Contoh Soal 5 (Segitiga dan Ketaksamaan)

$\begin{array}{l}\\ 21.& \text{Diketahui}a,b,c\text{sisi segitiga}\\ & \text{dengan}abc=1,\text{tunjukkan bahwa}\\ & a+b+c\ge 1\\ \\ & \mathbf{\text{Bukti}}\\ & \text{perhatikan bukti no.18 berikut}\\ & \begin{array}{rl}& a+b+c\ge 3\sqrt[3]{abc}\\ & \text{dengan mengganti nilai}abc=1,\\ & \text{maka akan didapatkan}\\ & a+b+c\ge 3\sqrt[3]{1}\\ & \iff a+b+c\ge 3.1\\ & \iff a+b+c\ge 3◼\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Jika}a\text{bilangan real positif}\\ & \text{tunjukkan bahwa}{a}^2+{\displaystyle \frac{2}{a}\ge 3}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui}{a}^2+{\displaystyle \frac{2}{a}\ge 3,\text{dengan}a>0}\\ & \text{maka dengan mengalikan dengan}a\\ & \text{bentuknya menjadi}{a}^3+2\ge 3a\\ & \text{Dengan AM-GM}\\ & {\displaystyle \frac{a^3+1+1}{3}\ge \sqrt[3]{a^3\times 1\times 1}}\\ & \iff {\displaystyle \frac{a^3+2}{3}\ge \sqrt[3]{a^3}}\\ & \iff {\displaystyle \frac{a^3+2}{3}\ge a}\\ & \iff a^3+2\ge 3a\\ & \iff a^3\ge 3a-2◼\end{array}\end{array}$.

$\begin{array}{l}\\ 23.& \text{Buktikan bahwa untuk bilangan}a\text{tidak}\\ & \text{negatif}\text{berlaku}{a}^3\ge 3a-2\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & a^3\ge 3a-2\\ & \iff a^3\ge a+2a-2\\ & \iff a^3-a\ge 2a-2\\ & \iff a(a^2-1)\ge 2(a-1)\\ & \iff a(a+1)(a-1)\ge 2(a-1)\\ & \iff (a^2+a)(a-1)\ge 2(a-1)\\ & \iff (a^2+a)(a-1)-2(a-1)\ge 0\\ & \iff (a-1)(a^2+a-2)\ge 0\\ & \iff (a-1)(a-1)(a+2)\ge 0\\ & \iff (a-1{)}^2(a+2)\ge 0◼\\ \\ & \text{Alternatif 2}\\ & \text{Lihat bukti no.22 di atas}\end{array}\end{array}$.

$\begin{array}{l}\\ 24.& \text{Jika}a,b,m,n\text{bilangan real yang}\\ & \text{memenuhi}{a}^2+b^2=1\text{dan}{m}^2+n^2=1\\ & \text{Tunjukkan bahwa}|am+bn|\le 1\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & (a^2+b^2)(m^2+n^2)-(am+bn{)}^2\\ & =a^2{m}^2+a^2{n}^2+b^2{m}^2+b^2{n}^2-a^2{m}^2-b^2{n}^2-2abmn\\ & =a^2{n}^2+b^2{m}^2-2abmn\\ & =(an-bm{)}^2\\ & \text{Sekarang kita punya}\\ & (an-bm{)}^2\ge 0\\ & \iff (a^2+b^2)(m^2+n^2)-(am+bn{)}^2\ge 0\\ & \iff 1\times 1-(am+bn{)}^2\ge 0\\ & \iff 1-(am+bn{)}^2\ge 0\\ & \iff (am+bn{)}^2-1\le 0\\ & \iff (am+bn{)}^2\le 1◼\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \text{Jika}a+b=1\text{dengan}a,b>0\\ & \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{a^2}{a+1}+{\displaystyle \frac{b^2}{b+1}\ge {\displaystyle \frac{1}{3}}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahi}a+b=1,a,b>0\\ & \text{akan ditunjukkan}\\ & {\displaystyle \frac{a^2}{a+1}+{\displaystyle \frac{b^2}{b+1}\ge {\displaystyle \frac{1}{3}}}}\\ & \text{Perhatikan bahwa}\\ & {\displaystyle \frac{a^2}{a+1}+{\displaystyle \frac{b^2}{b+1}}}\\ & ={\displaystyle \frac{a.a}{(1-b)+1}+{\displaystyle \frac{b.b}{(1-a)+1}}}\\ & ={\displaystyle \frac{a(1-b)}{2-b}+\frac{b(1-a)}{2-a}}\\ & ={\displaystyle \frac{a-ab}{2-b}+\frac{b-ab}{2-a}}\\ & ={\displaystyle \frac{(2-a)(a-ab)+(2-b)(b-ab)}{(2-a)(2-b)}}\\ & ={\displaystyle \frac{2a-2ab-a^2+a^{2}b+2b-2ab-b^2+a{b}^2}{4-2(a+b)+ab}}\\ & ={\displaystyle \frac{2(a+b)-4ab-(a^2+b^2)+ab(a+b)}{4-2(a+b)+ab}}\\ & ={\displaystyle \frac{2(a+b)-4ab-(1-2ab)+ab(a+b)}{4-2(a+b)+ab}}\\ & ={\displaystyle \frac{2(1)-4ab-1+2ab+ab(1)}{4-2(1)+ab}}\\ & ={\displaystyle \frac{1-ab}{2+ab}}\\ & \text{Selanjutnya dengan AM-GM kita peroleh}\\ & \begin{array}{|c|}\hline \begin{array}{rl}& a+b\ge 2\sqrt{ab}\\ & \iff 1\ge 2\sqrt{ab}\\ & \iff {\displaystyle \frac{1}{2}\ge \sqrt{ab}}\\ & \iff {\displaystyle \frac{1}{4}\ge ab}\end{array}\\ \hline\end{array}\\ & \text{Sehingga}\\ & \ge {\displaystyle \frac{1-{\displaystyle \frac{1}{4}}}{2+{\displaystyle \frac{1}{4}}}}\\ & \ge {\displaystyle \frac{{\displaystyle \frac{3}{4}}}{{\displaystyle \frac{9}{4}}}}\\ & \ge {\displaystyle \frac{1}{3}◼}\end{array}\end{array}$.

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.