Contoh Soal 14 (Segitiga dan Ketaksamaan)

$\begin{array}{l}\\ 66.& \text{Jika}x,y\text{bilangan positif, tunjukkan}\\ & x^2+y^2\ge {\displaystyle \frac{(x+y{)}^2}{2}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Alternatif 1}\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & (x-y{)}^2\ge 0\iff x^2-2xy+y^2\ge 0\\ & \iff x^2+y^2\ge 2xy\\ & \iff 2x^2+2y^2\ge x^2+y^2+2xy\\ & \iff 2(x^2+y^2)\ge (x+y{)}^2\\ & \iff x^2+y^2\ge {\displaystyle \frac{(x+y{)}^2}{2}◼}\end{array}\\ & \text{Alternatif 2}\\ & \text{Dengan Ketaksamaan Holder pilih}\\ & {\lambda }_1={\lambda }_2={\displaystyle \frac{1}{2}\left({\displaystyle \sum _{i=1}^2{\lambda }_i=1}\right)\text{akan diperoleh}}\\ & \begin{array}{rl}& {\left({\displaystyle \frac{x^2}{1}+\frac{y^2}{1}}\right)}^{\frac{1}{2}}(1+1{)}^{\frac{1}{2}}\ge x+y\\ & \iff (x^2+y^2)(1+1)\ge (x+y{)}^2\\ & \iff (x^2+y^2)\ge {\displaystyle \frac{(x+y{)}^2}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 67.& \text{Untuk}x,y\text{bilangan positif, tunjukkan}\\ & \text{kebenaran ketaksamaan Cauchy-Schwarz}\\ & {\displaystyle \frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\frac{x_3^2}{y_3}+\cdots +\frac{x_n^2}{y_n}\ge {\displaystyle \frac{(x_1+x_2+\cdots +x_n{)}^2}{y_1+y_2+\cdots +y_n}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Dengan Ketaksamaan Holder pilih}\\ & {\lambda }_1={\lambda }_2={\displaystyle \frac{1}{2}\left({\displaystyle \sum _{i=1}^2{\lambda }_i=1}\right)\text{akan diperoleh}}\\ & {\left({\displaystyle \frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\cdots +\frac{x_n^2}{y_n}}\right)}^{\frac{1}{2}}(y_1+y_2+\cdots +y_n{)}^{\frac{1}{2}}\ge (x_1+x_2+\cdots +x_n)\\ & \left({\displaystyle \frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\cdots +\frac{x_n^2}{y_n}}\right)(y_1+y_2+\cdots +y_n)\ge (\sqrt{x_1^2}+\sqrt{x_2^2}+\cdots +\sqrt{x_n^2}{)}^2\\ & \left({\displaystyle \frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\cdots +\frac{x_n^2}{y_n}}\right)(y_1+y_2+\cdots +y_n)\ge (x_1+x_2+\cdots +x_n{)}^2\\ & \left({\displaystyle \frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\cdots +\frac{x_n^2}{y_n}}\right)\ge {\displaystyle \frac{(x_1+x_2+\cdots +x_n{)}^2}{(y_1+y_2+\cdots +y_n)}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 68.& (\mathbf{\text{National Mathematical Contest, Belarus-2000}})\\ & \text{Jika}a,b,c,x,y\text{dan}z\text{bilangan real positif, tunjukkan}\\ & {\displaystyle \frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}\ge {\displaystyle \frac{(a+b+c{)}^3}{3(x+y+z)}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Dengan Ketaksamaan Holder pilih}\\ & {\lambda }_1={\lambda }_2={\lambda }_3={\displaystyle \frac{1}{3}\left({\displaystyle \sum _{i=1}^3{\lambda }_i=1}\right)\text{akan diperoleh}}\\ & {\left({\displaystyle \frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}}\right)}^{\frac{1}{3}}(1+1+1{)}^{\frac{1}{3}}(x+y+z{)}^{\frac{1}{3}}\ge (a+b+c)\\ & \iff \left({\displaystyle \frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}}\right)(3)(x+y+z)\ge (a+b+c{)}^3\\ & \iff \left({\displaystyle \frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}}\right)\ge {\displaystyle \frac{(a+b+c{)}^3}{3(x+y+z)}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 69.& \text{Jika}a,b,c,x,y\text{dan}z\text{bilangan real positif}\\ & \text{dengan}a+b+c=x+y+z\text{tunjukkan bahwa}\\ & {\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}\ge a+b+c}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Dengan Ketaksamaan Holder pilih}\\ & {\lambda }_1={\lambda }_2={\lambda }_3={\displaystyle \frac{1}{3}\left({\displaystyle \sum _{i=1}^3{\lambda }_i=1}\right)\text{akan diperoleh}}\\ & {\left({\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}}\right)}^{\frac{1}{3}}(x+y+z{)}^{\frac{1}{3}}(x+y+z{)}^{\frac{1}{3}}\ge (a+b+c)\\ & \iff \left({\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}}\right)(x+y+z{)}^2\ge (a+b+c{)}^3\\ & \iff \left({\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}}\right)(a+b+c{)}^2\ge (a+b+c{)}^3\\ & \iff \left({\displaystyle \frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}}\right)\ge {\displaystyle \frac{(a+b+c{)}^3}{(a+b+c{)}^2}}\\ & \iff {\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}\ge a+b+c◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 70.& \text{Jika}a,b,c\text{adalah bilangan real positif}\\ & \text{dengan}a+b+c=1,\text{tunjukkan}\\ & \text{bahwa}\left({\displaystyle \frac{1}{a}+1}\right)\left({\displaystyle \frac{1}{b}+1}\right)\left({\displaystyle \frac{1}{c}+1}\right)\ge 64\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Alternatif 1}\\ & \begin{array}{rl}& \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}\\ & \text{Dengan AM-GM kita mendapatkan}\\ & \bullet {\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 3\sqrt[3]{{\displaystyle \frac{1}{abc}}}}\\ & \bullet {\displaystyle \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge 3\sqrt[3]{{\displaystyle \frac{1}{(abc{)}^2}}}}\\ & \text{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}}\\ & \ge 1+3\sqrt[3]{{\displaystyle \frac{1}{abc}}}+3\sqrt[3]{{\displaystyle \frac{1}{(abc{)}^2}}}+\sqrt[3]{{\displaystyle \frac{1}{(abc{)}^3}}}\\ & \ge 1+3\sqrt[3]{{\displaystyle \frac{1}{abc}}}+3\sqrt[3]{{\displaystyle \frac{1}{(abc{)}^2}}}+\sqrt[3]{{\displaystyle \frac{1}{(abc{)}^3}}}\\ & ={(1+{\displaystyle \frac{1}{\sqrt[3]{abc}}})}^3\\ & \text{Karena}\sqrt[3]{abc}\le {\displaystyle \frac{a+b+c}{3}={\displaystyle \frac{1}{3},\text{maka}}}\\ & \left({\displaystyle \frac{1}{a}+1}\right)\left({\displaystyle \frac{1}{b}+1}\right)\left({\displaystyle \frac{1}{c}+1}\right)\ge {(1+{\displaystyle \frac{1}{\sqrt[3]{abc}}})}^3\\ & \ge {(1+{\displaystyle \frac{1}{\left(\frac{1}{3}\right)}})}^3\\ & \ge {(1+3)}^4\\ & \ge 4^4\\ & \ge 64◼\end{array}\\ & \begin{array}{rl}& \text{Alternatif 2}\\ & \text{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)\ge {(\sqrt[3]{{\displaystyle \frac{1}{a}.\frac{1}{b}.\frac{1}{c}}}+\sqrt[3]{1.1.1})}^3\\ & \ge {(\sqrt[3]{{\displaystyle \frac{1}{(abc)}}}+1)}^3\\ & \ge {\left({\displaystyle \frac{1}{\sqrt[3]{abc}}+1}\right)}^3\\ & \ge {(3+1)}^3\\ & \ge 64◼\end{array}\end{array}$.