Contoh Soal 13 (Segitiga dan Ketaksamaan)

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

$\begin{array}{l}\\ 61.& (\mathbf{\text{IMO 1995}})\\ & \text{Jika}a,b,c\text{bilangan-bilangan real positif}\\ & \text{dengan}abc=1,\text{maka tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\ge {\displaystyle \frac{3}{2}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Misalkan}x={\displaystyle \frac{1}{a},y={\displaystyle \frac{1}{b},\text{dan}z={\displaystyle \frac{1}{c},}}}\\ & \text{maka}\\ & {\displaystyle \frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}}\\ & ={\displaystyle \frac{x^{3}yz}{y+z}+\frac{y^{3}xz}{x+z}+\frac{z^{3}xy}{x+y},\text{karena}xyz=1}\\ & ={\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\\ & \text{Dengan ketaksamaan}\mathbf{\text{Cauchy-Schwarz}}\\ & \left(2{\displaystyle \sum _{\text{siklik}}^{.}y+z}\right)\left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge (x+y+z{)}^2\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{(x+y+z{)}^2}{2(x+y+z)}}\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{(a+b+c)}{2}}\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{3\sqrt[3]{xyz}}{2}}\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{3}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 62.& \text{Diketahui}a,b\text{bilangan real positif}\\ & \text{Tunjukkan bahwa}{\displaystyle \frac{a^2}{b}+\frac{b^2}{a}\ge a+b}\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Asumsikan bahwa}a\ge b,\text{maka}{a}^2\ge b^2\\ & \text{dan}{\displaystyle \frac{1}{b}\ge \frac{1}{a}.}\\ & \text{Perhatikan bahwa baik}(a^2,b^2)\text{dan}\left({\displaystyle \frac{1}{b},\frac{1}{a}}\right)\\ & \text{adalah kumpulan dua barisan yang monoton}\\ & \text{sama yaitu sama-sama naik. Sehingga}\\ & \text{dengan}\mathbf{\text{ketaksamaan Renata}}\text{diperoleh}\\ & a^2.{\displaystyle \frac{1}{b}+b^2.{\displaystyle \frac{1}{a}\ge a^2.{\displaystyle \frac{1}{a}+b^2.{\displaystyle \frac{1}{b}}}}}\\ & \iff {\displaystyle \frac{a^2}{b}+\frac{b^2}{a}\ge a+b◼}\end{array}$.

$\begin{array}{l}\\ 63.& \text{Diberikan}a,b,c>0,\text{tunjukkan bahwa}\\ & {\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Alternatif 1}\\ & \text{Dengan AM-GM diperoleh}\\ & \bullet {\displaystyle \frac{a}{c}+\frac{c}{a}\ge 2\iff {\displaystyle \frac{c}{a}\ge 2-{\displaystyle \frac{a}{c}}}}\\ & \bullet {\displaystyle \frac{b}{c}+\frac{c}{b}\ge 2\iff {\displaystyle \frac{b}{c}\ge 2-{\displaystyle \frac{c}{b}}}}\\ & \bullet {\displaystyle \frac{a}{b}+\frac{b}{a}\ge 2\iff {\displaystyle \frac{a}{b}\ge 2-{\displaystyle \frac{b}{a}}}}\\ & \text{Sehingga}\\ & \begin{array}{rl}{\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}}& \ge a(2-{\displaystyle \frac{c}{b}})+b(2-{\displaystyle \frac{a}{c}})+c(2-{\displaystyle \frac{b}{a}})\\ & =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}}& \ge 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)\ge 2(a+b+c)\\ {\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}}& \ge a+b+c◼\end{array}\\ & \text{Alternatif 2}\\ & \text{Dengan ketaksamaan Renata}\\ & \begin{array}{rl}& \text{Asumsikan}a\ge b\ge c,\text{maka}ab\ge ca\ge bc\\ & \text{dan}{\displaystyle \frac{1}{c}\ge \frac{1}{b}\ge \frac{1}{a}.}\\ & \text{Perhatikan bahwa}\\ & (ab\ge ca\ge bc)\text{dan}\left({\displaystyle \frac{1}{c}\ge \frac{1}{b}\ge \frac{1}{a}}\right)\\ & \text{memiliki kemonotonan yang sama}\\ & \text{maka dengan}\mathbf{\text{ketaksamaan Renata}}\\ & \text{dapat diperoleh bentuk}\\ & ab.{\displaystyle \frac{1}{c}+ac.{\displaystyle \frac{1}{b}+bc.{\displaystyle \frac{1}{a}\ge ab.{\displaystyle \frac{1}{b}+ac.{\displaystyle \frac{1}{a}+bc.{\displaystyle \frac{1}{c}}}}}}}\\ & \iff {\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+c+b}\\ & \iff {\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 64.& \text{Diberikan}a,b,c>0,\text{tunjukkan kebenaran}\\ & \mathbf{\text{ketaksamaan Nesbitt}}\text{berikut}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Alternatif 1}\\ & \text{Dengan AM-GM diperoleh}\\ & {\displaystyle \frac{(a+b)+(b+c)+(c+a)}{3}\ge {\displaystyle \frac{3}{{\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}}}}\\ & \iff ((a+b)+(b+c)+(c+a))\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge 9\\ & \iff 2\left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)+6\ge 9\\ & \iff 2\left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge 3\\ & \iff \left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge {\displaystyle \frac{3}{2}◼}\\ & \text{Alternatif 2}\\ & \text{Dengan ketaksamaan Renata}\\ & \begin{array}{rl}& \text{Asumsikan}a\le b\le c,\text{maka}a+b\le a+c\le b+c\\ & \text{dan}{\displaystyle \frac{1}{b+c}\le \frac{1}{a+c}\le \frac{1}{a+b}.}\\ & \text{Perhatikan bahwa}\\ & (a\le b\le c)\text{dan}{\displaystyle \frac{1}{b+c}\le \frac{1}{a+c}\le \frac{1}{a+b}}\\ & \text{memiliki kemonotonan yang sama}\\ & \text{maka dengan}\mathbf{\text{ketaksamaan Renata}}\\ & \text{dapat diperoleh bentuk}\\ & a.{\displaystyle \frac{1}{b+c}+b.{\displaystyle \frac{1}{a+c}+c.{\displaystyle \frac{1}{a+b}\ge b.{\displaystyle \frac{1}{b+c}+c.{\displaystyle \frac{1}{a+c}+a.{\displaystyle \frac{1}{a+b}}}}}}}\\ & \text{dan}\\ & a.{\displaystyle \frac{1}{b+c}+b.{\displaystyle \frac{1}{a+c}+c.{\displaystyle \frac{1}{a+b}\ge c.{\displaystyle \frac{1}{b+c}+a.{\displaystyle \frac{1}{a+c}+b.{\displaystyle \frac{1}{a+b}}}}}}}\\ & \text{Jika dijumlahkan keduanya, maka}\\ & 2\left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge 3\\ & \iff \left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge {\displaystyle \frac{3}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 65.& (\mathbf{\text{OSN 2015}})\\ & \text{Diberikan}a,b,c>0,\text{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}}}\ge 3\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Perhatikan bukti soal no. 4 di atas}\\ & \text{Dengan}\mathbf{\text{keksamaan Renata}}\text{dapat diperoleh}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{a+c}\ge \frac{b}{b+c}+\frac{a}{a+c}}\\ & \text{Misalkan}\\ & x=b+c,y=c+a,y=a+b,\text{maka}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{a+c}\ge \frac{b}{b+c}+\frac{a}{a+c}}\\ & \iff {\displaystyle \frac{a}{b+c}+\frac{b}{a+c}\ge \frac{y+z-x}{2x}+\frac{x+z-y}{2y}}\\ & \begin{array}{rl}& \iff \sqrt{{\displaystyle \frac{a}{b+c}+\frac{b}{a+c}}}\ge \sqrt{\frac{y+z-x}{2x}+\frac{x+z-y}{2y}}\\ & \iff =\sqrt{{\displaystyle \frac{1}{2}}}\sqrt{{\displaystyle \frac{y}{x}+\frac{z}{x}-1+\frac{x}{y}+\frac{z}{y}-1}}\\ & \iff \text{dengan AM-GM}\\ & \iff \ge {\displaystyle \frac{1}{\sqrt{2}}\sqrt{{\displaystyle \frac{z}{x}+\frac{z}{y}+2\sqrt{\frac{y}{x}.\frac{x}{y}}-2}}}\\ & \iff ={\displaystyle \frac{1}{\sqrt{2}}\sqrt{{\displaystyle \frac{z}{x}+\frac{z}{y}+2-2}}}\\ & \iff \ge {\displaystyle \frac{1}{\sqrt{2}}\sqrt{{\displaystyle \frac{z}{x}+\frac{z}{y}}}}\\ & \iff \ge {\displaystyle \frac{\sqrt{2}}{\sqrt{2}}\sqrt{{\displaystyle \frac{z^2}{xy}}}={\displaystyle \sqrt{{\displaystyle \frac{z^2}{xy}}}}}\end{array}\\ & \begin{array}{rl}& \bullet \sqrt{{\displaystyle \frac{a}{b+c}+\frac{b}{a+c}}}\ge {\displaystyle \sqrt{{\displaystyle \frac{z^2}{xy}}}}\\ & \bullet \sqrt{{\displaystyle \frac{b}{c+a}+\frac{c}{a+b}}}\ge \sqrt{{\displaystyle \frac{x^2}{yz}}}\\ & \bullet \sqrt{{\displaystyle \frac{c}{a+b}+\frac{a}{b+c}}}\ge \sqrt{{\displaystyle \frac{y^2}{xz}}}\end{array}\\ & \begin{array}{rl}& \text{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}}}\\ & \ge {\displaystyle \sqrt{{\displaystyle \frac{z^2}{xy}}}+\sqrt{{\displaystyle \frac{x^2}{yz}}}+\sqrt{{\displaystyle \frac{y^2}{xz}}}}\\ & \text{Dengan AM-GM lagi}\\ & \ge 3\sqrt[3]{\sqrt{{\displaystyle \frac{z^2}{xy}}}\times \sqrt{{\displaystyle \frac{x^2}{yz}}}\times \sqrt{{\displaystyle \frac{y^2}{xz}}}}\\ & \ge 3\sqrt[3]{\sqrt{{\displaystyle \frac{(xyz{)}^2}{(xyz{)}^2}}}}\\ & \ge 3◼\end{array}\end{array}$

DAFTAR PUSTAKA

1. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.

WEBSITE

1. https://holdenlee.github.io/high_school/omc/23-rearrange.pdf diakses 18 Januari 2022.
2. https://www.gotohaggstrom.com/Advanced%20inequality%20manipulations.pdf  diakses 20 Januari 2022