3. Ketaksamaan Holder
Misalkan $a_{1}, a_{2}, a_{3}, \cdots , a_{n}$ dan $b_{1}, b_{2}, b_{3}, \cdots , b_{n}$ adalah merupakan kumpulan bilangan real positif dan misalkan pula $\alpha,\beta >0$ dengan $\alpha +\beta =1$, maka
$\color{blue}\begin{aligned}&(a_{1}+\cdots +a_{n})^{\alpha }(b_{1}+\cdots +b_{n})^{\beta }\geq (a_{1}^{\alpha }b_{1}^{\beta }+\cdots +a_{n}^{\alpha }b_{n}^{\beta }) \end{aligned}$.
Kesamaan terjadi jika dan hanya jika
$\begin{aligned}&\displaystyle \frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}=\frac{a_{3}}{b_{3}}=\cdots =\displaystyle \frac{a_{n}}{b_{n}} \end{aligned}$.
Secara umum Ketaksamaan Holder dituliskan sebagai berikut:
Misalkan $\color{red}a_{1}, a_{2}, a_{3}, \cdots , a_{n}$, $\color{blue}b_{1}, b_{2}, b_{3}, \cdots , b_{n}$, ... , $\color{purple}z_{1}, z_{2}, z_{3}, \cdots , z_{n}$ adalah merupakan barisan bilangan tak negatif serta $\lambda _{1},\lambda _{2},\lambda _{3},\cdots ,\lambda _{n}$ adalah bilangan-bilangan real positif dengan $\lambda _{1}+\lambda _{2}+\lambda _{3}+\cdots +\lambda _{n}=1$, maka
$\begin{aligned}&(a_{1}+\cdots +a_{n})^{\lambda _{a}}(b_{1}+\cdots +b_{n})^{\lambda _{b}}\cdots (z_{1}+\cdots +z_{n})^{\lambda _{z}}\\ &\geq a_{1}^{\lambda _{a}}b_{1}^{\lambda _{b}}\cdots z_{1}^{\lambda _{z}}+a_{2}^{\lambda _{a}}b_{2}^{\lambda _{b}}\cdots z_{2}^{\lambda _{z}}+\cdots +a_{n}^{\lambda _{a}}b_{n}^{\lambda _{b}}\cdots z_{n}^{\lambda _{z}} \\ &(a_{1}+\cdots +a_{n})^{\lambda _{a}}(b_{1}+\cdots +b_{n})^{\lambda _{b}}\cdots (z_{1}+\cdots +z_{n})^{\lambda _{z}}\geq \displaystyle \sum_{i=1}^{n}a_{i}^{\lambda _{a}}b_{i}^{\lambda _{b}}\cdots z_{i}^{\lambda _{z}}\\ &\left ( \displaystyle \sum_{i=1}^{n}a_{i} \right )^{\lambda _{a}}\left ( \displaystyle \sum_{i=1}^{n}b_{i} \right )^{\lambda _{b}}\cdots \left ( \displaystyle \sum_{i=1}^{n}z_{i} \right )^{\lambda _{z}}\geq \displaystyle \sum_{i=1}^{n}a_{i}^{\lambda _{a}}b_{i}^{\lambda _{b}}\cdots z_{i}^{\lambda _{z}} \end{aligned}$.
$\begin{aligned}&\textrm{Ketaksamaan terjadi jika dan hanya jika}\\ &a_{1}:a_{2}:\cdots :a_{n}=b_{1}:b_{2}:\cdots :b_{n}=z_{1}:z_{2}:\cdots :z_{n} \end{aligned}$.
Bukti 1:
$\begin{aligned}&\textrm{Tanpa mengurangi keumuman misalkan untuk}\\ &a_{1}+a_{2}+\cdots +a_{n}=b_{1}+b_{2}+\cdots +b_{n}=\cdots =1\\ &\textrm{maka diperoleh}\\ &1\geq \displaystyle \sum_{i=1}^{n}a_{i}^{\lambda _{a}}b_{i}^{\lambda _{b}}\cdots z_{i}^{\lambda _{z}}\: \: \textrm{atau}\: \: \displaystyle \sum_{i=1}^{n}a_{i}^{\lambda _{a}}b_{i}^{\lambda _{b}}\cdots z_{i}^{\lambda _{z}}\leq 1\\ &\textrm{Dengan AM-GM akan diperoleh}\\ &\color{red}\displaystyle \sum_{i=1}^{n}a_{i}^{\lambda _{a}}b_{i}^{\lambda _{b}}\cdots z_{i}^{\lambda _{z}}\leq \displaystyle \sum_{i=1}^{n}(\lambda _{a}a_{1}+\lambda _{b}b_{1}+\cdots +\lambda _{z}z_{1})\color{black}=1\\ &\textrm{Misalkan lagi kita atur}\: \: \color{red}\lambda _{a}=\lambda _{b}\color{black}=\displaystyle \frac{1}{2}\\ &(a_{1}+a_{2}+\cdots +a_{n})(b_{1}+b_{2}+\cdots +b_{n})\geq (\sqrt{a_{1}b_{1}}+\sqrt{a_{2}b_{2}}+\cdots +\sqrt{a_{n}b_{n}})^{2}\\ &\textrm{bentuk di atas adalah Ketaksamaan Cauchy-Schwarz} \end{aligned}$.
Bentuk lain yang lebih luas untuk sekian-$k$ variabel, adalah
$\begin{aligned}&\underset{k}{\underbrace{(a_{1}+a_{2}+\cdots +a_{n})(b_{1}+b_{2}+\cdots +b_{n})\cdots (z_{1}+z_{2}+\cdots +z_{n})}}\\ &\geq \left (\sqrt[k]{a_{1}b_{1}\cdots z_{1}}+\sqrt[k]{a_{2}b_{2}\cdots z_{2}}+\cdots +\sqrt[k]{a_{n}b_{n}\cdots z_{n}} \right )^{k} \end{aligned}$.
Bukti 2:
$\begin{aligned} &\textrm{Diketahui}\\ &P(n)\equiv (a_{1}+\cdots +a_{n})^{\lambda _{a}}(b_{1}+\cdots +b_{n})^{\lambda _{b}}\cdots (z_{1}+\cdots +z_{n})^{\lambda _{z}}\geq \displaystyle \sum_{i=1}^{n}a_{i}^{\lambda _{a}}b_{i}^{\lambda _{b}}\cdots z_{i}^{\lambda _{z}}\\ &\textbf{Langkah basis}\\ &P(1)\: \: \textrm{benar},\: \: \textrm{karena}\: \: (a_{1} )^{\lambda _{a}}(b_{1})^{\lambda _{b}}\cdots (z_{1})^{\lambda _{z}}\geq a_{1}^{\lambda _{a}}b_{1}^{\lambda _{b}}\cdots z_{1}^{\lambda _{z}}\\ &\textbf{Langkah induksi}\\ &\textrm{Misalkan rumus berlaku untuk} \: \: n=k,\: \: \textrm{yaitu}\\&P(k)\equiv (a_{1}+\cdots +a_{k})^{\lambda _{a}}(b_{1}+\cdots +b_{k})^{\lambda _{b}}\cdots (z_{1}+\cdots +z_{k})^{\lambda _{z}}\\ &\quad \qquad \geq a_{1}^{\lambda _{a}}b_{1}^{\lambda _{b}}\cdots z_{1}^{\lambda _{z}}+a_{2}^{\lambda _{a}}b_{2}^{\lambda _{b}}\cdots z_{2}^{\lambda _{z}}+\cdots +a_{k}^{\lambda _{a}}b_{k}^{\lambda _{b}}\cdots z_{k}^{\lambda _{z}}\\ &\textrm{maka untuk}\: \: n=k+1\\ &(a_{1}+\cdots +a_{k}+a_{k+1})^{\lambda _{a}}(b_{1}+\cdots +b_{k}+b_{k+1})^{\lambda _{b}}\cdots (z_{1}+\cdots +z_{k}+z_{k+1})^{\lambda _{z}}\\ &\geq a_{1}^{\lambda _{a}}b_{1}^{\lambda _{b}}\cdots z_{1}^{\lambda _{z}}+a_{2}^{\lambda _{a}}b_{2}^{\lambda _{b}}\cdots z_{2}^{\lambda _{z}}+\cdots +a_{k}^{\lambda _{a}}b_{k}^{\lambda _{b}}\cdots z_{k}^{\lambda _{z}}+a_{k+1}^{\lambda _{a}}b_{k+1}^{\lambda _{b}}\cdots z_{k+1}^{\lambda _{z}}\\ &=\underset{k}{\underbrace{\displaystyle \sum_{i=1}^{k}a_{i}^{\lambda _{a}}b_{i}^{\lambda _{b}}\cdots z_{i}^{\lambda _{z}}}}+a_{k+1}^{\lambda _{a}}b_{k+1}^{\lambda _{b}}\cdots z_{k+1}^{\lambda _{z}}\\ &=\displaystyle \sum_{i=1}^{k+1}a_{i}^{\lambda _{a}}b_{i}^{\lambda _{b}}\cdots z_{i}^{\lambda _{z}}\equiv P(k+1)\\ &\textrm{Jadi},\: P(n)\: \: \textrm{benar untuk}\: \: n=k+1\\&\textbf{Kesimpulan}\\ &\textrm{Jadi},\: P(n)\: \: \textrm{benar untuk}\: \: \forall n\in \mathbb{N} \end{aligned}$.
$\LARGE\colorbox{yellow}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Jika}\: \: x,y\: \: \textrm{bilangan positif, tunjukkan}\\ &x^{2}+y^{2}\geq \displaystyle \frac{(x+y)^{2}}{2}\\\\ &\textbf{Bukti}:\\ &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\\ &(x-y)^{2}\geq 0\Leftrightarrow x^{2}-2xy+y^{2}\geq 0\\ &\Leftrightarrow x^{2}+y^{2}\geq 2xy\\ &\Leftrightarrow 2x^{2}+2y^{2}\geq x^{2}+y^{2}+2xy\\ &\Leftrightarrow 2(x^{2}+y^{2})\geq (x+y)^{2}\\ &\Leftrightarrow x^{2}+y^{2}\geq \displaystyle \frac{(x+y)^{2}}{2}\qquad \blacksquare \end{aligned}\\ &\color{blue}\textrm{Alternatif 2}\\ &\textrm{Dengan Ketaksamaan Holder pilih}\\ &\color{red}\lambda _{1}=\lambda _{2}=\displaystyle \frac{1}{2}\: \left ( \displaystyle \sum_{i=1}^{2}\lambda _{i}=1 \right )\: \: \color{black}\textrm{akan diperoleh}\\ &\begin{aligned}&\left ( \displaystyle \frac{x^{2}}{1}+\frac{y^{2}}{1} \right )^{\frac{1}{2}}(1+1)^{\frac{1}{2}}\geq x+y\\ &\Leftrightarrow (x^{2}+y^{2})(1+1)\geq (x+y)^{2}\\ &\Leftrightarrow (x^{2}+y^{2})\geq \displaystyle \frac{(x+y)^{2}}{2}\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Untuk}\: \: x,y\: \: \textrm{bilangan positif, tunjukkan}\\ &\textrm{kebenaran ketaksamaan Cauchy-Schwarz}\\ &\displaystyle \frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+\frac{x_{3}^{2}}{y_{3}}+\cdots +\frac{x_{n}^{2}}{y_{n}}\geq \displaystyle \frac{(x_{1}+x_{2}+\cdots +x_{n})^{2}}{y_{1}+y_{2}+\cdots +y_{n}}\\\\ &\textbf{Bukti}:\\ &\begin{aligned}&\textrm{Dengan Ketaksamaan Holder pilih}\\ &\color{red}\lambda _{1}=\lambda _{2}=\displaystyle \frac{1}{2}\: \left ( \displaystyle \sum_{i=1}^{2}\lambda _{i}=1 \right )\: \: \color{black}\textrm{akan diperoleh}\\ &\left ( \displaystyle \frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+\cdots +\frac{x_{n}^{2}}{y_{n}} \right )^{\frac{1}{2}}(y_{1}+y_{2}+\cdots +y_{n})^{\frac{1}{2}}\geq (x_{1}+x_{2}+\cdots +x_{n})\\ &\left ( \displaystyle \frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+\cdots +\frac{x_{n}^{2}}{y_{n}} \right )(y_{1}+y_{2}+\cdots +y_{n})\geq (\sqrt{x_{1}^{2}}+\sqrt{x_{2}^{2}}+\cdots +\sqrt{x_{n}^{2}})^{2}\\ &\left ( \displaystyle \frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+\cdots +\frac{x_{n}^{2}}{y_{n}} \right )(y_{1}+y_{2}+\cdots +y_{n})\geq (x_{1}+x_{2}+\cdots +x_{n})^{2}\\ &\left ( \displaystyle \frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+\cdots +\frac{x_{n}^{2}}{y_{n}} \right )\geq \displaystyle \frac{(x_{1}+x_{2}+\cdots +x_{n})^{2}}{(y_{1}+y_{2}+\cdots +y_{n})}\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 3.&(\textbf{National Mathematical Contest, Belarus-2000})\\ &\textrm{Jika}\: \: a,b,c,x,y\: \: \textrm{dan}\: \: z\: \: \textrm{bilangan real positif, tunjukkan}\\ &\displaystyle \frac{a^{3}}{x}+\frac{b^{3}}{y}+\frac{c^{3}}{z}\geq \displaystyle \frac{(a+b+c)^{3}}{3(x+y+z)}\\\\ &\textbf{Bukti}:\\ &\begin{aligned}&\textrm{Dengan Ketaksamaan Holder pilih}\\ &\color{red}\lambda _{1}=\lambda _{2}=\lambda _{3}=\displaystyle \frac{1}{3}\: \left ( \displaystyle \sum_{i=1}^{3}\lambda _{i}=1 \right )\: \: \color{black}\textrm{akan diperoleh}\\ &\left ( \displaystyle \frac{a^{3}}{x}+\frac{b^{3}}{y}+\frac{c^{3}}{z} \right )^{\frac{1}{3}}(1+1+1)^{\frac{1}{3}}(x+y+z)^{\frac{1}{3}}\geq (a+b+c)\\ &\Leftrightarrow \left ( \displaystyle \frac{a^{3}}{x}+\frac{b^{3}}{y}+\frac{c^{3}}{z} \right )(3)(x+y+z)\geq (a+b+c)^{3}\\ &\Leftrightarrow \left ( \displaystyle \frac{a^{3}}{x}+\frac{b^{3}}{y}+\frac{c^{3}}{z} \right )\geq \displaystyle \frac{(a+b+c)^{3}}{3(x+y+z)}\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: a,b,c,x,y\: \: \textrm{dan}\: \: z\: \: \textrm{bilangan real positif}\\ &\textrm{dengan}\: \: a+b+c=x+y+z\: \: \textrm{tunjukkan bahwa}\\ &\displaystyle \frac{a^{3}}{x^{2}}+\frac{b^{3}}{y^{2}}+\frac{c^{3}}{z^{2}}\geq a+b+c\\\\ &\textbf{Bukti}:\\ &\begin{aligned}&\textrm{Dengan Ketaksamaan Holder pilih}\\ &\color{red}\lambda _{1}=\lambda _{2}=\lambda _{3}=\displaystyle \frac{1}{3}\: \left ( \displaystyle \sum_{i=1}^{3}\lambda _{i}=1 \right )\: \: \color{black}\textrm{akan diperoleh}\\ &\left ( \displaystyle \frac{a^{3}}{x^{2}}+\frac{b^{3}}{y^{2}}+\frac{c^{3}}{z^{2}} \right )^{\frac{1}{3}}(x+y+z)^{\frac{1}{3}}(x+y+z)^{\frac{1}{3}}\geq (a+b+c)\\ &\Leftrightarrow \left ( \displaystyle \frac{a^{3}}{x^{2}}+\frac{b^{3}}{y^{2}}+\frac{c^{3}}{z^{2}} \right )(x+y+z)^{2}\geq (a+b+c)^{3}\\ &\Leftrightarrow \left ( \displaystyle \frac{a^{3}}{x^{2}}+\frac{b^{3}}{y^{2}}+\frac{c^{3}}{z^{2}} \right )(a+b+c)^{2}\geq (a+b+c)^{3}\\ &\Leftrightarrow \left ( \displaystyle \frac{a^{3}}{x}+\frac{b^{3}}{y}+\frac{c^{3}}{z} \right )\geq \displaystyle \frac{(a+b+c)^{3}}{(a+b+c)^{2}}\\ &\Leftrightarrow \displaystyle \frac{a^{3}}{x^{2}}+\frac{b^{3}}{y^{2}}+\frac{c^{3}}{z^{2}}\geq a+b+c\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \: a,b,c\: \: \textrm{adalah bilangan real positif}\\ &\textrm{dengan}\: \: a+b+c=1,\: \textrm{tunjukkan}\\ &\textrm{bahwa}\: \: \left ( \displaystyle \frac{1}{a}+1 \right )\left ( \displaystyle \frac{1}{b}+1 \right )\left ( \displaystyle \frac{1}{c}+1 \right )\geq 64\\\\ &\textbf{Bukti}:\\ &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned} &\left ( \displaystyle \frac{1}{a}+1 \right )\left ( \displaystyle \frac{1}{b}+1 \right )\left ( \displaystyle \frac{1}{c}+1 \right )\\ &=\displaystyle \frac{1}{abc}+\left (\displaystyle \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} \right )+\left (\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )+1\\ &\textrm{Dengan AM-GM kita mendapatkan}\\ &\bullet \: \: \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3\sqrt[3]{\displaystyle \frac{1}{abc}}\\ &\bullet \: \: \displaystyle \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq 3\sqrt[3]{\displaystyle \frac{1}{(abc)^{2}}}\\ &\textrm{Kita tulis sintak prosesnya di atas}\\ &=1+\left (\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )+\left (\displaystyle \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} \right )+\displaystyle \frac{1}{abc}\\ &\geq 1+3\sqrt[3]{\displaystyle \frac{1}{abc}}+3\sqrt[3]{\displaystyle \frac{1}{(abc)^{2}}}+\sqrt[3]{\displaystyle \frac{1}{(abc)^{3}}}\\ &\geq 1+3\sqrt[3]{\displaystyle \frac{1}{abc}}+3\sqrt[3]{\displaystyle \frac{1}{(abc)^{2}}}+\sqrt[3]{\displaystyle \frac{1}{(abc)^{3}}}\\ &=\left ( 1+\displaystyle \frac{1}{\sqrt[3]{abc}} \right )^{3}\\ &\textrm{Karena}\: \: \sqrt[3]{abc}\leq \displaystyle \frac{a+b+c}{3}=\displaystyle \frac{1}{3},\: \textrm{maka}\\ &\left ( \displaystyle \frac{1}{a}+1 \right )\left ( \displaystyle \frac{1}{b}+1 \right )\left ( \displaystyle \frac{1}{c}+1 \right )\geq \left ( 1+\displaystyle \frac{1}{\sqrt[3]{abc}} \right )^{3}\\ &\: \qquad\qquad\qquad\qquad\qquad\quad\quad \geq \left ( 1+\displaystyle \frac{1}{\left (\frac{1}{3} \right )} \right )^{3}\\ &\: \qquad\qquad\qquad\qquad\qquad\quad\quad \geq \left ( 1+3 \right )^{4}\\ &\: \qquad\qquad\qquad\qquad\qquad\quad\quad \geq 4^{4}\\ &\: \qquad\qquad\qquad\qquad\qquad\quad\quad \geq 64\qquad \blacksquare \end{aligned}\\ &\begin{aligned}&\color{blue}\textrm{Alternatif 2}\\ &\textrm{Dengan Ketaksamaan Holder diperoleh}\\ &\left ( \displaystyle \frac{1}{a}+1 \right )\left ( \displaystyle \frac{1}{b}+1 \right )\left ( \displaystyle \frac{1}{c}+1 \right )\geq \left ( \sqrt[3]{\displaystyle \frac{1}{a}.\frac{1}{b}.\frac{1}{c}}+\sqrt[3]{1.1.1} \right )^{3}\\ &\qquad\qquad\qquad\: \qquad\qquad\qquad \geq \left ( \sqrt[3]{\displaystyle \frac{1}{(abc)}}+1 \right )^{3}\\ &\qquad\qquad\qquad\: \qquad\qquad\qquad \geq \left ( \displaystyle \frac{1}{\sqrt[3]{abc}}+1 \right )^{3}\\ &\qquad\qquad\qquad\: \qquad\qquad\qquad \geq \left ( 3+1 \right )^{3}\\ &\qquad\qquad\qquad\: \qquad\qquad\qquad \geq 64\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 6.&(\textbf{IMO 2001})\\ &\textrm{Diberikan}\: \: a,b\: \: \textrm{dan}\: \: c\: \: \textrm{bilangan real positif, tunjukkan}\\ &\displaystyle \frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\geq 1\\\\ &\textbf{Bukti}:\\ &\begin{aligned}&\textrm{Dengan Ketaksamaan Holder diperoleh}\\ &\left ( \displaystyle \sum_{\textrm{siklik}}^{.} a(a^{2}+8bc)\right )^{\frac{1}{3}}\left ( \displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{a}{\sqrt{a^{2}+8bc}} \right )^{\frac{1}{3}}\left ( \displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{a}{\sqrt{a^{2}+8bc}} \right )^{\frac{1}{3}}\geq (\displaystyle \sum_{\textrm{siklik}}^{.}a)\\ &\textrm{dapat juga dituliskan lebih sederhana}\\ &\left ( \displaystyle \sum_{\textrm{siklik}}^{.} a(a^{2}+8bc)\right )^{\frac{1}{3}}\left ( \displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{a}{\sqrt{a^{2}+8bc}} \right )^{\frac{2}{3}}\geq \left ( \displaystyle \sum_{\textrm{siklik}}^{.}a \right )\\ &\Leftrightarrow \left ( \displaystyle \sum_{\textrm{siklik}}^{.} a(a^{2}+8bc)\right )\left ( \displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{a}{\sqrt{a^{2}+8bc}} \right )^{2}\geq \left ( \displaystyle \sum_{\textrm{siklik}}^{.}a \right )^{3}\\ &\Leftrightarrow \left ( \displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{a}{\sqrt{a^{2}+8bc}} \right )^{2}\geq \displaystyle \frac{\left ( \displaystyle \sum_{\textrm{siklik}}^{.}a \right )^{3}}{\left ( \displaystyle \sum_{\textrm{siklik}}^{.} a(a^{2}+8bc)\right )}\\ &\color{red}\textrm{Perhatikan bahwa}\\&\bullet \: \left ( \displaystyle \sum_{\textrm{siklik}}^{.}a \right )^{3}=(a+b+c)^{3}\geq \color{blue}a^{3}+b^{3}+c^{3}+24abc\\ &\bullet \: \left ( \displaystyle \sum_{\textrm{siklik}}^{.} a(a^{2}+8bc)\right )=\color{blue}a^{3}+b^{3}+c^{3}+24abc\\ &\textrm{maka}\\ &\left ( \displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{a}{\sqrt{a^{2}+8bc}} \right )^{2}\geq \displaystyle \frac{a^{3}+b^{3}+c^{3}+24abc}{a^{3}+b^{3}+c^{3}+24abc}=1\\ &\displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{a}{\sqrt{a^{2}+8bc}}\geq 1\qquad \blacksquare \end{aligned} \end{array}$.
$.\qquad\begin{aligned}&\color{blue}\textbf{Sebagai catatan}\\&(a+b+c)^{3}=\color{red}a^{3}+b^{3}+c^{3}+3a^{2}b+3ab^{2}\\&\qquad \color{red}+3b^{2}c+3bc^{2}+3a^{2}c+3ac^{2}+6abc\\ &(a+b+c)^{3}-(a^{3}+b^{3}+c^{3})=\color{red}3a^{2}b+3ab^{2}\\ &\qquad \color{red}+3b^{2}c+3bc^{2}+3a^{2}c+3ac^{2}+6abc\\&\textrm{Dengan AM-GM akan diperoleh bentuk}\\&(a+b+c)^{3}-(a^{3}+b^{3}+c^{3})\\ &\geq 3\left ( 3\sqrt[3]{(abc)^{3}}+3\sqrt[3]{(abc)^{3}} \right )+6abc\\&\geq 3\left ( 3abc+3abc \right )+6abc=24abc\\ &\textrm{Sehingga}\\ &(a+b+c)^{3}\geq a^{3}+b^{3}+c^{3}+24abc \end{aligned}$.
$\begin{array}{ll}\\ 7.&\textrm{Jika}\: \: a,b,\: \: \textrm{dan}\: \: c\: \: \textrm{bilangan real positif}\\ &\textrm{dengan}\: \: a+b+c=1\: \: \textrm{tunjukkan bahwa}\\ &\displaystyle \frac{1}{a(b+c)^{2}}+\frac{1}{b(a+c)^{2}}+\frac{1}{c(a+b)^{2}}\geq \displaystyle \frac{81}{4}\\\\ &\textbf{Bukti}:\\ &\begin{aligned}&\textrm{Dengan Ketaksamaan Holder pilih}\\ &\color{red}\lambda _{1}=\lambda _{2}=\lambda _{3}=\lambda _{4}=\displaystyle \frac{1}{4}\: \left ( \displaystyle \sum_{i=1}^{4}\lambda _{i}=1 \right )\: \: \color{black}\textrm{akan diperoleh}\\ &(a+b+c)(b+c+a+c+a+b)(b+c+a+c+a+b)\left (\displaystyle \frac{1}{a(b+c)^{2}}+\frac{1}{b(a+c)^{2}}+\frac{1}{c(a+b)^{2}} \right )\\ &\geq (1^{.^{\frac{1}{4}}}+1^{.^{\frac{1}{4}}}+1^{.^{\frac{1}{4}}})^{4}=3^{4}=81\\ &\Leftrightarrow (1)(2)(2)\left (\displaystyle \frac{1}{a(b+c)^{2}}+\frac{1}{b(a+c)^{2}}+\frac{1}{c(a+b)^{2}} \right )\geq 81\\ &\Leftrightarrow \displaystyle \frac{1}{a(b+c)^{2}}+\frac{1}{b(a+c)^{2}}+\frac{1}{c(a+b)^{2}}\geq \displaystyle \frac{81}{4}\qquad \blacksquare \end{aligned} \end{array}$.
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