Contoh Soal 7 (Segitiga dan Ketaksamaan)

$\begin{array}{ll}\\ 31.&(\textbf{Russia 1992})\\ &\textrm{Jika}\: \: a,b> 1\: ,\: \textrm{buktikan bahwa}\\ &\qquad  \displaystyle \frac{x^{2}}{y-1}+\displaystyle \frac{y^{2}}{x-1}\geq 8\\\\  &\textbf{Bukti}\\   &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}&\textrm{Gunakan AM-GM untuk mendapatkan}\\ &\displaystyle \frac{x^{2}}{y-1}+\displaystyle \frac{y^{2}}{x-1}\geq 2\sqrt{\displaystyle \frac{x^{2}y^{2}}{\sqrt{(x-1)(y-1)}}}\\ &\: \qquad\qquad\qquad \geq \displaystyle \frac{2xy}{\sqrt{(x-1)(y-1)}}\\ &\textrm{Sebelum kita lanjutkan, ingat bahwa}\\ &(x-2)^{2}\geq 0\Leftrightarrow x^{2}-4x+4\geq 0\\ &\Leftrightarrow x^{2}\geq 4x-4\Leftrightarrow x^{2}\geq 4(x-1)\\ &\Leftrightarrow \displaystyle \frac{x^{2}}{x-1}\geq 4\Leftrightarrow \displaystyle \frac{x}{\sqrt{x-1}}\geq 2\\ &\textrm{Demikian juga}\: \: \displaystyle \frac{y}{\sqrt{y-1}}\geq 2\\ &\textrm{Selanjutnya kembali ke semula yaitu}\\ &\: \qquad\qquad\qquad \geq \displaystyle \frac{2xy}{\sqrt{(x-1)(y-1)}}\\ &\: \qquad\qquad\qquad \geq \displaystyle 2.2.2\\ &\displaystyle \frac{x^{2}}{y-1}+\displaystyle \frac{y^{2}}{x-1}\geq 8\qquad \blacksquare  \end{aligned}\\ &\color{blue}\textrm{Alternatif 2}\\ &\begin{aligned}&\textrm{Misalkan saja}\\ &\begin{cases} a & =x-1 \Rightarrow x=a+1\\  b & =y-1 \Rightarrow y=b+1 \end{cases}\\ &\textrm{Maka}\\ &\displaystyle \frac{x^{2}}{y-1}+\displaystyle \frac{y^{2}}{x-1}=\displaystyle \frac{(a+1)^{2}}{b}+\displaystyle \frac{(b+1)^{2}}{a}\\ &\textrm{Dengan AM-GM akan diperoleh}\\ &\displaystyle \frac{(a+1)^{2}}{b}+\displaystyle \frac{(b+1)^{2}}{a}\geq 4\left ( \displaystyle \frac{a}{b}+\displaystyle \frac{b}{a} \right )\\ &\textrm{Bentuk di atas didapatkan dari}\\ &(a-1)^{2}\geq 0\Leftrightarrow a^{2}-2a+1\geq 0\\ &\Leftrightarrow a^{2}+2a-4a+1\geq 0\\ &\Leftrightarrow a^{2}+2a+1\geq 4a\Leftrightarrow (a+1)^{2}\geq 4a\\ &\textrm{Demikian juga yabf satunya, yaitu}\\ &(b+1)^{2}\geq 4b.\: \textrm{Serta bentuk}\\ &\left ( \displaystyle \frac{a}{b}+\displaystyle \frac{b}{a} \right )\geq 2\sqrt{\displaystyle \frac{a}{b}.\frac{b}{s}}=2\\ &\textrm{Selanjutnya kembali ke soal, yaitu}:\\ &\displaystyle \frac{(a+1)^{2}}{b}+\displaystyle \frac{(b+1)^{2}}{a}\geq 4\left ( \displaystyle \frac{a}{b}+\displaystyle \frac{b}{a} \right )\\ &\: \,  \qquad\qquad\qquad\qquad \geq 4(2)\\ &\: \,  \qquad\qquad\qquad\qquad \geq 8\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 32.&\textrm{Untuk}\: \: a>0 \: ,\: \textrm{tentukan nilai }\\ &\textrm{minimum dari bentuk}\: \: \displaystyle \frac{4a^{2}+8a+13}{6+6a}\\\\ &\textbf{Jawab}:\\   &\begin{aligned}&\textrm{Diketahui}\: \: a>0, \: \textrm{maka bentuk}\\ &\displaystyle \frac{4a^{2}+8a+13}{6+6a}=\displaystyle \frac{4(a+1)^{2}+9}{6(a+1)}\\ &=\displaystyle \frac{2(a+1)}{3}+\frac{3}{2(a+1)}\\ &\textrm{Dengan AM-GM akan diperoleh}\\ &\displaystyle \frac{2(a+1)}{3}+\frac{3}{2(a+1)}\\ &\geq 2\sqrt{\displaystyle \frac{2}{3}\times \frac{3}{2}\times \frac{(a+1)}{(a+1)}}=2\\ &\textrm{Jadi},\: \: \displaystyle \frac{2(a+1)}{3}+\frac{3}{2(a+1)}\geq 2  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 33.&\textrm{Untuk}\: \: 0\leq a< 6 \: ,\: \textrm{tentukan nilai }\\ &\textrm{minimum dari bentuk}\: :\: a\left ( 6-a \right )^{2}\\\\  &\textbf{Jawab}:\\   &\begin{aligned}&\textrm{Diketahui}\: \: 0\leq a< 6, \: \textrm{maka bentuk}\\ &a\left ( 6-a \right )^{2}=\displaystyle \frac{1}{2}(2a)(6-a)(6-a)\\ &\textrm{Dengan GM-AM akan kita peroleh}\\ &\displaystyle \frac{1}{2}(2a)(6-a)(6-a)\leq \displaystyle \frac{1}{2}\left ( \displaystyle \frac{2a+6-a+6-a}{3} \right )^{3}\\ &\: \: \qquad\qquad\qquad\quad\quad \leq \displaystyle \frac{1}{2}\left (\displaystyle \frac{12}{3}  \right )^{3}\\ &\: \: \qquad\qquad\qquad\quad\quad \leq \displaystyle \frac{1}{2}\left (4  \right )^{3}\\ &\: \: \qquad\qquad\qquad\quad\quad \leq \displaystyle \frac{1}{2}\left (64  \right )\\ &\: \: \qquad\qquad\qquad\quad\quad \leq 32\\  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 34.&\textrm{Jika}\: \: a,b> 0\: ,\: \textrm{tunjukkan bahwa}\\ &\qquad  (a+1)(b+1)(ab+1)\geq 8ab\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\textrm{Perhatikan bahwa dengan AM-GM}\\ &\bullet \: \: a+1\geq 2\sqrt{a}\\ &\bullet \: \: b+1\geq 2\sqrt{b},\: \: \textrm{dan}\\ &\bullet \: \: ab+1\geq 2\sqrt{ab}\\ &\textrm{Selanjutnya}\\ &(a+1)(b+1)(ab+1)\geq 2\sqrt{a}\times 2\sqrt{b}\times 2\sqrt{ab}\\ &\: \: \qquad\qquad\qquad\qquad\quad \geq 8\sqrt{a^{2}b^{2}}\\ &\: \: \qquad\qquad\qquad\qquad\quad \geq 8ab\qquad \blacksquare  \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 35.&\textrm{Jika}\: \: a,b,c> 0\: ,\: \textrm{tunjukkan bahwa}\\ &\qquad  (a+b)(b+c)(c+a)\geq 8abc\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Kurang lebih seperti pembuktian}\\ &\textrm{pada no.34 di atas dengan tetap}\\ &\textrm{menggunakan AM-GM akan}\\ &\textrm{didapatkan}\\ &\bullet \: \: a+b\geq 2\sqrt{ab}\\ &\bullet \: \: b+c\geq 2\sqrt{bc},\: \: \textrm{dan}\\ &\bullet \: \: c+a\geq 2\sqrt{ca}\\ &\textrm{Selanjutnya}\\ &(a+b)(b+c)(c+a)\geq 2\sqrt{ab}\times 2\sqrt{bc}\times 2\sqrt{ca}\\ & \: \: \: \qquad\qquad\qquad\quad\quad \geq 8\sqrt{a^{2}b^{2}c^{2}}\\ &\: \: \:  \qquad\qquad\qquad\quad\quad \geq 8abc\qquad \blacksquare  \end{aligned}   \end{array}$.

DAFTAR PUSTAKA

1. Bambang, S. 2012. Materi, Soal dan Penyelesaian Olimpiade Matematika Tingkat SMA/MA. Jakarta: BINA PRESTASI INSANI.
2. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
3. Manfrino, R.B., dkk. 2009. Inequalities A Mathematical Olympiad Approach. Basel: Birkhauser Verlag AG.