1. Ketaksamaan Pengaturan Ulang / Renata (Rearrangement)
$\begin{aligned}&a_{1}c_{1}+ a_{2}c_{2}+a_{3}c_{3}+ \cdots + a_{n}c_{n}\\ &\geq b_{1}c_{1}+b_{2}c_{2}+b_{3}c_{3}+...+b_{n}c_{n}\\ &\geq a_{n}c_{1}+a_{n-1}c_{2}+a_{n-2}c_{3}+...+a_{1}c_{n} \end{aligned}$.
Perhatikan bentuk $a_{1}\leq a_{2}\leq a_{3}\leq \cdots \leq a_{n}$ dengan $(b_{1},b_{2}, b_{3}, \cdots , b_{n})$ adalah sembarang permutasi dari $a_{1}\leq a_{2}\leq a_{3}\leq \cdots \leq a_{n}$, maka akan memiliki akibat
$\begin{aligned}1.\: \: &a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}\geq a_{1}b_{1}+ a_{2}b_{2}+ \cdots + a_{n}b_{n} \\ 2.\: \: &\displaystyle \frac{b_{1}}{a_{1}}+\frac{b_{2}}{a_{2}}+\frac{b_{3}}{a_{3}}+\cdots +\frac{b_{n}}{a_{n}}\geq n \end{aligned}$.
Penjelasan akibat no.1
Misalkan $a_{1}, a_{2}, a_{3}, \cdots , a_{n}$ adalah bilangan real positif asumsikan monoton naik, dengan akibat no. 1 akan diperoleh
$\begin{aligned}a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}&\geq a_{1}a_{2}+ a_{2}a_{3}+ \cdots + a_{n}a_{1}\\ a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}&\geq a_{1}a_{3}+ a_{2}a_{4}+ \cdots + a_{n}a_{2}\\ a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}&\geq a_{1}a_{4}+ a_{2}a_{5}+ \cdots + a_{n}a_{3}\\ \vdots &\qquad\vdots\\ a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}&\geq a_{1}a_{n}+ a_{2}a_{1}+ \cdots + a_{n}a_{n-1}\\ \end{aligned}$.
Jika kedua ruas setelah dijumlahkan ditambahkan dengan $a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}$, maka akan didapatkan bentuk:
$\begin{aligned} &n\left (a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2} \right )\geq \left ( a_{1}+a_{2}+\cdots +a_{n} \right )^{2}\\ &\Leftrightarrow \displaystyle \frac{n\left (a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2} \right )}{n^{2}}\geq \displaystyle \frac{\left ( a_{1}+a_{2}+\cdots +a_{n} \right )^{2}}{n^{2}}\\ &\Leftrightarrow \displaystyle \frac{\left (a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2} \right )}{n}\geq \displaystyle \frac{\left ( a_{1}+a_{2}+\cdots +a_{n} \right )^{2}}{n^{2}}\\ &\Leftrightarrow \sqrt{\displaystyle \frac{\left (a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2} \right )}{n}}\geq \sqrt{\displaystyle \frac{\left ( a_{1}+a_{2}+\cdots +a_{n} \right )^{2}}{n^{2}}}\\ &\Leftrightarrow \color{blue}\sqrt{\displaystyle \frac{\left (a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2} \right )}{n}}\color{black}\geq \color{blue}\displaystyle \frac{\left ( a_{1}+a_{2}+\cdots +a_{n} \right )}{n}\\ &\textrm{Dan bentuk terakhir di atas adalah bentuk}\: \color{red}\textrm{QM-AM} \end{aligned}$.
$\LARGE\colorbox{yellow}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Diketahui} \: \: a,b\: \: \textrm{bilangan real positif}\\&\textrm{Tunjukkan bahwa}\: \: a^{2}+b^{2}\geq 2ab\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\color{red}\textbf{Alternatif 1}\\ &\textrm{Perhatikan bahwa}\\ &(a-b)^{2}\geq 0\Leftrightarrow a^{2}-2ab+b^{2}\geq 0\\ &\Leftrightarrow a^{2}+b^{2}\geq 2ab\qquad \blacksquare \end{aligned} \\ &\begin{aligned}&\color{red}\textbf{Alternatif 2}\\ &\textrm{Dengan AM-GM diperoleh}\\ &\displaystyle \frac{a^{2}+b^{2}}{2}\geq \sqrt{(ab)^{2}}\\ &\Leftrightarrow \frac{a^{2}+b^{2}}{2}\geq ab\\ &\Leftrightarrow a^{2}+b^{2}\geq 2ab\qquad \blacksquare \end{aligned} \\ &\color{red}\textbf{Alternatif 3}\\ &\textrm{Asumsikan bahwa}\: \: a\geq b,\: \: \textrm{dengan}\\ & \textbf{ketaksamaan Renata}\: \textrm{diperoleh}\\ &a.a+b.b\geq a.b+b.a\\ &\Leftrightarrow a^{2}+b^{2}\geq ab+ab\\ &\Leftrightarrow a^{2}+b^{2}\geq 2ab\qquad \blacksquare \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Diketahui} \: \: a,b\: \: \textrm{bilangan real positif}\\&\textrm{Tunjukkan bahwa}\: \: \displaystyle \frac{a^{2}}{b}+\frac{b^{2}}{a}\geq a+b\\\\ &\textbf{Bukti}\\ &\textrm{Asumsikan bahwa}\: \: a\geq b,\: \textrm{maka}\: \: a^{2}\geq b^{2}\\&\textrm{dan}\: \: \displaystyle \frac{1}{b}\geq \frac{1}{a}.\\ &\textrm{Perhatikan bahwa baik}\: \left ( a^{2},b^{2} \right )\: \textrm{dan}\: \left ( \displaystyle \frac{1}{b}, \frac{1}{a} \right)\\ &\textrm{adalah kumpulan dua barisan yang monoton}\\ &\textrm{sama yaitu sama-sama naik. Sehingga}\\ &\textrm{dengan}\: \: \textbf{ketaksamaan Renata}\: \textrm{diperoleh}\\ &a^{2}.\displaystyle \frac{1}{b}+b^{2}.\displaystyle \frac{1}{a}\geq a^{2}.\displaystyle \frac{1}{a}+b^{2}.\displaystyle \frac{1}{b}\\ &\Leftrightarrow \displaystyle \frac{a^{2}}{b}+\frac{b^{2}}{a}\geq a+b\qquad \blacksquare \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Buktikan bahwa setiap bilangan real}\\ &\textrm{positif}\: \: a,\: b\: \: \textrm{dan}\: \: c\: \: \textrm{berlaku}\\ & a^{2}+b^{2}+c^{2}\geq ab+ac+bc\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\color{blue}\textrm{Alternatif 1}\\ &\textrm{Perhatikan bahwa}\: \: (a-b)^{2}\geq 0\\ &(a-c)^{2}\geq 0,\: \: \textrm{dan}\: \: (b-c)^{2}\geq 0\\ &\textrm{adalah benar, maka}\\ &(a-b)^{2}=a^{2}-2ab+b^{2}\geq 0\\ &\Leftrightarrow a^{2}+b^{2}\geq 2ab\: .....(1)\\ &\textrm{Dengan cara yang kurang lebih sama}\\ &\textrm{akan didapatkan}\\ &\bullet \quad a^{2}+c^{2}\geq 2ac\: .....(2)\\ &\bullet \quad b^{2}+c^{2}\geq 2bc\: .....(1)\\ &\textrm{Jika ketaksamaan}\quad (1),(2), \& \: (3)\: \: \textrm{dijumlahkan}\\ &\textrm{akan didapatkan bentuk}\\ &2a^{2}+2b^{2}+2c^{2}\geq 2ab+2ac+2bc\\ &\Leftrightarrow \: a^{2}+b^{2}+c^{2}\geq ab+ac+bc\quad \blacksquare\\ &\color{blue}\textrm{Alternatif 2}\\ &\textrm{Dengan ketaksamaan}\: \: \color{red}\textbf{Cauchy-Schwarz}\\ &(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+a^{2})\geq (ab+bc+ca)^{2}\\ &\Leftrightarrow a^{2}+b^{2}+c^{2}\geq ab+ac+bc\qquad \blacksquare\\ &\color{blue}\textrm{Alternatif 3}\\ &\textrm{Untuk barisan}\: \: (a,b,c),\: \textrm{asumsikan}\: a\geq b\geq c\\ &\textrm{maka dengan}\: \: \color{red}\textbf{Ketaksamaan Renata}\: \: \color{black}\textrm{diperoleh}\\ &a.a+b.b+c.c\geq ab+bc+ca\\ &a^{2}+b^{2}+c^{2}\geq ab+ac+bc\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&\textrm{Diberikan}\: \: a,b,c>0,\: \: \textrm{tunjukkan bahwa}\\ &\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+b+c\\\\ &\textbf{Bukti}:\\ &\color{red}\textrm{Alternatif 1}\\ &\textrm{Dengan AM-GM diperoleh}\\ &\bullet \: \: \displaystyle \frac{a}{c}+\frac{c}{a}\geq 2\Leftrightarrow \displaystyle \frac{c}{a}\geq 2-\displaystyle \frac{a}{c}\\ &\bullet \: \: \displaystyle \frac{b}{c}+\frac{c}{b}\geq 2\Leftrightarrow \displaystyle \frac{b}{c}\geq 2-\displaystyle \frac{c}{b}\\ &\bullet \: \: \displaystyle \frac{a}{b}+\frac{b}{a}\geq 2\Leftrightarrow \displaystyle \frac{a}{b}\geq 2-\displaystyle \frac{b}{a}\\ &\textrm{Sehingga}\\ &\begin{aligned}\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}&\geq a\left ( 2-\displaystyle \frac{c}{b} \right )+b\left ( 2-\displaystyle \frac{a}{c} \right )+c\left ( 2-\displaystyle \frac{b}{a} \right )\\ &=2a-\displaystyle \frac{ac}{b}+2b-\displaystyle \frac{ab}{c}+2c-\displaystyle \frac{bc}{a}\\ \displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}&\geq 2(a+b+c)-\left ( \displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} \right )\\ 2&\left ( \displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} \right )\geq 2(a+b+c)\\ \displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}&\geq a+b+c\qquad \blacksquare \end{aligned}\\ &\color{red}\textrm{Alternatif 2}\\ &\textrm{Dengan ketaksamaan Renata}\\ &\begin{aligned}&\textrm{Asumsikan}\: \: a\geq b\geq c,\: \textrm{maka}\: \: ab\geq ca\geq bc\\ &\textrm{dan}\: \: \displaystyle \frac{1}{c}\geq \frac{1}{b}\geq \frac{1}{a}.\\ &\textrm{Perhatikan bahwa}\\ & (ab\geq ca\geq bc)\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{1}{c}\geq \frac{1}{b}\geq \frac{1}{a} \right )\\ &\textrm{memiliki kemonotonan yang sama}\\ &\textrm{maka dengan}\: \: \textbf{ketaksamaan Renata}\\ &\textrm{dapat diperoleh bentuk}\\ &ab.\displaystyle \frac{1}{c}+ac.\displaystyle \frac{1}{b}+bc.\displaystyle \frac{1}{a}\geq ab.\displaystyle \frac{1}{b}+ac.\displaystyle \frac{1}{a}+bc.\displaystyle \frac{1}{c}\\ &\Leftrightarrow \displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+c+b\\ &\Leftrightarrow \displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+b+c\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 5.&\textrm{Diberikan}\: \: a,b,c>0,\: \: \textrm{tunjukkan kebenaran}\\ &\textbf{ketaksamaan Nesbitt}\: \textrm{berikut}\\ &\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}\\\\ &\textbf{Bukti}:\\ &\color{red}\textrm{Alternatif 1}\\ &\textrm{Dengan AM-GM diperoleh}\\ &\displaystyle \frac{(a+b)+(b+c)+(c+a)}{3}\geq \displaystyle \frac{3}{\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\\ &\Leftrightarrow ((a+b)+(b+c)+(c+a))\left ( \displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right )\geq 9\\ &\Leftrightarrow 2\left ( \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )+6\geq 9\\ &\Leftrightarrow 2\left ( \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )\geq 3\\ &\Leftrightarrow \left ( \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )\geq \displaystyle \frac{3}{2}\qquad \blacksquare \\ &\color{red}\textrm{Alternatif 2}\\ &\textrm{Dengan ketaksamaan Renata}\\ &\begin{aligned}&\textrm{Asumsikan}\: \: a\leq b\leq c,\: \textrm{maka}\: \: a+b\leq a+c\leq b+c\\ &\textrm{dan}\: \: \displaystyle \frac{1}{b+c}\leq \frac{1}{a+c}\leq \frac{1}{a+b}.\\ &\textrm{Perhatikan bahwa}\\ & (a\leq b\leq c)\: \: \textrm{dan}\: \: \displaystyle \frac{1}{b+c}\leq \frac{1}{a+c}\leq \frac{1}{a+b}\\ &\textrm{memiliki kemonotonan yang sama}\\ &\textrm{maka dengan}\: \: \textbf{ketaksamaan Renata}\\ &\textrm{dapat diperoleh bentuk}\\ &a.\displaystyle \frac{1}{b+c}+b.\displaystyle \frac{1}{a+c}+c.\displaystyle \frac{1}{a+b}\geq b.\displaystyle \frac{1}{b+c}+c.\displaystyle \frac{1}{a+c}+a.\displaystyle \frac{1}{a+b}\\ &\textrm{dan}\\ &a.\displaystyle \frac{1}{b+c}+b.\displaystyle \frac{1}{a+c}+c.\displaystyle \frac{1}{a+b}\geq c.\displaystyle \frac{1}{b+c}+a.\displaystyle \frac{1}{a+c}+b.\displaystyle \frac{1}{a+b}\\ &\textrm{Jika dijumlahkan keduanya, maka}\\ &2\left ( \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )\geq 3\\ &\Leftrightarrow \left ( \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )\geq \displaystyle \frac{3}{2}\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 6.&(\textbf{OSN 2015})\\ &\textrm{Diberikan}\: \: a,b,c>0,\: \: \textrm{Buktikan bahwa}\\ &\sqrt{\displaystyle \frac{a}{b+c}+\frac{b}{c+a}}+\sqrt{\displaystyle \frac{b}{c+a}+\frac{c}{a+b}}+\sqrt{\displaystyle\frac{c}{a+b}+ \frac{a}{b+c}}\geq 3\\\\ &\textbf{Bukti}:\\ &\textrm{Perhatikan bukti soal no. 4 di atas}\\ &\textrm{Dengan}\: \: \textbf{keksamaan Renata}\: \: \textrm{dapat diperoleh}\\ &\displaystyle \frac{a}{b+c}+\frac{b}{a+c}\geq \frac{b}{b+c}+\frac{a}{a+c}\\ &\textrm{Misalkan}\\ &x=b+c,\: y=c+a,\: y=a+b,\: \textrm{maka}\\ &\displaystyle \frac{a}{b+c}+\frac{b}{a+c}\geq \frac{b}{b+c}+\frac{a}{a+c}\\ &\Leftrightarrow \displaystyle \frac{a}{b+c}+\frac{b}{a+c}\geq \frac{y+z-x}{2x}+\frac{x+z-y}{2y}\\ &\begin{aligned} &\Leftrightarrow \sqrt{ \displaystyle \frac{a}{b+c}+\frac{b}{a+c}}\geq \sqrt{\frac{y+z-x}{2x}+\frac{x+z-y}{2y}}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \: = \sqrt{\displaystyle \frac{1}{2}}\sqrt{\displaystyle \frac{y}{x}+\frac{z}{x}-1+\frac{x}{y}+\frac{z}{y}-1}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \: \: \textrm{dengan AM-GM}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \: \geq \displaystyle \frac{1}{\sqrt{2}}\sqrt{\displaystyle \frac{z}{x}+\frac{z}{y}+2\sqrt{\frac{y}{x}.\frac{x}{y}}-2}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \: = \displaystyle \frac{1}{\sqrt{2}}\sqrt{\displaystyle \frac{z}{x}+\frac{z}{y}+2-2}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \: \geq \displaystyle \frac{1}{\sqrt{2}}\sqrt{\displaystyle \frac{z}{x}+\frac{z}{y}}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \: \geq \displaystyle \frac{\sqrt{2}}{\sqrt{2}}\sqrt{\displaystyle \frac{z^{2}}{xy}}=\displaystyle \sqrt{\displaystyle \frac{z^{2}}{xy}}\\ \end{aligned}\\ &\begin{aligned} &\bullet \: \sqrt{ \displaystyle \frac{a}{b+c}+\frac{b}{a+c}} \geq \displaystyle \sqrt{\displaystyle \frac{z^{2}}{xy}}\\ &\bullet \: \sqrt{ \displaystyle \frac{b}{c+a}+\frac{c}{a+b}}\geq \sqrt{\displaystyle \frac{x^{2}}{yz}}\\ &\bullet \: \sqrt{ \displaystyle \frac{c}{a+b}+\frac{a}{b+c}}\geq \sqrt{\displaystyle \frac{y^{2}}{xz}} \end{aligned}\\ &\begin{aligned} &\textrm{Selanjutnya}\\ &\sqrt{\displaystyle \frac{a}{b+c}+\frac{b}{c+a}}+\sqrt{\displaystyle \frac{b}{c+a}+\frac{c}{a+b}}+\sqrt{\displaystyle\frac{c}{a+b}+ \frac{a}{b+c}}\\ &\geq \displaystyle \sqrt{\displaystyle \frac{z^{2}}{xy}}+\sqrt{\displaystyle \frac{x^{2}}{yz}}+\sqrt{\displaystyle \frac{y^{2}}{xz}}\\ &\textrm{Dengan AM-GM lagi}\\ &\geq 3\sqrt[3]{\sqrt{\displaystyle \frac{z^{2}}{xy}}\times \sqrt{\displaystyle \frac{x^{2}}{yz}}\times \sqrt{\displaystyle \frac{y^{2}}{xz}}}\\ &\geq 3\sqrt[3]{\sqrt{\displaystyle \frac{(xyz)^{2}}{(xyz)^{2}}}}\\ &\geq 3\qquad \blacksquare \end{aligned} \end{array}$
$\begin{array}{ll}\\ 7.&\textrm{Diberikan}\: \: a,b,c>0,\: \: \textrm{tunjukkan bahwa}\\ &\displaystyle \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\geq 6\\\\ &\textbf{Bukti}:\\ &\color{red}\textrm{Alternatif 1}\\ &\textrm{Dengan mengaplikasikan AM-GM-HM}\\ &\textrm{pada}\: \: \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\: \: \textrm{kita dapat menemukan}\\ &\color{blue}\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\color{black}\geq \displaystyle \frac{3}{(abc)^{.^{\frac{1}{3}}}}\geq \color{blue}\displaystyle \frac{9}{a+b+c}\\ &\textrm{Jika kedua ruas dikalikan dengan}\: \: \color{red}a+b+c\color{black},\\ &\textrm{maka}\\ &\color{blue}3+\displaystyle \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\color{black}\geq \color{blue}\displaystyle \frac{9(a+b+c)}{a+b+c}\\ &\Leftrightarrow \displaystyle \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\geq 6\qquad \blacksquare \\ &\begin{aligned}&\color{red}\textrm{Alternatif 2}\\ &\textrm{Asumsikan}\: \: a\leq b\leq c,\: \textrm{maka}\: \: a+b\leq a+c\leq b+c\\ &\textrm{dan}\: \: \displaystyle \frac{1}{c}\leq \frac{1}{b}\leq \frac{1}{a}.\\ &\textrm{Perhatikan bahwa}\\ & (a+b\leq a+c\leq b+c)\: \: \textrm{dan}\: \: \displaystyle \frac{1}{c}\leq \frac{1}{b}\leq \frac{1}{a}\\ &\textrm{memiliki kemonotonan yang sama}\\ &\textrm{maka dengan}\: \: \textbf{ketaksamaan Renata}\\ &\textrm{dapat diperoleh bentuk}\\ &(b+c).\displaystyle \frac{1}{a}+(c+a).\displaystyle \frac{1}{b}+(a+b).\displaystyle \frac{1}{c}\geq (b+c).\displaystyle \frac{1}{b}+(c+a).\displaystyle \frac{1}{c}+(a+b).\displaystyle \frac{1}{a}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \doteq 1+\displaystyle \frac{c}{b}+1+\frac{a}{c}+1+\frac{b}{a}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \doteq 3+\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \doteq 3+3\left ( \displaystyle \frac{abc}{abc} \right )^{.^{\frac{1}{3}}}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \doteq 3+3\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \doteq 6\qquad \blacksquare \end{aligned} \end{array}$.
DAFTAR PUSTAKA
- Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.
WEBSITE
- https://holdenlee.github.io/high_school/omc/23-rearrange.pdf diakses 18 Januari 2022.
- https://www.gotohaggstrom.com/Advanced%20inequality%20manipulations.pdf diakses 20 Januari 2022
