\: a,b>0\\ &\textrm{akan ditunjukkan}\\\ &\displaystyle \frac{a^{2}}{a+1}+\displaystyle \frac{b^{2}}{b+1}\geq \displaystyle \frac{1}{3}\\ &\textrm{Perhatikan bahwa}\\ &\displaystyle \frac{a^{2}}{a+1}+\displaystyle \frac{b^{2}}{b+1}\\ &=\displaystyle \frac{a.a}{(1-b)+1}+\displaystyle \frac{b.b}{(1-a)+1}\\ &=\displaystyle \frac{a(1-b)}{2-b}+\frac{b(1-a)}{2-a}\\ &=\displaystyle \frac{a-ab}{2-b}+\frac{b-ab}{2-a}\\ &=\displaystyle \frac{(2-a)(a-ab)+(2-b)(b-ab)}{(2-a)(2-b)}\\ &=\displaystyle \frac{2a-2ab-a^{2}+a^{2}b+2b-2ab-b^{2}+ab^{2}}{4-2(a+b)+ab}\\ &=\displaystyle \frac{2(a+b)-4ab-(a^{2}+b^{2})+ab(a+b)}{4-2(a+b)+ab}\\ &=\displaystyle \frac{2(a+b)-4ab-(1-2ab)+ab(a+b)}{4-2(a+b)+ab}\\ &=\displaystyle \frac{2(1)-4ab-1+2ab+ab(1)}{4-2(1)+ab}\\ &=\displaystyle \frac{1-ab}{2+ab}\\ &\color{blue}\textrm{Selanjutnya dengan AM-GM kita peroleh}\\ &\begin{array}{|c|}\hline \begin{aligned}&a+b\geq 2\sqrt{ab}\\ &\Leftrightarrow 1\geq 2\sqrt{ab}\\ &\Leftrightarrow \displaystyle \frac{1}{2}\geq \sqrt{ab}\\ &\Leftrightarrow  \color{red}\displaystyle \frac{1}{4}\geq ab  \end{aligned}\\\hline \end{array}\\ &\textrm{Sehingga}\\ &\geq \displaystyle \frac{1-\displaystyle \frac{1}{4}}{2+\displaystyle \frac{1}{4}}\\ &\geq \displaystyle \frac{\displaystyle \frac{3}{4}}{\displaystyle \frac{9}{4}}\\ &\geq \displaystyle \frac{1}{3}\qquad \blacksquare  \end{aligned}  \end{array}$.

DAFTAR PUSTAKA

1. Bambang, S. 2012. Materi, Soal dan Penyelesaian Olimpiade Matematika Tingkat SMA/MA. Jakarta: BINA PRESTASI INSANI.
2. Susyanto, N. 2012. Tutor Senior Olimpiade Matematika Lima Benua Tingkat SMP. Yogyakarta: KENDI MAS MEDIA.
3. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.

Contoh Soal 4 (Segitiga dan Ketaksamaan)

 $\begin{array}{ll}\\ 16.&\textrm{Buktikan bahwa setiap bilangan real}\\ &\textrm{positif}\: \: a\: \: \textrm{dan}\: \: b\: \: \textrm{berlaku}\: \: a^{2}+b^{2}\geq 2ab\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\: \: (a-b)^{2}\geq 0\\ &\textrm{adalah benar, maka}\\ &(a-b)^{2}=a^{2}-2ab+b^{2}\geq 0\\ &\Leftrightarrow a^{2}+b^{2}\geq 2ab\qquad \blacksquare  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 17.&\textrm{Buktikan bahwa setiap bilangan real}\\ &\textrm{positif}\: \: a,\: b\: \: \textrm{dan}\: \: c\: \: \textrm{berlaku}\\ & a^{2}+b^{2}+c^{2}\geq ab+ac+bc\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\textrm{Perhatikan bahwa}\: \: (a-b)^{2}\geq 0\\  &(a-c)^{2}\geq 0,\: \: \textrm{dan}\: \: (b-c)^{2}\geq 0\\ &\textrm{adalah benar, maka}\\ &(a-b)^{2}=a^{2}-2ab+b^{2}\geq 0\\ &\Leftrightarrow a^{2}+b^{2}\geq 2ab\: .....(1)\\ &\textrm{Dengan cara yang kurang lebih sama}\\ &\textrm{akan didapatkan}\\ &\bullet \quad a^{2}+c^{2}\geq 2ac\: .....(2)\\ &\bullet \quad b^{2}+c^{2}\geq 2bc\: .....(1)\\ &\textrm{Jika ketaksamaan}\quad (1),(2), \& \: (3)\: \: \textrm{dijumlahkan}\\ &\textrm{akan didapatkan bentuk}\\ &2a^{2}+2b^{2}+2c^{2}\geq 2ab+2ac+2bc\\ &\Leftrightarrow \: a^{2}+b^{2}+c^{2}\geq ab+ac+bc\quad \blacksquare  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 18.&\textrm{Untuk}\: \: a,b,c\: \: \textrm{bilangan real, dengan}\\ &a^{2}+b^{2}+c^{2}=7\: \: \textrm{dan}\: \: ab+bc+ca=4\\  &\textrm{Tentukan nilai terbesar dari}\: \: ab\\\\  &\textbf{Jawab}:\\   &\begin{aligned}&\textrm{Pada}\: \: \color{red}\textrm{no. 17}\: \:  \color{black}\textrm{di atas telah ditunjukkan}\\ & 2(a^{2}+b^{2}+c^{2})\geq 2ab+2bc+2ca\\  &\textrm{dari bukti di atas, kita mendapatkan}\\ &\textrm{bentuk}\: \:  2(a^{2}+b^{2}+c^{2})\geq 2ab+\color{blue}bc+ca\\ &\textrm{Sehingga}\\ &2(a^{2}+b^{2}+c^{2})\geq ab+ab+bc+ca\\ &\Leftrightarrow \: 2(7)\geq ab+(4)\\ &\Leftrightarrow \: 14-4\geq ab\\ &\Leftrightarrow \: \quad\quad 10\geq ab\\ &\textrm{Jadi, maksimum nilai}\: \: ab\: \: \textrm{adalah 10} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Diketahui}\: \: a,\: b,\: c\: \: \textrm{sisi segitiga}\\ &\textrm{Tunjukkan bahwa}\: \: \displaystyle \frac{a+b+c}{3}\geq \sqrt[3]{abc}\\\\ &\textbf{Bukti}\\  &\begin{aligned}a+b+c+\sqrt[3]{abc}&=(a+b)+\left (c+\sqrt[3]{abc}  \right )\\ &\geq 2\sqrt{ab}+2\sqrt{c\times \sqrt[3]{abc}}\\ &\geq 2\left ( 2\sqrt{\sqrt{ab}\times\sqrt{c\times \sqrt[3]{abc}}}  \right )\\ &\geq 4\sqrt{\sqrt{abc}\times\sqrt{ \sqrt[3]{abc}}}\\ &\geq 4\sqrt{(abc)^{.^{\frac{1}{2}}}\times (abc)^{.^{\frac{1}{6}}}}\\ &\geq 4\sqrt{(abc)^{.^{\frac{1}{2}+\frac{1}{6}}}}\\ &\geq 4\sqrt{(abc)^{.^{\frac{2}{3}}}}\\ &\geq 4\sqrt{(abc)^{.^{\frac{1}{3}}}}\\ &\geq 4\sqrt[3]{abc}\\ a+b+c&\geq 4\sqrt[3]{abc}-\sqrt[3]{abc}\\ &\geq 3\sqrt[3]{abc}\\ \displaystyle \frac{a+b+c}{3}&\geq \sqrt[3]{abc}\qquad \blacksquare   \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 20.&\textrm{Diketahui}\: \: a,\: b,\: c\: \: \textrm{sisi segitiga}\\ &\textrm{Tunjukkan bahwa}\: \: \displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\textrm{Perhatikan bahwa sebelumnya}\\ &\textrm{telah ditunjukkan untuk sembarang}\\ &a,b,c\: \: \textrm{real positif akan berlaku}\\ &\displaystyle \frac{a+b+c}{3}\geq \sqrt[3]{abc}\\ &\color{red}\textrm{lihat bukti yang ditunjukkan no.18}\\ &\color{red}\textrm{di atas}\color{black},\: \textrm{maka}\\ &\displaystyle \frac{\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{3}\geq \sqrt[3]{\displaystyle \frac{a}{b}\times \frac{b}{c}\times \frac{c}{a}}\\ &\Leftrightarrow \: \displaystyle \frac{\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{3}\geq 1\\ &\Leftrightarrow \: \displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3\qquad \blacksquare  \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Bambang, S. 2012. Materi, Soal dan Penyelesaian Olimpiade Matematika Tingkat SMA/MA. Jakarta: BINA PRESTASI INSANI.
2. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
3. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.

Contoh Soal 3 (Segitiga dan Trigonometri)

 

$\begin{array}{ll}\\ 11.&\textrm{Nilai dari}\\ &\quad\quad \sin \displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \displaystyle \frac{1}{16} &&&\textrm{d}.&\displaystyle \frac{1}{2}\\\\ \textrm{b}.&\displaystyle \color{red}\frac{1}{8}&\quad \textrm{c}.&\displaystyle \frac{1}{4}\quad &\textrm{e}.&\displaystyle 1 \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\textrm{Perhat}&\textrm{ikan bahwa}\\ \sin \displaystyle \displaystyle \frac{\pi }{14}&=\sin \left (\displaystyle \frac{7\pi }{14}-\frac{6\pi }{14} \right )=\sin \left ( \displaystyle \frac{1}{2}\pi -\frac{6\pi }{14} \right )\\ &=\cos \displaystyle \frac{6\pi }{14} \\ \sin \displaystyle \frac{3\pi }{14}&=...=\cos \displaystyle \frac{4\pi }{14}\\ \sin \displaystyle \frac{9\pi }{14}&=...=\sin \displaystyle \frac{5\pi }{14}=\cos \displaystyle \frac{2\pi }{14} \end{aligned}\\ &...\\ &\sin \displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}\\ &=\cos \displaystyle \frac{6\pi }{14}\cos \displaystyle \frac{4\pi }{14}\cos \displaystyle \frac{2\pi }{14}\times \displaystyle \frac{2\sin \displaystyle \frac{2\pi }{14}}{2\sin \displaystyle \frac{2\pi }{14}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{6\pi }{14}\cos \displaystyle \frac{4\pi }{14}\sin \displaystyle \frac{4\pi }{14}}{2\sin \displaystyle \frac{2\pi }{14}}\\ &\textrm{silahkan dilanjutkan}\\ &...\\ &=\displaystyle \frac{1}{8} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Nilai dari}\\ & \cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \displaystyle \frac{8\pi }{5}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \color{red}-\displaystyle \frac{1}{16} &&&\textrm{d}.&\displaystyle \frac{1}{16}\\\\ \textrm{b}.&\displaystyle \frac{1}{8}&\quad \textrm{c}.&\displaystyle 0\quad &\textrm{e}.&\displaystyle \frac{1}{8} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \displaystyle \frac{8\pi }{5}\\ &=\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \left (\pi +\displaystyle \frac{3\pi }{5} \right )\\ &=\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\left (-\cos \displaystyle \frac{3\pi }{5} \right )\\ &=-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \displaystyle \frac{3\pi }{5}\\ &=-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}\cos \displaystyle \frac{4\pi }{5}\\ &=-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}\cos \displaystyle \frac{4\pi }{5}\times \displaystyle \frac{2\sin \displaystyle \frac{\pi }{5}}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\frac{-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}\left ( \sin \pi -\sin \displaystyle \frac{3\pi }{5} \right )}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5} \sin \frac{3\pi }{5}}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \displaystyle \frac{3\pi }{5}\left ( \cos \displaystyle \frac{2\pi }{5}\sin \displaystyle \frac{3\pi }{5} \right )}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \displaystyle \frac{3\pi }{5}\left ( \sin \pi -\sin \left ( -\displaystyle \frac{\pi }{5} \right ) \right )}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \displaystyle \frac{3\pi }{5}\sin \displaystyle \frac{\pi }{5}}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\cos \displaystyle \frac{\pi }{5}\sin \displaystyle \frac{\pi }{5}}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\left (\cos \displaystyle \frac{\pi }{5}\sin \displaystyle \frac{\pi }{5} \right )}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\left (\sin \displaystyle \frac{2\pi }{5}-\sin 0 \right )}{8\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\sin \displaystyle \frac{2\pi }{5}}{8\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\sin \pi -\sin \displaystyle \frac{\pi }{5}}{16\sin \displaystyle \frac{\pi }{5}}\\ &=-\displaystyle \frac{\sin \displaystyle \frac{\pi }{5}}{16\sin \displaystyle \frac{\pi }{5}}\\ &=\color{red}-\displaystyle \frac{1}{16} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Nilai dari}\: \: \sin \displaystyle \frac{\pi }{24}.\sin \frac{5\pi }{24}.\sin \frac{7\pi }{24}.\sin \frac{11\pi }{24}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{5}{16}&&\textrm{d}.\quad \displaystyle \frac{2}{16}\\\\ \textrm{b}.\quad \displaystyle \frac{4}{16}&\textrm{c}.\quad \displaystyle \frac{3}{16}&\textrm{e}.\quad \color{red}\displaystyle \frac{1}{16} \end{array}\\\\ &\textbf{(Olimpiade Sains PORSEMA NU 2012)}\\\\ &\textbf{Jawab}:\: \: \color{red}\textbf{e}\\ &\begin{aligned}&\sin \displaystyle \frac{\pi }{24}.\sin \frac{5\pi }{24}.\sin \frac{7\pi }{24}.\sin \frac{11\pi }{24}\\ &=\displaystyle \frac{1}{4}\left ( 2\sin \displaystyle \frac{11\pi }{24}.\sin \frac{\pi }{24}.2\sin \frac{7\pi }{24}.\sin \frac{5\pi }{24} \right )\\ &=\displaystyle \frac{1}{4}\left [ \left ( \cos \left ( \frac{10\pi }{24} \right )-\cos \left ( \frac{12\pi }{24} \right ) \right )\times \left ( \cos \left ( \frac{2\pi }{24} \right )-\cos \left ( \frac{12\pi }{24} \right ) \right ) \right ]\\ &=\displaystyle \frac{1}{4}\left [ \left ( \cos 75^{\circ}-\cos 90^{\circ} \right )\times \left ( \cos 15^{\circ}-\cos 90^{\circ} \right ) \right ]\\ &=\displaystyle \frac{1}{4}\left [ \cos 75^{\circ}.\cos 15^{\circ} \right ]\\ &=\displaystyle \frac{1}{8}\left [ \cos 90^{\circ}+\cos 60^{\circ} \right ]\\ &=\displaystyle \frac{1}{8}\left ( 0+\frac{1}{2} \right )\\ &=\color{red}\displaystyle \frac{1}{16} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Nilai dari}\\ & \qquad\qquad\sin 18^{\circ}\cos 36^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{6} &&&\textrm{d}.&\displaystyle \frac{1}{3}\\\\ \textrm{b}.&\displaystyle \frac{1}{5}&\quad \textrm{c}.&\displaystyle \color{red}\displaystyle \frac{1}{4}\quad &\textrm{e}.&\displaystyle \frac{1}{2} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\sin 18^{\circ}\cos 36^{\circ}\\ &=\sin 18^{\circ}\cos 36^{\circ}\times \displaystyle \frac{2\cos 18^{\circ}}{2\cos 18^{\circ}}\\ &=\displaystyle \frac{\cos 36^{\circ}\left ( \sin 36^{\circ}+\sin 0^{\circ} \right )}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\cos 36^{\circ}\sin 36^{\circ}}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\sin 72^{\circ}}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\sin \left ( 90^{\circ}-18^{\circ} \right )}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\cos 18^{\circ}}{4\cos 18^{\circ}}\\ &=\color{red}\displaystyle \frac{1}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Nilai eksak dari}\: \: \sin 36^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \displaystyle \frac{1}{4}\sqrt{10+2\sqrt{5}}&&&\textrm{d}.&\displaystyle \frac{\sqrt{5}-1}{4}\\ \textrm{b}.&\color{red}\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}}&&\quad &\textrm{e}.&\displaystyle \frac{\sqrt{5}-1}{2}\\ \textrm{c}.&\displaystyle \displaystyle \frac{\sqrt{5}+1}{4} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\textrm{Perhatikanlah ilustrasi gambar berikut} \end{array}$.

$.\qquad\begin{aligned}&\textrm{Perhatikan bahwa}\: \: \color{red}\bigtriangleup ABC\: \: \color{black}\textrm{sama kaki}\\ &\textrm{dengan}\: \: AD=DC=CB=1,\: AC=x\\ &\textrm{Diketahui pula}\: \: CD\: \: \textrm{adalah garis bagi}\\ &\textrm{serta}\: \: ABC\: \: \textrm{sebangun}\: \: \bigtriangleup BCD\\ &\textrm{akibatnya}:\\ &\color{red}\textrm{perbandingan sisi yang bersesuaian}\\ &\color{red}\textrm{akan sama},\: \: \color{black}\textrm{maka}\\ &\displaystyle \frac{AB}{BC}=\displaystyle \frac{BC}{AB-AD}\\ &\Leftrightarrow \displaystyle \frac{x}{1}=\frac{1}{x-1}\\ &\Leftrightarrow x(x-1)=1\\ &\Leftrightarrow x^{2}-x-1=0\\ &\Leftrightarrow x=\displaystyle \frac{1\pm \sqrt{5}}{2}\\ &\textrm{akibatnya}\: \: AB=AC=\displaystyle \frac{1+\sqrt{5}}{2}\\ &\textrm{Selanjutnya gunakan}\: \: \color{blue}\textrm{aturan sinus}\\ &\displaystyle \frac{AB}{\sin \angle C}=\frac{BC}{\sin \angle A}\\ &\Leftrightarrow \displaystyle \frac{AB}{BC}=\frac{\sin \angle C}{\sin \angle A}\\ &\Leftrightarrow \displaystyle \frac{ \left (\displaystyle \frac{1+\sqrt{5} }{2} \right )}{1}=\frac{\sin 72^{\circ}}{\sin 36^{\circ}}\\ &\Leftrightarrow \displaystyle \frac{1+\sqrt{5}}{2}=\displaystyle \frac{2\sin 36^{\circ}\cos 36^{\circ}}{\sin 36^{\circ}}\\ &\Leftrightarrow \displaystyle \frac{1+\sqrt{5}}{2}=2\cos 36^{\circ}\\ &\Leftrightarrow \cos 36^{\circ}=\Leftrightarrow \displaystyle \frac{1+\sqrt{5}}{4}\\ &\textrm{Dari fakta di atas kita akan dengan}\\ &\textrm{mudah menentukan nilai sinusnya}\\ &\textrm{yaitu dengan menggunakan}\\ &\textrm{identitas trigonometri berikut}:\\ &\sin ^{2}36^{\circ}+\cos ^{2}36^{\circ}=1\\ &\Leftrightarrow \sin ^{2}36^{\circ}=1-\cos ^{2}36^{\circ}\\ &\Leftrightarrow \sin 36^{\circ}=\sqrt{1-\cos ^{2}36^{\circ}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\sqrt{1-\left ( \displaystyle \frac{1+\sqrt{5}}{4} \right )^{2}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\sqrt{1-\displaystyle \frac{6+2\sqrt{5}}{16}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\sqrt{\displaystyle \frac{10-2\sqrt{5}}{16}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\color{red}\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}} \end{aligned}$.

DAFTAR PUSTAKA

1. Budhi, W.S. 2014. Matematika 4: Bahan Ajar Persiapan Menuju Olimpiade Matematika Sain Nasional/Internasional SMA. Jakarta: TRISULA ADISAKTI.
2. Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.
3. Sukino. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Contoh Soal 2 (Segitiga dan Trigonometri)

$\begin{array}{ll}\\ 6.&\textrm{Bentuk}\: \: \displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}\\ & \textrm{senilai dengan}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\tan 6x &&&\textrm{d}.&6\cot x\\ \textrm{b}.&\displaystyle -\cot 6x&&&\textrm{e}.&\displaystyle \color{red}\tan 6x\\ \textrm{c}.&\displaystyle 6\tan x \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}\\ &=\displaystyle \frac{\cos 3x-\cos 9x-\sin 6x}{\sin 9x-\sin 3x-\cos 6x}\\ &=\displaystyle \frac{-2\sin 6x\sin (-3x)-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}\\ &=\displaystyle \frac{2\sin 6x\sin 3x-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}\\ &=\displaystyle \frac{\sin 6x(2\sin 3x-1)}{\cos 6x(2\sin 3x-1)}\\ &=\tan 6x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Nilai dari}\\ &\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \tan 2x &&\textrm{d}.&\displaystyle \tan 8x\\ \textrm{b}.&\color{red}\displaystyle \tan 4x&\quad&\textrm{e}.&\displaystyle \tan 16x\\ \textrm{c}.&\displaystyle \displaystyle \tan 6x \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}\\ &=\displaystyle \frac{\sin 7x+\sin x+\sin 5x+\sin 3x}{\cos 7x+\cos x+\cos 5x+\cos 3x}\\ &=\displaystyle \frac{2\sin 4x\cos 3x+2\sin 4x\cos x}{2\cos 4x\cos 3x+2\cos 4x\cos x}\\ &=\displaystyle \frac{2\sin 4x\left ( \cos 3x+\cos x \right )}{2\cos 4x\left ( \cos 3x+\cos x \right )}\\ &=\tan 4x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Nilai dari}\: \: \cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 1&&\textrm{d}.\quad -\displaystyle \frac{1}{2}\\\\ \textrm{b}.\quad -1&\textrm{c}.\quad \displaystyle \frac{1}{2}&\textrm{e}.\quad \color{red}0 \end{array}\\\\ &\textbf{Jawab}:\: \: \color{red}\textbf{e}\\ &\begin{aligned}&\cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}\\ &=2\cos \displaystyle \frac{1}{2}\left ( 80^{\circ}+40^{\circ} \right )\cos \displaystyle \frac{1}{2}\left ( 80^{\circ}-40^{\circ} \right )-\cos 20^{\circ}\\ &=2\cos 60^{\circ}\cos 20^{\circ}-\cos 20^{\circ}\\ &=2.\displaystyle \frac{1}{2}.\cos 20^{\circ}-\cos 20^{\circ}\\ &=\cos 20^{\circ}-\cos 20^{\circ}\\ &=\color{red}0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Nilai dari}\\ &\quad\quad \sqrt{3}\sin 80^{\circ}\sin 160^{\circ}\sin 320^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\displaystyle \frac{3}{8} &&&\textrm{d}.&\displaystyle \color{red}\frac{3}{8}\\\\ \textrm{b}.&\displaystyle -\frac{1}{8}&&&\textrm{e}.&\displaystyle \frac{5}{8}\\ \textrm{c}.&\displaystyle \frac{1}{8} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\sqrt{3}\sin 80^{\circ}\sin 160^{\circ}\sin 320^{\circ}\\ &=\sqrt{3}\sin 80^{\circ}\sin 20^{\circ}\left (-\sin 40^{\circ} \right )\\ &=-\sqrt{3}\sin 80^{\circ}\sin 40^{\circ}\sin 20^{\circ}\\ &=-\sqrt{3}\sin 80^{\circ}\left ( -\displaystyle \frac{1}{2}\left ( \cos 60^{\circ}-\cos 20^{\circ} \right ) \right )\\ &=-\sqrt{3}\sin 80^{\circ}\left ( -\displaystyle \frac{1}{4}+\displaystyle \frac{\cos 20^{\circ}}{2} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{2}\sqrt{3}\sin 80^{\circ}\cos 20^{\circ}\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{4}\sqrt{3}\left ( \sin 100^{\circ}+\sin 60^{\circ} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{4}\sqrt{3}\left ( \sin 80^{\circ}+\displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}+\displaystyle \frac{1}{8}\sqrt{9}\\ &=\displaystyle \frac{3}{8} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Nilai dari}\\ &\quad\quad \cos \displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \color{red}-\displaystyle \frac{1}{8} &&&\textrm{d}.&\displaystyle \frac{1}{2}\\\\ \textrm{b}.&\displaystyle -\frac{1}{4}&\quad \textrm{c}.&0\quad &\textrm{e}.&\displaystyle \frac{1}{3} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\textbf{Alternatif 1}\\ &\begin{aligned}&\cos \displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\times \displaystyle \frac{2\sin \displaystyle \frac{2\pi }{7}}{2\sin \displaystyle \frac{2\pi }{7}}\\ &=\left (\sin \displaystyle \frac{4\pi }{7}-\sin 0 \right )\frac{\displaystyle \cos \frac{\pi }{7}\cos \displaystyle \frac{4\pi }{7}}{2\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin \displaystyle \frac{4\pi }{7}\displaystyle \cos \frac{\pi }{7}\cos \displaystyle \frac{4\pi }{7}}{2\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\left ( \sin \displaystyle \frac{5\pi }{7}+\sin \displaystyle \frac{3\pi }{7} \right )\cos \displaystyle \frac{4\pi }{7}}{4\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin \displaystyle \frac{5\pi }{7}\cos \displaystyle \frac{4\pi }{7}+\sin \displaystyle \frac{3\pi }{7}\cos \displaystyle \frac{4\pi }{7}}{4\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin \displaystyle \frac{9\pi }{7}+\sin \displaystyle \frac{\pi }{7}+\sin \displaystyle \frac{7\pi }{7}+\sin \left (-\displaystyle \frac{\pi }{7} \right )}{8\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{-\sin \displaystyle \frac{2\pi }{7}+\sin \displaystyle \frac{\pi }{7}+0-\sin \displaystyle \frac{\pi }{7}}{8\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{-\sin \displaystyle \frac{2\pi }{7}}{8\sin \displaystyle \frac{2\pi }{7}}\\ &=-\displaystyle \frac{1}{8} \end{aligned}\\ &\textbf{Alternatif 2}\\ &\begin{aligned}&\cos \displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\\ &=\cos \displaystyle \frac{4\pi }{7}\cos \frac{2\pi }{7}\cos \frac{\pi }{7}\\ &=\displaystyle \frac{1}{2}\left ( \cos \displaystyle \frac{6\pi }{7}+\cos \displaystyle \frac{2\pi }{7} \right )\cos \displaystyle \frac{\pi }{7}\\ &=\displaystyle \frac{1}{2}\left ( \cos \left ( \pi -\displaystyle \frac{\pi }{7} \right )+\cos \displaystyle \frac{2\pi }{7} \right )\cos \displaystyle \frac{\pi }{7}\\ &=\displaystyle \frac{1}{2}\left ( -\cos \displaystyle \frac{\pi }{7}+\cos \displaystyle \frac{2\pi }{7} \right )\cos \displaystyle \frac{\pi }{7}\\ &= \displaystyle \frac{1}{2}\left (-\cos ^{2}\displaystyle \frac{\pi }{7}+\cos \displaystyle \frac{2\pi }{7}\cos \displaystyle \frac{\pi }{7} \right )\\ &=\displaystyle \frac{1}{4}\left ( -\cos \displaystyle \frac{2\pi }{7}-\cos 0+\cos \displaystyle \frac{3\pi }{7}+\cos \displaystyle \frac{\pi }{7} \right )\\ &=\displaystyle \frac{1}{4}\left ( -\cos 0+\color{red}\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7} \color{black}\right )\\ &=\displaystyle \frac{1}{4}\left ( -1+\color{red}\displaystyle \frac{1}{2}\color{black} \right )\\ &=\displaystyle \frac{1}{4}\times \left (-\frac{1}{2} \right )\\ &=-\displaystyle \frac{1}{8} \end{aligned} \end{array}$.

Berikut penjelasan untuk  $\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7}=\color{red}\displaystyle \frac{1}{2}$.

$\begin{aligned}&\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7}\\ &=\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7}\times \displaystyle \frac{\left (2\sin\displaystyle \frac{2\pi }{7} \right ) }{\left (2\sin\displaystyle \frac{2\pi }{7} \right )}\\ &=\displaystyle \frac{2\cos\displaystyle \frac{\pi }{7}\sin\displaystyle \frac{2\pi }{7}-2\cos\displaystyle \frac{2\pi }{7}\sin\displaystyle \frac{2\pi }{7}+2\cos\displaystyle \frac{3\pi }{7}\sin\displaystyle \frac{2\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{3\pi }{7}-\sin\left (-\displaystyle \frac{\pi }{7} \right )-\left ( \sin\displaystyle \frac{4\pi }{7}-\sin\displaystyle \frac{0\pi }{7} \right )+\sin\displaystyle \frac{5\pi }{7}-\sin\displaystyle \frac{\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{3\pi }{7}+\sin\displaystyle \frac{\pi }{7}-\sin\displaystyle \frac{4\pi }{7}+\sin\displaystyle \frac{5\pi }{7}-\sin\displaystyle \frac{\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{3\pi }{7}-\sin\displaystyle \frac{4\pi }{7}+\sin\displaystyle \frac{5\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\left (\pi -\displaystyle \frac{4\pi }{7} \right )-\sin\displaystyle \frac{4\pi }{7}+\sin\left (\pi -\displaystyle \frac{2\pi }{7} \right )}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{4\pi }{7}-\sin\displaystyle \frac{4\pi }{7}+\sin\displaystyle \frac{2\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{2\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{1}{2}\qquad \blacksquare \end{aligned}$.