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