Contoh Soal 6 (Segitiga dan Ketaksamaan)

$\begin{array}{l}\\ 26.& \text{Diketahui}a,b\text{bilangan real positif}\\ & \text{dan}a+b=1.\text{Tunjukkan bahwa}\\ & {(a+{\displaystyle \frac{1}{a}})}^2+{(b+{\displaystyle \frac{1}{b}})}^2\ge {\displaystyle \frac{25}{2}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}a+b=1\\ & \text{Dengan QM-AM kita akan mendapatkan}\\ & \sqrt{{\displaystyle \frac{x^2+y^2}{2}}}\ge {\displaystyle \frac{x+y}{2}\iff {\displaystyle \frac{x^2+y^2}{2}\ge {\left({\displaystyle \frac{x+y}{2}}\right)}^2}}\\ & \text{Kita misalkan}\{\begin{array}{ll}x& =a+{\displaystyle \frac{1}{a}}\\ \\ y& =b+{\displaystyle \frac{1}{b}}\end{array}\\ & \text{maka}\\ & \begin{array}{rl}{\displaystyle \frac{{(a+{\displaystyle \frac{1}{a}})}^2+{(b+{\displaystyle \frac{1}{b}})}^2}{2}}& \ge {\left({\displaystyle \frac{a+{\displaystyle \frac{1}{a}+b+{\displaystyle \frac{1}{b}}}}{2}}\right)}^2\\ & \ge {\displaystyle \frac{1}{4}{(a+b+{\displaystyle \frac{1}{a}+\frac{1}{b}})}^2}\\ & \ge {\displaystyle \frac{1}{4}{(1+{\displaystyle \frac{1}{a}+\frac{1}{b}})}^2}\\ & \ge {\displaystyle \frac{1}{4}{(1+{\displaystyle \frac{a+b}{ab}})}^2}\\ & \ge {\displaystyle \frac{1}{4}{(1+{\displaystyle \frac{1}{ab}})}^2}\end{array}\\ & \text{Dari GM-AM kita akan mendapatkan}\\ & \sqrt{ab}\le {\displaystyle \frac{a+b}{2}\iff ab\le {\left({\displaystyle \frac{a+b}{2}}\right)}^2}\\ & \iff ab\le {\left({\displaystyle \frac{1}{2}}\right)}^2\iff ab\le {\displaystyle \frac{1}{4}}\\ & \text{Sehingga}\\ & \begin{array}{rl}{(a+{\displaystyle \frac{1}{a}})}^2+{(b+{\displaystyle \frac{1}{b}})}^2& \ge {\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{1}{ab}})}^2}\\ & \ge {\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{1}{{\displaystyle \frac{1}{4}}}})}^2}\\ & \ge {\displaystyle \frac{1}{2}{(1+4)}^2}\\ & \ge {\displaystyle \frac{1}{2}\times 25}\\ & \ge {\displaystyle \frac{25}{2}◼}\end{array}\end{array}\end{array}$.

$.\mathbf{\text{versi cara yang lain}}$ silahkan klik di sini

$\begin{array}{l}\\ 27.& \text{Jika}a,b>0,\text{tunjukkan bahwa}\\ & (a^2+1)(b^2+1)\ge (a+b{)}^2\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Dipilih}(ab-1{)}^2\ge 0\\ & \text{Selanjutnya}\\ & (ab-1{)}^2=a^2{b}^2-2ab+1\ge 0\\ & \iff a^2{b}^2+1\ge 2ab\\ & \iff a^2{b}^2+a^2+b^2+1\ge a^2+b^2+2ab\\ & \iff (a^2+1)(b^2+1)\ge (a+b{)}^2◼\end{array}\end{array}$.

$\begin{array}{l}\\ 28.& \text{Jika}a,b>0,\text{tunjukkan bahwa}\\ & {\displaystyle \frac{a^3+b^3}{2}\ge {\left({\displaystyle \frac{a+b}{2}}\right)}^3}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Dipilih}(a-b{)}^2\ge 0\\ & \text{Selanjutnya}\\ & (a+b)(a-b{)}^2\ge 0\\ & \iff (a+b)(a^2-2ab+b^2)\ge 0\\ & \iff a^3+b^3-a^{2}b-a{b}^2\ge 0\\ & \iff a^3+b^3\ge a^{2}b+a{b}^2\\ & \iff 3a^3+3b^3\ge 3a^{2}b+3a{b}^2\\ & \iff 4a^3+4b^3\ge a^3+b^3+3a^{2}b+3a{b}^2\\ & \iff 4a^3+4b^3\ge (a+b{)}^3\\ & \iff {\displaystyle \frac{1}{8}(4a^3+4b^3)\ge {\displaystyle \frac{1}{8}(a+b{)}^3}}\\ & \iff {\displaystyle \frac{a^3+b^3}{2}\ge {\left({\displaystyle \frac{a+b}{2}}\right)}^3◼}\end{array}\end{array}$ .

$.\mathbf{\text{versi cara yang lain}}$ silahkan klik di sini

$\begin{array}{l}\\ 29.& \text{Jika}a\ge b>0,\text{tunjukkan bahwa}\\ & a+{\displaystyle \frac{1}{b(a-b)}\ge 3}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}a+{\displaystyle \frac{1}{b(a-b)}}& =a-b+b+{\displaystyle \frac{1}{b(a-b)}}\\ \text{Dengan AM}& -\text{GM diperoleh}\\ & =a-b+b+{\displaystyle \frac{1}{b(a-b)}}\\ & \ge 3\sqrt[3]{(a-b)\times b\times {\displaystyle \frac{1}{b(a-b)}}}\\ & \ge 3\sqrt[3]{1}\\ & \ge 3◼\end{array}\end{array}$.

$\begin{array}{l}\\ 30.& \mathbf{\text{(OSN 2008)}}\\ & \text{Jika}a,b>0,\text{tunjukkan bahwa}\\ & {\displaystyle \frac{1}{{(1+\sqrt{a})}^2}+{\displaystyle \frac{1}{{(1+\sqrt{b})}^2}\ge {\displaystyle \frac{2}{a+b+2}}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Dipilih}\\ & \begin{array}{rl}{(1+\sqrt{a})}^2& =1+a+2\sqrt{a}\\ & =2+2a-1-a+2\sqrt{a}\\ & =2+2a-(1+a-2\sqrt{a})\\ & =2+2a-(1-\sqrt{a}{)}^2\\ \text{sehingga}& \\ {(1+\sqrt{a})}^2& =2+2a-(1-\sqrt{a}{)}^2\\ {(1+\sqrt{a})}^2& \le 2+2a\\ \text{atau}& \\ {\displaystyle \frac{1}{{(1+\sqrt{a})}^2}}& \ge {\displaystyle \frac{1}{2+2a}}\end{array}\\ & \text{Selanjutnya kembali ke bentuk soal}\\ & \text{yaitu}:\\ & \begin{array}{rl}{\displaystyle \frac{1}{{(1+\sqrt{a})}^2}+{\displaystyle \frac{1}{{(1+\sqrt{b})}^2}}}& \ge {\displaystyle \frac{1}{2+2a}+\frac{1}{2+2b}}\\ & \ge {\displaystyle \frac{1}{2}\left({\displaystyle \frac{1}{1+a}+\frac{1}{1+b}}\right)}\\ \text{Dengan AM-HM dipero}& \text{leh}\\ & \ge {\displaystyle \frac{1}{2}\left({\displaystyle \frac{2.2}{{\displaystyle \frac{1}{\frac{1}{1+a}}+{\displaystyle \frac{1}{\frac{1}{1+b}}}}}}\right)}\\ & \ge {\displaystyle \frac{2}{(1+a)+(1+b)}}\\ & \ge {\displaystyle \frac{2}{a+b+2}◼}\end{array}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Bambang, S. 2012. Materi, Soal dan Penyelesaian Olimpiade Matematika Tingkat SMA/MA. Jakarta: BINA PRESTASI INSANI.
2. Yohanes, S., Panji, R. 2008. Mahir Olimpiade Matematika SMA. Yogyakarta: KENDI MAS MEDIA.
3. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.