$.\quad\qquad\begin{aligned}&\color{red}\textrm{Mengenal penulisan pola}\: \textbf{Siklik dan Simetri}\\ &\textrm{Misal untuk}\: \: n=3,\: \: \textrm{pada penulisan unsur}\\ &x,y,\: \: \textrm{dan}\: \: z,\: \textrm{maka}\\ &\begin{array}{|l|l|}\hline \textbf{Pola Siklik}&\textbf{Pola Simetri}\\\hline\begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}=x^{2}+y^{2}+z^{2}\\ &\\ &\\ &\\ &\\ & \end{aligned} &\begin{aligned}\displaystyle\sum_{\textrm{sym}}^{.}x^{2}&=x^{2}+x^{2}\\ &+y^{2}+y^{2}\\ \\ &+z^{2}+z^{2}\\ \\ &=2\left (x^{2}+y^{2}+z^{2} \right ) \end{aligned}\\ \begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}x^{3}=x^{3}+y^{3}+z^{3}\end{aligned}&\begin{aligned}\displaystyle \sum_{\textrm{sym}}^{.}x^{3}=2\left (x^{3}+y^{3}+z^{3} \right ) \end{aligned}\\ \begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y=x^{2}y+y^{2}z+z^{2}x\\ &\\ &\\ & \end{aligned}&\begin{aligned}\displaystyle \sum_{\textrm{sym}}^{.}x^{2}y&=x^{2}y+x^{2}z\\ &+y^{2}x+y^{2}z\\\\ &+z^{2}x+z^{2}y \end{aligned}\\ \begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}xyz&=xyz+yzx+zxy\\ &=3xyz \end{aligned}&\begin{aligned}\displaystyle \sum_{\textrm{sym}}^{.}xyz&=xyz+xzy+\cdots \\ &=6xyz \end{aligned} \\\hline \end{array} \end{aligned}$.
$\begin{array}{ll}\\ 61.&(\textbf{IMO 1995})\\ &\textrm{Jika}\: \: a,b,c\: \: \textrm{bilangan-bilangan real positif}\\ &\textrm{dengan}\: \: abc=1,\: \: \textrm{maka tunjukkan bahwa}\\ &\displaystyle \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\geq \displaystyle \frac{3}{2}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Misalkan}\: \: x=\displaystyle \frac{1}{a},\: y=\displaystyle \frac{1}{b},\: \: \textrm{dan}\: \: z=\displaystyle \frac{1}{c},\\ & \textrm{maka}\\ &\displaystyle \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\\ &=\displaystyle \frac{x^{3}yz}{y+z}+ \frac{y^{3}xz}{x+z}+ \frac{z^{3}xy}{x+y},\: \: \textrm{karena}\: xyz=1\\ &=\displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y}\\ &\textrm{Dengan ketaksamaan}\: \textbf{Cauchy-Schwarz}\\ &\left ( 2\displaystyle \sum_{\textrm{siklik}}^{.}y+z \right )\left ( \displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right )\geq (x+y+z)^{2}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right )\geq \displaystyle \frac{(x+y+z)^{2}}{2(x+y+z)}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right )\geq \displaystyle \frac{(a+b+c)}{2}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right )\geq \displaystyle \frac{3\sqrt[3]{xyz}}{2}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right )\geq \displaystyle \frac{3}{2}\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 62.&\textrm{Diketahui} \: \: a,b\: \: \textrm{bilangan real positif}\\&\textrm{Tunjukkan bahwa}\: \: \displaystyle \frac{a^{2}}{b}+\frac{b^{2}}{a}\geq a+b\\\\ &\textbf{Bukti}\\ &\textrm{Asumsikan bahwa}\: \: a\geq b,\: \textrm{maka}\: \: a^{2}\geq b^{2}\\&\textrm{dan}\: \: \displaystyle \frac{1}{b}\geq \frac{1}{a}.\\ &\textrm{Perhatikan bahwa baik}\: \left ( a^{2},b^{2} \right )\: \textrm{dan}\: \left ( \displaystyle \frac{1}{b}, \frac{1}{a} \right)\\ &\textrm{adalah kumpulan dua barisan yang monoton}\\ &\textrm{sama yaitu sama-sama naik. Sehingga}\\ &\textrm{dengan}\: \: \textbf{ketaksamaan Renata}\: \textrm{diperoleh}\\ &a^{2}.\displaystyle \frac{1}{b}+b^{2}.\displaystyle \frac{1}{a}\geq a^{2}.\displaystyle \frac{1}{a}+b^{2}.\displaystyle \frac{1}{b}\\ &\Leftrightarrow \displaystyle \frac{a^{2}}{b}+\frac{b^{2}}{a}\geq a+b\qquad \blacksquare \end{array}$.
$\begin{array}{ll}\\ 63.&\textrm{Diberikan}\: \: a,b,c>0,\: \: \textrm{tunjukkan bahwa}\\ &\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+b+c\\\\ &\textbf{Bukti}:\\ &\color{red}\textrm{Alternatif 1}\\ &\textrm{Dengan AM-GM diperoleh}\\ &\bullet \: \: \displaystyle \frac{a}{c}+\frac{c}{a}\geq 2\Leftrightarrow \displaystyle \frac{c}{a}\geq 2-\displaystyle \frac{a}{c}\\ &\bullet \: \: \displaystyle \frac{b}{c}+\frac{c}{b}\geq 2\Leftrightarrow \displaystyle \frac{b}{c}\geq 2-\displaystyle \frac{c}{b}\\ &\bullet \: \: \displaystyle \frac{a}{b}+\frac{b}{a}\geq 2\Leftrightarrow \displaystyle \frac{a}{b}\geq 2-\displaystyle \frac{b}{a}\\ &\textrm{Sehingga}\\ &\begin{aligned}\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}&\geq a\left ( 2-\displaystyle \frac{c}{b} \right )+b\left ( 2-\displaystyle \frac{a}{c} \right )+c\left ( 2-\displaystyle \frac{b}{a} \right )\\ &=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}&\geq 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 )\geq 2(a+b+c)\\ \displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}&\geq a+b+c\qquad \blacksquare \end{aligned}\\ &\color{red}\textrm{Alternatif 2}\\ &\textrm{Dengan ketaksamaan Renata}\\ &\begin{aligned}&\textrm{Asumsikan}\: \: a\geq b\geq c,\: \textrm{maka}\: \: ab\geq ca\geq bc\\ &\textrm{dan}\: \: \displaystyle \frac{1}{c}\geq \frac{1}{b}\geq \frac{1}{a}.\\ &\textrm{Perhatikan bahwa}\\ & (ab\geq ca\geq bc)\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{1}{c}\geq \frac{1}{b}\geq \frac{1}{a} \right )\\ &\textrm{memiliki kemonotonan yang sama}\\ &\textrm{maka dengan}\: \: \textbf{ketaksamaan Renata}\\ &\textrm{dapat diperoleh bentuk}\\ &ab.\displaystyle \frac{1}{c}+ac.\displaystyle \frac{1}{b}+bc.\displaystyle \frac{1}{a}\geq ab.\displaystyle \frac{1}{b}+ac.\displaystyle \frac{1}{a}+bc.\displaystyle \frac{1}{c}\\ &\Leftrightarrow \displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+c+b\\ &\Leftrightarrow \displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+b+c\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 64.&\textrm{Diberikan}\: \: a,b,c>0,\: \: \textrm{tunjukkan kebenaran}\\ &\textbf{ketaksamaan Nesbitt}\: \textrm{berikut}\\ &\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}\\\\ &\textbf{Bukti}:\\ &\color{red}\textrm{Alternatif 1}\\ &\textrm{Dengan AM-GM diperoleh}\\ &\displaystyle \frac{(a+b)+(b+c)+(c+a)}{3}\geq \displaystyle \frac{3}{\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\\ &\Leftrightarrow ((a+b)+(b+c)+(c+a))\left ( \displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right )\geq 9\\ &\Leftrightarrow 2\left ( \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )+6\geq 9\\ &\Leftrightarrow 2\left ( \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )\geq 3\\ &\Leftrightarrow \left ( \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )\geq \displaystyle \frac{3}{2}\qquad \blacksquare \\ &\color{red}\textrm{Alternatif 2}\\ &\textrm{Dengan ketaksamaan Renata}\\ &\begin{aligned}&\textrm{Asumsikan}\: \: a\leq b\leq c,\: \textrm{maka}\: \: a+b\leq a+c\leq b+c\\ &\textrm{dan}\: \: \displaystyle \frac{1}{b+c}\leq \frac{1}{a+c}\leq \frac{1}{a+b}.\\ &\textrm{Perhatikan bahwa}\\ & (a\leq b\leq c)\: \: \textrm{dan}\: \: \displaystyle \frac{1}{b+c}\leq \frac{1}{a+c}\leq \frac{1}{a+b}\\ &\textrm{memiliki kemonotonan yang sama}\\ &\textrm{maka dengan}\: \: \textbf{ketaksamaan Renata}\\ &\textrm{dapat diperoleh bentuk}\\ &a.\displaystyle \frac{1}{b+c}+b.\displaystyle \frac{1}{a+c}+c.\displaystyle \frac{1}{a+b}\geq b.\displaystyle \frac{1}{b+c}+c.\displaystyle \frac{1}{a+c}+a.\displaystyle \frac{1}{a+b}\\ &\textrm{dan}\\ &a.\displaystyle \frac{1}{b+c}+b.\displaystyle \frac{1}{a+c}+c.\displaystyle \frac{1}{a+b}\geq c.\displaystyle \frac{1}{b+c}+a.\displaystyle \frac{1}{a+c}+b.\displaystyle \frac{1}{a+b}\\ &\textrm{Jika dijumlahkan keduanya, maka}\\ &2\left ( \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )\geq 3\\ &\Leftrightarrow \left ( \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )\geq \displaystyle \frac{3}{2}\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 65.&(\textbf{OSN 2015})\\ &\textrm{Diberikan}\: \: a,b,c>0,\: \: \textrm{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}}\geq 3\\\\ &\textbf{Bukti}:\\ &\textrm{Perhatikan bukti soal no. 4 di atas}\\ &\textrm{Dengan}\: \: \textbf{keksamaan Renata}\: \: \textrm{dapat diperoleh}\\ &\displaystyle \frac{a}{b+c}+\frac{b}{a+c}\geq \frac{b}{b+c}+\frac{a}{a+c}\\ &\textrm{Misalkan}\\ &x=b+c,\: y=c+a,\: y=a+b,\: \textrm{maka}\\ &\displaystyle \frac{a}{b+c}+\frac{b}{a+c}\geq \frac{b}{b+c}+\frac{a}{a+c}\\ &\Leftrightarrow \displaystyle \frac{a}{b+c}+\frac{b}{a+c}\geq \frac{y+z-x}{2x}+\frac{x+z-y}{2y}\\ &\begin{aligned} &\Leftrightarrow \sqrt{ \displaystyle \frac{a}{b+c}+\frac{b}{a+c}}\geq \sqrt{\frac{y+z-x}{2x}+\frac{x+z-y}{2y}}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \: = \sqrt{\displaystyle \frac{1}{2}}\sqrt{\displaystyle \frac{y}{x}+\frac{z}{x}-1+\frac{x}{y}+\frac{z}{y}-1}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \: \: \textrm{dengan AM-GM}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \: \geq \displaystyle \frac{1}{\sqrt{2}}\sqrt{\displaystyle \frac{z}{x}+\frac{z}{y}+2\sqrt{\frac{y}{x}.\frac{x}{y}}-2}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \: = \displaystyle \frac{1}{\sqrt{2}}\sqrt{\displaystyle \frac{z}{x}+\frac{z}{y}+2-2}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \: \geq \displaystyle \frac{1}{\sqrt{2}}\sqrt{\displaystyle \frac{z}{x}+\frac{z}{y}}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \: \geq \displaystyle \frac{\sqrt{2}}{\sqrt{2}}\sqrt{\displaystyle \frac{z^{2}}{xy}}=\displaystyle \sqrt{\displaystyle \frac{z^{2}}{xy}}\\ \end{aligned}\\ &\begin{aligned} &\bullet \: \sqrt{ \displaystyle \frac{a}{b+c}+\frac{b}{a+c}} \geq \displaystyle \sqrt{\displaystyle \frac{z^{2}}{xy}}\\ &\bullet \: \sqrt{ \displaystyle \frac{b}{c+a}+\frac{c}{a+b}}\geq \sqrt{\displaystyle \frac{x^{2}}{yz}}\\ &\bullet \: \sqrt{ \displaystyle \frac{c}{a+b}+\frac{a}{b+c}}\geq \sqrt{\displaystyle \frac{y^{2}}{xz}} \end{aligned}\\ &\begin{aligned} &\textrm{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}}\\ &\geq \displaystyle \sqrt{\displaystyle \frac{z^{2}}{xy}}+\sqrt{\displaystyle \frac{x^{2}}{yz}}+\sqrt{\displaystyle \frac{y^{2}}{xz}}\\ &\textrm{Dengan AM-GM lagi}\\ &\geq 3\sqrt[3]{\sqrt{\displaystyle \frac{z^{2}}{xy}}\times \sqrt{\displaystyle \frac{x^{2}}{yz}}\times \sqrt{\displaystyle \frac{y^{2}}{xz}}}\\ &\geq 3\sqrt[3]{\sqrt{\displaystyle \frac{(xyz)^{2}}{(xyz)^{2}}}}\\ &\geq 3\qquad \blacksquare \end{aligned} \end{array}$
DAFTAR PUSTAKA
- Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.
WEBSITE
- https://holdenlee.github.io/high_school/omc/23-rearrange.pdf diakses 18 Januari 2022.
- https://www.gotohaggstrom.com/Advanced%20inequality%20manipulations.pdf diakses 20 Januari 2022
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