Contoh Soal 9 (Segitiga dan Ketaksamaan)

$\begin{array}{ll}\\ 41.&\textrm{Jika}\: \: a,b,c> 0\: ,\: \textrm{tunjukkan bahwa}\\ &\qquad  \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \displaystyle \frac{3}{2}\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\left ( (a+b)+(b+c)+(c+a) \right )\left ( \displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right )\geq 9\\ &\Leftrightarrow \: 2(a+b+c)\left ( \displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right )\geq 9\\ &\Leftrightarrow \: (a+b+c)\left ( \displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right )\geq \displaystyle \frac{9}{2}\\ &\Leftrightarrow \: \displaystyle \frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\geq \displaystyle \frac{9}{2}\\ &\Leftrightarrow \: 1+\displaystyle \frac{c}{a+b}+1+\displaystyle \frac{a}{b+c}+1+\displaystyle \frac{b}{c+a}\geq \displaystyle \frac{9}{2}\\ &\Leftrightarrow \: 3+\displaystyle \frac{c}{a+b}+\displaystyle \frac{a}{b+c}+\displaystyle \frac{b}{c+a}\geq \displaystyle \frac{9}{2}\\ &\Leftrightarrow \: \displaystyle \frac{c}{a+b}+\displaystyle \frac{a}{b+c}+\displaystyle \frac{b}{c+a}\geq \displaystyle \frac{9}{2}-3\\ &\Leftrightarrow \: \displaystyle \frac{c}{a+b}+\displaystyle \frac{a}{b+c}+\displaystyle \frac{b}{c+a}\geq \displaystyle \frac{3}{2}\quad \blacksquare  \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 42.&\textrm{Jika}\: \: a,b,c\: \: \textrm{adalah sisi-sisi segitiga}\\  &\textrm{ABC, tunjukkan bahwa}\\ & \quad  \displaystyle \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\geq 3\\\\  &\textbf{Bukti}\\   &\begin{aligned}&\textrm{Perhatikan bahwa dengan AM-HM diperoleh}\\ &(a+b+c)\left (\displaystyle \frac{1}{b+c-a}+\frac{1}{c+a-b}+\frac{1}{a+b-c} \right )\geq 9\\ &\Leftrightarrow \: \left ( (b+c-a)+(c+a-b)+(a+b-c) \right )\\ &\quad \times \left (\displaystyle \frac{1}{b+c-a}+\frac{1}{c+a-b}+\frac{1}{a+b-c} \right )\geq 9\\ &\Leftrightarrow \:  \displaystyle \frac{a+b+c}{b+c-a}+\frac{a+b+c}{c+a-b}+\frac{a+b+c}{a+b-c}\geq 9\\ &\Leftrightarrow \: 1+\displaystyle \frac{2a}{b+c-a}+1+\displaystyle \frac{2b}{c+a-b}+1+\displaystyle \frac{2c}{a+b-c}\geq 9\\ &\Leftrightarrow \: 3+\displaystyle \frac{2a}{b+c-a}+\displaystyle \frac{2b}{c+a-b}+\displaystyle \frac{2c}{a+b-c}\geq 9\\ &\Leftrightarrow \: \displaystyle \frac{2a}{b+c-a}+\displaystyle \frac{2b}{c+a-b}+\displaystyle \frac{2c}{a+b-c}\geq 6\\ &\Leftrightarrow \: \displaystyle \frac{a}{b+c-a}+\displaystyle \frac{b}{c+a-b}+\displaystyle \frac{c}{a+b-c}\geq 3\quad \blacksquare  \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 43.&\textrm{Diketahui bilangan real positif}\: \: a,b,c\\ &\textrm{dengan}\: \:  abc=1\: .\: \textrm{Buktikan bahwa}\\  &\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3  \\\\   &\textbf{Bukti}\\    &\begin{aligned}&\textrm{Diketahui bahwa}\: \: abc=1\\ &\textrm{Perhatikan bahwa}\\ &\begin{cases} \bullet  & \displaystyle \frac{1+ab}{1+a}+\displaystyle \frac{abc+ab}{1+a}=ab\left ( \displaystyle \frac{1+c}{1+a} \right ) \\  \bullet  & \displaystyle \frac{1+bc}{1+b}+\displaystyle \frac{abc+bc}{1+b}=bc\left ( \displaystyle \frac{1+a}{1+b} \right ) \\  \bullet  & \displaystyle \frac{1+ca}{1+c}+\displaystyle \frac{abc+ca}{1+c}=ac\left ( \displaystyle \frac{1+b}{1+c} \right )  \end{cases}\\ &\color{red}\textrm{Jika hasilnya dijumlahkan}\color{black},\: \textrm{maka}\\ &\textrm{kita akan mendapatkan hasil}\\ &ab\left ( \displaystyle \frac{1+c}{1+a} \right )+bc\left ( \displaystyle \frac{1+a}{1+b} \right )+ac\left ( \displaystyle \frac{1+b}{1+c} \right )\\ &\color{blue}\textrm{Selanjutnya dengan AM-GM akan diperoleh}\\ &\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\\ &=ab\left ( \displaystyle \frac{1+c}{1+a} \right )+bc\left ( \displaystyle \frac{1+a}{1+b} \right )+ac\left ( \displaystyle \frac{1+b}{1+c} \right )\geq 3\sqrt[3]{(abc)^{2}}\\ &\Leftrightarrow \: \displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3\sqrt[3]{(abc)^{2}}\\ &\Leftrightarrow \: \displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3\sqrt[3]{1}\\ &\Leftrightarrow \: \displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3\qquad \blacksquare  \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 44.&(\textbf{SEAMO III-Malaysia})\\ &\textrm{Misalkan}\: \: a,b,c\: \: \textrm{bilangan real positif}\\ &\textrm{Tunjukkan bahwa}\\ &\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\geq \displaystyle \frac{3}{1+abc}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\\ &\displaystyle \frac{1}{a(1+b)}+\frac{1}{1+abc}=\frac{1}{1+abc}\left ( \displaystyle \frac{1+a}{a(1+b)}+\frac{b(1+c)}{(1+b)} \right )\\ &\textrm{Sehingga}\\ &\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc}\\ &=\displaystyle \frac{1}{1+abc}\left ( \displaystyle \frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a} \right )\\ &\color{blue}\textrm{Secara AM-GM kita mendapatkan}\\ &\left ( \displaystyle \frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a} \right )\geq 6\\ &\textrm{Kembali ke soal}\\ &\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc}\geq \displaystyle \frac{1}{1+abc}(6)\\ &\Leftrightarrow \: \displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\geq \displaystyle \frac{3}{1+abc}\qquad \blacksquare  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 45.&\textrm{Misalkan}\: \: a,b,c\: \: \textrm{bilangan real non negatif}\\ &\textrm{dengan}\: \: a+b+c=1,\: \: \textrm{Tunjukkan bahwa}\\ &\displaystyle \frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}\geq \displaystyle \frac{9}{10}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Diketahui}\: \: a+b+c=1\\ &\color{blue}\textrm{Dengan AM-GM kita memiliki}\\ &\displaystyle \frac{a+b+c}{3}\geq \sqrt[3]{abc}\Leftrightarrow \displaystyle \frac{1}{3}\geq \sqrt[3]{abc}\Leftrightarrow \displaystyle \frac{1}{27}\geq abc\\ &\textrm{Sehingga}\\ &\displaystyle \frac{1}{1+3abc}\geq \displaystyle \frac{1}{1+3\left ( \displaystyle \frac{1}{27} \right )}\geq \displaystyle \frac{1}{1+\displaystyle \frac{1}{9}}\geq \displaystyle \frac{9}{10}\\ &\color{red}\textrm{Selanjutnya dengan perluasan AM-HM}\\ &\textrm{Sebagaimana bentuk berikut}\\ &\left ( \displaystyle \frac{p_{1}a_{1}+p_{2}a_{2}+p_{3}a_{3}}{p_{1}+p_{2}+p_{3}} \right )\geq \left ( \displaystyle \frac{p_{1}+p_{2}+p_{3}}{p_{1}a_{1}^{-1}+p_{2}a_{2}^{-1}+p_{3}a_{3}^{-1}} \right )\\ &\textrm{maka}\\ &\displaystyle \frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}\geq \displaystyle \frac{1}{a(1+bc)+b(1+ac)+c(1+ab)}\\ &\qquad\qquad\qquad\qquad\qquad\quad  \geq \displaystyle \frac{1}{a+b+c+3abc}\\ &\qquad\qquad\qquad\qquad\qquad\quad  \geq \displaystyle \frac{1}{1+3abc}\\ &\qquad\qquad\qquad\qquad\qquad\quad  \geq \displaystyle \frac{9}{10}\qquad \blacksquare  \end{aligned}   \end{array}$.

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.