$\begin{array}{ll}\\ 66.&\textrm{Jika}\: \: x,y\: \: \textrm{bilangan positif, tunjukkan}\\ &x^{2}+y^{2}\geq \displaystyle \frac{(x+y)^{2}}{2}\\\\ &\textbf{Bukti}:\\ &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\\ &(x-y)^{2}\geq 0\Leftrightarrow x^{2}-2xy+y^{2}\geq 0\\ &\Leftrightarrow x^{2}+y^{2}\geq 2xy\\ &\Leftrightarrow 2x^{2}+2y^{2}\geq x^{2}+y^{2}+2xy\\ &\Leftrightarrow 2(x^{2}+y^{2})\geq (x+y)^{2}\\ &\Leftrightarrow x^{2}+y^{2}\geq \displaystyle \frac{(x+y)^{2}}{2}\qquad \blacksquare \end{aligned}\\ &\color{blue}\textrm{Alternatif 2}\\ &\textrm{Dengan Ketaksamaan Holder pilih}\\ &\color{red}\lambda _{1}=\lambda _{2}=\displaystyle \frac{1}{2}\: \left ( \displaystyle \sum_{i=1}^{2}\lambda _{i}=1 \right )\: \: \color{black}\textrm{akan diperoleh}\\ &\begin{aligned}&\left ( \displaystyle \frac{x^{2}}{1}+\frac{y^{2}}{1} \right )^{\frac{1}{2}}(1+1)^{\frac{1}{2}}\geq x+y\\ &\Leftrightarrow (x^{2}+y^{2})(1+1)\geq (x+y)^{2}\\ &\Leftrightarrow (x^{2}+y^{2})\geq \displaystyle \frac{(x+y)^{2}}{2}\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 67.&\textrm{Untuk}\: \: x,y\: \: \textrm{bilangan positif, tunjukkan}\\ &\textrm{kebenaran ketaksamaan Cauchy-Schwarz}\\ &\displaystyle \frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+\frac{x_{3}^{2}}{y_{3}}+\cdots +\frac{x_{n}^{2}}{y_{n}}\geq \displaystyle \frac{(x_{1}+x_{2}+\cdots +x_{n})^{2}}{y_{1}+y_{2}+\cdots +y_{n}}\\\\ &\textbf{Bukti}:\\ &\begin{aligned}&\textrm{Dengan Ketaksamaan Holder pilih}\\ &\color{red}\lambda _{1}=\lambda _{2}=\displaystyle \frac{1}{2}\: \left ( \displaystyle \sum_{i=1}^{2}\lambda _{i}=1 \right )\: \: \color{black}\textrm{akan diperoleh}\\ &\left ( \displaystyle \frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+\cdots +\frac{x_{n}^{2}}{y_{n}} \right )^{\frac{1}{2}}(y_{1}+y_{2}+\cdots +y_{n})^{\frac{1}{2}}\geq (x_{1}+x_{2}+\cdots +x_{n})\\ &\left ( \displaystyle \frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+\cdots +\frac{x_{n}^{2}}{y_{n}} \right )(y_{1}+y_{2}+\cdots +y_{n})\geq (\sqrt{x_{1}^{2}}+\sqrt{x_{2}^{2}}+\cdots +\sqrt{x_{n}^{2}})^{2}\\ &\left ( \displaystyle \frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+\cdots +\frac{x_{n}^{2}}{y_{n}} \right )(y_{1}+y_{2}+\cdots +y_{n})\geq (x_{1}+x_{2}+\cdots +x_{n})^{2}\\ &\left ( \displaystyle \frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+\cdots +\frac{x_{n}^{2}}{y_{n}} \right )\geq \displaystyle \frac{(x_{1}+x_{2}+\cdots +x_{n})^{2}}{(y_{1}+y_{2}+\cdots +y_{n})}\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 68.&(\textbf{National Mathematical Contest, Belarus-2000})\\ &\textrm{Jika}\: \: a,b,c,x,y\: \: \textrm{dan}\: \: z\: \: \textrm{bilangan real positif, tunjukkan}\\ &\displaystyle \frac{a^{3}}{x}+\frac{b^{3}}{y}+\frac{c^{3}}{z}\geq \displaystyle \frac{(a+b+c)^{3}}{3(x+y+z)}\\\\ &\textbf{Bukti}:\\ &\begin{aligned}&\textrm{Dengan Ketaksamaan Holder pilih}\\ &\color{red}\lambda _{1}=\lambda _{2}=\lambda _{3}=\displaystyle \frac{1}{3}\: \left ( \displaystyle \sum_{i=1}^{3}\lambda _{i}=1 \right )\: \: \color{black}\textrm{akan diperoleh}\\ &\left ( \displaystyle \frac{a^{3}}{x}+\frac{b^{3}}{y}+\frac{c^{3}}{z} \right )^{\frac{1}{3}}(1+1+1)^{\frac{1}{3}}(x+y+z)^{\frac{1}{3}}\geq (a+b+c)\\ &\Leftrightarrow \left ( \displaystyle \frac{a^{3}}{x}+\frac{b^{3}}{y}+\frac{c^{3}}{z} \right )(3)(x+y+z)\geq (a+b+c)^{3}\\ &\Leftrightarrow \left ( \displaystyle \frac{a^{3}}{x}+\frac{b^{3}}{y}+\frac{c^{3}}{z} \right )\geq \displaystyle \frac{(a+b+c)^{3}}{3(x+y+z)}\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 69.&\textrm{Jika}\: \: a,b,c,x,y\: \: \textrm{dan}\: \: z\: \: \textrm{bilangan real positif}\\ &\textrm{dengan}\: \: a+b+c=x+y+z\: \: \textrm{tunjukkan bahwa}\\ &\displaystyle \frac{a^{3}}{x^{2}}+\frac{b^{3}}{y^{2}}+\frac{c^{3}}{z^{2}}\geq a+b+c\\\\ &\textbf{Bukti}:\\ &\begin{aligned}&\textrm{Dengan Ketaksamaan Holder pilih}\\ &\color{red}\lambda _{1}=\lambda _{2}=\lambda _{3}=\displaystyle \frac{1}{3}\: \left ( \displaystyle \sum_{i=1}^{3}\lambda _{i}=1 \right )\: \: \color{black}\textrm{akan diperoleh}\\ &\left ( \displaystyle \frac{a^{3}}{x^{2}}+\frac{b^{3}}{y^{2}}+\frac{c^{3}}{z^{2}} \right )^{\frac{1}{3}}(x+y+z)^{\frac{1}{3}}(x+y+z)^{\frac{1}{3}}\geq (a+b+c)\\ &\Leftrightarrow \left ( \displaystyle \frac{a^{3}}{x^{2}}+\frac{b^{3}}{y^{2}}+\frac{c^{3}}{z^{2}} \right )(x+y+z)^{2}\geq (a+b+c)^{3}\\ &\Leftrightarrow \left ( \displaystyle \frac{a^{3}}{x^{2}}+\frac{b^{3}}{y^{2}}+\frac{c^{3}}{z^{2}} \right )(a+b+c)^{2}\geq (a+b+c)^{3}\\ &\Leftrightarrow \left ( \displaystyle \frac{a^{3}}{x}+\frac{b^{3}}{y}+\frac{c^{3}}{z} \right )\geq \displaystyle \frac{(a+b+c)^{3}}{(a+b+c)^{2}}\\ &\Leftrightarrow \displaystyle \frac{a^{3}}{x^{2}}+\frac{b^{3}}{y^{2}}+\frac{c^{3}}{z^{2}}\geq a+b+c\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 70.&\textrm{Jika}\: \: a,b,c\: \: \textrm{adalah bilangan real positif}\\ &\textrm{dengan}\: \: a+b+c=1,\: \textrm{tunjukkan}\\ &\textrm{bahwa}\: \: \left ( \displaystyle \frac{1}{a}+1 \right )\left ( \displaystyle \frac{1}{b}+1 \right )\left ( \displaystyle \frac{1}{c}+1 \right )\geq 64\\\\ &\textbf{Bukti}:\\ &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned} &\left ( \displaystyle \frac{1}{a}+1 \right )\left ( \displaystyle \frac{1}{b}+1 \right )\left ( \displaystyle \frac{1}{c}+1 \right )\\ &=\displaystyle \frac{1}{abc}+\left (\displaystyle \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} \right )+\left (\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )+1\\ &\textrm{Dengan AM-GM kita mendapatkan}\\ &\bullet \: \: \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3\sqrt[3]{\displaystyle \frac{1}{abc}}\\ &\bullet \: \: \displaystyle \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq 3\sqrt[3]{\displaystyle \frac{1}{(abc)^{2}}}\\ &\textrm{Kita tulis sintak prosesnya di atas}\\ &=1+\left (\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )+\left (\displaystyle \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} \right )+\displaystyle \frac{1}{abc}\\ &\geq 1+3\sqrt[3]{\displaystyle \frac{1}{abc}}+3\sqrt[3]{\displaystyle \frac{1}{(abc)^{2}}}+\sqrt[3]{\displaystyle \frac{1}{(abc)^{3}}}\\ &\geq 1+3\sqrt[3]{\displaystyle \frac{1}{abc}}+3\sqrt[3]{\displaystyle \frac{1}{(abc)^{2}}}+\sqrt[3]{\displaystyle \frac{1}{(abc)^{3}}}\\ &=\left ( 1+\displaystyle \frac{1}{\sqrt[3]{abc}} \right )^{3}\\ &\textrm{Karena}\: \: \sqrt[3]{abc}\leq \displaystyle \frac{a+b+c}{3}=\displaystyle \frac{1}{3},\: \textrm{maka}\\ &\left ( \displaystyle \frac{1}{a}+1 \right )\left ( \displaystyle \frac{1}{b}+1 \right )\left ( \displaystyle \frac{1}{c}+1 \right )\geq \left ( 1+\displaystyle \frac{1}{\sqrt[3]{abc}} \right )^{3}\\ &\: \qquad\qquad\qquad\qquad\qquad\quad\quad \geq \left ( 1+\displaystyle \frac{1}{\left (\frac{1}{3} \right )} \right )^{3}\\ &\: \qquad\qquad\qquad\qquad\qquad\quad\quad \geq \left ( 1+3 \right )^{4}\\ &\: \qquad\qquad\qquad\qquad\qquad\quad\quad \geq 4^{4}\\ &\: \qquad\qquad\qquad\qquad\qquad\quad\quad \geq 64\qquad \blacksquare \end{aligned}\\ &\begin{aligned}&\color{blue}\textrm{Alternatif 2}\\ &\textrm{Dengan Ketaksamaan Holder diperoleh}\\ &\left ( \displaystyle \frac{1}{a}+1 \right )\left ( \displaystyle \frac{1}{b}+1 \right )\left ( \displaystyle \frac{1}{c}+1 \right )\geq \left ( \sqrt[3]{\displaystyle \frac{1}{a}.\frac{1}{b}.\frac{1}{c}}+\sqrt[3]{1.1.1} \right )^{3}\\ &\qquad\qquad\qquad\: \qquad\qquad\qquad \geq \left ( \sqrt[3]{\displaystyle \frac{1}{(abc)}}+1 \right )^{3}\\ &\qquad\qquad\qquad\: \qquad\qquad\qquad \geq \left ( \displaystyle \frac{1}{\sqrt[3]{abc}}+1 \right )^{3}\\ &\qquad\qquad\qquad\: \qquad\qquad\qquad \geq \left ( 3+1 \right )^{3}\\ &\qquad\qquad\qquad\: \qquad\qquad\qquad \geq 64\qquad \blacksquare \end{aligned} \end{array}$.
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