$\begin{array}{ll}\\ 46.&\textrm{Misalkan}\: \: a,b,c\: \: \textrm{bilangan real non negatif}\\ &\textrm{dengan}\: \: a+b+c=1,\: \: \textrm{Tunjukkan bahwa}\\ &\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq \displaystyle \frac{9}{2}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Diketahui}\: \: a+b+c=1\\ &\color{blue}\textrm{Dengan AM-GM kita memiliki}\\ &2(a+b+c)\left (\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right )\\ &= ((a+b)+(b+c)+(c+a))\left (\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right )\\ &\geq 3\sqrt[3]{(a+b)(b+c)(c+a)}\times 3\sqrt[3]{\displaystyle \frac{1}{(a+b)(b+c)(c+a)}}\\ &=9\sqrt[3]{\displaystyle \frac{(a+b)(b+c)(c+a)}{(a+b)(b+c)(c+a)}}=9\sqrt[3]{1}=9\\ &\textrm{Sehingga}\\ &2\left (\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right )\geq 9\\ &\Leftrightarrow \: \left (\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right )\geq \displaystyle \frac{9}{2}\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 47.&(\textbf{OSN 2013})\\ &\textrm{Tentukan semua bilangan real}\: \: M\\ &\textrm{sedemikian sehingga untuk sebarang}\\ &\textrm{bilangan real}\: \: a,b,c\: \: \textrm{paling sedikit}\\ &\textrm{satu di antara tiga bilangan berikut}\\ &\qquad a+\displaystyle \frac{M}{ab},\: b+\displaystyle \frac{M}{bc},\: c+\displaystyle \frac{M}{ca}\\ &\textrm{bernilai lebih dari atau sama dengan}\\ & 1+M \\\\ &\textbf{Jawab}:\\ &\begin{aligned} &\textrm{Diketahui bahwa}\\ &\textrm{min}\left \{ a+\displaystyle \frac{M}{ab},\: b+\displaystyle \frac{M}{bc},\: c+\displaystyle \frac{M}{ca} \right \}\geq 1+M\\ &\textrm{Perhatikan bahwa}\\ &a+\displaystyle \frac{M}{ab}+ b+\displaystyle \frac{M}{bc}+ c+\displaystyle \frac{M}{ca}\\ &=a+b+c+\displaystyle \frac{M}{ab}+\displaystyle \frac{M}{bc}+\displaystyle \frac{M}{ca}\\ &=\displaystyle \frac{1}{2}\left ( 2(a+b+c)+\displaystyle \frac{2M}{ab}+\displaystyle \frac{2M}{bc}+\displaystyle \frac{2M}{ca} \right )\\ &=\displaystyle \frac{1}{2}\left ( a+b+\displaystyle \frac{2M}{ab}+b+c+\displaystyle \frac{2M}{bc}+c+a+\displaystyle \frac{2M}{ca} \right )\\ &\color{blue}\textrm{Dengan AM-GM akan diperoleh bentuk}\\ &\geq \displaystyle \frac{1}{2}\left ( 3\sqrt[3]{2M}+3\sqrt[3]{2M}+3\sqrt[3]{2M} \right )\\ &=\displaystyle \frac{9}{2} \sqrt[3]{2M}\\ &\color{red}\textrm{Pilih nilai minimum}\\ &\left \{ a+\displaystyle \frac{M}{ab},\: b+\displaystyle \frac{M}{bc},\: c+\displaystyle \frac{M}{ca} \right \}=1+M=\displaystyle \frac{3\sqrt[3]{2M}}{2}\\ &\textrm{Selanjutnya}\\ &1+M=\displaystyle \frac{3\sqrt[3]{2M}}{2}\\ &\Leftrightarrow \: M=\displaystyle \frac{1}{2} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 48.&(\textbf{OSK 2017})\\ &\textrm{Diketahui bilangan real positif}\: \: a,b,c\\ & \textrm{yang memenuhi}\: \: a+b+c=1.\: \textrm{Nilai}\\ &\textrm{minimum dari}\: \: \displaystyle \frac{a+b}{abc}\: \: \textrm{adalah}\: .\: ...\\\\ &\textbf{Jawab}\\ &\begin{aligned}&\textrm{Diketahui bahwa}\: \: a,b,c\in R^{+}\\ &\textrm{dengan}\: a+b+c=1.\: \textrm{kita dapat peroleh}\\ &a+b=1-c\\ &\textrm{Selanjutnya}\\ &\displaystyle \frac{a+b}{abc}=\displaystyle \frac{a}{abc}+\frac{b}{abc}=\displaystyle \frac{1}{bc}+\frac{1}{ac}\\ &\color{blue}\textrm{Sebelumnya ingat ketaksamaan AM-HM}\\ &\displaystyle \frac{m+n}{2}\geq \displaystyle \frac{2}{\displaystyle \frac{1}{m}+\frac{1}{n}}\Leftrightarrow \displaystyle \frac{1}{m}+\frac{1}{n}\geq \displaystyle \frac{4}{m+n}\\ &\textrm{Sehingga}\\ &\displaystyle \frac{a+b}{abc}=\displaystyle \frac{1}{bc}+\frac{1}{ac}\geq \displaystyle \frac{4}{ac+bc}= \displaystyle \frac{4}{c(a+b)}\\ &\qquad \geq \displaystyle \frac{4}{c(1-c)}=\displaystyle \frac{4}{c-c^{2}}=\displaystyle \frac{4}{\displaystyle \frac{1}{4}-\displaystyle \frac{1}{4}+c-c^{2}}\\ &\qquad =\displaystyle \frac{4}{\displaystyle \frac{1}{4}-\left ( c-\displaystyle \frac{1}{2} \right )^{2}}\\ &\textrm{Saat}\: \: c-\displaystyle \frac{1}{2}=0,\: \textrm{maka akan diperoleh}\\ &\textrm{nilai minimum yaitu}:\\ &\left ( \displaystyle \frac{a+b}{abc} \right )_{\textrm{minimum}}=\displaystyle \frac{4}{\displaystyle \frac{1}{4}-0}=16 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 49.&\textrm{Jika pada soal 48 diubah dengan}\\ &a,b,c\: \: \textrm{adalah sisi segitiga}\: \: ABC\\ & \textrm{yang memenuhi}\: \: a+b+c=1.\\ & \textrm{Tentukan nilai dari}\: \: \displaystyle \frac{a+b}{abc}\\\\ &\textbf{Jawab}\\ &\begin{aligned}&\textrm{Diketahui bahwa}\: \: a,b,c\: \: \textrm{sisi}\: \: \bigtriangleup ABC\\ &\textrm{dengan}\: a+b+c=1.\: \textrm{kita dapat peroleh}\\ &a+b=a+b\\ &\textrm{Ingat bahwa dalam}\: \: \bigtriangleup ABC,\: \textrm{berlaku}\\ &\begin{cases} \bullet & a+b>c \\ \bullet & a+c>b \\ \bullet & b+c>a \end{cases}\\ &\textrm{maka}\: \: a+b+a+b> a+b+c\\ &\Leftrightarrow \: 2(a+b)> 1\Leftrightarrow a+b> \displaystyle \frac{1}{2}\\ &\color{blue}\textrm{Dan dengan AM-GM diperoleh}\\ &\displaystyle \frac{a+b+c}{3}\geq \sqrt[3]{abc}\Leftrightarrow \displaystyle \frac{1}{3}\geq \sqrt[3]{abc}\\ &\Leftrightarrow \displaystyle \frac{1}{27}\geq abc\\ &\textrm{Selanjutnya}\\ &\displaystyle \frac{a+b}{abc}> \displaystyle \frac{\frac{1}{2}}{\frac{1}{27}}=\displaystyle \frac{27}{2}=13,5\\ &\textrm{Jadi, nilai}\: \: \displaystyle \frac{a+b}{abc}>13,5 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 50.&\textrm{Diketahui bilangan real positif}\: \: x_{1},x_{2},...,x_{k}\\ & \textrm{memenuhi persamaan}\: \: x_{1}+x_{2}+...+x_{k}=1\\ &\textrm{Buktikan bahwa}\\ &\displaystyle \frac{1}{(x_{1})^{2013}}+\frac{1}{(x_{2})^{2013}}+...+\frac{1}{(x_{k})^{2013}}\geq k^{2014} \\\\ &\textbf{Bukti}\\ &\begin{aligned}&\color{blue}\textrm{Dengan ketaksamaan AM-GM kita}\\ &\textrm{memiliki}\\ &\displaystyle \frac{x_{1}+x_{2}+...+x_{k}}{k}\geq \sqrt[k]{x_{1}.x_{2}...x_{k}}\\ &\Leftrightarrow \: \displaystyle \frac{1}{k} \geq \sqrt[k]{x_{1}.x_{2}...x_{k}}\\ &\Leftrightarrow \: \sqrt[k]{x_{1}.x_{2}...x_{k}}\leq \displaystyle \frac{1}{k}\\ & \Leftrightarrow \: x_{1}.x_{2}...x_{k}\leq \left ( \displaystyle \frac{1}{k} \right )^{k}\\ &\textrm{Selanjutnya}\\ &\displaystyle \frac{\displaystyle \frac{1}{(x_{1})^{2013}}+\frac{1}{(x_{2})^{2013}}+...+\frac{1}{(x_{k})^{2013}}}{k}\\ &\geq \sqrt[k]{\displaystyle \frac{1}{(x_{1})^{2013}.(x_{2})^{2013}...(x_{k})^{2013}}}\\ &\displaystyle \frac{1}{(x_{1})^{2013}}+\frac{1}{(x_{2})^{2013}}+...+\frac{1}{(x_{k})^{2013}}\\ &\geq k\sqrt[k]{\displaystyle \frac{1}{(x_{1})^{2013}.(x_{2})^{2013}...(x_{k})^{2013}}}\\ &\geq k\sqrt[k]{\displaystyle \frac{1}{(x_{1}.x_{2}...x_{k})^{2013}}}\geq k.\: \sqrt[k]{\displaystyle \frac{1}{\left ( \left ( \displaystyle \frac{1}{k} \right )^{k} \right )^{2013}}}\\ &\geq k.\sqrt[k]{k^{2013k}}\\ &\geq k.k^{2013}\\ &\geq k^{2014}\qquad \blacksquare \end{aligned} \end{array}$.
DAFTAR PUSTAKA
- Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
- Muslim, M.S. 2020. Kumpulan Soal dan Pembahasan Olimpiade Matematika SMA Tahun 2007-2019 Tingkat Kota/Kabupaten. Bandung: YRAMA WIDYA.
- Widodo, T. 2013. Pembahasan OSN Matematika SMA Tahun 2013 Seleksi Tingkat Nasional.
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