Contoh Soal 13 (Segitiga dan Ketaksamaan)

 $.\begin{array}{rl}& \text{Mengenal penulisan pola}\mathbf{\text{Siklik dan Simetri}}\\ & \text{Misal untuk}n=3,\text{pada penulisan unsur}\\ & x,y,\text{dan}z,\text{maka}\\ & \begin{array}{|ll|}\hline \mathbf{\text{Pola Siklik}}& \mathbf{\text{Pola Simetri}}\\ \begin{array}{r}{\displaystyle \sum _{\text{siklik}}^{.}{x}^2=x^2+y^2+z^2}\\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}{\displaystyle \sum _{\text{sym}}^{.}{x}^2}& =x^2+x^2\\ & +y^2+y^2\\ \\ & +z^2+z^2\\ \\ & =2(x^2+y^2+z^2)\end{array}\\ \begin{array}{r}{\displaystyle \sum _{\text{siklik}}^{.}{x}^3=x^3+y^3+z^3}\end{array}& \begin{array}{r}{\displaystyle \sum _{\text{sym}}^{.}{x}^3=2(x^3+y^3+z^3)}\end{array}\\ \begin{array}{r}{\displaystyle \sum _{\text{siklik}}^{.}{x}^{2}y=x^{2}y+y^{2}z+z^{2}x}\\ & \\ & \\ & \end{array}& \begin{array}{rl}{\displaystyle \sum _{\text{sym}}^{.}{x}^{2}y}& =x^{2}y+x^{2}z\\ & +y^{2}x+y^{2}z\\ \\ & +z^{2}x+z^{2}y\end{array}\\ \begin{array}{rl}{\displaystyle \sum _{\text{siklik}}^{.}xyz}& =xyz+yzx+zxy\\ & =3xyz\end{array}& \begin{array}{rl}{\displaystyle \sum _{\text{sym}}^{.}xyz}& =xyz+xzy+\cdots \\ & =6xyz\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 61.& (\mathbf{\text{IMO 1995}})\\ & \text{Jika}a,b,c\text{bilangan-bilangan real positif}\\ & \text{dengan}abc=1,\text{maka tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\ge {\displaystyle \frac{3}{2}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Misalkan}x={\displaystyle \frac{1}{a},y={\displaystyle \frac{1}{b},\text{dan}z={\displaystyle \frac{1}{c},}}}\\ & \text{maka}\\ & {\displaystyle \frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}}\\ & ={\displaystyle \frac{x^{3}yz}{y+z}+\frac{y^{3}xz}{x+z}+\frac{z^{3}xy}{x+y},\text{karena}xyz=1}\\ & ={\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\\ & \text{Dengan ketaksamaan}\mathbf{\text{Cauchy-Schwarz}}\\ & \left(2{\displaystyle \sum _{\text{siklik}}^{.}y+z}\right)\left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge (x+y+z{)}^2\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{(x+y+z{)}^2}{2(x+y+z)}}\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{(a+b+c)}{2}}\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{3\sqrt[3]{xyz}}{2}}\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{3}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 62.& \text{Diketahui}a,b\text{bilangan real positif}\\ & \text{Tunjukkan bahwa}{\displaystyle \frac{a^2}{b}+\frac{b^2}{a}\ge a+b}\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Asumsikan bahwa}a\ge b,\text{maka}{a}^2\ge b^2\\ & \text{dan}{\displaystyle \frac{1}{b}\ge \frac{1}{a}.}\\ & \text{Perhatikan bahwa baik}(a^2,b^2)\text{dan}\left({\displaystyle \frac{1}{b},\frac{1}{a}}\right)\\ & \text{adalah kumpulan dua barisan yang monoton}\\ & \text{sama yaitu sama-sama naik. Sehingga}\\ & \text{dengan}\mathbf{\text{ketaksamaan Renata}}\text{diperoleh}\\ & a^2.{\displaystyle \frac{1}{b}+b^2.{\displaystyle \frac{1}{a}\ge a^2.{\displaystyle \frac{1}{a}+b^2.{\displaystyle \frac{1}{b}}}}}\\ & \iff {\displaystyle \frac{a^2}{b}+\frac{b^2}{a}\ge a+b◼}\end{array}$.

$\begin{array}{l}\\ 63.& \text{Diberikan}a,b,c>0,\text{tunjukkan bahwa}\\ & {\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Alternatif 1}\\ & \text{Dengan AM-GM diperoleh}\\ & \bullet {\displaystyle \frac{a}{c}+\frac{c}{a}\ge 2\iff {\displaystyle \frac{c}{a}\ge 2-{\displaystyle \frac{a}{c}}}}\\ & \bullet {\displaystyle \frac{b}{c}+\frac{c}{b}\ge 2\iff {\displaystyle \frac{b}{c}\ge 2-{\displaystyle \frac{c}{b}}}}\\ & \bullet {\displaystyle \frac{a}{b}+\frac{b}{a}\ge 2\iff {\displaystyle \frac{a}{b}\ge 2-{\displaystyle \frac{b}{a}}}}\\ & \text{Sehingga}\\ & \begin{array}{rl}{\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}}& \ge a(2-{\displaystyle \frac{c}{b}})+b(2-{\displaystyle \frac{a}{c}})+c(2-{\displaystyle \frac{b}{a}})\\ & =2a-{\displaystyle \frac{ac}{b}+2b-{\displaystyle \frac{ab}{c}+2c-{\displaystyle \frac{bc}{a}}}}\\ {\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}}& \ge 2(a+b+c)-\left({\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}}\right)\\ 2& \left({\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}}\right)\ge 2(a+b+c)\\ {\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}}& \ge a+b+c◼\end{array}\\ & \text{Alternatif 2}\\ & \text{Dengan ketaksamaan Renata}\\ & \begin{array}{rl}& \text{Asumsikan}a\ge b\ge c,\text{maka}ab\ge ca\ge bc\\ & \text{dan}{\displaystyle \frac{1}{c}\ge \frac{1}{b}\ge \frac{1}{a}.}\\ & \text{Perhatikan bahwa}\\ & (ab\ge ca\ge bc)\text{dan}\left({\displaystyle \frac{1}{c}\ge \frac{1}{b}\ge \frac{1}{a}}\right)\\ & \text{memiliki kemonotonan yang sama}\\ & \text{maka dengan}\mathbf{\text{ketaksamaan Renata}}\\ & \text{dapat diperoleh bentuk}\\ & ab.{\displaystyle \frac{1}{c}+ac.{\displaystyle \frac{1}{b}+bc.{\displaystyle \frac{1}{a}\ge ab.{\displaystyle \frac{1}{b}+ac.{\displaystyle \frac{1}{a}+bc.{\displaystyle \frac{1}{c}}}}}}}\\ & \iff {\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+c+b}\\ & \iff {\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 64.& \text{Diberikan}a,b,c>0,\text{tunjukkan kebenaran}\\ & \mathbf{\text{ketaksamaan Nesbitt}}\text{berikut}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Alternatif 1}\\ & \text{Dengan AM-GM diperoleh}\\ & {\displaystyle \frac{(a+b)+(b+c)+(c+a)}{3}\ge {\displaystyle \frac{3}{{\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}}}}\\ & \iff ((a+b)+(b+c)+(c+a))\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge 9\\ & \iff 2\left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)+6\ge 9\\ & \iff 2\left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge 3\\ & \iff \left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge {\displaystyle \frac{3}{2}◼}\\ & \text{Alternatif 2}\\ & \text{Dengan ketaksamaan Renata}\\ & \begin{array}{rl}& \text{Asumsikan}a\le b\le c,\text{maka}a+b\le a+c\le b+c\\ & \text{dan}{\displaystyle \frac{1}{b+c}\le \frac{1}{a+c}\le \frac{1}{a+b}.}\\ & \text{Perhatikan bahwa}\\ & (a\le b\le c)\text{dan}{\displaystyle \frac{1}{b+c}\le \frac{1}{a+c}\le \frac{1}{a+b}}\\ & \text{memiliki kemonotonan yang sama}\\ & \text{maka dengan}\mathbf{\text{ketaksamaan Renata}}\\ & \text{dapat diperoleh bentuk}\\ & a.{\displaystyle \frac{1}{b+c}+b.{\displaystyle \frac{1}{a+c}+c.{\displaystyle \frac{1}{a+b}\ge b.{\displaystyle \frac{1}{b+c}+c.{\displaystyle \frac{1}{a+c}+a.{\displaystyle \frac{1}{a+b}}}}}}}\\ & \text{dan}\\ & a.{\displaystyle \frac{1}{b+c}+b.{\displaystyle \frac{1}{a+c}+c.{\displaystyle \frac{1}{a+b}\ge c.{\displaystyle \frac{1}{b+c}+a.{\displaystyle \frac{1}{a+c}+b.{\displaystyle \frac{1}{a+b}}}}}}}\\ & \text{Jika dijumlahkan keduanya, maka}\\ & 2\left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge 3\\ & \iff \left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge {\displaystyle \frac{3}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 65.& (\mathbf{\text{OSN 2015}})\\ & \text{Diberikan}a,b,c>0,\text{Buktikan bahwa}\\ & \sqrt{{\displaystyle \frac{a}{b+c}+\frac{b}{c+a}}}+\sqrt{{\displaystyle \frac{b}{c+a}+\frac{c}{a+b}}}+\sqrt{{\displaystyle \frac{c}{a+b}+\frac{a}{b+c}}}\ge 3\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Perhatikan bukti soal no. 4 di atas}\\ & \text{Dengan}\mathbf{\text{keksamaan Renata}}\text{dapat diperoleh}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{a+c}\ge \frac{b}{b+c}+\frac{a}{a+c}}\\ & \text{Misalkan}\\ & x=b+c,y=c+a,y=a+b,\text{maka}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{a+c}\ge \frac{b}{b+c}+\frac{a}{a+c}}\\ & \iff {\displaystyle \frac{a}{b+c}+\frac{b}{a+c}\ge \frac{y+z-x}{2x}+\frac{x+z-y}{2y}}\\ & \begin{array}{rl}& \iff \sqrt{{\displaystyle \frac{a}{b+c}+\frac{b}{a+c}}}\ge \sqrt{\frac{y+z-x}{2x}+\frac{x+z-y}{2y}}\\ & \iff =\sqrt{{\displaystyle \frac{1}{2}}}\sqrt{{\displaystyle \frac{y}{x}+\frac{z}{x}-1+\frac{x}{y}+\frac{z}{y}-1}}\\ & \iff \text{dengan AM-GM}\\ & \iff \ge {\displaystyle \frac{1}{\sqrt{2}}\sqrt{{\displaystyle \frac{z}{x}+\frac{z}{y}+2\sqrt{\frac{y}{x}.\frac{x}{y}}-2}}}\\ & \iff ={\displaystyle \frac{1}{\sqrt{2}}\sqrt{{\displaystyle \frac{z}{x}+\frac{z}{y}+2-2}}}\\ & \iff \ge {\displaystyle \frac{1}{\sqrt{2}}\sqrt{{\displaystyle \frac{z}{x}+\frac{z}{y}}}}\\ & \iff \ge {\displaystyle \frac{\sqrt{2}}{\sqrt{2}}\sqrt{{\displaystyle \frac{z^2}{xy}}}={\displaystyle \sqrt{{\displaystyle \frac{z^2}{xy}}}}}\end{array}\\ & \begin{array}{rl}& \bullet \sqrt{{\displaystyle \frac{a}{b+c}+\frac{b}{a+c}}}\ge {\displaystyle \sqrt{{\displaystyle \frac{z^2}{xy}}}}\\ & \bullet \sqrt{{\displaystyle \frac{b}{c+a}+\frac{c}{a+b}}}\ge \sqrt{{\displaystyle \frac{x^2}{yz}}}\\ & \bullet \sqrt{{\displaystyle \frac{c}{a+b}+\frac{a}{b+c}}}\ge \sqrt{{\displaystyle \frac{y^2}{xz}}}\end{array}\\ & \begin{array}{rl}& \text{Selanjutnya}\\ & \sqrt{{\displaystyle \frac{a}{b+c}+\frac{b}{c+a}}}+\sqrt{{\displaystyle \frac{b}{c+a}+\frac{c}{a+b}}}+\sqrt{{\displaystyle \frac{c}{a+b}+\frac{a}{b+c}}}\\ & \ge {\displaystyle \sqrt{{\displaystyle \frac{z^2}{xy}}}+\sqrt{{\displaystyle \frac{x^2}{yz}}}+\sqrt{{\displaystyle \frac{y^2}{xz}}}}\\ & \text{Dengan AM-GM lagi}\\ & \ge 3\sqrt[3]{\sqrt{{\displaystyle \frac{z^2}{xy}}}\times \sqrt{{\displaystyle \frac{x^2}{yz}}}\times \sqrt{{\displaystyle \frac{y^2}{xz}}}}\\ & \ge 3\sqrt[3]{\sqrt{{\displaystyle \frac{(xyz{)}^2}{(xyz{)}^2}}}}\\ & \ge 3◼\end{array}\end{array}$

DAFTAR PUSTAKA

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

WEBSITE

1. https://holdenlee.github.io/high_school/omc/23-rearrange.pdf diakses 18 Januari 2022.
2. https://www.gotohaggstrom.com/Advanced%20inequality%20manipulations.pdf  diakses 20 Januari 2022

Contoh Soal 12 (Segitiga dan Ketaksamaan)

$\begin{array}{l}\\ 56.& (\mathbf{\text{OSK 2018}})\\ & \text{Diketahui bilangan real}x\text{dan}y\\ & \text{yang memenuhi}{\displaystyle \frac{1}{2}<\frac{x}{y}<2}\\ & \text{Nilai minimum}{\displaystyle \frac{x}{2y-x}+\frac{2y}{2x-y}\text{adalah}....}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & \text{Misal}t={\displaystyle \frac{x}{y}}\\ & \text{Misalkan juga}f(t)={\displaystyle \frac{x}{2y-x}+{\displaystyle \frac{2y}{2x-y}}}\\ & \text{maka}f(t)={\displaystyle \frac{2t^2-3t+4}{-2t^2+5t-2}}\\ & \bullet \text{Agar minimum, maka}{f}^{\prime }(t)=0\\ & \text{Sehingga}\\ & f^{\prime }(t)=4t^2+8t-14=0\\ & \iff t_{1,2}=-1\pm {\displaystyle \frac{3}{2}\sqrt{2}}\\ & \text{Pilih yang positif, yaitu}t=-1+{\displaystyle \frac{3}{2}\sqrt{2}}\\ & \bullet \text{Dengan proses substistusi harga}t\\ & \text{di atas, maka akan didapatkan }\\ & \text{nilai}f(t)=1+{\displaystyle \frac{4}{3}\sqrt{2}}\end{array}\\ & \begin{array}{rl}& \text{Alternatif 2}\\ & \text{Menurut bentuk}{\displaystyle \frac{1}{2}<\frac{x}{y}<2}\\ & \text{jelas bahwa baik}2y-x\text{dan}2x-y\\ & \text{keduanya}\mathbf{\text{positif}}\\ & \text{Lihat tabel berikut}\\ & \begin{array}{|ccc|}\hline \text{Bentuk}& \text{Pengecekan 1}& \text{Pengecekan 2}\\ \begin{array}{rl}& {\displaystyle \frac{1}{2}<\frac{x}{y}<2}\\ & \text{Jelas bahwa}\\ & x,y\ne 0\\ & \\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}& \text{Saat}(\times y)\\ & \text{Yaitu}:\\ & {\displaystyle \frac{1}{2}(y)<\frac{x}{y}(y)<2(y)}\\ & \iff {\displaystyle \frac{1}{2}y<x<2y}\\ & \text{Jelas bahwa}\\ & 2y-x>0\\ & \\ & \\ & \end{array}& \begin{array}{rl}& \text{Saat dibali posisinya}\\ & {\displaystyle \frac{1}{2}<\frac{y}{x}<2}\\ & \text{Saat}(\times x)\\ & \text{Yaitu}:\\ & {\displaystyle \frac{1}{2}(x)<\frac{y}{x}(x)<2(x)}\\ & \iff {\displaystyle \frac{1}{2}x<y<2x}\\ & \text{Jelas bahwa}\\ & 2x-y>0\end{array}\\ \hline\end{array}\\ & \text{Saat masing-masing}\\ & \bullet {\displaystyle \frac{x}{2y-x}={\displaystyle \frac{1}{3}+\frac{2}{3}\left({\displaystyle \frac{2x-y}{2y-x}}\right)\text{dan}}}\\ & \bullet {\displaystyle \frac{2y}{2x-y}={\displaystyle \frac{2}{3}+\frac{4}{3}\left({\displaystyle \frac{2y-x}{2x-y}}\right)}}\\ & \text{Dengan ketaksamaan AM-GM diperoleh}\\ & \begin{array}{rl}{\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)\\ & \ge 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{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 57.& \text{Diketahui}a,b\text{bilangan real positif}\\ & \text{dan}a+b=1.\text{Tunjukkan bahwa}\\ & {(a+{\displaystyle \frac{1}{a}})}^2+{(b+{\displaystyle \frac{1}{b}})}^2\ge {\displaystyle \frac{25}{2}}\\ \\ & \mathbf{\text{Bukti}}\end{array}$.

$\begin{array}{rl}& \text{Dengan AM-GM kita akan mendapatkan}\\ & 1=a+b\ge 2\sqrt{ab}\iff {\displaystyle \frac{1}{2}\ge \sqrt{ab}\iff 2\le {\displaystyle \frac{1}{\sqrt{ab}}}}\\ & \iff {\displaystyle \frac{1}{\sqrt{ab}}\ge 2\iff {\displaystyle \frac{1}{ab}\ge 4}}\\ & \text{Perhatikan soal, dengan CS-Engel}\text{akan}\\ & \text{didapatkan}\\ & \begin{array}{rl}& (1+1)\left({\displaystyle \frac{{(a+\frac{1}{a})}^2}{1}+{\displaystyle \frac{{(b+\frac{1}{b})}^2}{1}}}\right)\ge {(a+\frac{1}{a}+b+\frac{1}{b})}^2\\ & \iff 2({\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2)\ge {(a+b+\frac{1}{a}+\frac{1}{b})}^2\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{a+b}{ab}})}^2}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{1}{ab}})}^2}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}(1+4{)}^2}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}\times 25}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{25}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 58.& \text{Jika}a,b\text{bilangan positif, buktikan}\\ & \text{a}.2(a^2+b^2)\ge (a+b{)}^2\\ & \text{b}.4(a^3+b^3)\ge (a+b{)}^3\\ & \text{c}.8(a^4+b^4)\ge (a+b{)}^4\\ & \text{d}.16(a^5+b^5)\ge (a+b{)}^5\\ & \text{e}.32(a^6+b^6)\ge (a+b{)}^6\\ & \text{f}.64(a^7+b^7)\ge (a+b{)}^7\\ & \text{g}.128(a^8+b^8)\ge (a+b{)}^8\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Akan ditunjukkan bukti poin 6.c saja}\\ & \text{untuk poin yang lain, silahkan pembaca}\\ & \text{sekalian untuk dibuktikan sendiri sebagai}\\ & \text{bahan latihan mandiri}.\\ & \text{Adapun bukti poin 6.c adalah sebagaimana}\\ & \text{berikut ini}\\ & \begin{array}{rl}& \text{Dengan ketaksamaan CS-Engel}\text{akan}\\ & \text{didapatkan}\\ & \bullet (1+1)(a^4+b^4)\ge (a^2+b^2{)}^2\\ & \iff 2(a^4+b^4)\ge (a^2+b^2{)}^2..........(1)\\ & \bullet (1+1)(a^2+b^2)\ge (a+b{)}^2\\ & \iff 2(a^2+b^2)\ge {\displaystyle (a+b{)}^2,(\text{kuadratkan})}\\ & \iff 4{\displaystyle {(a^2+b^2)}^2\ge (a+b{)}^4}\\ & \iff {(a^2+b^2)}^2\ge {\displaystyle \frac{(a+b{)}^4}{4}...........(2)}\\ & \text{Dari (1) dan (2) didapatkan hubungan}\\ & 2(a^4+b^4)\ge {(a^2+b^2)}^2\ge {\displaystyle \frac{(a+b{)}^4}{4}}\\ & \iff 8(a^4+b^4)\ge 4{(a^2+b^2)}^2\ge (a+b{)}^4\\ & \iff 8(a^4+b^4)\ge (a+b{)}^4◼\end{array}\end{array}$.

$\begin{array}{l}\\ 59.& \text{Jika}x,y,z\text{adalah bilangan real positif}\\ & \text{dengan}{x}^2+y^2+z^2=27.\text{Tunjukkan}\\ & \text{bahwa}{x}^3+y^3+z^3\ge 81\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Pada contoh soal no.2 terdapat}\\ & (ax+by+cz{)}^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)\\ & \text{ganti mengganti}a=b=c=1,\text{maka menjadi}\\ & (x+y+z{)}^2\le (1^2+1^2+1^2)(x^2+y^2+z^2)\\ & \iff (x+y+z{)}^2\le (3)(x^2+y^2+z^2)........(1)\\ & \text{Selanjutnya dengan mengganti dengan}\\ & x^{{.}^{\frac{3}{2}}},y^{{.}^{\frac{3}{2}}},z^{{.}^{\frac{3}{2}}}\text{dan}{x}^{{.}^{\frac{1}{2}}},y^{{.}^{\frac{1}{2}}},z^{{.}^{\frac{1}{2}}},\text{pada}\\ & (ax+by+cz{)}^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)\\ & \text{Kita akan dapatkan}\\ & (x^2+y^2+z^2)\le (x^3+y^3+z^3)(x+y+z)........(2)\\ & \text{Jika masing-masing ruas dikuadratkan, maka}\\ & (x^2+y^2+z^2{)}^4\le (x^3+y^3+z^3{)}^2(x+y+z{)}^2\\ & \iff (x^2+y^2+z^2{)}^4\le 3(x^3+y^3+z^3{)}^2(x^2+y^2+z^2)\\ & \iff (x^2+y^2+z^2{)}^3\le 3(x^3+y^3+z^3{)}^2\\ & \iff (27{)}^3\le 3(x^3+y^3+z^3{)}^2\\ & \iff 3^9\le 3(x^3+y^3+z^3{)}^2\\ & \iff 3^8\le (x^3+y^3+z^3{)}^2\\ & \iff (x^3+y^3+z^3{)}^2\ge 3^8\\ & \iff (x^3+y^3+z^3)\ge 3^{{.}^{\frac{8}{2}}}\\ & \iff (x^3+y^3+z^3)\ge 3^4=81◼\end{array}$.

$\begin{array}{l}\\ 60.& \text{Untuk}a,b,c,d\text{adalah bilangan real }\\ & \text{positif, tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\ge {\displaystyle \frac{64}{a+b+c+d}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Dengan ketaksamaan CS-Engel}\\ & (a+b+c+d)\left({\displaystyle \frac{1^2}{a}+{\displaystyle \frac{1^2}{b}+{\displaystyle \frac{2^2}{c}+{\displaystyle \frac{4^2}{d}}}}}\right)\ge (1+1+2+4{)}^2\\ & \iff {\displaystyle \frac{1^2}{a}+{\displaystyle \frac{1^2}{b}+{\displaystyle \frac{2^2}{c}+{\displaystyle \frac{4^2}{d}\ge {\displaystyle \frac{(1+1+2+4{)}^2}{a+b+c+d}}}}}}\\ & \iff {\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\ge {\displaystyle \frac{64}{a+b+c+d}◼}}\end{array}\end{array}$.

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}{l}\\ 51.& \text{Diketahui rata-rata bilangan positif}\\ & m_1,m_2,...,m_k\text{adalah}A\\ & \text{Buktikan bahwa}\\ & {(m_1+{\displaystyle \frac{1}{m_1}})}^2+{(m_2+{\displaystyle \frac{1}{m_2}})}^2+...+{(m_k+{\displaystyle \frac{1}{m_k}})}^2\ge k{(A+{\displaystyle \frac{1}{A}})}^2\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & {\displaystyle \frac{m_1+m_2+m_3+...+m_k}{k}=A}\\ & \text{Dengan ketaksamaan QM-AM akan diperoleh}\\ & \sqrt{{\displaystyle \frac{(m_1{)}^2+(m_2{)}^2+...+(m_k{)}^2}{k}}}\ge {\displaystyle \frac{m_1+m_2+...+m_k}{k}}\\ & \iff \sqrt{{\displaystyle \frac{(m_1{)}^2+(m_2{)}^2+...+(m_k{)}^2}{k}}}\ge A\\ & \iff {\displaystyle \frac{(m_1{)}^2+(m_2{)}^2+...+(m_k{)}^2}{k}\ge A^2}\\ & \iff (m_1{)}^2+(m_2{)}^2+...+(m_k{)}^2\ge k{A}^2\\ & \iff {\displaystyle \sum _{i=1}^k(m_i{)}^2\ge k{A}^2........(1)}\end{array}\\ & \begin{array}{rl}& \text{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}}}\ge {\displaystyle \frac{k}{m_1+m_2+...+m_k}}\\ & \iff \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}}}\ge {\displaystyle \frac{1}{A}}\\ & \iff {\displaystyle \frac{{\left(\frac{1}{m_1}\right)}^2+{\left(\frac{1}{m_2}\right)}^2+...+{\left(\frac{1}{m_k}\right)}^2}{k}\ge {\displaystyle \frac{1}{A^2}}}\\ & \iff {\left(\frac{1}{m_1}\right)}^2+{\left(\frac{1}{m_2}\right)}^2+...+{\left(\frac{1}{m_k}\right)}^2\ge {\displaystyle \frac{k}{A^2}}\\ & \iff \sum _{i=1}^k{\left({\displaystyle \frac{1}{m_i}}\right)}^2\ge {\displaystyle \frac{k}{A^2}........(2)}\end{array}\\ & \begin{array}{rl}& \text{Dengan (1) dan (2) akan diperoleh}\\ & {\displaystyle \sum _{i=1}^k{(m_i+{\displaystyle \frac{1}{m_i}})}^2={\displaystyle \sum _{i=1}^k((m_i{)}^2+2+{\left({\displaystyle \frac{1}{m_i}}\right)}^2)}}\\ & ={\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}}}\\ & \ge k{A}^2+2k+{\displaystyle \frac{k}{A^2}}\\ & =k(A^2+2+{\displaystyle \frac{1}{A^2}})\\ & =k{(A+{\displaystyle \frac{1}{A}})}^2◼\end{array}\end{array}$.

$\begin{array}{l}\\ 52.& \text{Buktikan bahwa untuk bilangan asli}n>1\\ & \text{berlaku}\sqrt[n]{1+{\displaystyle \frac{\sqrt[n]{n}}{n}}}+\sqrt[n]{1-{\displaystyle \frac{\sqrt[n]{n}}{n}}}<2\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa untuk}\sqrt[n]{1-{\displaystyle \frac{\sqrt[n]{n}}{n}}}\\ & \text{dengan}n>1,\text{maka}0<{\displaystyle \frac{\sqrt[n]{n}}{n}<1}\\ & \text{Dengan ketaksamaan AM-GM dapat diperoleh}\\ & \begin{array}{rl}{\displaystyle \frac{(1+{\displaystyle \frac{\sqrt[n]{n}}{n}})+\underset{\text{sebanyak}(n-1)}{\underbrace{1+1+1+...+1}}}{n}}& >\sqrt[n]{(1+{\displaystyle \frac{\sqrt[n]{n}}{n}})111...1}\\ & =\sqrt[n]{(1+{\displaystyle \frac{\sqrt[n]{n}}{n}})}......(1)\\ {\displaystyle \frac{(1-{\displaystyle \frac{\sqrt[n]{n}}{n}})+\underset{\text{sebanyak}(n-1)}{\underbrace{1+1+1+...+1}}}{n}}& >\sqrt[n]{(1-{\displaystyle \frac{\sqrt[n]{n}}{n}})111...1}\\ & =\sqrt[n]{(1-{\displaystyle \frac{\sqrt[n]{n}}{n}})}......(2)\end{array}\\ & \text{Jika ketaksamaan (1) dan (2) dijumlahkna, maka}\\ & \begin{array}{rl}{\displaystyle \sqrt[n]{1+{\displaystyle \frac{\sqrt[n]{n}}{n}}}+{\displaystyle \sqrt[n]{1-{\displaystyle \frac{\sqrt[n]{n}}{n}}}}}& <{\displaystyle \frac{\underset{\text{sebanyak}(2n-2)}{\underbrace{1+1+1+...+1}}+(1+{\displaystyle \frac{\sqrt[n]{n}}{n}})+(1-{\displaystyle \frac{\sqrt[n]{n}}{n}})}{n}}\\ & ={\displaystyle \frac{\underset{\text{sebanyak}(2n-2)}{\underbrace{1+1+1+...+1}}+1+1}{n}}\\ & =2◼\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 53.& \text{Jika bilangan real positif}x,y,z\text{dengan}\\ & xyz=1.\text{tentukan nilai minimum dari}\\ & 2x^3+12y^2+24z\\ \\ & \mathbf{\text{Jawab}}\\ & \text{Misal}\\ & A=2x^3+12y^2+24z\\ & \iff A=x^3+x^3+4y^2+4y^2+4y^2+\underset{\text{sebanyak}6\text{kali}}{\underbrace{4z+4z+...+4z}}\\ & \begin{array}{rl}& \text{Dengan ketaksamaan AM-GM dapat diperoleh}\\ & {\displaystyle \frac{A}{11}\ge \sqrt[11]{(x^3)(x^3)(4y^2)(4y^2)(4y^2)\underset{\text{sebanyak}6}{\underbrace{(4z)...(4z)}}}}\\ & \iff {\displaystyle \frac{A}{11}\ge \sqrt[11]{(xyz{)}^6(4{)}^9}}\\ & \iff A\ge 11\sqrt[11]{4^9}\end{array}\end{array}$.

$\begin{array}{l}\\ 54.& (\mathbf{\text{OSK 2019}})\\ & \text{Tentukan bilangan real terbesar}M,\\ & \text{sehingga untuk setiap}x\text{positif berlaku}\\ & (x+1)(x+3)(x+5)(x+11)\ge Mx\\ \\ & \mathbf{\text{Jawab}}\\ & \text{Diketahui}\\ & (x+1)(x+3)(x+5)(x+11)\ge Mx\\ & \iff x^4+20x^3+122x^2+268x+165\ge Mx\\ & \text{Dengan AM-GM kita akan mendapatkan}\\ & {\displaystyle \frac{x^4+\stackrel{\text{sebanyak}20}{\overbrace{x^3+x^3+...+x^3}}+\stackrel{\text{sebanyak}122}{\overbrace{x^2+x^2+...+x^2}}+\stackrel{\text{sebanyak}268}{\overbrace{x+x+...+x}}+\stackrel{\text{sebanyak}165}{\overbrace{1+1+...+1}}}{1+20+122+268+165}}\\ & \ge \sqrt[576]{x^4\left(\underset{\text{sebanyak}20}{\underbrace{x^3{x}^3...x^3}}\right)\left(\underset{\text{sebanyak}268}{\underbrace{x^2{x}^2...x^2}}\right)\left(\underset{\text{sebanyak}165}{\underbrace{11...1}}\right)}\\ & \iff {\displaystyle \frac{x^4+20x^3+122x^2+268x+165}{576}\ge \sqrt[576]{x^{576}}}\\ & \iff x^4+20x^3+122x^2+268x+165\ge 576x\\ & \text{Jadi, nilai}M=576\end{array}$.

$\begin{array}{l}\\ 55.& (\mathbf{\text{OMITS 2012}})\\ & \text{Jika diketahui}{x}_1,x_2,x_3,...,x_{2012}\in \left({\displaystyle \frac{1}{4},1}\right),\\ & \text{maka nilai minimum dari bentuk}\\ & {}^{{.}^{x_1}}\log (x_2-{\displaystyle \frac{1}{4}}){+}^{{.}^{x_2}}\log (x_3-{\displaystyle \frac{1}{4}})+...{+}^{{.}^{x_{2012}}}\log (x_1-{\displaystyle \frac{1}{4}})\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & \bullet \text{untuk}:x\ge 1,\text{berlaku}{(x-{\displaystyle \frac{1}{2}})}^2\ge 0\\ & \text{sehingga berlaku juga}(x-{\displaystyle \frac{1}{4}})\le x^2\\ & \bullet \text{untuk}:x_1,x_2,x_3,...,x_{2012}\in \left({\displaystyle \frac{1}{4},1}\right)\\ & \text{kita akan memperoleh fakta bahwa}\\ & {}^{{.}^{x_i}}\log (x_{i+1}-{\displaystyle \frac{1}{4}}){\ge }^{{.}^{x_1}}\log x_{x_{i+1}}^2={2.}^{{.}^{x_1}}\log x_{x_{i+1}}\\ & \text{Selanjutnya}\\ & {}^{{.}^{x_1}}\log (x_2-{\displaystyle \frac{1}{4}}){+}^{{.}^{x_2}}\log (x_3-{\displaystyle \frac{1}{4}})+...{+}^{{.}^{x_n}}\log (x_1-{\displaystyle \frac{1}{4}})\\ & ={\displaystyle \sum _{i=1}^n{.}^{{.}^{x_i}}\log (x_{i+1}-{\displaystyle \frac{1}{4}})\ge 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}}}}}\\ & \text{Dengan AM-GM kita mendapatkan}\\ & {\displaystyle \sum _{i=1}^n{.}^{{.}^{x_i}}\log (x_{i+1}-{\displaystyle \frac{1}{4}})}\\ & \ge 2{\displaystyle \sum _{i=1}^n{\displaystyle \frac{\log x_{i+1}}{\log x_i}\ge 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\\ & \text{Sehingga nilai minimum dari}\\ & {}^{{.}^{x_1}}\log (x_2-{\displaystyle \frac{1}{4}}){+}^{{.}^{x_2}}\log (x_3-{\displaystyle \frac{1}{4}})+...{+}^{{.}^{x_{2012}}}\log (x_1-{\displaystyle \frac{1}{4}})\\ & =2n=2(2012)=4024\end{array}\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)

 $\begin{array}{l}\\ 46.& \text{Misalkan}a,b,c\text{bilangan real non negatif}\\ & \text{dengan}a+b+c=1,\text{Tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge {\displaystyle \frac{9}{2}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui}a+b+c=1\\ & \text{Dengan AM-GM kita memiliki}\\ & 2(a+b+c)\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\\ & =((a+b)+(b+c)+(c+a))\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\\ & \ge 3\sqrt[3]{(a+b)(b+c)(c+a)}\times 3\sqrt[3]{{\displaystyle \frac{1}{(a+b)(b+c)(c+a)}}}\\ & =9\sqrt[3]{{\displaystyle \frac{(a+b)(b+c)(c+a)}{(a+b)(b+c)(c+a)}}}=9\sqrt[3]{1}=9\\ & \text{Sehingga}\\ & 2\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge 9\\ & \iff \left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge {\displaystyle \frac{9}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 47.& (\mathbf{\text{OSN 2013}})\\ & \text{Tentukan semua bilangan real}M\\ & \text{sedemikian sehingga untuk sebarang}\\ & \text{bilangan real}a,b,c\text{paling sedikit}\\ & \text{satu di antara tiga bilangan berikut}\\ & a+{\displaystyle \frac{M}{ab},b+{\displaystyle \frac{M}{bc},c+{\displaystyle \frac{M}{ca}}}}\\ & \text{bernilai lebih dari atau sama dengan}\\ & 1+M\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \text{min}\{a+{\displaystyle \frac{M}{ab},b+{\displaystyle \frac{M}{bc},c+{\displaystyle \frac{M}{ca}}}}\}\ge 1+M\\ & \text{Perhatikan bahwa}\\ & a+{\displaystyle \frac{M}{ab}+b+{\displaystyle \frac{M}{bc}+c+{\displaystyle \frac{M}{ca}}}}\\ & =a+b+c+{\displaystyle \frac{M}{ab}+{\displaystyle \frac{M}{bc}+{\displaystyle \frac{M}{ca}}}}\\ & ={\displaystyle \frac{1}{2}(2(a+b+c)+{\displaystyle \frac{2M}{ab}+{\displaystyle \frac{2M}{bc}+{\displaystyle \frac{2M}{ca}}}})}\\ & ={\displaystyle \frac{1}{2}(a+b+{\displaystyle \frac{2M}{ab}+b+c+{\displaystyle \frac{2M}{bc}+c+a+{\displaystyle \frac{2M}{ca}}}})}\\ & \text{Dengan AM-GM akan diperoleh bentuk}\\ & \ge {\displaystyle \frac{1}{2}(3\sqrt[3]{2M}+3\sqrt[3]{2M}+3\sqrt[3]{2M})}\\ & ={\displaystyle \frac{9}{2}\sqrt[3]{2M}}\\ & \text{Pilih nilai minimum}\\ & \{a+{\displaystyle \frac{M}{ab},b+{\displaystyle \frac{M}{bc},c+{\displaystyle \frac{M}{ca}}}}\}=1+M={\displaystyle \frac{3\sqrt[3]{2M}}{2}}\\ & \text{Selanjutnya}\\ & 1+M={\displaystyle \frac{3\sqrt[3]{2M}}{2}}\\ & \iff M={\displaystyle \frac{1}{2}}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 49.& \text{Jika pada soal 48 diubah dengan}\\ & a,b,c\text{adalah sisi segitiga}ABC\\ & \text{yang memenuhi}a+b+c=1.\\ & \text{Tentukan nilai dari}{\displaystyle \frac{a+b}{abc}}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}a,b,c\text{sisi}△ABC\\ & \text{dengan}a+b+c=1.\text{kita dapat peroleh}\\ & a+b=a+b\\ & \text{Ingat bahwa dalam}△ABC,\text{berlaku}\\ & \{\begin{array}{ll}\bullet & a+b>c\\ \bullet & a+c>b\\ \bullet & b+c>a\end{array}\\ & \text{maka}a+b+a+b>a+b+c\\ & \iff 2(a+b)>1\iff a+b>{\displaystyle \frac{1}{2}}\\ & \text{Dan dengan AM-GM diperoleh}\\ & {\displaystyle \frac{a+b+c}{3}\ge \sqrt[3]{abc}\iff {\displaystyle \frac{1}{3}\ge \sqrt[3]{abc}}}\\ & \iff {\displaystyle \frac{1}{27}\ge abc}\\ & \text{Selanjutnya}\\ & {\displaystyle \frac{a+b}{abc}>{\displaystyle \frac{\frac{1}{2}}{\frac{1}{27}}={\displaystyle \frac{27}{2}=13,5}}}\\ & \text{Jadi, nilai}{\displaystyle \frac{a+b}{abc}>13,5}\end{array}\end{array}$.

$\begin{array}{l}\\ 50.& \text{Diketahui bilangan real positif}{x}_1,x_2,...,x_k\\ & \text{memenuhi persamaan}{x}_1+x_2+...+x_k=1\\ & \text{Buktikan bahwa}\\ & {\displaystyle \frac{1}{(x_1{)}^{2013}}+\frac{1}{(x_2{)}^{2013}}+...+\frac{1}{(x_k{)}^{2013}}\ge k^{2014}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Dengan ketaksamaan AM-GM kita}\\ & \text{memiliki}\\ & {\displaystyle \frac{x_1+x_2+...+x_k}{k}\ge \sqrt[k]{x_1.x_2...x_k}}\\ & \iff {\displaystyle \frac{1}{k}\ge \sqrt[k]{x_1.x_2...x_k}}\\ & \iff \sqrt[k]{x_1.x_2...x_k}\le {\displaystyle \frac{1}{k}}\\ & \iff x_1.x_2...x_k\le {\left({\displaystyle \frac{1}{k}}\right)}^k\\ & \text{Selanjutnya}\\ & {\displaystyle \frac{{\displaystyle \frac{1}{(x_1{)}^{2013}}+\frac{1}{(x_2{)}^{2013}}+...+\frac{1}{(x_k{)}^{2013}}}}{k}}\\ & \ge \sqrt[k]{{\displaystyle \frac{1}{(x_1{)}^{2013}.(x_2{)}^{2013}...(x_k{)}^{2013}}}}\\ & {\displaystyle \frac{1}{(x_1{)}^{2013}}+\frac{1}{(x_2{)}^{2013}}+...+\frac{1}{(x_k{)}^{2013}}}\\ & \ge k\sqrt[k]{{\displaystyle \frac{1}{(x_1{)}^{2013}.(x_2{)}^{2013}...(x_k{)}^{2013}}}}\\ & \ge k\sqrt[k]{{\displaystyle \frac{1}{(x_1.x_2...x_k{)}^{2013}}}}\ge k.\sqrt[k]{{\displaystyle \frac{1}{{\left({\left({\displaystyle \frac{1}{k}}\right)}^k\right)}^{2013}}}}\\ & \ge k.\sqrt[k]{k^{2013k}}\\ & \ge k.k^{2013}\\ & \ge k^{2014}◼\end{array}\end{array}$.

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 $a_1\le a_2\le a_3\le \cdots \le a_n$ dan $b_1\le b_2\le b_3\le \cdots \le b_n$ adalah merupakan kumpulan bilangan yang monoton naik atau $a_1\ge a_2\ge a_3\ge \cdots \ge a_n$ dan $b_1\ge b_2\ge b_3\ge \cdots \ge b_n$ adalah merupakan kumpulan bilangan yang monoton turun, maka

$\begin{array}{rl}& \left({\displaystyle \frac{a_1{b}_1+a_2{b}_2+a_3{b}_3+\cdots +a_n{b}_n}{n}}\right)\\ & \ge \left({\displaystyle \frac{a_1+a_2+a_3+...+a_n}{n}}\right)\\ & \times \left({\displaystyle \frac{b_1+b_2+b_3+...+b_n}{n}}\right)\end{array}$.

$\begin{array}{rl}& \mathbf{\text{atau}}\\ \\ & n(a_1{b}_1+a_2{b}_2+a_3{b}_3+\cdots +a_n{b}_n)\\ & \ge (a_1+a_2+a_3+...+a_n)\times (b_1+b_2+b_3+...+b_n)\end{array}$.

Tetapi jika kumpulan bilangan di atas memiliki kemonotonan yang berbeda, yaitu $a_1\le a_2\le a_3\le \cdots \le a_n$  dan  $b_1\ge b_2\ge b_3\ge \cdots \ge b_n$ atau  $a_1\ge a_2\ge a_3\ge \cdots \ge a_n$  dan $b_1\le b_2\le b_3\le \cdots \le b_n$, maka ketaksamaan akan menjadi

$\begin{array}{rl}& \left({\displaystyle \frac{a_1{b}_1+a_2{b}_2+a_3{b}_3+\cdots +a_n{b}_n}{n}}\right)\\ & \le \left({\displaystyle \frac{a_1+a_2+a_3+...+a_n}{n}}\right)\\ & \times \left({\displaystyle \frac{b_1+b_2+b_3+...+b_n}{n}}\right)\end{array}$.

$\begin{array}{rl}& \mathbf{\text{atau}}\\ \\ & n(a_1{b}_1+a_2{b}_2+a_3{b}_3+\cdots +a_n{b}_n)\\ & \le (a_1+a_2+a_3+...+a_n)\times (b_1+b_2+b_3+...+b_n)\end{array}$.

Bukti

Pada kumpulan bilangan yang memiliki kemonotonan yang sama yaitu:

$a_1\le a_2\le a_3\le \cdots \le a_n$ dan $b_1\le b_2\le b_3\le \cdots \le b_n$, dengan mengaplikasikan ketaksamaan Renata akan diperoleh

$\begin{array}{rl}{a}_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n& =a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n\\ a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n& \ge a_1{b}_2+a_2{b}_3+\cdots +a_n{b}_1\\ a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n& \ge a_1{b}_3+a_2{b}_4+\cdots +a_n{b}_2\\ & ⋮\\ a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n& \ge a_1{b}_n+a_2{b}_1+\cdots +a_n{b}_{n-1}\\ \text{Jika ketaksamaan di at}& \text{as ditmabahkan, maka}\\ n(a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n)& \ge (a_1+a_2+\cdots +a_n)(b_1+b_2+\cdots +b_n)\\ \text{Bentuk terakhir adalah}& \text{bukti dari ketaksamaan ini}\end{array}$.

$\begin{array}{rl}\text{Sebagai}& \text{misal, andaikan}\\ \bullet n=2& \iff 2(a_1{b}_1+a_2{b}_2)\ge (a_1+a_2)(b_1+b_2)\\ \bullet n=3& \iff 3(a_1{b}_1+a_2{b}_2+a_3{b}_3)\ge (a_1+a_2+a_3)(b_1+b_2+b_3)\\ & ⋮\\ \bullet n=k& \iff k(a_1{b}_1+a_2{b}_2+\cdots +a_k{b}_k)\ge (a_1+a_2+\cdots +a_k)(b_1+b_2+\cdots +b_k)\end{array}$ .

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika}a,b,x,y\text{bilangan real positif}\\ & \text{sehingga}a\ge b\text{dan}x\ge y.\\ & \text{Tunjukkan bahwa}\\ & (ax+by)\ge {\displaystyle \frac{1}{2}(a+b)(x+y)}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{dapat diperoleh bentuk}\\ & 2(ax+by)\ge (a+b)(x+y)\\ & \iff (ax+by)\ge {\displaystyle \frac{1}{2}(a+b)(x+y)◼}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika}a,b>0,\text{tunjukkan bahwa}\\ & 2(a^2+b^2)\ge (a+b{)}^2\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Alternatif 1}\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{untuk}a\ge b,\text{dapat diperoleh bentuk}\\ & 2(a.a+b.b)\ge (a+b)(a+b)\\ & \iff 2(a^2+b^2)\ge (a+b{)}^2◼\\ & \text{Alternatif 2}\\ & \text{Dengan}\mathbf{\text{ketaksamaan CS-Engel}}\\ & (1+1)(a^2+b^2)\ge (1.a+1.b{)}^2\\ & \iff 2(a^2+b^2)\ge (a+b{)}^2◼\\ & \text{Alternatif 3}\\ & \text{Perhatikan bahwa}(a-b{)}^2\ge 0\\ & \iff a^2+b^2-2ab\ge 0\\ & \iff a^2+b^2\ge 2ab\\ & \iff 2(a^2+b^2)\ge a^2+b^2+2ab\\ & \iff 2(a^2+b^2)\ge (a+b{)}^2◼\end{array}$.

$\begin{array}{l}\\ 3.& \text{Jika}a,b,c>0,\text{tunjukkan bahwa}\\ & 3(a^2+b^2+c^2)\ge (a+b+c{)}^2\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{untuk}a\ge b\ge c,\text{dapat diperoleh bentuk}\\ & 3(a.a+b.b+c.c)\ge (a+b+c)(a+b+c)\\ & \iff 3(a^2+b^2+c^2)\ge (a+b+c{)}^2◼\end{array}$.

$\begin{array}{l}\\ 4.& \text{Jika}a,b,c>0,\text{dengan}{a}^2+b^2+c^2=1\\ & \text{tunjukkan bahwa}a+b+c\le \sqrt{3}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{Kurang lebih caranya sama dengan no.3 di atas}\\ & \text{untuk}a\ge b\ge c,\text{dapat diperoleh bentuk}\\ & 3(a.a+b.b+c.c)\ge (a+b+c)(a+b+c)\\ & \iff 3(a^2+b^2+c^2)\ge (a+b+c{)}^2\\ & \iff 3(1)\ge (a+b+c{)}^2\\ & \iff \sqrt{3}\ge (a+b+c)\\ & \iff (a+b+c)\le \sqrt{3}◼\end{array}$.

$\begin{array}{l}\\ 5.& \text{Diberikan}a,b,c>0,\text{tunjukkan kebenaran}\\ & \mathbf{\text{ketaksamaan Nesbitt}}\text{berikut}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Alternatif 1}\\ & \begin{array}{rl}& \text{Asumsikan},\\ & \{\begin{array}{ll}& a\ge b\ge c\\ & {\displaystyle \frac{1}{b+c}\ge \frac{1}{a+c}\ge \frac{1}{a+b}}\end{array}\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Chebyshev}}\\ & {\displaystyle \frac{{\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}}{3}\ge {\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)}}\\ & \text{Dengan AM-HM dan}K={\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\\ & \iff {\displaystyle \frac{K}{3}\ge {\displaystyle \frac{(a+b+c)}{3}\left({\displaystyle \frac{3}{(b+c)+(a+c)+(a+b)}}\right)}}\\ & \iff K\ge {\displaystyle \frac{3(a+b+c)}{2(a+b+c)}}\\ & \iff K\ge {\displaystyle \frac{3}{2}}\\ & \iff {\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge {\displaystyle \frac{3}{2}◼}}\end{array}\\ & \text{Alternatif 2}\\ & \begin{array}{rl}& \mathbf{\text{Pertama}},\text{asumsikan}\\ & \{\begin{array}{ll}& a\ge b\ge c\\ & {\displaystyle \frac{1}{b+c}\ge \frac{1}{a+c}\ge \frac{1}{a+b}}\end{array}\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Chebyshev}}\\ & 3\left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge (a+b+c)\left({\displaystyle \frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}}\right)\\ & \mathbf{\text{Kedua}},\text{asumsikan}\\ & \{\begin{array}{ll}& a+b\ge a+c\ge b+c\\ & {\displaystyle \frac{1}{a+b}\le \frac{1}{a+c}\le \frac{1}{b+c}}\end{array}\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Chebyshev}}\\ & 3\left({\displaystyle \frac{a+b}{a+b}+\frac{a+c}{a+c}+\frac{b+c}{b+c}}\right)\le (a+b+a+c+b+c)\left({\displaystyle \frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}}\right)\\ & \iff 3(1+1+1)\le 2(a+b+c)\left({\displaystyle \frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}}\right)\\ & \iff {\displaystyle \frac{9}{2}\le (a+b+c)\left({\displaystyle \frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}}\right)}\\ & \text{Dari dua ketaksamaan di atas didapatkan}\\ & 3\left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge {\displaystyle \frac{9}{2}}\\ & \iff \left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge {\displaystyle \frac{3}{2}◼}\end{array}\end{array}$ 

$\begin{array}{l}\\ 6.& \text{Diberikan}a,b,c,d>0,\text{tunjukkan bahwa}\\ & (a+b+c+d)\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}\right)\ge 16\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Asumsikan}\\ & \{\begin{array}{ll}& a\ge b\ge c\ge d\\ & {\displaystyle \frac{1}{d}\ge \frac{1}{c}\ge \frac{1}{b}\ge \frac{1}{a}}\end{array}\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Chebyshev}}\\ & {\displaystyle \frac{{\displaystyle \frac{a}{a}+\frac{b}{b}+\frac{c}{c}+\frac{d}{d}}}{4}\le {\displaystyle \left({\displaystyle \frac{(a+b+c+d)}{4}}\right)\left({\displaystyle \frac{{\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}}{4}}\right)}}\\ & \iff 1\le {\displaystyle \left({\displaystyle \frac{(a+b+c+d)}{4}}\right)\left({\displaystyle \frac{{\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}}{4}}\right)}\\ & \iff 16\le (a+b+c+d)\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}\right)\\ & \iff (a+b+c+d)\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}\right)\ge 16◼\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika}a,b,c>0,\text{dengan}a\ne b\ne c\\ & \text{tunjukkan bahwa}\\ & (a^3+b^3+c^3)>{\displaystyle \frac{(a+b+c{)}^3}{9}>3abc}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{untuk}a\ge b\ge c,\text{dapat diperoleh bentuk}\\ & {\displaystyle \frac{(a^3+b^3+c^3)}{3}>\left({\displaystyle \frac{a+b+c}{3}}\right)\left({\displaystyle \frac{a+b+c}{3}}\right)\left({\displaystyle \frac{a+b+c}{3}}\right)}\\ & \iff {\displaystyle \frac{(a^3+b^3+c^3)}{3}>{\left({\displaystyle \frac{a+b+c}{3}}\right)}^3>{\left({\displaystyle \frac{3\sqrt[3]{abc}}{3}}\right)}^3}\\ & \iff {\displaystyle \frac{(a^3+b^3+c^3)}{3}>{\displaystyle \frac{(a+b+c{)}^3}{27}>abc}}\\ & \iff (a^3+b^3+c^3)>{\displaystyle \frac{(a+b+c{)}^3}{9}>3abc◼}\end{array}$.

$\begin{array}{l}\\ 8.& \text{tunjukkan bahwa untuk}n\text{bilangan asli}\\ & \text{berlaku}\\ & 1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n}\le n\sqrt{{\displaystyle \frac{n+1}{2}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{dapat diperoleh bentuk berikut}\\ & {\displaystyle \frac{1+2+3+\cdots +n}{n}\ge {\displaystyle \frac{(1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n})}{n}\times {\displaystyle \frac{(1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n})}{n}}}}\\ & \iff {\displaystyle \frac{1+2+3+\cdots +n}{n}\ge {\displaystyle \frac{{(1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n})}^2}{n^2}}}\\ & \iff {\displaystyle \frac{{(1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n})}^2}{n^2}\le {\displaystyle \frac{1+2+3+\cdots +n}{n}}}\\ & \iff {(1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n})}^2\le n(1+2+3+\cdots +n)\\ & \iff {(1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n})}^2\le n\left({\displaystyle \frac{n(n+1)}{2}}\right)\\ & \iff (1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n})\le n\sqrt{{\displaystyle \frac{n+1}{2}}}◼\end{array}$.

$\begin{array}{l}\\ 9.& \text{tunjukkan bahwa untuk}n\text{bilangan asli}\\ & \text{berlaku}\\ & {\displaystyle \frac{1}{\sqrt{n}}(1+\sqrt{{\displaystyle \frac{1}{2}}}+\sqrt{{\displaystyle \frac{1}{3}}}+\cdots +\sqrt{{\displaystyle \frac{1}{n}}})\le (2n-1{)}^{{.}^{\frac{1}{4}}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{untuk}:(1\ge {\displaystyle \frac{1}{2}\ge \frac{1}{3}\ge \cdots \ge \frac{1}{n}})\\ & \text{dapat diperoleh bentuk berikut}\\ & {(1+{\displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}})}^2\le n(1+{\displaystyle \frac{1}{2.2}+\frac{1}{3.3}+\cdots +\frac{1}{n.n}})\\ & \iff {(1+{\displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}})}^2\le n(1+{\displaystyle \frac{1}{1.2}+\frac{1}{2.3}+\cdots +\frac{1}{(n-1).n}})\\ & \iff {(1+{\displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}})}^2\le n(1+(1-{\displaystyle \frac{1}{2}})+\left({\displaystyle \frac{1}{2}-\frac{1}{3}}\right)+\cdots +\left({\displaystyle \frac{1}{(n-1)}-\frac{1}{n}}\right))\\ & \iff {(1+{\displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}})}^2\le n(1+1-{\displaystyle \frac{1}{n}})=n(2-{\displaystyle \frac{1}{n}})\\ & \iff (1+{\displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}})\le \sqrt{2n-1}..........(1)\\ & \text{Gunakan lagi}\mathbf{\text{ketaksamaan Chebyshev}}\\ & {(1+\sqrt{{\displaystyle \frac{1}{2}}}+\sqrt{{\displaystyle \frac{1}{3}}}+\cdots +\sqrt{{\displaystyle \frac{1}{n}}})}^2\le n(1+{\displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}})\\ & \iff {\displaystyle \frac{1}{n}{(1+\sqrt{{\displaystyle \frac{1}{2}}}+\sqrt{{\displaystyle \frac{1}{3}}}+\cdots +\sqrt{{\displaystyle \frac{1}{n}}})}^2\le (1+{\displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}})...(2)}\\ & \text{Dari ketaksamaan (1) dan (2), dapat diperoleh}\\ & {\displaystyle \frac{1}{n}{(1+\sqrt{{\displaystyle \frac{1}{2}}}+\sqrt{{\displaystyle \frac{1}{3}}}+\cdots +\sqrt{{\displaystyle \frac{1}{n}}})}^2\le \sqrt{2n-1}=(2n-1{)}^{{.}^{\frac{1}{2}}}}\\ & \iff {\displaystyle \frac{1}{\sqrt{n}}(1+\sqrt{{\displaystyle \frac{1}{2}}}+\sqrt{{\displaystyle \frac{1}{3}}}+\cdots +\sqrt{{\displaystyle \frac{1}{n}}})\le (2n-1{)}^{{.}^{\frac{1}{4}}}◼}\end{array}$.

$\begin{array}{l}\\ 10.& (\mathbf{\text{OSN 2011}})\\ & \text{Jika}a,b,c>0,\text{dengan}abc=1\\ & \text{Jika diketahui}\\ & a^{2011}+b^{2011}+c^{2011}<{\displaystyle \frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}}\\ & \text{tunjukkan bahwa}\\ & (a+b+c)>{\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Asumsikan}\\ & \{\begin{array}{ll}& a\ge b\ge c\\ & {\displaystyle \frac{1}{c}\ge \frac{1}{b}\ge \frac{1}{a}}\end{array}\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{Perhatikan}\\ & \begin{array}{rl}& {\displaystyle \frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}\ge {\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)}}\\ & \iff {\displaystyle \frac{1}{a^{2010}}+\frac{1}{b^{2010}}+\frac{1}{c^{2010}}\ge {\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)}}\\ & \iff {\displaystyle \frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}\ge {\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)}}\\ & ⋮\\ & \iff {\displaystyle \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\ge {\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)}}\\ & \iff {\displaystyle \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge {\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)}}\\ & \text{Sehingga}\\ & {\displaystyle \frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}\ge {\displaystyle \frac{1}{3^{2010}}{\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)}^{2011}.....(1)}}\end{array}\\ & \text{dan}\\ & \begin{array}{rl}& a^{2011}+b^{2011}+c^{2011}\ge {\displaystyle \frac{1}{3}(a+b+c)(a^{2010}+b^{2010}+c^{2010})}\\ & \iff a^{2010}+b^{2010}+c^{2010}\ge {\displaystyle \frac{1}{3}(a+b+c)(a^{2009}+b^{2009}+c^{2009})}\\ & \iff a^{2009}+b^{2009}+c^{2009}\ge {\displaystyle \frac{1}{3}(a+b+c)(a^{2008}+b^{2008}+c^{2008})}\\ & ⋮\\ & \iff a^3+b^3+c^3\ge {\displaystyle \frac{1}{3}(a+b+c)(a^2+b^2+c^2)}\\ & \iff a^2+b^2+c^2\ge {\displaystyle \frac{1}{3}(a+b+c)(a+b+c)}\\ & \text{Sehingga}\\ & \iff a^{2011}+b^{2011}+c^{2011}\ge {\displaystyle \frac{1}{3^{2010}}(a+b+c{)}^{2011}........(2)}\end{array}\\ & \begin{array}{rl}& \text{Dari ketaksamaan (1) dan (2) didapatkan}\\ & \frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}\ge {\displaystyle \frac{1}{3^{2010}}{\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)}^{2011}}\\ & >a^{2011}+b^{2011}+c^{2011}\ge {\displaystyle \frac{1}{3^{2010}}(a+b+c{)}^{2011}}\\ & \text{atau}\\ & {\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{1}{a^{2011}}\ge {\displaystyle \frac{1}{3^{2010}}{\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{1}{a}}}\right)}^{2011}>{\displaystyle \sum _{\text{siklik}}^{.}{a}^{2011}\ge {\displaystyle \frac{1}{3^{2010}}{\left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)}^{2011}}}}}}\\ & \iff {\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}}}\\ & \iff \left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)>(a+b+c)\\ & \iff a+b+c<{\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}◼}\end{array}\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)

$\begin{array}{l}\\ 41.& \text{Jika}a,b,c>0,\text{tunjukkan bahwa}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge {\displaystyle \frac{3}{2}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& ((a+b)+(b+c)+(c+a))\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge 9\\ & \iff 2(a+b+c)\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge 9\\ & \iff (a+b+c)\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge {\displaystyle \frac{9}{2}}\\ & \iff {\displaystyle \frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\ge {\displaystyle \frac{9}{2}}}\\ & \iff 1+{\displaystyle \frac{c}{a+b}+1+{\displaystyle \frac{a}{b+c}+1+{\displaystyle \frac{b}{c+a}\ge {\displaystyle \frac{9}{2}}}}}\\ & \iff 3+{\displaystyle \frac{c}{a+b}+{\displaystyle \frac{a}{b+c}+{\displaystyle \frac{b}{c+a}\ge {\displaystyle \frac{9}{2}}}}}\\ & \iff {\displaystyle \frac{c}{a+b}+{\displaystyle \frac{a}{b+c}+{\displaystyle \frac{b}{c+a}\ge {\displaystyle \frac{9}{2}-3}}}}\\ & \iff {\displaystyle \frac{c}{a+b}+{\displaystyle \frac{a}{b+c}+{\displaystyle \frac{b}{c+a}\ge {\displaystyle \frac{3}{2}◼}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 42.& \text{Jika}a,b,c\text{adalah sisi-sisi segitiga}\\ & \text{ABC, tunjukkan bahwa}\\ & {\displaystyle \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\ge 3}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa dengan AM-HM diperoleh}\\ & (a+b+c)\left({\displaystyle \frac{1}{b+c-a}+\frac{1}{c+a-b}+\frac{1}{a+b-c}}\right)\ge 9\\ & \iff ((b+c-a)+(c+a-b)+(a+b-c))\\ & \times \left({\displaystyle \frac{1}{b+c-a}+\frac{1}{c+a-b}+\frac{1}{a+b-c}}\right)\ge 9\\ & \iff {\displaystyle \frac{a+b+c}{b+c-a}+\frac{a+b+c}{c+a-b}+\frac{a+b+c}{a+b-c}\ge 9}\\ & \iff 1+{\displaystyle \frac{2a}{b+c-a}+1+{\displaystyle \frac{2b}{c+a-b}+1+{\displaystyle \frac{2c}{a+b-c}\ge 9}}}\\ & \iff 3+{\displaystyle \frac{2a}{b+c-a}+{\displaystyle \frac{2b}{c+a-b}+{\displaystyle \frac{2c}{a+b-c}\ge 9}}}\\ & \iff {\displaystyle \frac{2a}{b+c-a}+{\displaystyle \frac{2b}{c+a-b}+{\displaystyle \frac{2c}{a+b-c}\ge 6}}}\\ & \iff {\displaystyle \frac{a}{b+c-a}+{\displaystyle \frac{b}{c+a-b}+{\displaystyle \frac{c}{a+b-c}\ge 3◼}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 43.& \text{Diketahui bilangan real positif}a,b,c\\ & \text{dengan}abc=1.\text{Buktikan bahwa}\\ & {\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\ge 3}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}abc=1\\ & \text{Perhatikan bahwa}\\ & \{\begin{array}{ll}\bullet & {\displaystyle \frac{1+ab}{1+a}+{\displaystyle \frac{abc+ab}{1+a}=ab\left({\displaystyle \frac{1+c}{1+a}}\right)}}\\ \bullet & {\displaystyle \frac{1+bc}{1+b}+{\displaystyle \frac{abc+bc}{1+b}=bc\left({\displaystyle \frac{1+a}{1+b}}\right)}}\\ \bullet & {\displaystyle \frac{1+ca}{1+c}+{\displaystyle \frac{abc+ca}{1+c}=ac\left({\displaystyle \frac{1+b}{1+c}}\right)}}\end{array}\\ & \text{Jika hasilnya dijumlahkan},\text{maka}\\ & \text{kita akan mendapatkan hasil}\\ & ab\left({\displaystyle \frac{1+c}{1+a}}\right)+bc\left({\displaystyle \frac{1+a}{1+b}}\right)+ac\left({\displaystyle \frac{1+b}{1+c}}\right)\\ & \text{Selanjutnya dengan AM-GM akan diperoleh}\\ & {\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}}\\ & =ab\left({\displaystyle \frac{1+c}{1+a}}\right)+bc\left({\displaystyle \frac{1+a}{1+b}}\right)+ac\left({\displaystyle \frac{1+b}{1+c}}\right)\ge 3\sqrt[3]{(abc{)}^2}\\ & \iff {\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\ge 3\sqrt[3]{(abc{)}^2}}\\ & \iff {\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\ge 3\sqrt[3]{1}}\\ & \iff {\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\ge 3◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 44.& (\mathbf{\text{SEAMO III-Malaysia}})\\ & \text{Misalkan}a,b,c\text{bilangan real positif}\\ & \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge {\displaystyle \frac{3}{1+abc}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & {\displaystyle \frac{1}{a(1+b)}+\frac{1}{1+abc}=\frac{1}{1+abc}\left({\displaystyle \frac{1+a}{a(1+b)}+\frac{b(1+c)}{(1+b)}}\right)}\\ & \text{Sehingga}\\ & {\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc}}\\ & ={\displaystyle \frac{1}{1+abc}\left({\displaystyle \frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a}}\right)}\\ & \text{Secara AM-GM kita mendapatkan}\\ & \left({\displaystyle \frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a}}\right)\ge 6\\ & \text{Kembali ke soal}\\ & {\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc}\ge {\displaystyle \frac{1}{1+abc}(6)}}\\ & \iff {\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge {\displaystyle \frac{3}{1+abc}◼}}\end{array}\end{array}$.

$\begin{array}{l}\\ 45.& \text{Misalkan}a,b,c\text{bilangan real non negatif}\\ & \text{dengan}a+b+c=1,\text{Tunjukkan bahwa}\\ & {\displaystyle \frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}\ge {\displaystyle \frac{9}{10}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui}a+b+c=1\\ & \text{Dengan AM-GM kita memiliki}\\ & {\displaystyle \frac{a+b+c}{3}\ge \sqrt[3]{abc}\iff {\displaystyle \frac{1}{3}\ge \sqrt[3]{abc}\iff {\displaystyle \frac{1}{27}\ge abc}}}\\ & \text{Sehingga}\\ & {\displaystyle \frac{1}{1+3abc}\ge {\displaystyle \frac{1}{1+3\left({\displaystyle \frac{1}{27}}\right)}\ge {\displaystyle \frac{1}{1+{\displaystyle \frac{1}{9}}}\ge {\displaystyle \frac{9}{10}}}}}\\ & \text{Selanjutnya dengan perluasan AM-HM}\\ & \text{Sebagaimana bentuk berikut}\\ & \left({\displaystyle \frac{p_1{a}_1+p_2{a}_2+p_3{a}_3}{p_1+p_2+p_3}}\right)\ge \left({\displaystyle \frac{p_1+p_2+p_3}{p_1{a}_1^{-1}+p_2{a}_2^{-1}+p_3{a}_3^{-1}}}\right)\\ & \text{maka}\\ & {\displaystyle \frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}\ge {\displaystyle \frac{1}{a(1+bc)+b(1+ac)+c(1+ab)}}}\\ & \ge {\displaystyle \frac{1}{a+b+c+3abc}}\\ & \ge {\displaystyle \frac{1}{1+3abc}}\\ & \ge {\displaystyle \frac{9}{10}◼}\end{array}\end{array}$.

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)

 $\begin{array}{l}\\ 36.& \text{Jika}a,b,c>0,\text{dengan}a+b+c=1\\ & \text{tunjukkan bahwa}\\ & (1-a)(1-b)(1-c)\ge 8abc\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & \{\begin{array}{ll}1-a& =b+c\\ 1-b& =a+c\\ 1-c& =a+b\end{array}\\ & \text{Kurang lebih seperti pembuktian}\\ & \text{pada no.35 di atas dengan tetap}\\ & \text{menggunakan AM-GM akan}\\ & \text{didapatkan}\\ & \bullet a+b\ge 2\sqrt{ab}\\ & \bullet b+c\ge 2\sqrt{bc},\text{dan}\\ & \bullet c+a\ge 2\sqrt{ca}\\ & \text{Selanjutnya}\\ & (1-a)(1-b)(1-c)=(b+c)(a+c)(a+b)\\ & \ge 2\sqrt{bc}\times 2\sqrt{ac}\times 2\sqrt{ab}\\ & \ge 8\sqrt{a^2{b}^2{c}^2}\\ & \ge 8abc◼\end{array}\end{array}$.

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

$\begin{array}{l}\\ 38.& \text{Diketahui}a,b,c\text{bilangan real positif}\\ & \text{dan}(a+1)(b+1)(c+1)=8.\\ & \text{tunjukkan bahwa}abc<1\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & (a+1)(b+1)(c+1)=8\\ & abc+(ab+bc+ca)+(a+b+c)+1=8\\ & \text{Dengan AM-GM kita akan mendapatkan}\\ & abc+3(abc{)}^{{.}^{\frac{2}{3}}}+3(abc{)}^{{.}^{\frac{1}{3}}}+1<8\\ & \iff {((abc{)}^{{.}^{\frac{1}{3}}}+1)}^3<2^3\\ & \iff (abc{)}^{{.}^{\frac{1}{3}}}+1<2\\ & \iff (abc{)}^{{.}^{\frac{1}{3}}}<1◼\end{array}\end{array}$.

$\begin{array}{l}\\ 39.& \text{Jika}a,b,c>0\\ & \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{a}{a^2+1}+{\displaystyle \frac{b}{b^2+1}+{\displaystyle \frac{c}{c^2+1}\le {\displaystyle \frac{3}{2}}}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & (a-1{)}^2\ge 0\\ & \iff a^2-2a+1\ge 0\\ & \iff a^2+1\ge 2a\\ & {\displaystyle \frac{1}{2}\ge {\displaystyle \frac{a}{a^2+1}}}\\ & \text{atau}\\ & {\displaystyle \frac{a}{a^2+1}\le {\displaystyle \frac{1}{2}..................(1)}}\\ & \text{dengan cara yang sama akan}\\ & \text{pula jiika}b\text{dan}c\text{dikondisikan}\\ & \text{akan didapatkan ketaksamaan}\\ & {\displaystyle \frac{b}{b^2+1}\le {\displaystyle \frac{1}{2}..................(2)}}\\ & {\displaystyle \frac{c}{c^2+1}\le {\displaystyle \frac{1}{2}..................(3)}}\\ & \text{Jika ketaksamaan}(1),(2),\mathrm{\&}(3)\\ & \text{dijumlahkan akan menghasilkan}\\ & {\displaystyle \frac{a}{a^2+1}+{\displaystyle \frac{b}{b^2+1}+{\displaystyle \frac{c}{c^2+1}\le {\displaystyle \frac{1}{2}+{\displaystyle \frac{1}{2}+{\displaystyle \frac{1}{2}}}}}}}\\ & \le 3\left({\displaystyle \frac{1}{2}}\right)\\ & \le {\displaystyle \frac{3}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 40.& \text{Jika}a,b,c\text{adalah sisi-sisi segitiga}\\ & \text{ABC, tunjukkan bahwa}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa pada segitiga}\\ & \text{ABC berlaku}\\ & \{\begin{array}{ll}a+b& >c\\ a+c& >b\\ b+c& >a\end{array}\\ & \text{Misalkan}2s=a+b+c,\text{maka}\\ & a+b=a+b\\ & \iff a+b+a+b>a+b+c\\ & \iff 2(a+b)>2s\\ & \iff (a+b)>s\\ & \text{Demikian juga akan berlaku}\\ & \{\begin{array}{ll}b+c& >s\\ c+a& >s\end{array}\\ & \text{Sehingga}\\ & \{\begin{array}{ll}{\displaystyle \frac{a}{b+c}}& <{\displaystyle \frac{a}{s}}\\ {\displaystyle \frac{b}{c+a}}& <{\displaystyle \frac{b}{s}}\\ {\displaystyle \frac{c}{a+b}}& <{\displaystyle \frac{c}{s}}\end{array}\\ & \text{Selanjutnya kita kembali ke soal}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<{\displaystyle \frac{a+b+c}{s}}}\\ & <{\displaystyle \frac{2s}{s}}\\ & <2◼\end{array}\end{array}$.

Contoh Soal 7 (Segitiga dan Ketaksamaan)

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

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

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

$\begin{array}{l}\\ 34.& \text{Jika}a,b>0,\text{tunjukkan bahwa}\\ & (a+1)(b+1)(ab+1)\ge 8ab\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa dengan AM-GM}\\ & \bullet a+1\ge 2\sqrt{a}\\ & \bullet b+1\ge 2\sqrt{b},\text{dan}\\ & \bullet ab+1\ge 2\sqrt{ab}\\ & \text{Selanjutnya}\\ & (a+1)(b+1)(ab+1)\ge 2\sqrt{a}\times 2\sqrt{b}\times 2\sqrt{ab}\\ & \ge 8\sqrt{a^2{b}^2}\\ & \ge 8ab◼\end{array}\end{array}$.

$\begin{array}{l}\\ 35.& \text{Jika}a,b,c>0,\text{tunjukkan bahwa}\\ & (a+b)(b+c)(c+a)\ge 8abc\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Kurang lebih seperti pembuktian}\\ & \text{pada no.34 di atas dengan tetap}\\ & \text{menggunakan AM-GM akan}\\ & \text{didapatkan}\\ & \bullet a+b\ge 2\sqrt{ab}\\ & \bullet b+c\ge 2\sqrt{bc},\text{dan}\\ & \bullet c+a\ge 2\sqrt{ca}\\ & \text{Selanjutnya}\\ & (a+b)(b+c)(c+a)\ge 2\sqrt{ab}\times 2\sqrt{bc}\times 2\sqrt{ca}\\ & \ge 8\sqrt{a^2{b}^2{c}^2}\\ & \ge 8abc◼\end{array}\end{array}$.

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)

$\begin{array}{l}\\ 26.& \text{Diketahui}a,b\text{bilangan real positif}\\ & \text{dan}a+b=1.\text{Tunjukkan bahwa}\\ & {(a+{\displaystyle \frac{1}{a}})}^2+{(b+{\displaystyle \frac{1}{b}})}^2\ge {\displaystyle \frac{25}{2}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}a+b=1\\ & \text{Dengan QM-AM kita akan mendapatkan}\\ & \sqrt{{\displaystyle \frac{x^2+y^2}{2}}}\ge {\displaystyle \frac{x+y}{2}\iff {\displaystyle \frac{x^2+y^2}{2}\ge {\left({\displaystyle \frac{x+y}{2}}\right)}^2}}\\ & \text{Kita misalkan}\{\begin{array}{ll}x& =a+{\displaystyle \frac{1}{a}}\\ \\ y& =b+{\displaystyle \frac{1}{b}}\end{array}\\ & \text{maka}\\ & \begin{array}{rl}{\displaystyle \frac{{(a+{\displaystyle \frac{1}{a}})}^2+{(b+{\displaystyle \frac{1}{b}})}^2}{2}}& \ge {\left({\displaystyle \frac{a+{\displaystyle \frac{1}{a}+b+{\displaystyle \frac{1}{b}}}}{2}}\right)}^2\\ & \ge {\displaystyle \frac{1}{4}{(a+b+{\displaystyle \frac{1}{a}+\frac{1}{b}})}^2}\\ & \ge {\displaystyle \frac{1}{4}{(1+{\displaystyle \frac{1}{a}+\frac{1}{b}})}^2}\\ & \ge {\displaystyle \frac{1}{4}{(1+{\displaystyle \frac{a+b}{ab}})}^2}\\ & \ge {\displaystyle \frac{1}{4}{(1+{\displaystyle \frac{1}{ab}})}^2}\end{array}\\ & \text{Dari GM-AM kita akan mendapatkan}\\ & \sqrt{ab}\le {\displaystyle \frac{a+b}{2}\iff ab\le {\left({\displaystyle \frac{a+b}{2}}\right)}^2}\\ & \iff ab\le {\left({\displaystyle \frac{1}{2}}\right)}^2\iff ab\le {\displaystyle \frac{1}{4}}\\ & \text{Sehingga}\\ & \begin{array}{rl}{(a+{\displaystyle \frac{1}{a}})}^2+{(b+{\displaystyle \frac{1}{b}})}^2& \ge {\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{1}{ab}})}^2}\\ & \ge {\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{1}{{\displaystyle \frac{1}{4}}}})}^2}\\ & \ge {\displaystyle \frac{1}{2}{(1+4)}^2}\\ & \ge {\displaystyle \frac{1}{2}\times 25}\\ & \ge {\displaystyle \frac{25}{2}◼}\end{array}\end{array}\end{array}$.

$.\mathbf{\text{versi cara yang lain}}$ silahkan klik di sini

$\begin{array}{l}\\ 27.& \text{Jika}a,b>0,\text{tunjukkan bahwa}\\ & (a^2+1)(b^2+1)\ge (a+b{)}^2\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Dipilih}(ab-1{)}^2\ge 0\\ & \text{Selanjutnya}\\ & (ab-1{)}^2=a^2{b}^2-2ab+1\ge 0\\ & \iff a^2{b}^2+1\ge 2ab\\ & \iff a^2{b}^2+a^2+b^2+1\ge a^2+b^2+2ab\\ & \iff (a^2+1)(b^2+1)\ge (a+b{)}^2◼\end{array}\end{array}$.

$\begin{array}{l}\\ 28.& \text{Jika}a,b>0,\text{tunjukkan bahwa}\\ & {\displaystyle \frac{a^3+b^3}{2}\ge {\left({\displaystyle \frac{a+b}{2}}\right)}^3}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Dipilih}(a-b{)}^2\ge 0\\ & \text{Selanjutnya}\\ & (a+b)(a-b{)}^2\ge 0\\ & \iff (a+b)(a^2-2ab+b^2)\ge 0\\ & \iff a^3+b^3-a^{2}b-a{b}^2\ge 0\\ & \iff a^3+b^3\ge a^{2}b+a{b}^2\\ & \iff 3a^3+3b^3\ge 3a^{2}b+3a{b}^2\\ & \iff 4a^3+4b^3\ge a^3+b^3+3a^{2}b+3a{b}^2\\ & \iff 4a^3+4b^3\ge (a+b{)}^3\\ & \iff {\displaystyle \frac{1}{8}(4a^3+4b^3)\ge {\displaystyle \frac{1}{8}(a+b{)}^3}}\\ & \iff {\displaystyle \frac{a^3+b^3}{2}\ge {\left({\displaystyle \frac{a+b}{2}}\right)}^3◼}\end{array}\end{array}$ .

$.\mathbf{\text{versi cara yang lain}}$ silahkan klik di sini

$\begin{array}{l}\\ 29.& \text{Jika}a\ge b>0,\text{tunjukkan bahwa}\\ & a+{\displaystyle \frac{1}{b(a-b)}\ge 3}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}a+{\displaystyle \frac{1}{b(a-b)}}& =a-b+b+{\displaystyle \frac{1}{b(a-b)}}\\ \text{Dengan AM}& -\text{GM diperoleh}\\ & =a-b+b+{\displaystyle \frac{1}{b(a-b)}}\\ & \ge 3\sqrt[3]{(a-b)\times b\times {\displaystyle \frac{1}{b(a-b)}}}\\ & \ge 3\sqrt[3]{1}\\ & \ge 3◼\end{array}\end{array}$.

$\begin{array}{l}\\ 30.& \mathbf{\text{(OSN 2008)}}\\ & \text{Jika}a,b>0,\text{tunjukkan bahwa}\\ & {\displaystyle \frac{1}{{(1+\sqrt{a})}^2}+{\displaystyle \frac{1}{{(1+\sqrt{b})}^2}\ge {\displaystyle \frac{2}{a+b+2}}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Dipilih}\\ & \begin{array}{rl}{(1+\sqrt{a})}^2& =1+a+2\sqrt{a}\\ & =2+2a-1-a+2\sqrt{a}\\ & =2+2a-(1+a-2\sqrt{a})\\ & =2+2a-(1-\sqrt{a}{)}^2\\ \text{sehingga}& \\ {(1+\sqrt{a})}^2& =2+2a-(1-\sqrt{a}{)}^2\\ {(1+\sqrt{a})}^2& \le 2+2a\\ \text{atau}& \\ {\displaystyle \frac{1}{{(1+\sqrt{a})}^2}}& \ge {\displaystyle \frac{1}{2+2a}}\end{array}\\ & \text{Selanjutnya kembali ke bentuk soal}\\ & \text{yaitu}:\\ & \begin{array}{rl}{\displaystyle \frac{1}{{(1+\sqrt{a})}^2}+{\displaystyle \frac{1}{{(1+\sqrt{b})}^2}}}& \ge {\displaystyle \frac{1}{2+2a}+\frac{1}{2+2b}}\\ & \ge {\displaystyle \frac{1}{2}\left({\displaystyle \frac{1}{1+a}+\frac{1}{1+b}}\right)}\\ \text{Dengan AM-HM dipero}& \text{leh}\\ & \ge {\displaystyle \frac{1}{2}\left({\displaystyle \frac{2.2}{{\displaystyle \frac{1}{\frac{1}{1+a}}+{\displaystyle \frac{1}{\frac{1}{1+b}}}}}}\right)}\\ & \ge {\displaystyle \frac{2}{(1+a)+(1+b)}}\\ & \ge {\displaystyle \frac{2}{a+b+2}◼}\end{array}\end{array}\end{array}$.

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)

$\begin{array}{l}\\ 21.& \text{Diketahui}a,b,c\text{sisi segitiga}\\ & \text{dengan}abc=1,\text{tunjukkan bahwa}\\ & a+b+c\ge 1\\ \\ & \mathbf{\text{Bukti}}\\ & \text{perhatikan bukti no.18 berikut}\\ & \begin{array}{rl}& a+b+c\ge 3\sqrt[3]{abc}\\ & \text{dengan mengganti nilai}abc=1,\\ & \text{maka akan didapatkan}\\ & a+b+c\ge 3\sqrt[3]{1}\\ & \iff a+b+c\ge 3.1\\ & \iff a+b+c\ge 3◼\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Jika}a\text{bilangan real positif}\\ & \text{tunjukkan bahwa}{a}^2+{\displaystyle \frac{2}{a}\ge 3}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui}{a}^2+{\displaystyle \frac{2}{a}\ge 3,\text{dengan}a>0}\\ & \text{maka dengan mengalikan dengan}a\\ & \text{bentuknya menjadi}{a}^3+2\ge 3a\\ & \text{Dengan AM-GM}\\ & {\displaystyle \frac{a^3+1+1}{3}\ge \sqrt[3]{a^3\times 1\times 1}}\\ & \iff {\displaystyle \frac{a^3+2}{3}\ge \sqrt[3]{a^3}}\\ & \iff {\displaystyle \frac{a^3+2}{3}\ge a}\\ & \iff a^3+2\ge 3a\\ & \iff a^3\ge 3a-2◼\end{array}\end{array}$.

$\begin{array}{l}\\ 23.& \text{Buktikan bahwa untuk bilangan}a\text{tidak}\\ & \text{negatif}\text{berlaku}{a}^3\ge 3a-2\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & a^3\ge 3a-2\\ & \iff a^3\ge a+2a-2\\ & \iff a^3-a\ge 2a-2\\ & \iff a(a^2-1)\ge 2(a-1)\\ & \iff a(a+1)(a-1)\ge 2(a-1)\\ & \iff (a^2+a)(a-1)\ge 2(a-1)\\ & \iff (a^2+a)(a-1)-2(a-1)\ge 0\\ & \iff (a-1)(a^2+a-2)\ge 0\\ & \iff (a-1)(a-1)(a+2)\ge 0\\ & \iff (a-1{)}^2(a+2)\ge 0◼\\ \\ & \text{Alternatif 2}\\ & \text{Lihat bukti no.22 di atas}\end{array}\end{array}$.

$\begin{array}{l}\\ 24.& \text{Jika}a,b,m,n\text{bilangan real yang}\\ & \text{memenuhi}{a}^2+b^2=1\text{dan}{m}^2+n^2=1\\ & \text{Tunjukkan bahwa}|am+bn|\le 1\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & (a^2+b^2)(m^2+n^2)-(am+bn{)}^2\\ & =a^2{m}^2+a^2{n}^2+b^2{m}^2+b^2{n}^2-a^2{m}^2-b^2{n}^2-2abmn\\ & =a^2{n}^2+b^2{m}^2-2abmn\\ & =(an-bm{)}^2\\ & \text{Sekarang kita punya}\\ & (an-bm{)}^2\ge 0\\ & \iff (a^2+b^2)(m^2+n^2)-(am+bn{)}^2\ge 0\\ & \iff 1\times 1-(am+bn{)}^2\ge 0\\ & \iff 1-(am+bn{)}^2\ge 0\\ & \iff (am+bn{)}^2-1\le 0\\ & \iff (am+bn{)}^2\le 1◼\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \text{Jika}a+b=1\text{dengan}a,b>0\\ & \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{a^2}{a+1}+{\displaystyle \frac{b^2}{b+1}\ge {\displaystyle \frac{1}{3}}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahi}a+b=1,a,b>0\\ & \text{akan ditunjukkan}\\ & {\displaystyle \frac{a^2}{a+1}+{\displaystyle \frac{b^2}{b+1}\ge {\displaystyle \frac{1}{3}}}}\\ & \text{Perhatikan bahwa}\\ & {\displaystyle \frac{a^2}{a+1}+{\displaystyle \frac{b^2}{b+1}}}\\ & ={\displaystyle \frac{a.a}{(1-b)+1}+{\displaystyle \frac{b.b}{(1-a)+1}}}\\ & ={\displaystyle \frac{a(1-b)}{2-b}+\frac{b(1-a)}{2-a}}\\ & ={\displaystyle \frac{a-ab}{2-b}+\frac{b-ab}{2-a}}\\ & ={\displaystyle \frac{(2-a)(a-ab)+(2-b)(b-ab)}{(2-a)(2-b)}}\\ & ={\displaystyle \frac{2a-2ab-a^2+a^{2}b+2b-2ab-b^2+a{b}^2}{4-2(a+b)+ab}}\\ & ={\displaystyle \frac{2(a+b)-4ab-(a^2+b^2)+ab(a+b)}{4-2(a+b)+ab}}\\ & ={\displaystyle \frac{2(a+b)-4ab-(1-2ab)+ab(a+b)}{4-2(a+b)+ab}}\\ & ={\displaystyle \frac{2(1)-4ab-1+2ab+ab(1)}{4-2(1)+ab}}\\ & ={\displaystyle \frac{1-ab}{2+ab}}\\ & \text{Selanjutnya dengan AM-GM kita peroleh}\\ & \begin{array}{|c|}\hline \begin{array}{rl}& a+b\ge 2\sqrt{ab}\\ & \iff 1\ge 2\sqrt{ab}\\ & \iff {\displaystyle \frac{1}{2}\ge \sqrt{ab}}\\ & \iff {\displaystyle \frac{1}{4}\ge ab}\end{array}\\ \hline\end{array}\\ & \text{Sehingga}\\ & \ge {\displaystyle \frac{1-{\displaystyle \frac{1}{4}}}{2+{\displaystyle \frac{1}{4}}}}\\ & \ge {\displaystyle \frac{{\displaystyle \frac{3}{4}}}{{\displaystyle \frac{9}{4}}}}\\ & \ge {\displaystyle \frac{1}{3}◼}\end{array}\end{array}$.

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)

 $\begin{array}{l}\\ 16.& \text{Buktikan bahwa setiap bilangan real}\\ & \text{positif}a\text{dan}b\text{berlaku}{a}^2+b^2\ge 2ab\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa}(a-b{)}^2\ge 0\\ & \text{adalah benar, maka}\\ & (a-b{)}^2=a^2-2ab+b^2\ge 0\\ & \iff a^2+b^2\ge 2ab◼\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Buktikan bahwa setiap bilangan real}\\ & \text{positif}a,b\text{dan}c\text{berlaku}\\ & a^2+b^2+c^2\ge ab+ac+bc\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa}(a-b{)}^2\ge 0\\ & (a-c{)}^2\ge 0,\text{dan}(b-c{)}^2\ge 0\\ & \text{adalah benar, maka}\\ & (a-b{)}^2=a^2-2ab+b^2\ge 0\\ & \iff a^2+b^2\ge 2ab.....(1)\\ & \text{Dengan cara yang kurang lebih sama}\\ & \text{akan didapatkan}\\ & \bullet a^2+c^2\ge 2ac.....(2)\\ & \bullet b^2+c^2\ge 2bc.....(1)\\ & \text{Jika ketaksamaan}(1),(2),\mathrm{\&}(3)\text{dijumlahkan}\\ & \text{akan didapatkan bentuk}\\ & 2a^2+2b^2+2c^2\ge 2ab+2ac+2bc\\ & \iff a^2+b^2+c^2\ge ab+ac+bc◼\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Untuk}a,b,c\text{bilangan real, dengan}\\ & a^2+b^2+c^2=7\text{dan}ab+bc+ca=4\\ & \text{Tentukan nilai terbesar dari}ab\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Pada}\text{no. 17}\text{di atas telah ditunjukkan}\\ & 2(a^2+b^2+c^2)\ge 2ab+2bc+2ca\\ & \text{dari bukti di atas, kita mendapatkan}\\ & \text{bentuk}2(a^2+b^2+c^2)\ge 2ab+bc+ca\\ & \text{Sehingga}\\ & 2(a^2+b^2+c^2)\ge ab+ab+bc+ca\\ & \iff 2(7)\ge ab+(4)\\ & \iff 14-4\ge ab\\ & \iff 10\ge ab\\ & \text{Jadi, maksimum nilai}ab\text{adalah 10}\end{array}\end{array}$.

$\begin{array}{l}\\ 19.& \text{Diketahui}a,b,c\text{sisi segitiga}\\ & \text{Tunjukkan bahwa}{\displaystyle \frac{a+b+c}{3}\ge \sqrt[3]{abc}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}a+b+c+\sqrt[3]{abc}& =(a+b)+(c+\sqrt[3]{abc})\\ & \ge 2\sqrt{ab}+2\sqrt{c\times \sqrt[3]{abc}}\\ & \ge 2\left(2\sqrt{\sqrt{ab}\times \sqrt{c\times \sqrt[3]{abc}}}\right)\\ & \ge 4\sqrt{\sqrt{abc}\times \sqrt{\sqrt[3]{abc}}}\\ & \ge 4\sqrt{(abc{)}^{{.}^{\frac{1}{2}}}\times (abc{)}^{{.}^{\frac{1}{6}}}}\\ & \ge 4\sqrt{(abc{)}^{{.}^{\frac{1}{2}+\frac{1}{6}}}}\\ & \ge 4\sqrt{(abc{)}^{{.}^{\frac{2}{3}}}}\\ & \ge 4\sqrt{(abc{)}^{{.}^{\frac{1}{3}}}}\\ & \ge 4\sqrt[3]{abc}\\ a+b+c& \ge 4\sqrt[3]{abc}-\sqrt[3]{abc}\\ & \ge 3\sqrt[3]{abc}\\ {\displaystyle \frac{a+b+c}{3}}& \ge \sqrt[3]{abc}◼\end{array}\end{array}$.

$\begin{array}{l}\\ 20.& \text{Diketahui}a,b,c\text{sisi segitiga}\\ & \text{Tunjukkan bahwa}{\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 3}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa sebelumnya}\\ & \text{telah ditunjukkan untuk sembarang}\\ & a,b,c\text{real positif akan berlaku}\\ & {\displaystyle \frac{a+b+c}{3}\ge \sqrt[3]{abc}}\\ & \text{lihat bukti yang ditunjukkan no.18}\\ & \text{di atas},\text{maka}\\ & {\displaystyle \frac{{\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}}}{3}\ge \sqrt[3]{{\displaystyle \frac{a}{b}\times \frac{b}{c}\times \frac{c}{a}}}}\\ & \iff {\displaystyle \frac{{\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}}}{3}\ge 1}\\ & \iff {\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 3◼}\end{array}\end{array}$.

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)

 

$\begin{array}{l}\\ 11.& \text{Nilai dari}\\ & \sin {\displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {\displaystyle \frac{1}{16}}}& & & \text{d}.& {\displaystyle \frac{1}{2}}\\ \\ \text{b}.& {\displaystyle \frac{1}{8}}& \text{c}.& {\displaystyle \frac{1}{4}}& \text{e}.& {\displaystyle 1}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Perhat}& \text{ikan bahwa}\\ \sin {\displaystyle {\displaystyle \frac{\pi }{14}}}& =\sin \left({\displaystyle \frac{7\pi }{14}-\frac{6\pi }{14}}\right)=\sin \left({\displaystyle \frac{1}{2}\pi -\frac{6\pi }{14}}\right)\\ & =\cos {\displaystyle \frac{6\pi }{14}}\\ \sin {\displaystyle \frac{3\pi }{14}}& =...=\cos {\displaystyle \frac{4\pi }{14}}\\ \sin {\displaystyle \frac{9\pi }{14}}& =...=\sin {\displaystyle \frac{5\pi }{14}=\cos {\displaystyle \frac{2\pi }{14}}}\end{array}\\ & ...\\ & \sin {\displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}}\\ & =\cos {\displaystyle \frac{6\pi }{14}\cos {\displaystyle \frac{4\pi }{14}\cos {\displaystyle \frac{2\pi }{14}\times {\displaystyle \frac{2\sin {\displaystyle \frac{2\pi }{14}}}{2\sin {\displaystyle \frac{2\pi }{14}}}}}}}\\ & ={\displaystyle \frac{\cos {\displaystyle \frac{6\pi }{14}\cos {\displaystyle \frac{4\pi }{14}\sin {\displaystyle \frac{4\pi }{14}}}}}{2\sin {\displaystyle \frac{2\pi }{14}}}}\\ & \text{silahkan dilanjutkan}\\ & ...\\ & ={\displaystyle \frac{1}{8}}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Nilai dari}\\ & \cos {\displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos {\displaystyle \frac{8\pi }{5}}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -{\displaystyle \frac{1}{16}}}& & & \text{d}.& {\displaystyle \frac{1}{16}}\\ \\ \text{b}.& {\displaystyle \frac{1}{8}}& \text{c}.& {\displaystyle 0}& \text{e}.& {\displaystyle \frac{1}{8}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \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 (\pi +{\displaystyle \frac{3\pi }{5}})}\\ & =\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{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}(\sin \pi -\sin {\displaystyle \frac{3\pi }{5}})}}{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}(\sin \pi -\sin (-{\displaystyle \frac{\pi }{5}}))}}}{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}}}}\\ & =-{\displaystyle \frac{1}{16}}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 14.& \text{Nilai dari}\\ & \sin {18}^{\circ }\cos {36}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{6}}& & & \text{d}.& {\displaystyle \frac{1}{3}}\\ \\ \text{b}.& {\displaystyle \frac{1}{5}}& \text{c}.& {\displaystyle {\displaystyle \frac{1}{4}}}& \text{e}.& {\displaystyle \frac{1}{2}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \sin {18}^{\circ }\cos {36}^{\circ }\\ & =\sin {18}^{\circ }\cos {36}^{\circ }\times {\displaystyle \frac{2\cos {18}^{\circ }}{2\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\cos {36}^{\circ }(\sin {36}^{\circ }+\sin 0^{\circ })}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\cos {36}^{\circ }\sin {36}^{\circ }}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\sin {72}^{\circ }}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\sin ({90}^{\circ }-{18}^{\circ })}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\cos {18}^{\circ }}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{1}{4}}\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Nilai eksak dari}\sin {36}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {\displaystyle \frac{1}{4}\sqrt{10+2\sqrt{5}}}}& & & \text{d}.& {\displaystyle \frac{\sqrt{5}-1}{4}}\\ \text{b}.& {\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}}}& & & \text{e}.& {\displaystyle \frac{\sqrt{5}-1}{2}}\\ \text{c}.& {\displaystyle {\displaystyle \frac{\sqrt{5}+1}{4}}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikanlah ilustrasi gambar berikut}\end{array}$.

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

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.