Ketaksamaan Holder

3. Ketaksamaan Holder

Misalkan $a_1,a_2,a_3,\cdots ,a_n$ dan $b_1,b_2,b_3,\cdots ,b_n$ adalah merupakan kumpulan bilangan real positif dan misalkan pula $\alpha ,\beta >0$ dengan $\alpha +\beta =1$, maka 

$\begin{array}{rl}& (a_1+\cdots +a_n{)}^{\alpha }(b_1+\cdots +b_n{)}^{\beta }\ge (a_1^{\alpha }{b}_1^{\beta }+\cdots +a_n^{\alpha }{b}_n^{\beta })\end{array}$.

Kesamaan terjadi jika dan hanya jika

$\begin{array}{rl}& {\displaystyle \frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}=\cdots ={\displaystyle \frac{a_n}{b_n}}}\end{array}$.

Secara umum Ketaksamaan Holder dituliskan sebagai berikut:

Misalkan $a_1,a_2,a_3,\cdots ,a_n$, $b_1,b_2,b_3,\cdots ,b_n$, ... , $z_1,z_2,z_3,\cdots ,z_n$ adalah merupakan barisan bilangan tak negatif serta ${\lambda }_1,{\lambda }_2,{\lambda }_3,\cdots ,{\lambda }_n$ adalah bilangan-bilangan real positif dengan ${\lambda }_1+{\lambda }_2+{\lambda }_3+\cdots +{\lambda }_n=1$, maka

$\begin{array}{rl}& (a_1+\cdots +a_n{)}^{{\lambda }_a}(b_1+\cdots +b_n{)}^{{\lambda }_b}\cdots (z_1+\cdots +z_n{)}^{{\lambda }_z}\\ & \ge a_1^{{\lambda }_a}{b}_1^{{\lambda }_b}\cdots z_1^{{\lambda }_z}+a_2^{{\lambda }_a}{b}_2^{{\lambda }_b}\cdots z_2^{{\lambda }_z}+\cdots +a_n^{{\lambda }_a}{b}_n^{{\lambda }_b}\cdots z_n^{{\lambda }_z}\\ & (a_1+\cdots +a_n{)}^{{\lambda }_a}(b_1+\cdots +b_n{)}^{{\lambda }_b}\cdots (z_1+\cdots +z_n{)}^{{\lambda }_z}\ge {\displaystyle \sum _{i=1}^n{a}_i^{{\lambda }_a}{b}_i^{{\lambda }_b}\cdots z_i^{{\lambda }_z}}\\ & {\left({\displaystyle \sum _{i=1}^n{a}_i}\right)}^{{\lambda }_a}{\left({\displaystyle \sum _{i=1}^n{b}_i}\right)}^{{\lambda }_b}\cdots {\left({\displaystyle \sum _{i=1}^n{z}_i}\right)}^{{\lambda }_z}\ge {\displaystyle \sum _{i=1}^n{a}_i^{{\lambda }_a}{b}_i^{{\lambda }_b}\cdots z_i^{{\lambda }_z}}\end{array}$.

$\begin{array}{rl}& \text{Ketaksamaan terjadi jika dan hanya jika}\\ & a_1:a_2:\cdots :a_n=b_1:b_2:\cdots :b_n=z_1:z_2:\cdots :z_n\end{array}$.

Bukti 1:

$\begin{array}{rl}& \text{Tanpa mengurangi keumuman misalkan untuk}\\ & a_1+a_2+\cdots +a_n=b_1+b_2+\cdots +b_n=\cdots =1\\ & \text{maka diperoleh}\\ & 1\ge {\displaystyle \sum _{i=1}^n{a}_i^{{\lambda }_a}{b}_i^{{\lambda }_b}\cdots z_i^{{\lambda }_z}\text{atau}{\displaystyle \sum _{i=1}^n{a}_i^{{\lambda }_a}{b}_i^{{\lambda }_b}\cdots z_i^{{\lambda }_z}\le 1}}\\ & \text{Dengan AM-GM akan diperoleh}\\ & {\displaystyle \sum _{i=1}^n{a}_i^{{\lambda }_a}{b}_i^{{\lambda }_b}\cdots z_i^{{\lambda }_z}\le {\displaystyle \sum _{i=1}^n({\lambda }_a{a}_1+{\lambda }_b{b}_1+\cdots +{\lambda }_z{z}_1)=1}}\\ & \text{Misalkan lagi kita atur}{\lambda }_a={\lambda }_b={\displaystyle \frac{1}{2}}\\ & (a_1+a_2+\cdots +a_n)(b_1+b_2+\cdots +b_n)\ge (\sqrt{a_1{b}_1}+\sqrt{a_2{b}_2}+\cdots +\sqrt{a_n{b}_n}{)}^2\\ & \text{bentuk di atas adalah Ketaksamaan Cauchy-Schwarz}\end{array}$.

Bentuk lain yang lebih luas untuk sekian-$k$ variabel, adalah

$\begin{array}{rl}& \underset{k}{\underbrace{(a_1+a_2+\cdots +a_n)(b_1+b_2+\cdots +b_n)\cdots (z_1+z_2+\cdots +z_n)}}\\ & \ge {(\sqrt[k]{a_1{b}_1\cdots z_1}+\sqrt[k]{a_2{b}_2\cdots z_2}+\cdots +\sqrt[k]{a_n{b}_n\cdots z_n})}^k\end{array}$.

Bukti 2:

Dengan induksi matematika sebagaimana berikut ini

$\begin{array}{rl}& \text{Diketahui}\\ & P(n)\equiv (a_1+\cdots +a_n{)}^{{\lambda }_a}(b_1+\cdots +b_n{)}^{{\lambda }_b}\cdots (z_1+\cdots +z_n{)}^{{\lambda }_z}\ge {\displaystyle \sum _{i=1}^n{a}_i^{{\lambda }_a}{b}_i^{{\lambda }_b}\cdots z_i^{{\lambda }_z}}\\ & \mathbf{\text{Langkah basis}}\\ & P(1)\text{benar},\text{karena}(a_1{)}^{{\lambda }_a}(b_1{)}^{{\lambda }_b}\cdots (z_1{)}^{{\lambda }_z}\ge a_1^{{\lambda }_a}{b}_1^{{\lambda }_b}\cdots z_1^{{\lambda }_z}\\ & \mathbf{\text{Langkah induksi}}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv (a_1+\cdots +a_k{)}^{{\lambda }_a}(b_1+\cdots +b_k{)}^{{\lambda }_b}\cdots (z_1+\cdots +z_k{)}^{{\lambda }_z}\\ & \ge a_1^{{\lambda }_a}{b}_1^{{\lambda }_b}\cdots z_1^{{\lambda }_z}+a_2^{{\lambda }_a}{b}_2^{{\lambda }_b}\cdots z_2^{{\lambda }_z}+\cdots +a_k^{{\lambda }_a}{b}_k^{{\lambda }_b}\cdots z_k^{{\lambda }_z}\\ & \text{maka untuk}n=k+1\\ & (a_1+\cdots +a_k+a_{k+1}{)}^{{\lambda }_a}(b_1+\cdots +b_k+b_{k+1}{)}^{{\lambda }_b}\cdots (z_1+\cdots +z_k+z_{k+1}{)}^{{\lambda }_z}\\ & \ge a_1^{{\lambda }_a}{b}_1^{{\lambda }_b}\cdots z_1^{{\lambda }_z}+a_2^{{\lambda }_a}{b}_2^{{\lambda }_b}\cdots z_2^{{\lambda }_z}+\cdots +a_k^{{\lambda }_a}{b}_k^{{\lambda }_b}\cdots z_k^{{\lambda }_z}+a_{k+1}^{{\lambda }_a}{b}_{k+1}^{{\lambda }_b}\cdots z_{k+1}^{{\lambda }_z}\\ & =\underset{k}{\underbrace{{\displaystyle \sum _{i=1}^k{a}_i^{{\lambda }_a}{b}_i^{{\lambda }_b}\cdots z_i^{{\lambda }_z}}}}+a_{k+1}^{{\lambda }_a}{b}_{k+1}^{{\lambda }_b}\cdots z_{k+1}^{{\lambda }_z}\\ & ={\displaystyle \sum _{i=1}^{k+1}{a}_i^{{\lambda }_a}{b}_i^{{\lambda }_b}\cdots z_i^{{\lambda }_z}\equiv P(k+1)}\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ & \mathbf{\text{Kesimpulan}}\\ & \text{Jadi},P(n)\text{benar untuk}\mathrm{\forall }n\in \mathbb{N}\end{array}$.

$\text{CONTOH SOAL}$.

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

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

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

$\begin{array}{l}\\ 4.& \text{Jika}a,b,c,x,y\text{dan}z\text{bilangan real positif}\\ & \text{dengan}a+b+c=x+y+z\text{tunjukkan bahwa}\\ & {\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}\ge a+b+c}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Dengan Ketaksamaan Holder pilih}\\ & {\lambda }_1={\lambda }_2={\lambda }_3={\displaystyle \frac{1}{3}\left({\displaystyle \sum _{i=1}^3{\lambda }_i=1}\right)\text{akan diperoleh}}\\ & {\left({\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}}\right)}^{\frac{1}{3}}(x+y+z{)}^{\frac{1}{3}}(x+y+z{)}^{\frac{1}{3}}\ge (a+b+c)\\ & \iff \left({\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}}\right)(x+y+z{)}^2\ge (a+b+c{)}^3\\ & \iff \left({\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}}\right)(a+b+c{)}^2\ge (a+b+c{)}^3\\ & \iff \left({\displaystyle \frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}}\right)\ge {\displaystyle \frac{(a+b+c{)}^3}{(a+b+c{)}^2}}\\ & \iff {\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}\ge a+b+c◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \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}$.

$\begin{array}{l}\\ 6.& (\mathbf{\text{IMO 2001}})\\ & \text{Diberikan}a,b\text{dan}c\text{bilangan real positif, tunjukkan}\\ & {\displaystyle \frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ca}}+\frac{c}{\sqrt{c^2+8ab}}\ge 1}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Dengan Ketaksamaan Holder diperoleh}\\ & {\left({\displaystyle \sum _{\text{siklik}}^{.}a(a^2+8bc)}\right)}^{\frac{1}{3}}{\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^{\frac{1}{3}}{\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^{\frac{1}{3}}\ge ({\displaystyle \sum _{\text{siklik}}^{.}a)}\\ & \text{dapat juga dituliskan lebih sederhana}\\ & {\left({\displaystyle \sum _{\text{siklik}}^{.}a(a^2+8bc)}\right)}^{\frac{1}{3}}{\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^{\frac{2}{3}}\ge \left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)\\ & \iff \left({\displaystyle \sum _{\text{siklik}}^{.}a(a^2+8bc)}\right){\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^2\ge {\left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)}^3\\ & \iff {\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^2\ge {\displaystyle \frac{{\left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)}^3}{\left({\displaystyle \sum _{\text{siklik}}^{.}a(a^2+8bc)}\right)}}\\ & \text{Perhatikan bahwa}\\ & \bullet {\left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)}^3=(a+b+c{)}^3\ge a^3+b^3+c^3+24abc\\ & \bullet \left({\displaystyle \sum _{\text{siklik}}^{.}a(a^2+8bc)}\right)=a^3+b^3+c^3+24abc\\ & \text{maka}\\ & {\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^2\ge {\displaystyle \frac{a^3+b^3+c^3+24abc}{a^3+b^3+c^3+24abc}=1}\\ & {\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}\ge 1◼}}\end{array}\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Sebagai catatan}}\\ & (a+b+c{)}^3=a^3+b^3+c^3+3a^{2}b+3a{b}^2\\ & +3b^{2}c+3b{c}^2+3a^{2}c+3a{c}^2+6abc\\ & (a+b+c{)}^3-(a^3+b^3+c^3)=3a^{2}b+3a{b}^2\\ & +3b^{2}c+3b{c}^2+3a^{2}c+3a{c}^2+6abc\\ & \text{Dengan AM-GM akan diperoleh bentuk}\\ & (a+b+c{)}^3-(a^3+b^3+c^3)\\ & \ge 3(3\sqrt[3]{(abc{)}^3}+3\sqrt[3]{(abc{)}^3})+6abc\\ & \ge 3(3abc+3abc)+6abc=24abc\\ & \text{Sehingga}\\ & (a+b+c{)}^3\ge a^3+b^3+c^3+24abc\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika}a,b,\text{dan}c\text{bilangan real positif}\\ & \text{dengan}a+b+c=1\text{tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}\ge {\displaystyle \frac{81}{4}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Dengan Ketaksamaan Holder pilih}\\ & {\lambda }_1={\lambda }_2={\lambda }_3={\lambda }_4={\displaystyle \frac{1}{4}\left({\displaystyle \sum _{i=1}^4{\lambda }_i=1}\right)\text{akan diperoleh}}\\ & (a+b+c)(b+c+a+c+a+b)(b+c+a+c+a+b)\left({\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}}\right)\\ & \ge (1^{{.}^{\frac{1}{4}}}+1^{{.}^{\frac{1}{4}}}+1^{{.}^{\frac{1}{4}}}{)}^4=3^4=81\\ & \iff (1)(2)(2)\left({\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}}\right)\ge 81\\ & \iff {\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}\ge {\displaystyle \frac{81}{4}◼}}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Chen, E. 2014. A Brief Introduction to Olympiad Inequalities.
2. Riasat, S. 2008. Basics of Olympiad Inequalities.
3. Todinov, M.T. 2020. Risk And Uncertaintly Reduction by Using Algebraic Inequalities. Boca Raton: CRC Press.
4. Tung. K.Y. 2013. Ayo Raih Medali Emas Olimpiade Matematika SMA. Yogyakarta: ANDI.
5. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.