Lanjutan 2 Materi Segitiga dan Ketaksamaan Cauchy-Schwarz-Engel

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

$\begin{array}{rl}& (x_1{y}_1+x_2{y}_2+x_3{y}_3+...+x_n{y}_n{)}^2\\ & \le (x_1^2+x_2^2+x_3^2+...+x_n^2)(y_1^2+y_2^2+y_3^2+...+y_n^2)\end{array}$.

Jika dituliskan dengan notasi sigma adalah:

${\left({\displaystyle \sum _{i=1}^n{x}_i{y}_i}\right)}^2\le \left({\displaystyle \sum _{i=1}^n{x}_i^2}\right)\left({\displaystyle \sum _{i=1}^n{y}_i^2}\right)$.

$\begin{array}{rl}& \mathbf{\text{Bukti}}\text{dari ketaksamaan ini adalah}:\\ & \text{Diberikan}\\ & {\alpha }_1,{\alpha }_2,{\alpha }_3,\cdots ,{\alpha }_n\text{dan}{\beta }_1+{\beta }_2+{\beta }_3+\cdots +{\beta }_n\\ & \text{Untuk setiap bilangan real}x,\text{pada polinom kuadrat}\\ & \text{dapat berlaku}\\ & f(x)={\displaystyle \sum _{i=1}^n{(\alpha x+\beta )}^2\ge 0}\\ & \iff \left({\displaystyle \sum _{i=1}^n{\alpha }_i^2}\right)x^2+2\left({\displaystyle \sum _{i=1}^n{\alpha }_i{\beta }_i}\right)x+\left({\displaystyle \sum _{i=1}^n{\beta }_i^2}\right)\ge 0\\ & \text{akibatnya diskriminan (D)}\le 0,\text{maka}\\ & D=b^2-4ac\le 0\\ & \iff {\left(2\left({\displaystyle \sum _{i=1}^n{\alpha }_i{\beta }_i}\right)\right)}^2-4\left({\displaystyle \sum _{i=1}^n{\alpha }_i^2}\right)\left({\displaystyle \sum _{i=1}^n{\beta }_i^2}\right)\le 0\\ & \iff 4\left({\displaystyle \sum _{i=1}^n{\alpha }_i{\beta }_i}\right)\le 4\left({\displaystyle \sum _{i=1}^n{\alpha }_i^2}\right)\left({\displaystyle \sum _{i=1}^n{\beta }_i^2}\right)\\ & \iff \left({\displaystyle \sum _{i=1}^n{\alpha }_i{\beta }_i}\right)\le \left({\displaystyle \sum _{i=1}^n{\alpha }_i^2}\right)\left({\displaystyle \sum _{i=1}^n{\beta }_i^2}\right)◼\end{array}$.

Ada hal yang sangat menarik ketika substitusi bentuk  $x_i={\displaystyle \frac{p_i}{\sqrt{q_i}}}$  dan  $y_i=\sqrt{q_i}$, yaitu:

$\begin{array}{rl}& \text{Perhatikan bentuk berikut}\\ & (x_1{y}_1+x_2{y}_2+x_3{y}_3+...+x_n{y}_n{)}^2\\ & \le (x_1^2+x_2^2+x_3^2+...+x_n^2)(y_1^2+y_2^2+y_3^2+...+y_n^2)\end{array}$.

$\begin{array}{rl}& \text{dengan substitusi}\\ & \{\begin{array}{ll}{x}_i& ={\displaystyle \frac{p_i}{\sqrt{q_i}}}\\ y_i& =\sqrt{q_i}\end{array}\\ & \text{maka menjadi bentuk}\end{array}$.

$\begin{array}{rl}& (p_1+p_2+p_3+...+p_n{)}^2\\ & \le \left({\displaystyle \frac{p_1^2}{q_1}+\frac{p_2^2}{q_2}+\frac{p_3^2}{q_3}+...+\frac{p_n^2}{q_n}}\right)(q_1+q_2+q_3+...+q_n)\\ & \iff {\displaystyle \frac{(p_1+p_2+p_3+...+p_n{)}^2}{(q_1+q_2+q_3+...+q_n)}\le \left({\displaystyle \frac{p_1^2}{q_1}+\frac{p_2^2}{q_2}+\frac{p_3^2}{q_3}+...+\frac{p_n^2}{q_n}}\right)}\end{array}$.

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

${\displaystyle \frac{a^2}{x}+\frac{b^2}{y}\ge {\displaystyle \frac{(a+b{)}^2}{x+y}}}$.

$\begin{array}{rl}& \mathbf{\text{Bukti}}\\ & \text{Perhatikan bahwa}(ay-bx{)}^2\ge 0\\ & \iff a^2{y}^2+b^2{y}^2-2abxy\ge 0\\ & \iff a^2{y}^2+b^2{y}^2\ge 2abxy\\ & \iff a^{2}xy+a^2{y}^2+b^2{y}^2+b^{2}xy\ge a^{2}xy+2abxy+b^{2}xy\\ & \iff (a^{2}y+b^{2}x)(x+y)\ge (a^2+2ab+b^2)xy\\ & \iff {\displaystyle \frac{(a^{2}y+b^{2}x)}{xy}\ge {\displaystyle \frac{(a^2+2ab+b^2)}{(x+y)}}}\\ & \iff {\displaystyle \frac{a^2}{x}+\frac{b^2}{y}\ge {\displaystyle \frac{(a+b{)}^2}{x+y}◼}}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika}x,y\text{bilangan positif, tunjukkan}\\ & x^2+y^2\ge {\displaystyle \frac{(x+y{)}^2}{2}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Alternatif 1}\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & (x-y{)}^2\ge 0\iff x^2-2xy+y^2\ge 0\\ & \iff x^2+y^2\ge 2xy\\ & \iff 2x^2+2y^2\ge x^2+y^2+2xy\\ & \iff 2(x^2+y^2)\ge (x+y{)}^2\\ & \iff x^2+y^2\ge {\displaystyle \frac{(x+y{)}^2}{2}◼}\end{array}\\ & \begin{array}{rl}& \text{Alternatif 2}\\ & \text{Dengan QM-AM kita akan mendapatkan}\\ & \sqrt{{\displaystyle \frac{x^2+y^2}{2}}}\ge {\displaystyle \frac{x+y}{2}}\\ & \iff \left({\displaystyle \frac{x^2+y^2}{2}}\right)\ge {\left({\displaystyle \frac{x+y}{2}}\right)}^2\\ & \iff \left({\displaystyle \frac{x^2+y^2}{2}}\right)\ge {\displaystyle \frac{(x+y{)}^2}{4}}\\ & \iff x^2+y^2\ge {\displaystyle \frac{(x+y{)}^2}{2}◼}\end{array}\\ & \text{Alternatif 3}\\ & \text{Dengan ketaksamaan CS-Engel}\\ & \text{kita dapat peroleh}\\ & {\displaystyle \frac{x^2}{1}+\frac{y^2}{1}\ge {\displaystyle \frac{(x+y{)}^2}{2}}}\\ & \iff x^2+y^2\ge {\displaystyle \frac{(x+y{)}^2}{2}◼}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Buktikan bahwa setiap bilangan real}\\ & \text{positif}a,b\text{dan}c\text{berlaku}\\ & a^2+b^2+c^2\ge ab+ac+bc\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & \text{Perhatikan bahwa}(a-b{)}^2\ge 0\\ & (a-c{)}^2\ge 0,\text{dan}(b-c{)}^2\ge 0\\ & \text{adalah benar, maka}\\ & (a-b{)}^2=a^2-2ab+b^2\ge 0\\ & \iff a^2+b^2\ge 2ab.....(1)\\ & \text{Dengan cara yang kurang lebih sama}\\ & \text{akan didapatkan}\\ & \bullet a^2+c^2\ge 2ac.....(2)\\ & \bullet b^2+c^2\ge 2bc.....(1)\\ & \text{Jika ketaksamaan}(1),(2),\mathrm{\&}(3)\text{dijumlahkan}\\ & \text{akan didapatkan bentuk}\\ & 2a^2+2b^2+2c^2\ge 2ab+2ac+2bc\\ & \iff a^2+b^2+c^2\ge ab+ac+bc◼\\ & \text{Alternatif 2}\\ & \text{Dengan ketaksamaan}\mathbf{\text{Cauchy-Schwarz}}\\ & (a^2+b^2+c^2)(b^2+c^2+a^2)\ge (ab+bc+ca{)}^2\\ & \iff a^2+b^2+c^2\ge ab+ac+bc◼\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Buktikan bahwa setiap bilangan real}\\ & \text{positif}a,b\text{dan}c\text{berlaku}\\ & a^2+b^2+c^2\ge 3(ab+ac+bc)\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Lihat jawaban no.2 di atas, yaitu}\\ & a^2+b^2+c^2\ge ab+ac+bc\\ & \text{karena nilai dari}(a+b+c{)}^2\\ & =a^2+b^2+c^2+2(ab+ac+bc)\\ & \text{maka nilai}\\ & a^2+b^2+c^2=(a+b+c{)}^2-2(ab+ac+bc)\\ & \text{Sehingga nilai untuk}\\ & a^2+b^2+c^2\ge ab+ac+bc\text{menjadi}\\ & (a+b+c{)}^2-2(ab+ac+bc)\ge ab+ac+bc\\ & \iff (a+b+c{)}^2\ge 3(ab+ac+bc)◼\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Diketahui}x,y,z\text{adalah bilangan real }\\ & \text{positif. Tunjukkan bahwa}\\ & (x+y+z{)}^2\le 3(x^2+y^2+z^2)\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Alternatif 1}\\ & \begin{array}{rl}& \text{Dengan ketaksamaan Cauchy-Schwarz}\\ & \text{Kita dapat peroleh}\\ & (ax+by+cz{)}^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)\\ & \text{untuk}a=b=c=1,\text{akan diperoleh}\\ & (ax+by+cz{)}^2\le (1^2+1^2+1^2)(x^2+y^2+z^2)\\ & \iff (ax+by+cz{)}^2\le (3)(x^2+y^2+z^2)◼\end{array}\end{array}$.

$\begin{array}{rl}& \text{Alternatif 2}\\ & \text{Dengan ketaksamaan CS-Engel}\\ & \text{Kita dapat peroleh}\\ & {\displaystyle \frac{x^2}{1}+\frac{y^2}{1}+\frac{z^2}{1}\ge {\displaystyle \frac{(x+y+z{)}^2}{1+1+1}}}\\ & \iff x^2+y^2+z^2\ge {\displaystyle \frac{(x+y+z{)}^2}{3}}\\ & \iff 3(x^2+y^2+z^2)\ge (x+y+z{)}^2,\\ & \text{atau}\\ & \iff (x+y+z{)}^2\le 3(x^2+y^2+z^2)◼\end{array}$.

$\begin{array}{rl}& \text{Alternatif 3}\\ & \text{Dengan QM-AM kita akan mendapatkan}\\ & \sqrt{{\displaystyle \frac{x^2+y^2+z^2}{3}}}\ge {\displaystyle \frac{x+y+z}{3}}\\ & \iff \left({\displaystyle \frac{x^2+y^2+z^2}{3}}\right)\ge {\left({\displaystyle \frac{x+y+z}{3}}\right)}^2\\ & \iff \left({\displaystyle \frac{x^2+y^2+z^2}{3}}\right)\ge {\displaystyle \frac{(x+y+z{)}^2}{9}}\\ & \iff x^2+y^2+z^2\ge {\displaystyle \frac{(x+y+z{)}^2}{3}}\\ & \iff 3(x^2+y^2+z^2)\ge (x+y+z{)}^2\text{atau}\\ & \iff (x+y+z{)}^2\le 3(x^2+y^2+z^2)◼\end{array}$.

$\begin{array}{l}\\ 5.& \text{Jika}x,y,z\text{adalah bilangan real positif}\\ & \text{dengan}{x}^2+y^2+z^2=27.\text{Tunjukkan}\\ & \text{bahwa}{x}^3+y^3+z^3\ge 81\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Pada contoh soal no.2 terdapat}\\ & (ax+by+cz{)}^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)\\ & \text{ganti mengganti}a=b=c=1,\text{maka menjadi}\\ & (x+y+z{)}^2\le (1^2+1^2+1^2)(x^2+y^2+z^2)\\ & \iff (x+y+z{)}^2\le (3)(x^2+y^2+z^2)........(1)\\ & \text{Selanjutnya dengan mengganti dengan}\\ & x^{{.}^{\frac{3}{2}}},y^{{.}^{\frac{3}{2}}},z^{{.}^{\frac{3}{2}}}\text{dan}{x}^{{.}^{\frac{1}{2}}},y^{{.}^{\frac{1}{2}}},z^{{.}^{\frac{1}{2}}},\text{pada}\\ & (ax+by+cz{)}^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)\\ & \text{Kita akan dapatkan}\\ & (x^2+y^2+z^2)\le (x^3+y^3+z^3)(x+y+z)........(2)\\ & \text{Jika masing-masing ruas dikuadratkan, maka}\\ & (x^2+y^2+z^2{)}^4\le (x^3+y^3+z^3{)}^2(x+y+z{)}^2\\ & \iff (x^2+y^2+z^2{)}^4\le 3(x^3+y^3+z^3{)}^2(x^2+y^2+z^2)\\ & \iff (x^2+y^2+z^2{)}^3\le 3(x^3+y^3+z^3{)}^2\\ & \iff (27{)}^3\le 3(x^3+y^3+z^3{)}^2\\ & \iff 3^9\le 3(x^3+y^3+z^3{)}^2\\ & \iff 3^8\le (x^3+y^3+z^3{)}^2\\ & \iff (x^3+y^3+z^3{)}^2\ge 3^8\\ & \iff (x^3+y^3+z^3)\ge 3^{{.}^{\frac{8}{2}}}\\ & \iff (x^3+y^3+z^3)\ge 3^4=81◼\end{array}$.

$\begin{array}{l}\\ 6.& \text{Untuk}a,b,c,d\text{adalah bilangan real }\\ & \text{positif, tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\ge {\displaystyle \frac{64}{a+b+c+d}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Dengan ketaksamaan CS-Engel}\\ & (a+b+c+d)\left({\displaystyle \frac{1^2}{a}+{\displaystyle \frac{1^2}{b}+{\displaystyle \frac{2^2}{c}+{\displaystyle \frac{4^2}{d}}}}}\right)\ge (1+1+2+4{)}^2\\ & \iff {\displaystyle \frac{1^2}{a}+{\displaystyle \frac{1^2}{b}+{\displaystyle \frac{2^2}{c}+{\displaystyle \frac{4^2}{d}\ge {\displaystyle \frac{(1+1+2+4{)}^2}{a+b+c+d}}}}}}\\ & \iff {\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\ge {\displaystyle \frac{64}{a+b+c+d}◼}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Diketahui}a,b\text{bilangan real positif}\\ & \text{dan}a+b=1.\text{Tunjukkan bahwa}\\ & {(a+{\displaystyle \frac{1}{a}})}^2+{(b+{\displaystyle \frac{1}{b}})}^2\ge {\displaystyle \frac{25}{2}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & \text{Diketahui bahwa}a+b=1\\ & \text{Dengan QM-AM kita akan mendapatkan}\\ & \sqrt{{\displaystyle \frac{x^2+y^2}{2}}}\ge {\displaystyle \frac{x+y}{2}\iff {\displaystyle \frac{x^2+y^2}{2}\ge {\left({\displaystyle \frac{x+y}{2}}\right)}^2}}\\ & \text{Kita misalkan}\{\begin{array}{ll}x& =a+{\displaystyle \frac{1}{a}}\\ \\ y& =b+{\displaystyle \frac{1}{b}}\end{array}\\ & \text{maka}\\ & \begin{array}{rl}{\displaystyle \frac{{(a+{\displaystyle \frac{1}{a}})}^2+{(b+{\displaystyle \frac{1}{b}})}^2}{2}}& \ge {\left({\displaystyle \frac{a+{\displaystyle \frac{1}{a}+b+{\displaystyle \frac{1}{b}}}}{2}}\right)}^2\\ & \ge {\displaystyle \frac{1}{4}{(a+b+{\displaystyle \frac{1}{a}+\frac{1}{b}})}^2}\\ & \ge {\displaystyle \frac{1}{4}{(1+{\displaystyle \frac{1}{a}+\frac{1}{b}})}^2}\\ & \ge {\displaystyle \frac{1}{4}{(1+{\displaystyle \frac{a+b}{ab}})}^2}\\ & \ge {\displaystyle \frac{1}{4}{(1+{\displaystyle \frac{1}{ab}})}^2}\end{array}\\ & \text{Dari GM-AM kita akan mendapatkan}\\ & \sqrt{ab}\le {\displaystyle \frac{a+b}{2}\iff ab\le {\left({\displaystyle \frac{a+b}{2}}\right)}^2}\\ & \iff ab\le {\left({\displaystyle \frac{1}{2}}\right)}^2\iff ab\le {\displaystyle \frac{1}{4}}\\ & \text{Sehingga}\\ & \begin{array}{rl}{(a+{\displaystyle \frac{1}{a}})}^2+{(b+{\displaystyle \frac{1}{b}})}^2& \ge {\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{1}{ab}})}^2}\\ & \ge {\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{1}{{\displaystyle \frac{1}{4}}}})}^2}\\ & \ge {\displaystyle \frac{1}{2}{(1+4)}^2}\\ & \ge {\displaystyle \frac{1}{2}\times 25}\\ & \ge {\displaystyle \frac{25}{2}◼}\end{array}\end{array}\end{array}$.

$\begin{array}{rl}& \text{Alternatif 2}\\ & \text{Dengan AM-GM kita akan mendapatkan}\\ & 1=a+b\ge 2\sqrt{ab}\iff {\displaystyle \frac{1}{2}\ge \sqrt{ab}\iff 2\le {\displaystyle \frac{1}{\sqrt{ab}}}}\\ & \iff {\displaystyle \frac{1}{\sqrt{ab}}\ge 2\iff {\displaystyle \frac{1}{ab}\ge 4}}\\ & \text{Perhatikan soal, dengan QM-AM}\text{akan}\\ & \text{didapatkan}\\ & \begin{array}{rl}{\displaystyle \frac{{(a+\frac{1}{a})}^2+{(b+\frac{1}{b})}^2}{2}}& \ge {\left({\displaystyle \frac{a+\frac{1}{a}+b+\frac{1}{b}}{2}}\right)}^2\\ {(a+{\displaystyle \frac{1}{a}})}^2+{(b+{\displaystyle \frac{1}{b}})}^2& \ge {\displaystyle \frac{1}{2}{(a+b+{\displaystyle \frac{1}{a}+\frac{1}{b}})}^2}\\ & ={\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{a+b}{ab}})}^2}\\ & ={\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{1}{ab}})}^2}\\ & ={\displaystyle \frac{1}{2}(1+4{)}^2}\\ & ={\displaystyle \frac{1}{2}\times 25}\\ & \ge {\displaystyle \frac{25}{2}◼}\end{array}\end{array}$.

$\begin{array}{rl}& \text{Alternatif 3}\\ & \text{Dengan AM-GM kita akan mendapatkan}\\ & 1=a+b\ge 2\sqrt{ab}\iff {\displaystyle \frac{1}{2}\ge \sqrt{ab}\iff 2\le {\displaystyle \frac{1}{\sqrt{ab}}}}\\ & \iff {\displaystyle \frac{1}{\sqrt{ab}}\ge 2\iff {\displaystyle \frac{1}{ab}\ge 4}}\\ & \text{Perhatikan soal, dengan CS-Engel}\text{akan}\\ & \text{didapatkan}\\ & \begin{array}{rl}& (1+1)\left({\displaystyle \frac{{(a+\frac{1}{a})}^2}{1}+{\displaystyle \frac{{(b+\frac{1}{b})}^2}{1}}}\right)\ge {(a+\frac{1}{a}+b+\frac{1}{b})}^2\\ & \iff 2({\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2)\ge {(a+b+\frac{1}{a}+\frac{1}{b})}^2\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{a+b}{ab}})}^2}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{1}{ab}})}^2}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}(1+4{)}^2}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}\times 25}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{25}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Jika}a,b\text{bilangan positif, buktikan}\\ & \text{a}.2(a^2+b^2)\ge (a+b{)}^2\\ & \text{b}.4(a^3+b^3)\ge (a+b{)}^3\\ & \text{c}.8(a^4+b^4)\ge (a+b{)}^4\\ & \text{d}.16(a^5+b^5)\ge (a+b{)}^5\\ & \text{e}.32(a^6+b^6)\ge (a+b{)}^6\\ & \text{f}.64(a^7+b^7)\ge (a+b{)}^7\\ & \text{g}.128(a^8+b^8)\ge (a+b{)}^8\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Akan ditunjukkan bukti poin 8.c saja}\\ & \text{untuk poin yang lain, silahkan pembaca}\\ & \text{sekalian untuk dibuktikan sendiri sebagai}\\ & \text{bahan latihan mandiri}.\\ & \text{Adapun bukti poin 8.c adalah sebagaimana}\\ & \text{berikut ini}\\ & \begin{array}{rl}& \text{Dengan ketaksamaan CS-Engel}\text{akan}\\ & \text{didapatkan}\\ & \bullet (1+1)(a^4+b^4)\ge (a^2+b^2{)}^2\\ & \iff 2(a^4+b^4)\ge (a^2+b^2{)}^2..........(1)\\ & \bullet (1+1)(a^2+b^2)\ge (a+b{)}^2\\ & \iff 2(a^2+b^2)\ge {\displaystyle (a+b{)}^2,(\text{kuadratkan})}\\ & \iff 4{\displaystyle {(a^2+b^2)}^2\ge (a+b{)}^4}\\ & \iff {(a^2+b^2)}^2\ge {\displaystyle \frac{(a+b{)}^4}{4}...........(2)}\\ & \text{Dari (1) dan (2) didapatkan hubungan}\\ & 2(a^4+b^4)\ge {(a^2+b^2)}^2\ge {\displaystyle \frac{(a+b{)}^4}{4}}\\ & \iff 8(a^4+b^4)\ge 4{(a^2+b^2)}^2\ge (a+b{)}^4\\ & \iff 8(a^4+b^4)\ge (a+b{)}^4◼\end{array}\end{array}$.

$.\begin{array}{rl}& \text{Mengenal penulisan pola}\mathbf{\text{Siklik dan Simetri}}\\ & \text{Misal untuk}n=3,\text{pada penulisan unsur}\\ & x,y,\text{dan}z,\text{maka}\\ & \begin{array}{|ll|}\hline \mathbf{\text{Pola Siklik}}& \mathbf{\text{Pola Simetri}}\\ \begin{array}{r}{\displaystyle \sum _{\text{siklik}}^{.}{x}^2=x^2+y^2+z^2}\\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}{\displaystyle \sum _{\text{sym}}^{.}{x}^2}& =x^2+x^2\\ & +y^2+y^2\\ \\ & +z^2+z^2\\ \\ & =2(x^2+y^2+z^2)\end{array}\\ \begin{array}{r}{\displaystyle \sum _{\text{siklik}}^{.}{x}^3=x^3+y^3+z^3}\end{array}& \begin{array}{r}{\displaystyle \sum _{\text{sym}}^{.}{x}^3=2(x^3+y^3+z^3)}\end{array}\\ \begin{array}{r}{\displaystyle \sum _{\text{siklik}}^{.}{x}^{2}y=x^{2}y+y^{2}z+z^{2}x}\\ & \\ & \\ & \end{array}& \begin{array}{rl}{\displaystyle \sum _{\text{sym}}^{.}{x}^{2}y}& =x^{2}y+x^{2}z\\ & +y^{2}x+y^{2}z\\ \\ & +z^{2}x+z^{2}y\end{array}\\ \begin{array}{rl}{\displaystyle \sum _{\text{siklik}}^{.}xyz}& =xyz+yzx+zxy\\ & =3xyz\end{array}& \begin{array}{rl}{\displaystyle \sum _{\text{sym}}^{.}xyz}& =xyz+xzy+\cdots \\ & =6xyz\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& (\mathbf{\text{IMO 1995}})\\ & \text{Jika}a,b,c\text{bilangan-bilangan real positif}\\ & \text{dengan}abc=1,\text{maka tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\ge {\displaystyle \frac{3}{2}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Misalkan}x={\displaystyle \frac{1}{a},y={\displaystyle \frac{1}{b},\text{dan}z={\displaystyle \frac{1}{c},}}}\\ & \text{maka}\\ & {\displaystyle \frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}}\\ & ={\displaystyle \frac{x^{3}yz}{y+z}+\frac{y^{3}xz}{x+z}+\frac{z^{3}xy}{x+y},\text{karena}xyz=1}\\ & ={\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\\ & \text{Dengan ketaksamaan}\mathbf{\text{Cauchy-Schwarz}}\\ & \left(2{\displaystyle \sum _{\text{siklik}}^{.}y+z}\right)\left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge (x+y+z{)}^2\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{(x+y+z{)}^2}{2(x+y+z)}}\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{(a+b+c)}{2}}\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{3\sqrt[3]{xyz}}{2}}\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{3}{2}◼}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Tung. K.Y. 2013. Ayo Raih Medali Emas Olimpiade Matematika SMA. Yogyakarta: ANDI.
2. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.