$\begin{array}{ll}\\ 66.&\textrm{Diberikan}\: \: a,b,c>0,\: \: \textrm{tunjukkan bahwa}\\ &\displaystyle \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\geq 6\\\\ &\textbf{Bukti}:\\ &\color{red}\textrm{Alternatif 1}\\ &\textrm{Dengan mengaplikasikan AM-GM-HM}\\ &\textrm{pada}\: \: \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\: \: \textrm{kita dapat menemukan}\\ &\color{blue}\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\color{black}\geq \displaystyle \frac{3}{(abc)^{.^{\frac{1}{3}}}}\geq \color{blue}\displaystyle \frac{9}{a+b+c}\\ &\textrm{Jika kedua ruas dikalikan dengan}\: \: \color{red}a+b+c\color{black},\\ &\textrm{maka}\\ &\color{blue}3+\displaystyle \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\color{black}\geq \color{blue}\displaystyle \frac{9(a+b+c)}{a+b+c}\\ &\Leftrightarrow \displaystyle \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\geq 6\qquad \blacksquare \\ &\begin{aligned}&\color{red}\textrm{Alternatif 2}\\ &\textrm{Asumsikan}\: \: a\leq b\leq c,\: \textrm{maka}\: \: a+b\leq a+c\leq b+c\\ &\textrm{dan}\: \: \displaystyle \frac{1}{c}\leq \frac{1}{b}\leq \frac{1}{a}.\\ &\textrm{Perhatikan bahwa}\\ & (a+b\leq a+c\leq b+c)\: \: \textrm{dan}\: \: \displaystyle \frac{1}{c}\leq \frac{1}{b}\leq \frac{1}{a}\\ &\textrm{memiliki kemonotonan yang sama}\\ &\textrm{maka dengan}\: \: \textbf{ketaksamaan Renata}\\ &\textrm{dapat diperoleh bentuk}\\ &(b+c).\displaystyle \frac{1}{a}+(c+a).\displaystyle \frac{1}{b}+(a+b).\displaystyle \frac{1}{c}\geq (b+c).\displaystyle \frac{1}{b}+(c+a).\displaystyle \frac{1}{c}+(a+b).\displaystyle \frac{1}{a}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \doteq 1+\displaystyle \frac{c}{b}+1+\frac{a}{c}+1+\frac{b}{a}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \doteq 3+\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \doteq 3+3\left ( \displaystyle \frac{abc}{abc} \right )^{.^{\frac{1}{3}}}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \doteq 3+3\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \doteq 6\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 67.&\textrm{Diberikan}\: \: a,b,c>0,\: \: \textrm{tunjukkan kebenaran}\\ &\textbf{ketaksamaan Nesbitt}\: \textrm{berikut}\\ &\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}\\\\ &\textbf{Bukti}:\\ &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}&\textrm{Asumsikan},\\ &\color{red}\begin{cases} & a\geq b\geq c \\ & \displaystyle \frac{1}{b+c}\geq \frac{1}{a+c}\geq \frac{1}{a+b} \end{cases} \\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan Chebyshev}\\ &\displaystyle \frac{\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}{3}\geq \displaystyle \frac{(a+b+c)}{3}\left ( \displaystyle \frac{\left (\displaystyle \frac{1}{b+c}+ \frac{1}{a+c}+ \frac{1}{a+b} \right )}{3} \right )\\ &\textrm{Dengan AM-HM dan}\: K=\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\\ &\Leftrightarrow \displaystyle \frac{K}{3}\geq \displaystyle \frac{(a+b+c)}{3}\left ( \displaystyle \frac{3}{(b+c)+(a+c)+(a+b)} \right )\\ &\Leftrightarrow K\geq \displaystyle \frac{3(a+b+c)}{2(a+b+c)}\\ &\Leftrightarrow K\geq \displaystyle \frac{3}{2}\\ &\Leftrightarrow \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \displaystyle \frac{3}{2}\qquad \blacksquare \end{aligned}\\ &\color{blue}\textrm{Alternatif 2}\\ &\begin{aligned}&\color{magenta}\textbf{Pertama},\: \color{black}\textrm{asumsikan}\\ &\color{red}\begin{cases} & a\geq b\geq c \\ & \displaystyle \frac{1}{b+c}\geq \frac{1}{a+c}\geq \frac{1}{a+b} \end{cases} \\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan Chebyshev}\\ &3\left (\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )\geq (a+b+c)\left (\displaystyle \frac{1}{b+c}+ \frac{1}{a+c}+ \frac{1}{a+b} \right )\\ &\color{magenta}\textbf{Kedua},\: \color{black}\textrm{asumsikan}\\ &\color{red}\begin{cases} & a+b\geq a+c\geq b+c \\ & \displaystyle \frac{1}{a+b}\leq \frac{1}{a+c}\leq \frac{1}{b+c} \end{cases} \\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan Chebyshev}\\ &3\left (\displaystyle \frac{a+b}{a+b}+\frac{a+c}{a+c}+\frac{b+c}{b+c} \right )\leq (a+b+a+c+b+c)\left (\displaystyle \frac{1}{a+b}+ \frac{1}{a+c}+ \frac{1}{b+c} \right )\\ &\Leftrightarrow 3(1+1+1)\leq 2(a+b+c)\left (\displaystyle \frac{1}{a+b}+ \frac{1}{a+c}+ \frac{1}{b+c} \right )\\ &\Leftrightarrow \displaystyle \frac{9}{2}\leq (a+b+c)\left (\displaystyle \frac{1}{a+b}+ \frac{1}{a+c}+ \frac{1}{b+c} \right )\\ &\textrm{Dari dua ketaksamaan di atas didapatkan}\\ &3\left (\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )\geq \displaystyle \frac{9}{2}\\ &\Leftrightarrow \left (\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )\geq \displaystyle \frac{3}{2}\qquad \blacksquare \end{aligned} \end{array}$
$\begin{array}{ll}\\ 68.&\textrm{Jika}\: \: a,b,c>0\: ,\: \textrm{dengan}\: \: a\neq b\neq c\\ &\textrm{tunjukkan bahwa}\\ &\left ( a^{3}+b^{3}+c^{3} \right )> \displaystyle \frac{(a+b+c)^{3}}{9}> 3abc\\\\ &\textbf{Bukti}:\\ &\textrm{Dengan}\: \textbf{ketaksamaan Chebyshev}\\ &\textrm{untuk}\: \: a\geq b\geq c,\: \textrm{dapat diperoleh bentuk}\\ &\displaystyle \frac{\left ( a^{3}+b^{3}+c^{3} \right )}{3}> \left ( \displaystyle \frac{a+b+c}{3} \right )\left ( \displaystyle \frac{a+b+c}{3} \right )\left ( \displaystyle \frac{a+b+c}{3} \right )\\ &\Leftrightarrow \displaystyle \frac{\left ( a^{3}+b^{3}+c^{3} \right )}{3}> \left ( \displaystyle \frac{a+b+c}{3} \right )^{3}> \left ( \displaystyle \frac{3\sqrt[3]{abc}}{3} \right )^{3}\\ &\Leftrightarrow \displaystyle \frac{\left ( a^{3}+b^{3}+c^{3} \right )}{3}> \displaystyle \frac{(a+b+c)^{3}}{27}> abc\\ &\Leftrightarrow \left ( a^{3}+b^{3}+c^{3} \right )> \displaystyle \frac{(a+b+c)^{3}}{9}> 3abc\quad \blacksquare \end{array}$.
$\begin{array}{ll}\\ 69.&\textrm{tunjukkan bahwa untuk}\: \: n\: \: \textrm{bilangan asli}\\&\textrm{berlaku}\\ &\displaystyle \frac{1}{\sqrt{n}}\left (1+\sqrt{\displaystyle \frac{1}{2}}+\sqrt{\displaystyle \frac{1}{3}}+\cdots +\sqrt{\displaystyle \frac{1}{n}} \right )\leq (2n-1)^{.^{\frac{1}{4}}}\\\\ &\textbf{Bukti}:\\ &\textrm{Dengan}\: \textbf{ketaksamaan Chebyshev}\\&\textrm{untuk}:\: \left ( 1\geq \displaystyle \frac{1}{2}\geq \frac{1}{3}\geq \cdots \geq \frac{1}{n} \right )\\ &\textrm{dapat diperoleh bentuk berikut}\\ &\left (1+ \displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right )^{2}\leq n\left ( 1+\displaystyle \frac{1}{2.2}+\frac{1}{3.3}+\cdots +\frac{1}{n.n} \right )\\ &\Leftrightarrow \left (1+ \displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right )^{2}\leq n\left ( 1+\displaystyle \frac{1}{1.2}+\frac{1}{2.3}+\cdots +\frac{1}{(n-1).n} \right )\\ &\Leftrightarrow \left (1+ \displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right )^{2}\leq n\left ( 1+\left (1-\displaystyle \frac{1}{2} \right )+\left (\displaystyle \frac{1}{2}-\frac{1}{3} \right )+\cdots +\left (\displaystyle \frac{1}{(n-1)}-\frac{1}{n} \right ) \right )\\ &\Leftrightarrow \left (1+ \displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right )^{2}\leq n\left ( 1+1-\displaystyle \frac{1}{n} \right )=n\left ( 2-\displaystyle \frac{1}{n} \right )\\&\Leftrightarrow \left (1+ \displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right )\leq \sqrt{2n-1}\quad \color{red}..........(1)\\ &\textrm{Gunakan lagi}\: \textbf{ketaksamaan Chebyshev}\\ &\left (1+\sqrt{\displaystyle \frac{1}{2}}+\sqrt{\displaystyle \frac{1}{3}}+\cdots +\sqrt{\displaystyle \frac{1}{n}} \right )^{2}\leq n\left (1+ \displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right )\\ &\Leftrightarrow \displaystyle \frac{1}{n}\left (1+\sqrt{\displaystyle \frac{1}{2}}+\sqrt{\displaystyle \frac{1}{3}}+\cdots +\sqrt{\displaystyle \frac{1}{n}} \right )^{2}\leq \left (1+ \displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right )\: \: \color{red}...(2)\\&\textrm{Dari ketaksamaan (1) dan (2), dapat diperoleh}\\ & \displaystyle \frac{1}{n}\left (1+\sqrt{\displaystyle \frac{1}{2}}+\sqrt{\displaystyle \frac{1}{3}}+\cdots +\sqrt{\displaystyle \frac{1}{n}} \right )^{2}\leq \sqrt{2n-1}=(2n-1)^{.^{\frac{1}{2}}}\\ &\Leftrightarrow \displaystyle \frac{1}{\sqrt{n}}\left (1+\sqrt{\displaystyle \frac{1}{2}}+\sqrt{\displaystyle \frac{1}{3}}+\cdots +\sqrt{\displaystyle \frac{1}{n}} \right )\leq (2n-1)^{.^{\frac{1}{4}}}\qquad \blacksquare \end{array}$.
$\begin{array}{ll}\\ 70.&(\textbf{OSN 2011})\\ &\textrm{Jika}\: \: a,b,c>0\: ,\: \textrm{dengan}\: \: abc=1\\ &\textrm{Jika diketahui}\\ &a^{2011}+b^{2011}+c^{2011}< \displaystyle \frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}\\ &\textrm{tunjukkan bahwa}\\ &(a+b+c)> \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\\\\ &\textbf{Bukti}:\\ &\textrm{Asumsikan}\\&\color{red}\begin{cases} &a\geq b\geq c \\ &\displaystyle \frac{1}{c}\geq \frac{1}{b}\geq \frac{1}{a} \end{cases}\\ &\textrm{Dengan}\: \textbf{ketaksamaan Chebyshev}\\ &\textrm{Perhatikan}\\ &\begin{aligned}&\displaystyle \frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}\geq \displaystyle \frac{1}{3}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\left ( \displaystyle \frac{1}{a^{2010}}+\frac{1}{b^{2010}}+\frac{1}{c^{2010}} \right )\\ &\Leftrightarrow \displaystyle \frac{1}{a^{2010}}+\frac{1}{b^{2010}}+\frac{1}{c^{2010}}\geq \displaystyle \frac{1}{3}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\left ( \displaystyle \frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}} \right )\\ &\Leftrightarrow \displaystyle \frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}\geq \displaystyle \frac{1}{3}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\left ( \displaystyle \frac{1}{a^{2008}}+\frac{1}{b^{2008}}+\frac{1}{c^{2008}} \right )\\ &\qquad\qquad \vdots \\ &\Leftrightarrow \displaystyle \frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}\geq \displaystyle \frac{1}{3}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\left ( \displaystyle \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \right )\\ &\Leftrightarrow \displaystyle \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\geq \displaystyle \frac{1}{3}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\\ &\textrm{Sehingga}\\ &\displaystyle \frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}\geq \displaystyle \frac{1}{3^{2010}}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^{2011}\: \: \color{red}.....(1) \end{aligned} \\&\color{red}\textrm{dan}\\ &\begin{aligned}&a^{2011}+b^{2011}+c^{2011}\geq \displaystyle \frac{1}{3}(a+b+c)(a^{2010}+b^{2010}+c^{2010})\\ &\Leftrightarrow a^{2010}+b^{2010}+c^{2010}\geq \displaystyle \frac{1}{3}(a+b+c)(a^{2009}+b^{2009}+c^{2009})\\ &\Leftrightarrow a^{2009}+b^{2009}+c^{2009}\geq \displaystyle \frac{1}{3}(a+b+c)(a^{2008}+b^{2008}+c^{2008})\\&\qquad\qquad \vdots \\&\Leftrightarrow a^{3}+b^{3}+c^{3}\geq \displaystyle \frac{1}{3}(a+b+c)(a^{2}+b^{2}+c^{2})\\&\Leftrightarrow a^{2}+b^{2}+c^{2}\geq \displaystyle \frac{1}{3}(a+b+c)(a+b+c)\\ &\textrm{Sehingga}\\ &\Leftrightarrow a^{2011}+b^{2011}+c^{2011}\geq \displaystyle \frac{1}{3^{2010}}(a+b+c)^{2011}\: \: \color{red}........(2) \end{aligned}\\ &\begin{aligned}&\color{purple}\textrm{Dari ketaksamaan (1) dan (2) didapatkan}\\ &\frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}\geq \displaystyle \frac{1}{3^{2010}}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^{2011}\\ &>a^{2011}+b^{2011}+c^{2011}\geq \displaystyle \frac{1}{3^{2010}}(a+b+c)^{2011}\\ &\color{blue}\textrm{atau}\\ &\displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{1}{a^{2011}}\geq \displaystyle \frac{1}{3^{2010}}\left ( \displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{1}{a}\right )^{2011}> \displaystyle \sum_{\textrm{siklik}}^{.}a^{2011}\geq \displaystyle \frac{1}{3^{2010}}\left ( \displaystyle \sum_{\textrm{siklik}}^{.} \right )^{2011}\\ &\Leftrightarrow \displaystyle \frac{1}{3^{2010}}\left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^{2011}>\displaystyle \frac{1}{3^{2010}}(a+b+c)^{2011}\\ &\Leftrightarrow \left ( \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )> (a+b+c)\\&\Leftrightarrow \: a+b+c< \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\qquad \blacksquare \end{aligned} \end{array}$.
DAFTAR PUSTAKA
- Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.
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