Contoh Soal 11 (Segitiga dan Ketaksamaan)

$\begin{array}{l}\\ 51.& \text{Diketahui rata-rata bilangan positif}\\ & m_1,m_2,...,m_k\text{adalah}A\\ & \text{Buktikan bahwa}\\ & {(m_1+{\displaystyle \frac{1}{m_1}})}^2+{(m_2+{\displaystyle \frac{1}{m_2}})}^2+...+{(m_k+{\displaystyle \frac{1}{m_k}})}^2\ge k{(A+{\displaystyle \frac{1}{A}})}^2\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & {\displaystyle \frac{m_1+m_2+m_3+...+m_k}{k}=A}\\ & \text{Dengan ketaksamaan QM-AM akan diperoleh}\\ & \sqrt{{\displaystyle \frac{(m_1{)}^2+(m_2{)}^2+...+(m_k{)}^2}{k}}}\ge {\displaystyle \frac{m_1+m_2+...+m_k}{k}}\\ & \iff \sqrt{{\displaystyle \frac{(m_1{)}^2+(m_2{)}^2+...+(m_k{)}^2}{k}}}\ge A\\ & \iff {\displaystyle \frac{(m_1{)}^2+(m_2{)}^2+...+(m_k{)}^2}{k}\ge A^2}\\ & \iff (m_1{)}^2+(m_2{)}^2+...+(m_k{)}^2\ge k{A}^2\\ & \iff {\displaystyle \sum _{i=1}^k(m_i{)}^2\ge k{A}^2........(1)}\end{array}\\ & \begin{array}{rl}& \text{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}}}\ge {\displaystyle \frac{k}{m_1+m_2+...+m_k}}\\ & \iff \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}}}\ge {\displaystyle \frac{1}{A}}\\ & \iff {\displaystyle \frac{{\left(\frac{1}{m_1}\right)}^2+{\left(\frac{1}{m_2}\right)}^2+...+{\left(\frac{1}{m_k}\right)}^2}{k}\ge {\displaystyle \frac{1}{A^2}}}\\ & \iff {\left(\frac{1}{m_1}\right)}^2+{\left(\frac{1}{m_2}\right)}^2+...+{\left(\frac{1}{m_k}\right)}^2\ge {\displaystyle \frac{k}{A^2}}\\ & \iff \sum _{i=1}^k{\left({\displaystyle \frac{1}{m_i}}\right)}^2\ge {\displaystyle \frac{k}{A^2}........(2)}\end{array}\\ & \begin{array}{rl}& \text{Dengan (1) dan (2) akan diperoleh}\\ & {\displaystyle \sum _{i=1}^k{(m_i+{\displaystyle \frac{1}{m_i}})}^2={\displaystyle \sum _{i=1}^k((m_i{)}^2+2+{\left({\displaystyle \frac{1}{m_i}}\right)}^2)}}\\ & ={\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}}}\\ & \ge k{A}^2+2k+{\displaystyle \frac{k}{A^2}}\\ & =k(A^2+2+{\displaystyle \frac{1}{A^2}})\\ & =k{(A+{\displaystyle \frac{1}{A}})}^2◼\end{array}\end{array}$.

$\begin{array}{l}\\ 52.& \text{Buktikan bahwa untuk bilangan asli}n>1\\ & \text{berlaku}\sqrt[n]{1+{\displaystyle \frac{\sqrt[n]{n}}{n}}}+\sqrt[n]{1-{\displaystyle \frac{\sqrt[n]{n}}{n}}}<2\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa untuk}\sqrt[n]{1-{\displaystyle \frac{\sqrt[n]{n}}{n}}}\\ & \text{dengan}n>1,\text{maka}0<{\displaystyle \frac{\sqrt[n]{n}}{n}<1}\\ & \text{Dengan ketaksamaan AM-GM dapat diperoleh}\\ & \begin{array}{rl}{\displaystyle \frac{(1+{\displaystyle \frac{\sqrt[n]{n}}{n}})+\underset{\text{sebanyak}(n-1)}{\underbrace{1+1+1+...+1}}}{n}}& >\sqrt[n]{(1+{\displaystyle \frac{\sqrt[n]{n}}{n}})111...1}\\ & =\sqrt[n]{(1+{\displaystyle \frac{\sqrt[n]{n}}{n}})}......(1)\\ {\displaystyle \frac{(1-{\displaystyle \frac{\sqrt[n]{n}}{n}})+\underset{\text{sebanyak}(n-1)}{\underbrace{1+1+1+...+1}}}{n}}& >\sqrt[n]{(1-{\displaystyle \frac{\sqrt[n]{n}}{n}})111...1}\\ & =\sqrt[n]{(1-{\displaystyle \frac{\sqrt[n]{n}}{n}})}......(2)\end{array}\\ & \text{Jika ketaksamaan (1) dan (2) dijumlahkna, maka}\\ & \begin{array}{rl}{\displaystyle \sqrt[n]{1+{\displaystyle \frac{\sqrt[n]{n}}{n}}}+{\displaystyle \sqrt[n]{1-{\displaystyle \frac{\sqrt[n]{n}}{n}}}}}& <{\displaystyle \frac{\underset{\text{sebanyak}(2n-2)}{\underbrace{1+1+1+...+1}}+(1+{\displaystyle \frac{\sqrt[n]{n}}{n}})+(1-{\displaystyle \frac{\sqrt[n]{n}}{n}})}{n}}\\ & ={\displaystyle \frac{\underset{\text{sebanyak}(2n-2)}{\underbrace{1+1+1+...+1}}+1+1}{n}}\\ & =2◼\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 53.& \text{Jika bilangan real positif}x,y,z\text{dengan}\\ & xyz=1.\text{tentukan nilai minimum dari}\\ & 2x^3+12y^2+24z\\ \\ & \mathbf{\text{Jawab}}\\ & \text{Misal}\\ & A=2x^3+12y^2+24z\\ & \iff A=x^3+x^3+4y^2+4y^2+4y^2+\underset{\text{sebanyak}6\text{kali}}{\underbrace{4z+4z+...+4z}}\\ & \begin{array}{rl}& \text{Dengan ketaksamaan AM-GM dapat diperoleh}\\ & {\displaystyle \frac{A}{11}\ge \sqrt[11]{(x^3)(x^3)(4y^2)(4y^2)(4y^2)\underset{\text{sebanyak}6}{\underbrace{(4z)...(4z)}}}}\\ & \iff {\displaystyle \frac{A}{11}\ge \sqrt[11]{(xyz{)}^6(4{)}^9}}\\ & \iff A\ge 11\sqrt[11]{4^9}\end{array}\end{array}$.

$\begin{array}{l}\\ 54.& (\mathbf{\text{OSK 2019}})\\ & \text{Tentukan bilangan real terbesar}M,\\ & \text{sehingga untuk setiap}x\text{positif berlaku}\\ & (x+1)(x+3)(x+5)(x+11)\ge Mx\\ \\ & \mathbf{\text{Jawab}}\\ & \text{Diketahui}\\ & (x+1)(x+3)(x+5)(x+11)\ge Mx\\ & \iff x^4+20x^3+122x^2+268x+165\ge Mx\\ & \text{Dengan AM-GM kita akan mendapatkan}\\ & {\displaystyle \frac{x^4+\stackrel{\text{sebanyak}20}{\overbrace{x^3+x^3+...+x^3}}+\stackrel{\text{sebanyak}122}{\overbrace{x^2+x^2+...+x^2}}+\stackrel{\text{sebanyak}268}{\overbrace{x+x+...+x}}+\stackrel{\text{sebanyak}165}{\overbrace{1+1+...+1}}}{1+20+122+268+165}}\\ & \ge \sqrt[576]{x^4\left(\underset{\text{sebanyak}20}{\underbrace{x^3{x}^3...x^3}}\right)\left(\underset{\text{sebanyak}268}{\underbrace{x^2{x}^2...x^2}}\right)\left(\underset{\text{sebanyak}165}{\underbrace{11...1}}\right)}\\ & \iff {\displaystyle \frac{x^4+20x^3+122x^2+268x+165}{576}\ge \sqrt[576]{x^{576}}}\\ & \iff x^4+20x^3+122x^2+268x+165\ge 576x\\ & \text{Jadi, nilai}M=576\end{array}$.

$\begin{array}{l}\\ 55.& (\mathbf{\text{OMITS 2012}})\\ & \text{Jika diketahui}{x}_1,x_2,x_3,...,x_{2012}\in \left({\displaystyle \frac{1}{4},1}\right),\\ & \text{maka nilai minimum dari bentuk}\\ & {}^{{.}^{x_1}}\log (x_2-{\displaystyle \frac{1}{4}}){+}^{{.}^{x_2}}\log (x_3-{\displaystyle \frac{1}{4}})+...{+}^{{.}^{x_{2012}}}\log (x_1-{\displaystyle \frac{1}{4}})\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & \bullet \text{untuk}:x\ge 1,\text{berlaku}{(x-{\displaystyle \frac{1}{2}})}^2\ge 0\\ & \text{sehingga berlaku juga}(x-{\displaystyle \frac{1}{4}})\le x^2\\ & \bullet \text{untuk}:x_1,x_2,x_3,...,x_{2012}\in \left({\displaystyle \frac{1}{4},1}\right)\\ & \text{kita akan memperoleh fakta bahwa}\\ & {}^{{.}^{x_i}}\log (x_{i+1}-{\displaystyle \frac{1}{4}}){\ge }^{{.}^{x_1}}\log x_{x_{i+1}}^2={2.}^{{.}^{x_1}}\log x_{x_{i+1}}\\ & \text{Selanjutnya}\\ & {}^{{.}^{x_1}}\log (x_2-{\displaystyle \frac{1}{4}}){+}^{{.}^{x_2}}\log (x_3-{\displaystyle \frac{1}{4}})+...{+}^{{.}^{x_n}}\log (x_1-{\displaystyle \frac{1}{4}})\\ & ={\displaystyle \sum _{i=1}^n{.}^{{.}^{x_i}}\log (x_{i+1}-{\displaystyle \frac{1}{4}})\ge 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}}}}}\\ & \text{Dengan AM-GM kita mendapatkan}\\ & {\displaystyle \sum _{i=1}^n{.}^{{.}^{x_i}}\log (x_{i+1}-{\displaystyle \frac{1}{4}})}\\ & \ge 2{\displaystyle \sum _{i=1}^n{\displaystyle \frac{\log x_{i+1}}{\log x_i}\ge 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\\ & \text{Sehingga nilai minimum dari}\\ & {}^{{.}^{x_1}}\log (x_2-{\displaystyle \frac{1}{4}}){+}^{{.}^{x_2}}\log (x_3-{\displaystyle \frac{1}{4}})+...{+}^{{.}^{x_{2012}}}\log (x_1-{\displaystyle \frac{1}{4}})\\ & =2n=2(2012)=4024\end{array}\end{array}$. 

DAFTAR PUSTAKA

1. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
2. Muslim, M.S. 2020. Kumpulan Soal dan Pembahasan Olimpiade Matematika SMA Tahun 2007-2019 Tingkat Kota/Kabupaten. Bandung: YRAMA WIDYA.
3. Sidi, A.B. 2010. Aljabar: Alhaqibiyyah Attadribiyyah lita'hil Attullab litasfiyat Oulimbiyat Arriyadliyyat bi Hazakhistan. Saudi Arabia.