3. 2 Ketaksamaan Cauchy-Schwars
Jika diberikan sebarang bilangan real $x_{1},x_{2},x_{3},...,x_{n}$ dan $y_{1},y_{2},y_{3},...,y_{n}$, maka akan berlaku
$\begin{aligned}&(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}+...+x_{n}y_{n})^{\color{red}2}\\ &\leq (x_{1}^{\color{red}2}+x_{2}^{\color{red}2}+x_{3}^{\color{red}2}+...+x_{n}^{\color{red}2})(y_{1}^{\color{red}2}+y_{2}^{\color{red}2}+y_{3}^{\color{red}2}+...+y_{n}^{\color{red}2}) \end{aligned}$.
Jika dituliskan dengan notasi sigma adalah:
$\left ( \displaystyle \sum_{i=1}^{n}x_{i}y_{i} \right )^{\color{red}2}\leq \left ( \displaystyle \sum_{i=1}^{n}x_{i}^{\color{red}2} \right )\left ( \displaystyle \sum_{i=1}^{n}y_{i}^{\color{red}2} \right )$.
$\begin{aligned}&\color{blue}\textbf{Bukti}\: \: \color{black}\textrm{dari ketaksamaan ini adalah}:\\ &\textrm{Diberikan}\\&\alpha _{1},\alpha _{2},\alpha _{3},\cdots ,\alpha _{n}\: \: \textrm{dan}\: \: \beta_{1}+\beta_{2}+\beta _{3}+\cdots +\beta _{n}\\ &\textrm{Untuk setiap bilangan real}\: \: x\: ,\: \textrm{pada polinom kuadrat}\\ &\textrm{dapat berlaku}\\ &f(x)=\displaystyle \sum_{i=1}^{n}\left ( \alpha x+\beta \right )^{2}\geq 0\\ &\Leftrightarrow \left ( \displaystyle \sum_{i=1}^{n}\alpha _{i}^{2} \right )x^{2}+2\left (\displaystyle \sum_{i=1}^{n}\alpha _{i}\beta _{i} \right )x+\left ( \displaystyle \sum_{i=1}^{n}\beta _{i}^{2} \right )\geq 0\\ &\textrm{akibatnya diskriminan (D)}\leq 0,\: \textrm{maka}\\&D=b^{2}-4ac\leq 0\\ &\Leftrightarrow \left ( 2\left ( \displaystyle \sum_{i=1}^{n}\alpha _{i}\beta _{i} \right ) \right )^{2}-4\left ( \displaystyle \sum_{i=1}^{n}\alpha _{i}^{2} \right )\left ( \displaystyle \sum_{i=1}^{n}\beta _{i}^{2} \right )\leq 0\\ &\Leftrightarrow 4\left ( \displaystyle \sum_{i=1}^{n}\alpha _{i}\beta _{i} \right )\leq 4\left ( \displaystyle \sum_{i=1}^{n}\alpha _{i}^{2} \right )\left ( \displaystyle \sum_{i=1}^{n}\beta _{i}^{2} \right )\\ &\Leftrightarrow \left ( \displaystyle \sum_{i=1}^{n}\alpha _{i}\beta _{i} \right )\leq \left ( \displaystyle \sum_{i=1}^{n}\alpha _{i}^{2} \right )\left ( \displaystyle \sum_{i=1}^{n}\beta _{i}^{2} \right )\qquad \blacksquare \end{aligned}$.
Ada hal yang sangat menarik ketika substitusi bentuk $x_{i}=\displaystyle \frac{p_{i}}{\sqrt{q_{i}}}$ dan $y_{i}=\sqrt{q_{i}}$, yaitu:
$\begin{aligned}&\color{blue}\textrm{Perhatikan bentuk berikut}\\ &(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}+...+x_{n}y_{n})^{\color{red}2}\\ &\leq (x_{1}^{\color{red}2}+x_{2}^{\color{red}2}+x_{3}^{\color{red}2}+...+x_{n}^{\color{red}2})(y_{1}^{\color{red}2}+y_{2}^{\color{red}2}+y_{3}^{\color{red}2}+...+y_{n}^{\color{red}2})\\ \end{aligned}$.
$\begin{aligned} &\textrm{dengan substitusi}\\ &\begin{cases} x_{i} &= \displaystyle \frac{p_{i}}{\sqrt{q_{i}}} \\ y_{i} &= \sqrt{q_{i}} \end{cases}\\ &\textrm{maka menjadi bentuk}\\ \end{aligned}$.
$\begin{aligned} &(p_{1}+p_{2}+p_{3}+...+p_{n})^{2}\\ &\leq \left ( \displaystyle \frac{p_{1}^{2}}{q_{1}}+\frac{p_{2}^{2}}{q_{2}}+\frac{p_{3}^{2}}{q_{3}}+...+\frac{p_{n}^{2}}{q_{n}} \right )\left ( q_{1}+q_{2}+q_{3}+...+q_{n} \right )\\ &\Leftrightarrow \displaystyle \frac{(p_{1}+p_{2}+p_{3}+...+p_{n})^{2}}{\left ( q_{1}+q_{2}+q_{3}+...+q_{n} \right )}\leq \left ( \displaystyle \frac{p_{1}^{2}}{q_{1}}+\frac{p_{2}^{2}}{q_{2}}+\frac{p_{3}^{2}}{q_{3}}+...+\frac{p_{n}^{2}}{q_{n}} \right ) \end{aligned}$.
Bentuk di atas selanjutnya lebih dikenal dengan sebutan Cauchy-Schwarz Engel, karena bentuk ketaksamaan ini dipopulerkan oleh Arthur Engel dan biasa disebut dengan sebutan CS-Engel.
Perhatikan lemma berikut berkaitan dengan ketaksamaan CS-Engel di atas.
Jika diberikan $a,b$ bilangan real dan $x,y$ real positif akan ditunjukkan berlaku
$\displaystyle \frac{a^{2}}{x}+\frac{b^{2}}{y}\geq \displaystyle \frac{(a+b)^{2}}{x+y}$.
$\begin{aligned}&\color{blue}\textbf{Bukti}\\ &\textrm{Perhatikan bahwa}\: \: (ay-bx)^{2}\geq 0\\ &\Leftrightarrow a^{2}y^{2}+b^{2}y^{2}-2abxy\geq 0\\ &\Leftrightarrow a^{2}y^{2}+b^{2}y^{2}\geq 2abxy\\ &\Leftrightarrow \color{red}a^{2}xy\color{black}+a^{2}y^{2}+b^{2}y^{2}+\color{red}b^{2}xy\color{black}\geq \color{red}a^{2}xy+\color{black}2abxy+\color{red}b^{2}xy\\ &\Leftrightarrow (a^{2}y+b^{2}x)(x+y)\geq (a^{2}+2ab+b^{2})xy\\ &\Leftrightarrow \displaystyle \frac{(a^{2}y+b^{2}x)}{xy}\geq \displaystyle \frac{(a^{2}+2ab+b^{2})}{(x+y)}\\ &\Leftrightarrow \displaystyle \frac{a^{2}}{x}+\frac{b^{2}}{y}\geq \displaystyle \frac{(a+b)^{2}}{x+y}\qquad \blacksquare \end{aligned}$.
$\LARGE\colorbox{yellow}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Jika}\: \: x,y\: \: \textrm{bilangan positif, tunjukkan}\\ &x^{2}+y^{2}\geq \displaystyle \frac{(x+y)^{2}}{2}\\\\ &\textbf{Bukti}:\\ &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\\ &(x-y)^{2}\geq 0\Leftrightarrow x^{2}-2xy+y^{2}\geq 0\\ &\Leftrightarrow x^{2}+y^{2}\geq 2xy\\ &\Leftrightarrow 2x^{2}+2y^{2}\geq x^{2}+y^{2}+2xy\\ &\Leftrightarrow 2(x^{2}+y^{2})\geq (x+y)^{2}\\ &\Leftrightarrow x^{2}+y^{2}\geq \displaystyle \frac{(x+y)^{2}}{2}\qquad \blacksquare \end{aligned}\\ &\begin{aligned}&\color{blue}\textrm{Alternatif 2}\\ &\color{red}\textrm{Dengan QM-AM kita akan mendapatkan}\\ &\sqrt{\displaystyle \frac{x^{2}+y^{2}}{2}}\geq \displaystyle \frac{x+y}{2}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}+y^{2}}{2} \right )\geq \left ( \displaystyle \frac{x+y}{2} \right )^{2}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}+y^{2}}{2} \right )\geq \displaystyle \frac{(x+y)^{2}}{4}\\ &\Leftrightarrow x^{2}+y^{2}\geq \displaystyle \frac{(x+y)^{2}}{2}\qquad \blacksquare \end{aligned}\\ &\color{blue}\textrm{Alternatif 3}\\ &\color{purple}\textrm{Dengan ketaksamaan CS-Engel}\\ &\textrm{kita dapat peroleh}\\ &\displaystyle \frac{x^{2}}{1}+\frac{y^{2}}{1}\geq \displaystyle \frac{(x+y)^{2}}{2}\\ &\Leftrightarrow x^{2}+y^{2}\geq \displaystyle \frac{(x+y)^{2}}{2}\qquad \blacksquare \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Buktikan bahwa setiap bilangan real}\\ &\textrm{positif}\: \: a,\: b\: \: \textrm{dan}\: \: c\: \: \textrm{berlaku}\\ & a^{2}+b^{2}+c^{2}\geq ab+ac+bc\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\color{blue}\textrm{Alternatif 1}\\ &\textrm{Perhatikan bahwa}\: \: (a-b)^{2}\geq 0\\ &(a-c)^{2}\geq 0,\: \: \textrm{dan}\: \: (b-c)^{2}\geq 0\\ &\textrm{adalah benar, maka}\\ &(a-b)^{2}=a^{2}-2ab+b^{2}\geq 0\\ &\Leftrightarrow a^{2}+b^{2}\geq 2ab\: .....(1)\\ &\textrm{Dengan cara yang kurang lebih sama}\\ &\textrm{akan didapatkan}\\ &\bullet \quad a^{2}+c^{2}\geq 2ac\: .....(2)\\ &\bullet \quad b^{2}+c^{2}\geq 2bc\: .....(1)\\ &\textrm{Jika ketaksamaan}\quad (1),(2), \& \: (3)\: \: \textrm{dijumlahkan}\\ &\textrm{akan didapatkan bentuk}\\ &2a^{2}+2b^{2}+2c^{2}\geq 2ab+2ac+2bc\\ &\Leftrightarrow \: a^{2}+b^{2}+c^{2}\geq ab+ac+bc\quad \blacksquare\\ &\color{blue}\textrm{Alternatif 2}\\ &\textrm{Dengan ketaksamaan}\: \: \color{red}\textbf{Cauchy-Schwarz}\\ &(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+a^{2})\geq (ab+bc+ca)^{2}\\ &\Leftrightarrow a^{2}+b^{2}+c^{2}\geq ab+ac+bc\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Buktikan bahwa setiap bilangan real}\\ &\textrm{positif}\: \: a,\: b\: \: \textrm{dan}\: \: c\: \: \textrm{berlaku}\\ & a^{2}+b^{2}+c^{2}\geq 3(ab+ac+bc)\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Lihat jawaban no.2 di atas, yaitu}\\ &\color{red}a^{2}+b^{2}+c^{2}\geq ab+ac+bc\\ &\textrm{karena nilai dari}\: \: \: (a+b+c)^{2}\\ &= a^{2}+b^{2}+c^{2}+2(ab+ac+bc)\\ &\textrm{maka nilai}\\ &a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+ac+bc)\\ &\textrm{Sehingga nilai untuk}\\ &\color{red}a^{2}+b^{2}+c^{2}\geq ab+ac+bc\: \: \: \color{black}\textrm{menjadi}\\ &(a+b+c)^{2}-2(ab+ac+bc)\geq ab+ac+bc\\ &\Leftrightarrow (a+b+c)^{2}\geq 3(ab+ac+bc)\qquad \blacksquare \end{aligned} \end{array}$
$\begin{array}{ll}\\ 4.&\textrm{Diketahui}\: \: x,y,z\: \: \textrm{adalah bilangan real }\\ &\textrm{positif. Tunjukkan bahwa}\\ &\qquad\qquad (x+y+z)^{2}\leq 3(x^{2}+y^{2}+z^{2})\\\\ &\textbf{Bukti}\\ &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}&\color{red}\textrm{Dengan ketaksamaan Cauchy-Schwarz}\\ &\textrm{Kita dapat peroleh}\\ &(ax+by+cz)^{2}\leq (a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})\\ &\textrm{untuk}\: \: \color{red}a=b=c=1,\: \color{black}\textrm{akan diperoleh}\\ &(ax+by+cz)^{2}\leq (1^{2}+1^{2}+1^{2})(x^{2}+y^{2}+z^{2})\\ &\Leftrightarrow (ax+by+cz)^{2}\leq (3)(x^{2}+y^{2}+z^{2})\quad \blacksquare \\ \end{aligned} \end{array}$.
$\: \: \: \quad\begin{aligned}&\color{blue}\textrm{Alternatif 2}\\ &\color{purple}\textrm{Dengan ketaksamaan CS-Engel}\\ &\textrm{Kita dapat peroleh}\\ &\displaystyle \frac{x^{2}}{1}+\frac{y^{2}}{1}+\frac{z^{2}}{1}\geq \displaystyle \frac{(x+y+z)^{2}}{1+1+1}\\ &\Leftrightarrow x^{2}+y^{2}+z^{2}\geq \displaystyle \frac{(x+y+z)^{2}}{3}\\ &\Leftrightarrow 3(x^{2}+y^{2}+z^{2})\geq (x+y+z)^{2},\\ &\quad \textrm{atau}\\ &\Leftrightarrow (x+y+z)^{2}\leq 3(x^{2}+y^{2}+z^{2})\qquad \blacksquare \end{aligned}$.
$\: \: \: \quad\begin{aligned}&\color{blue}\textrm{Alternatif 3}\\ &\color{red}\textrm{Dengan QM-AM kita akan mendapatkan}\\ &\sqrt{\displaystyle \frac{x^{2}+y^{2}+z^{2}}{3}}\geq \displaystyle \frac{x+y+z}{3}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}+y^{2}+z^{2}}{3} \right )\geq \left ( \displaystyle \frac{x+y+z}{3} \right )^{2}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}+y^{2}+z^{2}}{3} \right )\geq \displaystyle \frac{(x+y+z)^{2}}{9}\\ &\Leftrightarrow x^{2}+y^{2}+z^{2}\geq \displaystyle \frac{(x+y+z)^{2}}{3}\\ &\Leftrightarrow 3(x^{2}+y^{2}+z^{2})\geq (x+y+z)^{2}\quad \textrm{atau}\\ &\Leftrightarrow (x+y+z)^{2}\leq 3(x^{2}+y^{2}+z^{2}) \qquad \blacksquare \end{aligned}$.
$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \: x,y,z\: \: \textrm{adalah bilangan real positif}\\ &\textrm{dengan}\: \: x^{2}+y^{2}+z^{2}=27.\: \textrm{Tunjukkan}\\ & \textrm{bahwa}\: \: x^{3}+y^{3}+z^{3}\geq 81\\\\ &\textbf{Bukti}\\ &\color{red}\textrm{Pada contoh soal no.2 terdapat}\\ &(ax+by+cz)^{2}\leq (a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})\\ &\textrm{ganti mengganti}\: \: a=b=c=1,\: \textrm{maka menjadi}\\ &(x+y+z)^{2}\leq (1^{2}+1^{2}+1^{2})(x^{2}+y^{2}+z^{2})\\ &\Leftrightarrow (x+y+z)^{2}\leq (3)(x^{2}+y^{2}+z^{2})\: \color{red}........(1)\\ &\textrm{Selanjutnya dengan mengganti dengan}\\ &x^{.^{\frac{3}{2}}},y^{.^{\frac{3}{2}}},z^{.^{\frac{3}{2}}}\: \: \textrm{dan}\: \: x^{.^{\frac{1}{2}}},y^{.^{\frac{1}{2}}},z^{.^{\frac{1}{2}}},\: \textrm{pada}\\ &(ax+by+cz)^{2}\leq (a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})\\ &\textrm{Kita akan dapatkan}\\ &(x^{2}+y^{2}+z^{2})\leq (x^{3}+y^{3}+z^{3})(x+y+z)\: \color{red}........(2)\\ &\textrm{Jika masing-masing ruas dikuadratkan, maka}\\ &(x^{2}+y^{2}+z^{2})^{4}\leq (x^{3}+y^{3}+z^{3})^{2}(x+y+z)^{2}\\ &\Leftrightarrow (x^{2}+y^{2}+z^{2})^{4} \leq 3(x^{3}+y^{3}+z^{3})^{2}\left ( x^{2}+y^{2}+z^{2} \right )\\ &\Leftrightarrow (x^{2}+y^{2}+z^{2})^{3} \leq 3(x^{3}+y^{3}+z^{3})^{2}\\ &\Leftrightarrow (27)^{3}\leq 3(x^{3}+y^{3}+z^{3})^{2}\\ &\Leftrightarrow 3^{9}\leq 3(x^{3}+y^{3}+z^{3})^{2}\\ &\Leftrightarrow 3^{8}\leq (x^{3}+y^{3}+z^{3})^{2}\\ &\Leftrightarrow (x^{3}+y^{3}+z^{3})^{2}\geq 3^{8}\\ &\Leftrightarrow (x^{3}+y^{3}+z^{3})\geq 3^{.^{\frac{8}{2}}}\\ &\Leftrightarrow (x^{3}+y^{3}+z^{3})\geq 3^{4}=81\qquad \blacksquare \end{array}$.
$\begin{array}{ll}\\ 6.&\textrm{Untuk}\: \: a,b,c,d\: \: \textrm{adalah bilangan real }\\ &\textrm{positif, tunjukkan bahwa}\\ &\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\geq \displaystyle \frac{64}{a+b+c+d}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\color{red}\textrm{Dengan ketaksamaan CS-Engel}\\ &(a+b+c+d)\left (\displaystyle \frac{1^{2}}{a}+\displaystyle \frac{1^{2}}{b}+\displaystyle \frac{2^{2}}{c}+\displaystyle \frac{4^{2}}{d} \right )\geq (1+1+2+4)^{2}\\ &\Leftrightarrow \displaystyle \frac{1^{2}}{a}+\displaystyle \frac{1^{2}}{b}+\displaystyle \frac{2^{2}}{c}+\displaystyle \frac{4^{2}}{d}\geq \displaystyle \frac{(1+1+2+4)^{2}}{a+b+c+d}\\ &\Leftrightarrow \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\geq \displaystyle \frac{64}{a+b+c+d}\quad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 7.&\textrm{Diketahui}\: \: a,b\: \: \textrm{bilangan real positif}\\ &\textrm{dan}\: \: a+b=1.\: \textrm{Tunjukkan bahwa}\\ &\qquad \left ( a+\displaystyle \frac{1}{a} \right )^{2}+\left ( b+\displaystyle \frac{1}{b} \right )^{2}\geq \displaystyle \frac{25}{2}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\color{blue}\textrm{Alternatif 1}\\ &\textrm{Diketahui bahwa}\: \: a+b=1\\ &\color{red}\textrm{Dengan QM-AM kita akan mendapatkan}\\ &\sqrt{\displaystyle \frac{x^{2}+y^{2}}{2}}\geq \displaystyle \frac{x+y}{2}\Leftrightarrow \displaystyle \frac{x^{2}+y^{2}}{2}\geq \left ( \displaystyle \frac{x+y}{2} \right )^{2}\\ &\textrm{Kita misalkan}\: \: \color{purple}\begin{cases} x & =a+\displaystyle \frac{1}{a} \\\\ y & =b+\displaystyle \frac{1}{b} \end{cases}\\ &\textrm{maka}\\ &\begin{aligned}\displaystyle \frac{\left ( a+\displaystyle \frac{1}{a} \right )^{2}+\left (b+\displaystyle \frac{1}{b} \right )^{2}}{2}&\geq \left ( \displaystyle \frac{a+\displaystyle \frac{1}{a}+b+\displaystyle \frac{1}{b}}{2} \right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( a+b+\displaystyle \frac{1}{a}+\frac{1}{b} \right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( 1+\displaystyle \frac{1}{a}+\frac{1}{b} \right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( 1+ \displaystyle \frac{a+b}{ab}\right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( 1+\displaystyle \frac{1}{ab} \right )^{2} \end{aligned}\\ &\color{red}\textrm{Dari GM-AM kita akan mendapatkan}\\ &\sqrt{ab}\leq \displaystyle \frac{a+b}{2}\Leftrightarrow ab\leq \left ( \displaystyle \frac{a+b}{2} \right )^{2}\\ &\Leftrightarrow ab\leq \left ( \displaystyle \frac{1}{2} \right )^{2}\Leftrightarrow ab\leq \displaystyle \frac{1}{4} \\ &\textrm{Sehingga}\\ &\begin{aligned}\left ( a+\displaystyle \frac{1}{a} \right )^{2}+\left (b+\displaystyle \frac{1}{b} \right )^{2}&\geq \displaystyle \frac{1}{2}\left ( 1+\displaystyle \frac{1}{ab} \right )^{2}\\ &\geq \displaystyle \frac{1}{2}\left ( 1+ \displaystyle \frac{1}{\displaystyle \frac{1}{4}}\right )^{2}\\ &\geq \displaystyle \frac{1}{2}\left ( 1+4 \right )^{2}\\ &\geq \displaystyle \frac{1}{2}\times 25\\ &\geq \displaystyle \frac{25}{2}\qquad \blacksquare \end{aligned} \end{aligned} \end{array}$.
$\: \: \: \quad\begin{aligned}&\color{blue}\textrm{Alternatif 2}\\ &\color{red}\textrm{Dengan AM-GM kita akan mendapatkan}\\ &1=a+b\geq 2\sqrt{ab}\Leftrightarrow \displaystyle \frac{1}{2}\geq \sqrt{ab}\Leftrightarrow 2\leq \displaystyle \frac{1}{\sqrt{ab}}\\ &\Leftrightarrow \displaystyle \frac{1}{\sqrt{ab}}\geq 2\Leftrightarrow \displaystyle \frac{1}{ab}\geq 4\\ &\color{red}\textrm{Perhatikan soal, dengan QM-AM}\: \color{black}\textrm{akan}\\ &\textrm{didapatkan}\\ &\begin{aligned}\displaystyle \frac{\left ( a+\frac{1}{a} \right )^{2}+\left (b+\frac{1}{b} \right )^{2}}{2}&\geq \left ( \displaystyle \frac{a+\frac{1}{a}+b+\frac{1}{b}}{2} \right )^{2}\\ \left ( a+\displaystyle \frac{1}{a} \right )^{2}+\left (b+\displaystyle \frac{1}{b} \right )^{2}&\geq \displaystyle \frac{1}{2}\left ( a+b+\displaystyle \frac{1}{a}+\frac{1}{b} \right )^{2}\\ &=\displaystyle \frac{1}{2}\left ( 1+\displaystyle \frac{a+b}{ab} \right )^{2}\\ &=\displaystyle \frac{1}{2}\left ( 1+\displaystyle \frac{1}{ab} \right )^{2}\\ &=\displaystyle \frac{1}{2}(1+4)^{2}\\ &= \displaystyle \frac{1}{2}\times 25\\ &\geq \displaystyle \frac{25}{2}\qquad \blacksquare \end{aligned} \end{aligned}$.
$\: \: \: \quad\begin{aligned}&\color{blue}\textrm{Alternatif 3}\\ &\color{red}\textrm{Dengan AM-GM kita akan mendapatkan}\\ &1=a+b\geq 2\sqrt{ab}\Leftrightarrow \displaystyle \frac{1}{2}\geq \sqrt{ab}\Leftrightarrow 2\leq \displaystyle \frac{1}{\sqrt{ab}}\\ &\Leftrightarrow \displaystyle \frac{1}{\sqrt{ab}}\geq 2\Leftrightarrow \displaystyle \frac{1}{ab}\geq 4\\ &\color{red}\textrm{Perhatikan soal, dengan CS-Engel}\: \color{black}\textrm{akan}\\ &\textrm{didapatkan}\\ &\begin{aligned} &(1+1)\left (\displaystyle \frac{\left ( a+\frac{1}{a} \right )^{2}}{1} +\displaystyle \frac{\left ( b+\frac{1}{b} \right )^{2}}{1} \right )\geq \left (a+\frac{1}{a}+b+\frac{1}{b} \right )^{2}\\ &\Leftrightarrow 2\left (\left (\displaystyle a+\frac{1}{a} \right)^{2} +\left (b+\frac{1}{b} \right )^{2} \right )\geq \left (a+b+\frac{1}{a}+\frac{1}{b} \right )^{2}\\ &\Leftrightarrow \left (\displaystyle a+\frac{1}{a} \right )^{2} +\left (b+\frac{1}{b} \right )^{2}\geq \displaystyle \frac{1}{2}\left ( 1+\displaystyle \frac{a+b}{ab} \right )^{2}\\ &\Leftrightarrow \left (\displaystyle a+\frac{1}{a} \right )^{2} +\left (b+\frac{1}{b} \right )^{2}\geq\displaystyle \frac{1}{2}\left ( 1+\displaystyle \frac{1}{ab} \right )^{2}\\ &\Leftrightarrow \left (\displaystyle a+\frac{1}{a} \right )^{2} +\left (b+\frac{1}{b} \right )^{2}\geq\displaystyle \frac{1}{2}(1+4)^{2}\\ &\Leftrightarrow \left (\displaystyle a+\frac{1}{a} \right )^{2} +\left (b+\frac{1}{b} \right )^{2}\geq \displaystyle \frac{1}{2}\times 25\\ &\Leftrightarrow \left (\displaystyle a+\frac{1}{a} \right )^{2} +\left (b+\frac{1}{b} \right )^{2}\geq \displaystyle \frac{25}{2}\qquad \blacksquare \end{aligned} \end{aligned}$.
$\begin{array}{ll}\\ 8.&\textrm{Jika}\: \: a,b\: \: \textrm{bilangan positif, buktikan}\\ &\textrm{a}.\quad 2(a^{2}+b^{2})\geq (a+b)^{2}\\ &\textrm{b}.\quad 4(a^{3}+b^{3})\geq (a+b)^{3}\\ &\textrm{c}.\quad 8(a^{4}+b^{4})\geq (a+b)^{4}\\ &\textrm{d}.\quad 16(a^{5}+b^{5})\geq (a+b)^{5}\\ &\textrm{e}.\quad 32(a^{6}+b^{6})\geq (a+b)^{6}\\ &\textrm{f}.\quad 64(a^{7}+b^{7})\geq (a+b)^{7}\\ &\textrm{g}.\quad 128(a^{8}+b^{8})\geq (a+b)^{8}\\\\ &\textbf{Bukti}:\\ &\textrm{Akan ditunjukkan bukti poin 8.c saja}\\ &\textrm{untuk poin yang lain, silahkan pembaca}\\ &\textrm{sekalian untuk dibuktikan sendiri sebagai}\\ &\textrm{bahan latihan mandiri}.\\ &\textrm{Adapun bukti poin 8.c adalah sebagaimana}\\ &\textrm{berikut ini}\\ &\begin{aligned} &\color{red}\textrm{Dengan ketaksamaan CS-Engel}\: \color{black}\textrm{akan}\\ &\textrm{didapatkan}\\ &\bullet \: \: (1+1)(a^{4}+b^{4})\geq (a^{2}+b^{2})^{2}\\ &\: \quad\Leftrightarrow 2\left (a^{4}+b^{4} \right )\geq (a^{2}+b^{2})^{2}\: \color{red}..........(1)\\ &\bullet \: \: (1+1)(a^{2}+b^{2})\geq (a+b)^{2}\\ &\: \quad\Leftrightarrow 2\left (a^{2}+b^{2} \right )\geq \displaystyle (a+b)^{2},\quad (\textrm{kuadratkan})\\ &\: \quad \Leftrightarrow 4\displaystyle \left (a^{2}+b^{2} \right )^{2}\geq (a+b)^{4}\\ &\: \quad \Leftrightarrow \left (a^{2}+b^{2} \right )^{2}\geq \displaystyle \frac{(a+b)^{4}}{4}\: \color{red}...........(2)\\ &\textrm{Dari (1) dan (2) didapatkan hubungan}\\ &2\left (a^{4}+b^{4} \right )\geq \left (a^{2}+b^{2} \right )^{2}\geq \displaystyle \frac{(a+b)^{4}}{4}\\ &\Leftrightarrow 8\left ( a^{4}+b^{4} \right )\geq 4\left ( a^{2}+b^{2} \right )^{2}\geq (a+b)^{4}\\ &\Leftrightarrow 8\left ( a^{4}+b^{4} \right )\geq (a+b)^{4}\qquad \blacksquare \end{aligned} \end{array}$.
$.\quad\qquad\begin{aligned}&\color{red}\textrm{Mengenal penulisan pola}\: \textbf{Siklik dan Simetri}\\ &\textrm{Misal untuk}\: \: n=3,\: \: \textrm{pada penulisan unsur}\\ &x,y,\: \: \textrm{dan}\: \: z,\: \textrm{maka}\\ &\begin{array}{|l|l|}\hline \textbf{Pola Siklik}&\textbf{Pola Simetri}\\\hline\begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}=x^{2}+y^{2}+z^{2}\\ &\\ &\\ &\\ &\\ & \end{aligned} &\begin{aligned}\displaystyle\sum_{\textrm{sym}}^{.}x^{2}&=x^{2}+x^{2}\\ &+y^{2}+y^{2}\\ \\ &+z^{2}+z^{2}\\ \\ &=2\left (x^{2}+y^{2}+z^{2} \right ) \end{aligned}\\ \begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}x^{3}=x^{3}+y^{3}+z^{3}\end{aligned}&\begin{aligned}\displaystyle \sum_{\textrm{sym}}^{.}x^{3}=2\left (x^{3}+y^{3}+z^{3} \right ) \end{aligned}\\ \begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y=x^{2}y+y^{2}z+z^{2}x\\ &\\ &\\ & \end{aligned}&\begin{aligned}\displaystyle \sum_{\textrm{sym}}^{.}x^{2}y&=x^{2}y+x^{2}z\\ &+y^{2}x+y^{2}z\\\\ &+z^{2}x+z^{2}y \end{aligned}\\ \begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}xyz&=xyz+yzx+zxy\\ &=3xyz \end{aligned}&\begin{aligned}\displaystyle \sum_{\textrm{sym}}^{.}xyz&=xyz+xzy+\cdots \\ &=6xyz \end{aligned} \\\hline \end{array} \end{aligned}$.
$\begin{array}{ll}\\ 9.&(\textbf{IMO 1995})\\ &\textrm{Jika}\: \: a,b,c\: \: \textrm{bilangan-bilangan real positif}\\ &\textrm{dengan}\: \: abc=1,\: \: \textrm{maka tunjukkan bahwa}\\ &\displaystyle \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\geq \displaystyle \frac{3}{2}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Misalkan}\: \: x=\displaystyle \frac{1}{a},\: y=\displaystyle \frac{1}{b},\: \: \textrm{dan}\: \: z=\displaystyle \frac{1}{c},\\ & \textrm{maka}\\ &\displaystyle \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\\ &=\displaystyle \frac{x^{3}yz}{y+z}+ \frac{y^{3}xz}{x+z}+ \frac{z^{3}xy}{x+y},\: \: \textrm{karena}\: xyz=1\\ &=\displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y}\\ &\textrm{Dengan ketaksamaan}\: \textbf{Cauchy-Schwarz}\\ &\left ( 2\displaystyle \sum_{\textrm{siklik}}^{.}y+z \right )\left ( \displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right )\geq (x+y+z)^{2}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right )\geq \displaystyle \frac{(x+y+z)^{2}}{2(x+y+z)}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right )\geq \displaystyle \frac{(a+b+c)}{2}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right )\geq \displaystyle \frac{3\sqrt[3]{xyz}}{2}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right )\geq \displaystyle \frac{3}{2}\qquad \blacksquare \end{aligned} \end{array}$.
DAFTAR PUSTAKA
- Tung. K.Y. 2013. Ayo Raih Medali Emas Olimpiade Matematika SMA. Yogyakarta: ANDI.
- Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.
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