Contoh Soal 5 (Segitiga dan Ketaksamaan)

$\begin{array}{ll}\\ 21.&\textrm{Diketahui}\: \: a,\: b,\: c\: \: \textrm{sisi segitiga}\\ &\textrm{dengan}\: \: abc=1,\: \textrm{tunjukkan bahwa}\\  &a+b+c\geq 1\\\\ &\textbf{Bukti}\\ &\textrm{perhatikan bukti no.18 berikut}\\   &\begin{aligned}&a+b+c\geq 3\sqrt[3]{abc}\\ &\textrm{dengan mengganti nilai}\: \: abc=1,\\ &\textrm{maka akan didapatkan}\\ &a+b+c\geq 3\sqrt[3]{1} \\ &\Leftrightarrow \: a+b+c\geq 3.1\\ &\Leftrightarrow \: a+b+c\geq 3\qquad \blacksquare   \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textrm{Jika}\: \: a\: \: \textrm{bilangan real positif}\\  &\textrm{tunjukkan bahwa}\: \: a^{2}+\displaystyle \frac{2}{a}\geq 3\\\\ &\textbf{Bukti}\\   &\begin{aligned}&\textrm{Diketahui}\: \: a^{2}+\displaystyle \frac{2}{a}\geq 3,\: \textrm{dengan}\: a> 0\\ &\textrm{maka dengan mengalikan dengan}\: \: a\\ &\textrm{bentuknya menjadi}\: \: a^{3}+2\geq 3a\\ &\textrm{Dengan AM-GM}\\ &\displaystyle \frac{a^{3}+1+1}{3}\geq \sqrt[3]{a^{3}\times 1\times 1}\\ &\Leftrightarrow \: \displaystyle \frac{a^{3}+2}{3}\geq \sqrt[3]{a^{3}}\\ &\Leftrightarrow \: \displaystyle \frac{a^{3}+2}{3}\geq a\\ &\Leftrightarrow \: \: a^{3}+2\geq 3a\\ &\Leftrightarrow \: \qquad a^{3}\geq 3a-2 \qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 23.&\textrm{Buktikan bahwa untuk bilangan}\: \: a\: \: \textrm{tidak}\\ &\textrm{negatif}\: \:  \textrm{berlaku}\: \: a^{3}\geq 3a-2\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\color{blue}\textrm{Alternatif 1}\\ &a^{3}\geq 3a-2\\ &\Leftrightarrow \: a^{3}\geq a+2a-2\\ &\Leftrightarrow \: a^{3}-a\geq 2a-2\\ &\Leftrightarrow \: a(a^{2}-1)\geq 2(a-1)\\ &\Leftrightarrow \: a(a+1)(a-1)\geq 2(a-1)\\ &\Leftrightarrow \: (a^{2}+a)(a-1)\geq 2(a-1)\\ &\Leftrightarrow \: (a^{2}+a)(a-1)- 2(a-1)\geq 0\\ &\Leftrightarrow \: (a-1)(a^{2}+a-2)\geq 0\\ &\Leftrightarrow \: (a-1)(a-1)(a+2)\geq 0\\ &\Leftrightarrow \: (a-1)^{2}(a+2)\geq 0\qquad \blacksquare \\\\  &\color{blue}\textrm{Alternatif 2}\\ &\textrm{Lihat bukti no.22 di atas} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 24.&\textrm{Jika}\: \: a,b,m,n\: \: \textrm{bilangan real yang}\\  &\textrm{memenuhi}\: \: a^{2}+b^{2}=1\: \: \textrm{dan}\: \: m^{2}+n^{2}=1\\ &\textrm{Tunjukkan bahwa}\quad  \left | am+bn \right |\leq 1\\\\ &\textbf{Bukti}\\   &\begin{aligned}&\color{blue}\textrm{Perhatikan bahwa}\\ &(a^{2}+b^{2})(m^{2}+n^{2})-(am+bn)^{2}\\ &=a^{2}m^{2}+a^{2}n^{2}+b^{2}m^{2} +b^{2}n^{2}-a^{2}m^{2}-b^{2}n^{2}-2abmn\\ &=a^{2}n^{2}+b^{2}m^{2}-2abmn\\ &=(an-bm)^{2}\\ &\color{red}\textrm{Sekarang kita punya}\\ &(an-bm)^{2}\geq 0\\ &\Leftrightarrow \: (a^{2}+b^{2})(m^{2}+n^{2})-(am+bn)^{2}\geq 0\\ &\Leftrightarrow \: \quad 1\times 1-(am+bn)^{2}\geq 0\\ &\Leftrightarrow \: \quad\qquad 1-(am+bn)^{2}\geq 0\\ &\Leftrightarrow \: \quad\qquad (am+bn)^{2}-1\leq  0\\ &\Leftrightarrow \: \: \: \: \: \qquad\qquad (am+bn)^{2}\leq  1\qquad \blacksquare  \end{aligned}  \end{array}$

$\begin{array}{ll}\\ 25.&\textrm{Jika}\: \: a+b=1\: \: \textrm{dengan}\: \: a,b> 0\\ &\textrm{Tunjukkan bahwa}\\ &\qquad\qquad  \displaystyle \frac{a^{2}}{a+1}+\displaystyle \frac{b^{2}}{b+1}\geq \displaystyle \frac{1}{3}\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\textrm{Diketahi}\: \: a+b=1,\: a,b>0\\ &\textrm{akan ditunjukkan}\\\ &\displaystyle \frac{a^{2}}{a+1}+\displaystyle \frac{b^{2}}{b+1}\geq \displaystyle \frac{1}{3}\\ &\textrm{Perhatikan bahwa}\\ &\displaystyle \frac{a^{2}}{a+1}+\displaystyle \frac{b^{2}}{b+1}\\ &=\displaystyle \frac{a.a}{(1-b)+1}+\displaystyle \frac{b.b}{(1-a)+1}\\ &=\displaystyle \frac{a(1-b)}{2-b}+\frac{b(1-a)}{2-a}\\ &=\displaystyle \frac{a-ab}{2-b}+\frac{b-ab}{2-a}\\ &=\displaystyle \frac{(2-a)(a-ab)+(2-b)(b-ab)}{(2-a)(2-b)}\\ &=\displaystyle \frac{2a-2ab-a^{2}+a^{2}b+2b-2ab-b^{2}+ab^{2}}{4-2(a+b)+ab}\\ &=\displaystyle \frac{2(a+b)-4ab-(a^{2}+b^{2})+ab(a+b)}{4-2(a+b)+ab}\\ &=\displaystyle \frac{2(a+b)-4ab-(1-2ab)+ab(a+b)}{4-2(a+b)+ab}\\ &=\displaystyle \frac{2(1)-4ab-1+2ab+ab(1)}{4-2(1)+ab}\\ &=\displaystyle \frac{1-ab}{2+ab}\\ &\color{blue}\textrm{Selanjutnya dengan AM-GM kita peroleh}\\ &\begin{array}{|c|}\hline \begin{aligned}&a+b\geq 2\sqrt{ab}\\ &\Leftrightarrow 1\geq 2\sqrt{ab}\\ &\Leftrightarrow \displaystyle \frac{1}{2}\geq \sqrt{ab}\\ &\Leftrightarrow  \color{red}\displaystyle \frac{1}{4}\geq ab  \end{aligned}\\\hline \end{array}\\ &\textrm{Sehingga}\\ &\geq \displaystyle \frac{1-\displaystyle \frac{1}{4}}{2+\displaystyle \frac{1}{4}}\\ &\geq \displaystyle \frac{\displaystyle \frac{3}{4}}{\displaystyle \frac{9}{4}}\\ &\geq \displaystyle \frac{1}{3}\qquad \blacksquare  \end{aligned}  \end{array}$.

DAFTAR PUSTAKA

1. Bambang, S. 2012. Materi, Soal dan Penyelesaian Olimpiade Matematika Tingkat SMA/MA. Jakarta: BINA PRESTASI INSANI.
2. Susyanto, N. 2012. Tutor Senior Olimpiade Matematika Lima Benua Tingkat SMP. Yogyakarta: KENDI MAS MEDIA.
3. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.