Contoh Soal 9 (Segitiga dan Ketaksamaan)

$\begin{array}{l}\\ 41.& \text{Jika}a,b,c>0,\text{tunjukkan bahwa}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge {\displaystyle \frac{3}{2}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& ((a+b)+(b+c)+(c+a))\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge 9\\ & \iff 2(a+b+c)\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge 9\\ & \iff (a+b+c)\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge {\displaystyle \frac{9}{2}}\\ & \iff {\displaystyle \frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\ge {\displaystyle \frac{9}{2}}}\\ & \iff 1+{\displaystyle \frac{c}{a+b}+1+{\displaystyle \frac{a}{b+c}+1+{\displaystyle \frac{b}{c+a}\ge {\displaystyle \frac{9}{2}}}}}\\ & \iff 3+{\displaystyle \frac{c}{a+b}+{\displaystyle \frac{a}{b+c}+{\displaystyle \frac{b}{c+a}\ge {\displaystyle \frac{9}{2}}}}}\\ & \iff {\displaystyle \frac{c}{a+b}+{\displaystyle \frac{a}{b+c}+{\displaystyle \frac{b}{c+a}\ge {\displaystyle \frac{9}{2}-3}}}}\\ & \iff {\displaystyle \frac{c}{a+b}+{\displaystyle \frac{a}{b+c}+{\displaystyle \frac{b}{c+a}\ge {\displaystyle \frac{3}{2}◼}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 42.& \text{Jika}a,b,c\text{adalah sisi-sisi segitiga}\\ & \text{ABC, tunjukkan bahwa}\\ & {\displaystyle \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\ge 3}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa dengan AM-HM diperoleh}\\ & (a+b+c)\left({\displaystyle \frac{1}{b+c-a}+\frac{1}{c+a-b}+\frac{1}{a+b-c}}\right)\ge 9\\ & \iff ((b+c-a)+(c+a-b)+(a+b-c))\\ & \times \left({\displaystyle \frac{1}{b+c-a}+\frac{1}{c+a-b}+\frac{1}{a+b-c}}\right)\ge 9\\ & \iff {\displaystyle \frac{a+b+c}{b+c-a}+\frac{a+b+c}{c+a-b}+\frac{a+b+c}{a+b-c}\ge 9}\\ & \iff 1+{\displaystyle \frac{2a}{b+c-a}+1+{\displaystyle \frac{2b}{c+a-b}+1+{\displaystyle \frac{2c}{a+b-c}\ge 9}}}\\ & \iff 3+{\displaystyle \frac{2a}{b+c-a}+{\displaystyle \frac{2b}{c+a-b}+{\displaystyle \frac{2c}{a+b-c}\ge 9}}}\\ & \iff {\displaystyle \frac{2a}{b+c-a}+{\displaystyle \frac{2b}{c+a-b}+{\displaystyle \frac{2c}{a+b-c}\ge 6}}}\\ & \iff {\displaystyle \frac{a}{b+c-a}+{\displaystyle \frac{b}{c+a-b}+{\displaystyle \frac{c}{a+b-c}\ge 3◼}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 43.& \text{Diketahui bilangan real positif}a,b,c\\ & \text{dengan}abc=1.\text{Buktikan bahwa}\\ & {\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\ge 3}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}abc=1\\ & \text{Perhatikan bahwa}\\ & \{\begin{array}{ll}\bullet & {\displaystyle \frac{1+ab}{1+a}+{\displaystyle \frac{abc+ab}{1+a}=ab\left({\displaystyle \frac{1+c}{1+a}}\right)}}\\ \bullet & {\displaystyle \frac{1+bc}{1+b}+{\displaystyle \frac{abc+bc}{1+b}=bc\left({\displaystyle \frac{1+a}{1+b}}\right)}}\\ \bullet & {\displaystyle \frac{1+ca}{1+c}+{\displaystyle \frac{abc+ca}{1+c}=ac\left({\displaystyle \frac{1+b}{1+c}}\right)}}\end{array}\\ & \text{Jika hasilnya dijumlahkan},\text{maka}\\ & \text{kita akan mendapatkan hasil}\\ & ab\left({\displaystyle \frac{1+c}{1+a}}\right)+bc\left({\displaystyle \frac{1+a}{1+b}}\right)+ac\left({\displaystyle \frac{1+b}{1+c}}\right)\\ & \text{Selanjutnya dengan AM-GM akan diperoleh}\\ & {\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}}\\ & =ab\left({\displaystyle \frac{1+c}{1+a}}\right)+bc\left({\displaystyle \frac{1+a}{1+b}}\right)+ac\left({\displaystyle \frac{1+b}{1+c}}\right)\ge 3\sqrt[3]{(abc{)}^2}\\ & \iff {\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\ge 3\sqrt[3]{(abc{)}^2}}\\ & \iff {\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\ge 3\sqrt[3]{1}}\\ & \iff {\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\ge 3◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 44.& (\mathbf{\text{SEAMO III-Malaysia}})\\ & \text{Misalkan}a,b,c\text{bilangan real positif}\\ & \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge {\displaystyle \frac{3}{1+abc}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & {\displaystyle \frac{1}{a(1+b)}+\frac{1}{1+abc}=\frac{1}{1+abc}\left({\displaystyle \frac{1+a}{a(1+b)}+\frac{b(1+c)}{(1+b)}}\right)}\\ & \text{Sehingga}\\ & {\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc}}\\ & ={\displaystyle \frac{1}{1+abc}\left({\displaystyle \frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a}}\right)}\\ & \text{Secara AM-GM kita mendapatkan}\\ & \left({\displaystyle \frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a}}\right)\ge 6\\ & \text{Kembali ke soal}\\ & {\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc}\ge {\displaystyle \frac{1}{1+abc}(6)}}\\ & \iff {\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge {\displaystyle \frac{3}{1+abc}◼}}\end{array}\end{array}$.

$\begin{array}{l}\\ 45.& \text{Misalkan}a,b,c\text{bilangan real non negatif}\\ & \text{dengan}a+b+c=1,\text{Tunjukkan bahwa}\\ & {\displaystyle \frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}\ge {\displaystyle \frac{9}{10}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui}a+b+c=1\\ & \text{Dengan AM-GM kita memiliki}\\ & {\displaystyle \frac{a+b+c}{3}\ge \sqrt[3]{abc}\iff {\displaystyle \frac{1}{3}\ge \sqrt[3]{abc}\iff {\displaystyle \frac{1}{27}\ge abc}}}\\ & \text{Sehingga}\\ & {\displaystyle \frac{1}{1+3abc}\ge {\displaystyle \frac{1}{1+3\left({\displaystyle \frac{1}{27}}\right)}\ge {\displaystyle \frac{1}{1+{\displaystyle \frac{1}{9}}}\ge {\displaystyle \frac{9}{10}}}}}\\ & \text{Selanjutnya dengan perluasan AM-HM}\\ & \text{Sebagaimana bentuk berikut}\\ & \left({\displaystyle \frac{p_1{a}_1+p_2{a}_2+p_3{a}_3}{p_1+p_2+p_3}}\right)\ge \left({\displaystyle \frac{p_1+p_2+p_3}{p_1{a}_1^{-1}+p_2{a}_2^{-1}+p_3{a}_3^{-1}}}\right)\\ & \text{maka}\\ & {\displaystyle \frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}\ge {\displaystyle \frac{1}{a(1+bc)+b(1+ac)+c(1+ab)}}}\\ & \ge {\displaystyle \frac{1}{a+b+c+3abc}}\\ & \ge {\displaystyle \frac{1}{1+3abc}}\\ & \ge {\displaystyle \frac{9}{10}◼}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Bambang, S. 2012. Materi, Soal dan Penyelesaian Olimpiade Matematika Tingkat SMA/MA. Jakarta: BINA PRESTASI INSANI.
2. Bintari, N., Gunarto, D. 2007. Panduan Menguasai Soal-Soal Olimpiade Matematika Nasional dan Internasional. Yogyakarta: INDONESIA CERDAS.
3. Tung, K.Y. 2013. Ayo Raih Medali Emas Olimpiade Matematika SMA. Yogyakarta: ANDI.