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.