Contoh Soal 6 (Segitiga dan Ketaksamaan)

$\begin{array}{ll}\\ 26.&\textrm{Diketahui}\: \: a,b\: \: \textrm{bilangan real positif}\\ &\textrm{dan}\: \: a+b=1.\: \textrm{Tunjukkan bahwa}\\ &\qquad  \left ( a+\displaystyle \frac{1}{a} \right )^{2}+\left ( b+\displaystyle \frac{1}{b} \right )^{2}\geq \displaystyle \frac{25}{2}\\\\  &\textbf{Bukti}\\   &\begin{aligned}&\textrm{Diketahui bahwa}\: \: a+b=1\\ &\color{red}\textrm{Dengan QM-AM kita akan mendapatkan}\\ &\sqrt{\displaystyle \frac{x^{2}+y^{2}}{2}}\geq \displaystyle \frac{x+y}{2}\Leftrightarrow \displaystyle \frac{x^{2}+y^{2}}{2}\geq \left ( \displaystyle \frac{x+y}{2} \right )^{2}\\ &\textrm{Kita misalkan}\: \: \color{purple}\begin{cases} x & =a+\displaystyle \frac{1}{a} \\\\  y & =b+\displaystyle \frac{1}{b}  \end{cases}\\ &\textrm{maka}\\ &\begin{aligned}\displaystyle \frac{\left ( a+\displaystyle \frac{1}{a} \right )^{2}+\left (b+\displaystyle \frac{1}{b}  \right )^{2}}{2}&\geq \left ( \displaystyle \frac{a+\displaystyle \frac{1}{a}+b+\displaystyle \frac{1}{b}}{2} \right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( a+b+\displaystyle \frac{1}{a}+\frac{1}{b} \right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( 1+\displaystyle \frac{1}{a}+\frac{1}{b} \right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( 1+ \displaystyle \frac{a+b}{ab}\right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( 1+\displaystyle \frac{1}{ab} \right )^{2} \end{aligned}\\ &\color{blue}\textrm{Dari GM-AM kita akan mendapatkan}\\ &\sqrt{ab}\leq \displaystyle \frac{a+b}{2}\Leftrightarrow ab\leq \left ( \displaystyle \frac{a+b}{2} \right )^{2}\\ &\Leftrightarrow ab\leq \left ( \displaystyle \frac{1}{2} \right )^{2}\Leftrightarrow ab\leq \displaystyle \frac{1}{4} \\ &\textrm{Sehingga}\\ &\begin{aligned}\left ( a+\displaystyle \frac{1}{a} \right )^{2}+\left (b+\displaystyle \frac{1}{b}  \right )^{2}&\geq \displaystyle \frac{1}{2}\left ( 1+\displaystyle \frac{1}{ab} \right )^{2}\\ &\geq \displaystyle \frac{1}{2}\left ( 1+ \displaystyle \frac{1}{\displaystyle \frac{1}{4}}\right )^{2}\\ &\geq \displaystyle \frac{1}{2}\left ( 1+4 \right )^{2}\\ &\geq \displaystyle \frac{1}{2}\times 25\\ &\geq \displaystyle \frac{25}{2}\qquad \blacksquare  \end{aligned}  \end{aligned}  \end{array}$.

$.\: \: \: \quad \textbf{versi cara yang lain}\: $ silahkan klik di sini

$\begin{array}{ll}\\ 27.&\textrm{Jika}\: \: a,b> 0\: ,\: \textrm{tunjukkan bahwa}\\ &\qquad  (a^{2}+1)(b^{2}+1)\geq (a+b)^{2}\\\\  &\textbf{Bukti}\\   &\begin{aligned}&\textrm{Dipilih}\: \: (ab-1)^{2}\geq 0\\ &\textrm{Selanjutnya}\\ &(ab-1)^{2}=a^{2}b^{2}-2ab+1\geq 0\\ &\Leftrightarrow \: a^{2}b^{2}+1\geq 2ab\\ &\Leftrightarrow \: a^{2}b^{2}+a^{2}+b^{2}+1\geq a^{2}+b^{2}+2ab\\ &\Leftrightarrow \: (a^{2}+1)(b^{2}+1)\geq (a+b)^{2}\qquad \blacksquare   \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 28.&\textrm{Jika}\: \: a,b> 0\: ,\: \textrm{tunjukkan bahwa}\\ &\qquad  \displaystyle \frac{a^{3}+b^{3}}{2}\geq \left ( \displaystyle \frac{a+b}{2} \right )^{3}\\\\  &\textbf{Bukti}\\   &\begin{aligned}&\textrm{Dipilih}\: \: (a-b)^{2}\geq 0\\ &\textrm{Selanjutnya}\\ &(a+b)(a-b)^{2}\geq 0\\ &\Leftrightarrow \: (a+b)(a^{2}-2ab+b^{2})\geq 0\\ &\Leftrightarrow \: a^{3}+b^{3}-a^{2}b-ab^{2}\geq 0\\ &\Leftrightarrow \: a^{3}+b^{3}\geq a^{2}b+ab^{2}\\ &\Leftrightarrow \: 3a^{3}+3b^{3}\geq 3a^{2}b+3ab^{2}\\ &\Leftrightarrow \: 4a^{3}+4b^{3}\geq a^{3}+b^{3}+3a^{2}b+3ab^{2}\\ &\Leftrightarrow \: 4a^{3}+4b^{3}\geq (a+b)^{3}\\ &\Leftrightarrow \: \displaystyle \frac{1}{8}\left (4a^{3}+4b^{3}  \right )\geq \displaystyle \frac{1}{8}(a+b)^{3}\\ &\Leftrightarrow \: \displaystyle \frac{a^{3}+b^{3}}{2}\geq \left ( \displaystyle \frac{a+b}{2} \right )^{3}\qquad \blacksquare   \end{aligned}    \end{array}$ .

$.\: \: \: \quad \textbf{versi cara yang lain}\: $ silahkan klik di sini

$\begin{array}{ll}\\ 29.&\textrm{Jika}\: \: a\geq b> 0\: ,\: \textrm{tunjukkan bahwa}\\  &\qquad  a+\displaystyle \frac{1}{b(a-b)}\geq 3\\\\  &\textbf{Bukti}\\   &\begin{aligned}a+\displaystyle \frac{1}{b(a-b)}&=a-b+b+\displaystyle \frac{1}{b(a-b)}\\ \color{blue}\textrm{Dengan AM}&\color{blue}-\textrm{GM diperoleh}\\ &=a-b+b+\displaystyle \frac{1}{b(a-b)}\\ &\geq 3\sqrt[3]{(a-b)\times b\times \displaystyle \frac{1}{b(a-b)}}\\ &\geq 3\sqrt[3]{1}\\ &\geq 3\qquad \blacksquare   \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 30.&\textbf{(OSN 2008)}\\ &\textrm{Jika}\: \: a,b> 0\: ,\: \textrm{tunjukkan bahwa}\\  &\quad  \displaystyle \frac{1}{\left ( 1+\sqrt{a} \right )^{2}}+\displaystyle \frac{1}{\left ( 1+\sqrt{b} \right )^{2}}\geq \displaystyle \frac{2}{a+b+2}\\\\  &\textbf{Bukti}\\   &\begin{aligned}&\textrm{Dipilih}\\ &\begin{aligned}\left ( 1+\sqrt{a} \right )^{2}&=1+a+2\sqrt{a}\\ &=2+2a-1-a+2\sqrt{a}\\ &=2+2a-(1+a-2\sqrt{a})\\ &=2+2a-(1-\sqrt{a})^{2}\\ \textrm{sehingga}\: \: &\\ \left ( 1+\sqrt{a} \right )^{2}&=2+2a-(1-\sqrt{a})^{2}\\ \left ( 1+\sqrt{a} \right )^{2}&\leq 2+2a\\ \textrm{atau}\qquad\: &\\ \displaystyle \frac{1}{\left (1+\sqrt{a}  \right )^{2}}&\geq \displaystyle \frac{1}{2+2a} \end{aligned}\\ &\textrm{Selanjutnya kembali ke bentuk soal}\\ &\textrm{yaitu}:\\ &\begin{aligned}\displaystyle \frac{1}{\left ( 1+\sqrt{a} \right )^{2}}+\displaystyle \frac{1}{\left ( 1+\sqrt{b} \right )^{2}}&\geq \displaystyle \frac{1}{2+2a}+\frac{1}{2+2b}\\ &\geq \displaystyle \frac{1}{2}\left ( \displaystyle \frac{1}{1+a}+\frac{1}{1+b} \right )\\ \textrm{Dengan AM-HM dipero}&\textrm{leh}\\ &\geq \displaystyle \frac{1}{2}\left (\displaystyle \frac{2.2}{\displaystyle \frac{1}{\frac{1}{1+a}}+\displaystyle \frac{1}{\frac{1}{1+b}}} \right )\\ &\geq \displaystyle \frac{2}{(1+a)+(1+b)}\\ &\geq \displaystyle \frac{2}{a+b+2}\qquad \blacksquare  \end{aligned}   \end{aligned}  \end{array}$.

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.