$\begin{array}{ll}\\ 51.&\textrm{Diketahui rata-rata bilangan positif}\\ & m_{1},m_{2},...,m_{k}\: \: \textrm{adalah}\: \: A\\ &\textrm{Buktikan bahwa}\\ & \left (m_{1}+\displaystyle \frac{1}{m_{1}} \right )^{2}+\left (m_{2}+\displaystyle \frac{1}{m_{2}} \right )^{2}+...+\left (m_{k}+\displaystyle \frac{1}{m_{k}} \right )^{2}\geq k\left ( A+\displaystyle \frac{1}{A} \right )^{2} \\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Diketahui}\\ &\displaystyle \frac{m_{1}+m_{2}+m_{3}+...+m_{k}}{k}=A\\ &\color{blue}\textrm{Dengan ketaksamaan QM-AM akan diperoleh}\\ &\sqrt{\displaystyle \frac{(m_{1})^{2}+(m_{2})^{2}+...+(m_{k})^{2}}{k}}\geq \displaystyle \frac{m_{1}+m_{2}+...+m_{k}}{k}\\ &\Leftrightarrow \sqrt{\displaystyle \frac{(m_{1})^{2}+(m_{2})^{2}+...+(m_{k})^{2}}{k}}\geq A\\ &\Leftrightarrow \displaystyle \frac{(m_{1})^{2}+(m_{2})^{2}+...+(m_{k})^{2}}{k}\geq A^{2}\\ &\Leftrightarrow (m_{1})^{2}+(m_{2})^{2}+...+(m_{k})^{2}\geq kA^{2}\\ &\Leftrightarrow \displaystyle \sum_{i=1}^{k}(m_{i})^{2}\geq kA^{2}\: \color{red}........(1) \end{aligned}\\ &\begin{aligned} &\color{blue}\textrm{Dengan ketaksamaan QM-HM diperoleh juga}\\ &\sqrt{\displaystyle \frac{\left ( \frac{1}{m_{1}} \right )^{2}+\left ( \frac{1}{m_{2}} \right )^{2}+...+\left ( \frac{1}{m_{k}} \right )^{2}}{k}}\geq \displaystyle \frac{k}{m_{1}+m_{2}+...+m_{k}}\\ &\Leftrightarrow \sqrt{\displaystyle \frac{\left ( \frac{1}{m_{1}} \right )^{2}+\left ( \frac{1}{m_{2}} \right )^{2}+...+\left ( \frac{1}{m_{k}} \right )^{2}}{k}}\geq \displaystyle \frac{1}{A}\\ &\Leftrightarrow \displaystyle \frac{\left ( \frac{1}{m_{1}} \right )^{2}+\left ( \frac{1}{m_{2}} \right )^{2}+...+\left ( \frac{1}{m_{k}} \right )^{2}}{k}\geq \displaystyle \frac{1}{A^{2}}\\ &\Leftrightarrow \left ( \frac{1}{m_{1}} \right )^{2}+\left ( \frac{1}{m_{2}} \right )^{2}+...+\left ( \frac{1}{m_{k}} \right )^{2}\geq \displaystyle \frac{k}{A^{2}}\\ &\Leftrightarrow \sum_{i=1}^{k}\left ( \displaystyle \frac{1}{m_{i}} \right )^{2}\geq \displaystyle \frac{k}{A^{2}}\: \color{red}........(2) \end{aligned}\\ &\begin{aligned} &\color{blue}\textrm{Dengan (1) dan (2) akan diperoleh}\\ &\displaystyle \sum_{i=1}^{k}\left ( m_{i}+\displaystyle \frac{1}{m_{i}} \right )^{2}=\displaystyle \sum_{i=1}^{k}\left ( (m_{i})^{2}+2+\left ( \displaystyle \frac{1}{m_{i}} \right )^{2} \right )\\ &=\displaystyle \sum_{i=1}^{k}(m_{i})^{2}+\displaystyle \sum_{i=1}^{k}2+\displaystyle \sum_{i=1}^{k}\left ( \frac{1}{m_{i}} \right )^{2}\\ &\geq kA^{2}+2k+\displaystyle \frac{k}{A^{2}}\\ &= k\left ( A^{2}+2+\displaystyle \frac{1}{A^{2}} \right )\\ &=k\left ( A+\displaystyle \frac{1}{A} \right )^{2}\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 52.&\textrm{Buktikan bahwa untuk bilangan asli}\: n>1\\ &\textrm{berlaku}\quad \sqrt[n]{1+\displaystyle \frac{\sqrt[n]{n}}{n}}+\sqrt[n]{1-\displaystyle \frac{\sqrt[n]{n}}{n}}<2\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Perhatikan bahwa untuk}\: \: \sqrt[n]{1-\color{red}\displaystyle \frac{\sqrt[n]{n}}{n}}\\ &\textrm{dengan}\: \: n>1,\: \: \textrm{maka}\: \: 0<\color{red}\displaystyle \frac{\sqrt[n]{n}}{n}\color{black}<1\\ &\color{blue}\textrm{Dengan ketaksamaan AM-GM dapat diperoleh}\\ &\begin{aligned} \displaystyle \frac{\left ( 1+\displaystyle \frac{\sqrt[n]{n}}{n} \right )+\underset{\textrm{sebanyak}\: \: (n-1)}{\underbrace{1+1+1+...+1}}}{n}&>\sqrt[n]{\left ( 1+\displaystyle \frac{\sqrt[n]{n}}{n} \right )111...1}\\ &=\sqrt[n]{\left ( 1+\displaystyle \frac{\sqrt[n]{n}}{n} \right )}\:\color{blue} ......(1)\\ \displaystyle \frac{\left ( 1-\displaystyle \frac{\sqrt[n]{n}}{n} \right )+\underset{\textrm{sebanyak}\: \: (n-1)}{\underbrace{1+1+1+...+1}}}{n}&>\sqrt[n]{\left ( 1-\displaystyle \frac{\sqrt[n]{n}}{n} \right )111...1}\\ &=\sqrt[n]{\left ( 1-\displaystyle \frac{\sqrt[n]{n}}{n} \right )}\: \color{blue} ......(2)\\ \end{aligned} \\ &\textrm{Jika ketaksamaan (1) dan (2) dijumlahkna, maka}\\ &\begin{aligned} \displaystyle \sqrt[n]{1+\displaystyle \frac{\sqrt[n]{n}}{n}}+\displaystyle \sqrt[n]{1-\displaystyle \frac{\sqrt[n]{n}}{n}}&<\displaystyle \frac{\underset{\textrm{sebanyak}\: \: (2n-2)}{\underbrace{1+1+1+...+1}}+\left ( 1+\displaystyle \frac{\sqrt[n]{n}}{n} \right )+\left ( 1-\displaystyle \frac{\sqrt[n]{n}}{n} \right )}{n}\\ &=\displaystyle \frac{\underset{\textrm{sebanyak}\: \: (2n-2)}{\underbrace{1+1+1+...+1}} +1+1}{n}\\ &=2\qquad \blacksquare \end{aligned} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 53.&\textrm{Jika bilangan real positif}\: \: x,y,z\: \: \textrm{dengan}\\ &xyz=1\: .\: \textrm{tentukan nilai minimum dari}\\ &2x^{3}+12y^{2}+24z\\\\ &\textbf{Jawab}\\ &\textrm{Misal}\\ & A=2x^{3}+12y^{2}+24z\\ &\Leftrightarrow A=x^{3}+x^{3}+4y^{2}+4y^{2}+4y^{2}+\underset{\textrm{sebanyak}\: \: 6\: \: \textrm{kali}}{\underbrace{4z+4z+...+4z}}\\ &\begin{aligned}&\color{blue}\textrm{Dengan ketaksamaan AM-GM dapat diperoleh}\\ &\displaystyle \frac{A}{11}\geq \sqrt[11]{(x^{3})(x^{3})(4y^{2})(4y^{2})(4y^{2})\underset{\textrm{sebanyak}\: \: 6}{\underbrace{(4z)...(4z)}} }\\ &\Leftrightarrow \displaystyle \frac{A}{11}\geq \sqrt[11]{(xyz)^{6}(4)^{9}}\\ &\Leftrightarrow A\geq 11\sqrt[11]{4^{9}} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 54.&(\textbf{OSK 2019})\\ &\textrm{Tentukan bilangan real terbesar}\: \: M,\\ & \textrm{sehingga untuk setiap}\: \: x\: \: \textrm{positif berlaku}\\ &(x+1)(x+3)(x+5)(x+11)\geq Mx\\\\ &\textbf{Jawab}\\ &\textrm{Diketahui}\\ &(x+1)(x+3)(x+5)(x+11)\geq Mx\\ &\Leftrightarrow x^{4}+20x^{3}+122x^{2}+268x+165\geq Mx\\ &\color{blue}\textrm{Dengan AM-GM kita akan mendapatkan}\\ &\displaystyle \frac{x^{4}+\overset{\textrm{sebanyak}\: \: 20}{\overbrace{x^{3}+x^{3}+...+x^{3}}}+\overset{\textrm{sebanyak}\: \: 122}{\overbrace{x^{2}+x^{2}+...+x^{2}}}+\overset{\textrm{sebanyak}\: \: 268}{\overbrace{x+x+...+x}}+\overset{\textrm{sebanyak}\: \: 165}{\overbrace{1+1+...+1}}}{1+20+122+268+165}\\ &\geq \sqrt[576]{x^{4}\left (\underset{\textrm{sebanyak}\: \: 20}{\underbrace{x^{3}x^{3}...x^{3}}} \right )\left (\underset{\textrm{sebanyak}\: \: 268}{\underbrace{x^{2}x^{2}...x^{2}}} \right )\left (\underset{\textrm{sebanyak}\: \: 165}{\underbrace{11...1}} \right )}\\ &\Leftrightarrow \displaystyle \frac{x^{4}+20x^{3}+122x^{2}+268x+165}{576}\geq \sqrt[576]{x^{576}}\\ &\Leftrightarrow x^{4}+20x^{3}+122x^{2}+268x+165\geq 576x\\ &\textrm{Jadi, nilai}\: \: M=\color{red}576 \end{array}$.
$\begin{array}{ll}\\ 55.&(\textbf{OMITS 2012})\\ &\textrm{Jika diketahui}\: \: x_{1},x_{2},x_{3},...,x_{2012}\in \left ( \displaystyle \frac{1}{4}\: ,1 \right ),\\ & \textrm{maka nilai minimum dari bentuk}\\ &^{.^{x_{1}}}\log \left ( x_{2}-\displaystyle \frac{1}{4} \right )+^{.^{x_{2}}}\log \left ( x_{3}-\displaystyle \frac{1}{4} \right )+...+^{.^{x_{2012}}}\log \left ( x_{1}-\displaystyle \frac{1}{4} \right ) \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\\ &\bullet \: \textrm{untuk}:\: x\geq 1,\: \textrm{berlaku}\: \left ( x-\displaystyle \frac{1}{2} \right )^{2}\geq 0\\ &\: \quad \textrm{sehingga berlaku juga}\: \left (x-\displaystyle \frac{1}{4} \right )\leq x^{2}\\ &\bullet \: \textrm{untuk}:\: x_{1},x_{2},x_{3},...,x_{2012}\in \left ( \displaystyle \frac{1}{4}\: ,1 \right )\\ &\: \quad \textrm{kita akan memperoleh fakta bahwa}\\ &\: \quad ^{.^{x_{i}}}\log \left ( x_{i+1}-\displaystyle \frac{1}{4} \right )\geq ^{.^{x_{1}}}\log x_{x_{i+1}}^{2}=2.^{.^{x_{1}}}\log x_{x_{i+1}}\\ &\textrm{Selanjutnya}\\ &^{.^{x_{1}}}\log \left ( x_{2}-\displaystyle \frac{1}{4} \right )+^{.^{x_{2}}}\log \left ( x_{3}-\displaystyle \frac{1}{4} \right )+...+^{.^{x_{n}}}\log \left ( x_{1}-\displaystyle \frac{1}{4} \right )\\ &=\displaystyle \sum_{i=1}^{n} .^{.^{x_{i}}}\log \left ( x_{i+1}-\displaystyle \frac{1}{4} \right )\geq 2\displaystyle \sum_{i=1}^{n}.^{.^{x_{i}}}\log (x_{i+1})=2\displaystyle \sum_{i=1}^{n}\displaystyle \frac{\log x_{i+1}}{\log x_{i}}\\ &\color{blue}\textrm{Dengan AM-GM kita mendapatkan}\\ &\displaystyle \sum_{i=1}^{n} .^{.^{x_{i}}}\log \left ( x_{i+1}-\displaystyle \frac{1}{4} \right )\\ &\geq 2\displaystyle \sum_{i=1}^{n}\displaystyle \frac{\log x_{i+1}}{\log x_{i}}\geq 2n.\sqrt[n]{\prod_{i=1}^{n}\displaystyle \frac{\log x_{i+1}}{\log x_{i}}}\\ &= 2n.\sqrt[n]{\displaystyle \frac{\log x_{2}}{\log x_{1}}.\frac{\log x_{3}}{\log x_{2}}.\frac{\log x_{4}}{\log x_{3}}...\frac{\log x_{n}}{\log x_{n-1}}.\frac{\log x_{1}}{\log x_{n}}}\\ &=2n.1=2n\\ &\textrm{Sehingga nilai minimum dari}\\ &^{.^{x_{1}}}\log \left ( x_{2}-\displaystyle \frac{1}{4} \right )+^{.^{x_{2}}}\log \left ( x_{3}-\displaystyle \frac{1}{4} \right )+...+^{.^{x_{2012}}}\log \left ( x_{1}-\displaystyle \frac{1}{4} \right )\\ &=2n=2(2012)=4024 \end{aligned} \end{array}$.
DAFTAR PUSTAKA
- Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
- Muslim, M.S. 2020. Kumpulan Soal dan Pembahasan Olimpiade Matematika SMA Tahun 2007-2019 Tingkat Kota/Kabupaten. Bandung: YRAMA WIDYA.
- Sidi, A.B. 2010. Aljabar: Alhaqibiyyah Attadribiyyah lita'hil Attullab litasfiyat Oulimbiyat Arriyadliyyat bi Hazakhistan. Saudi Arabia.
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