Contoh Soal 7 (Segitiga dan Ketaksamaan)

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

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

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

$\begin{array}{l}\\ 34.& \text{Jika}a,b>0,\text{tunjukkan bahwa}\\ & (a+1)(b+1)(ab+1)\ge 8ab\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa dengan AM-GM}\\ & \bullet a+1\ge 2\sqrt{a}\\ & \bullet b+1\ge 2\sqrt{b},\text{dan}\\ & \bullet ab+1\ge 2\sqrt{ab}\\ & \text{Selanjutnya}\\ & (a+1)(b+1)(ab+1)\ge 2\sqrt{a}\times 2\sqrt{b}\times 2\sqrt{ab}\\ & \ge 8\sqrt{a^2{b}^2}\\ & \ge 8ab◼\end{array}\end{array}$.

$\begin{array}{l}\\ 35.& \text{Jika}a,b,c>0,\text{tunjukkan bahwa}\\ & (a+b)(b+c)(c+a)\ge 8abc\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Kurang lebih seperti pembuktian}\\ & \text{pada no.34 di atas dengan tetap}\\ & \text{menggunakan AM-GM akan}\\ & \text{didapatkan}\\ & \bullet a+b\ge 2\sqrt{ab}\\ & \bullet b+c\ge 2\sqrt{bc},\text{dan}\\ & \bullet c+a\ge 2\sqrt{ca}\\ & \text{Selanjutnya}\\ & (a+b)(b+c)(c+a)\ge 2\sqrt{ab}\times 2\sqrt{bc}\times 2\sqrt{ca}\\ & \ge 8\sqrt{a^2{b}^2{c}^2}\\ & \ge 8abc◼\end{array}\end{array}$.

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.