Lanjutan 4 Materi Ketaksamaan : Ketaksamaan Chebyshev

2. Ketaksamaan Chebyshev

Jika $a_1\le a_2\le a_3\le \cdots \le a_n$ dan $b_1\le b_2\le b_3\le \cdots \le b_n$ adalah merupakan kumpulan bilangan yang monoton naik atau $a_1\ge a_2\ge a_3\ge \cdots \ge a_n$ dan $b_1\ge b_2\ge b_3\ge \cdots \ge b_n$ adalah merupakan kumpulan bilangan yang monoton turun, maka

$\begin{array}{rl}& \left({\displaystyle \frac{a_1{b}_1+a_2{b}_2+a_3{b}_3+\cdots +a_n{b}_n}{n}}\right)\\ & \ge \left({\displaystyle \frac{a_1+a_2+a_3+...+a_n}{n}}\right)\\ & \times \left({\displaystyle \frac{b_1+b_2+b_3+...+b_n}{n}}\right)\end{array}$.

$\begin{array}{rl}& \mathbf{\text{atau}}\\ \\ & n(a_1{b}_1+a_2{b}_2+a_3{b}_3+\cdots +a_n{b}_n)\\ & \ge (a_1+a_2+a_3+...+a_n)\times (b_1+b_2+b_3+...+b_n)\end{array}$.

Tetapi jika kumpulan bilangan di atas memiliki kemonotonan yang berbeda, yaitu $a_1\le a_2\le a_3\le \cdots \le a_n$  dan  $b_1\ge b_2\ge b_3\ge \cdots \ge b_n$ atau  $a_1\ge a_2\ge a_3\ge \cdots \ge a_n$  dan $b_1\le b_2\le b_3\le \cdots \le b_n$, maka ketaksamaan akan menjadi

$\begin{array}{rl}& \left({\displaystyle \frac{a_1{b}_1+a_2{b}_2+a_3{b}_3+\cdots +a_n{b}_n}{n}}\right)\\ & \le \left({\displaystyle \frac{a_1+a_2+a_3+...+a_n}{n}}\right)\\ & \times \left({\displaystyle \frac{b_1+b_2+b_3+...+b_n}{n}}\right)\end{array}$.

$\begin{array}{rl}& \mathbf{\text{atau}}\\ \\ & n(a_1{b}_1+a_2{b}_2+a_3{b}_3+\cdots +a_n{b}_n)\\ & \le (a_1+a_2+a_3+...+a_n)\times (b_1+b_2+b_3+...+b_n)\end{array}$.

Bukti

Pada kumpulan bilangan yang memiliki kemonotonan yang sama yaitu:

$a_1\le a_2\le a_3\le \cdots \le a_n$ dan $b_1\le b_2\le b_3\le \cdots \le b_n$, dengan mengaplikasikan ketaksamaan Renata akan diperoleh

$\begin{array}{rl}{a}_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n& =a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n\\ a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n& \ge a_1{b}_2+a_2{b}_3+\cdots +a_n{b}_1\\ a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n& \ge a_1{b}_3+a_2{b}_4+\cdots +a_n{b}_2\\ & ⋮\\ a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n& \ge a_1{b}_n+a_2{b}_1+\cdots +a_n{b}_{n-1}\\ \text{Jika ketaksamaan di at}& \text{as ditmabahkan, maka}\\ n(a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n)& \ge (a_1+a_2+\cdots +a_n)(b_1+b_2+\cdots +b_n)\\ \text{Bentuk terakhir adalah}& \text{bukti dari ketaksamaan ini}\end{array}$.

$\begin{array}{rl}\text{Sebagai}& \text{misal, andaikan}\\ \bullet n=2& \iff 2(a_1{b}_1+a_2{b}_2)\ge (a_1+a_2)(b_1+b_2)\\ \bullet n=3& \iff 3(a_1{b}_1+a_2{b}_2+a_3{b}_3)\ge (a_1+a_2+a_3)(b_1+b_2+b_3)\\ & ⋮\\ \bullet n=k& \iff k(a_1{b}_1+a_2{b}_2+\cdots +a_k{b}_k)\ge (a_1+a_2+\cdots +a_k)(b_1+b_2+\cdots +b_k)\end{array}$ .

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika}a,b,x,y\text{bilangan real positif}\\ & \text{sehingga}a\ge b\text{dan}x\ge y.\\ & \text{Tunjukkan bahwa}\\ & (ax+by)\ge {\displaystyle \frac{1}{2}(a+b)(x+y)}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{dapat diperoleh bentuk}\\ & 2(ax+by)\ge (a+b)(x+y)\\ & \iff (ax+by)\ge {\displaystyle \frac{1}{2}(a+b)(x+y)◼}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika}a,b>0,\text{tunjukkan bahwa}\\ & 2(a^2+b^2)\ge (a+b{)}^2\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Alternatif 1}\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{untuk}a\ge b,\text{dapat diperoleh bentuk}\\ & 2(a.a+b.b)\ge (a+b)(a+b)\\ & \iff 2(a^2+b^2)\ge (a+b{)}^2◼\\ & \text{Alternatif 2}\\ & \text{Dengan}\mathbf{\text{ketaksamaan CS-Engel}}\\ & (1+1)(a^2+b^2)\ge (1.a+1.b{)}^2\\ & \iff 2(a^2+b^2)\ge (a+b{)}^2◼\\ & \text{Alternatif 3}\\ & \text{Perhatikan bahwa}(a-b{)}^2\ge 0\\ & \iff a^2+b^2-2ab\ge 0\\ & \iff a^2+b^2\ge 2ab\\ & \iff 2(a^2+b^2)\ge a^2+b^2+2ab\\ & \iff 2(a^2+b^2)\ge (a+b{)}^2◼\end{array}$.

$\begin{array}{l}\\ 3.& \text{Jika}a,b,c>0,\text{tunjukkan bahwa}\\ & 3(a^2+b^2+c^2)\ge (a+b+c{)}^2\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{untuk}a\ge b\ge c,\text{dapat diperoleh bentuk}\\ & 3(a.a+b.b+c.c)\ge (a+b+c)(a+b+c)\\ & \iff 3(a^2+b^2+c^2)\ge (a+b+c{)}^2◼\end{array}$.

$\begin{array}{l}\\ 4.& \text{Jika}a,b,c>0,\text{dengan}{a}^2+b^2+c^2=1\\ & \text{tunjukkan bahwa}a+b+c\le \sqrt{3}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{Kurang lebih caranya sama dengan no.3 di atas}\\ & \text{untuk}a\ge b\ge c,\text{dapat diperoleh bentuk}\\ & 3(a.a+b.b+c.c)\ge (a+b+c)(a+b+c)\\ & \iff 3(a^2+b^2+c^2)\ge (a+b+c{)}^2\\ & \iff 3(1)\ge (a+b+c{)}^2\\ & \iff \sqrt{3}\ge (a+b+c)\\ & \iff (a+b+c)\le \sqrt{3}◼\end{array}$.

$\begin{array}{l}\\ 5.& \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}\\ & \begin{array}{rl}& \text{Asumsikan},\\ & \{\begin{array}{ll}& a\ge b\ge c\\ & {\displaystyle \frac{1}{b+c}\ge \frac{1}{a+c}\ge \frac{1}{a+b}}\end{array}\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Chebyshev}}\\ & {\displaystyle \frac{{\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}}{3}\ge {\displaystyle \frac{(a+b+c)}{3}\left({\displaystyle \frac{\left({\displaystyle \frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}}\right)}{3}}\right)}}\\ & \text{Dengan AM-HM dan}K={\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\\ & \iff {\displaystyle \frac{K}{3}\ge {\displaystyle \frac{(a+b+c)}{3}\left({\displaystyle \frac{3}{(b+c)+(a+c)+(a+b)}}\right)}}\\ & \iff K\ge {\displaystyle \frac{3(a+b+c)}{2(a+b+c)}}\\ & \iff K\ge {\displaystyle \frac{3}{2}}\\ & \iff {\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge {\displaystyle \frac{3}{2}◼}}\end{array}\\ & \text{Alternatif 2}\\ & \begin{array}{rl}& \mathbf{\text{Pertama}},\text{asumsikan}\\ & \{\begin{array}{ll}& a\ge b\ge c\\ & {\displaystyle \frac{1}{b+c}\ge \frac{1}{a+c}\ge \frac{1}{a+b}}\end{array}\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Chebyshev}}\\ & 3\left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge (a+b+c)\left({\displaystyle \frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}}\right)\\ & \mathbf{\text{Kedua}},\text{asumsikan}\\ & \{\begin{array}{ll}& a+b\ge a+c\ge b+c\\ & {\displaystyle \frac{1}{a+b}\le \frac{1}{a+c}\le \frac{1}{b+c}}\end{array}\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Chebyshev}}\\ & 3\left({\displaystyle \frac{a+b}{a+b}+\frac{a+c}{a+c}+\frac{b+c}{b+c}}\right)\le (a+b+a+c+b+c)\left({\displaystyle \frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}}\right)\\ & \iff 3(1+1+1)\le 2(a+b+c)\left({\displaystyle \frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}}\right)\\ & \iff {\displaystyle \frac{9}{2}\le (a+b+c)\left({\displaystyle \frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}}\right)}\\ & \text{Dari dua ketaksamaan di atas didapatkan}\\ & 3\left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge {\displaystyle \frac{9}{2}}\\ & \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}$ 

$\begin{array}{l}\\ 6.& \text{Diberikan}a,b,c,d>0,\text{tunjukkan bahwa}\\ & (a+b+c+d)\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}\right)\ge 16\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Asumsikan}\\ & \{\begin{array}{ll}& a\ge b\ge c\ge d\\ & {\displaystyle \frac{1}{d}\ge \frac{1}{c}\ge \frac{1}{b}\ge \frac{1}{a}}\end{array}\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Chebyshev}}\\ & {\displaystyle \frac{{\displaystyle \frac{a}{a}+\frac{b}{b}+\frac{c}{c}+\frac{d}{d}}}{4}\le {\displaystyle \left({\displaystyle \frac{(a+b+c+d)}{4}}\right)\left({\displaystyle \frac{{\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}}{4}}\right)}}\\ & \iff 1\le {\displaystyle \left({\displaystyle \frac{(a+b+c+d)}{4}}\right)\left({\displaystyle \frac{{\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}}{4}}\right)}\\ & \iff 16\le (a+b+c+d)\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}\right)\\ & \iff (a+b+c+d)\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}\right)\ge 16◼\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika}a,b,c>0,\text{dengan}a\ne b\ne c\\ & \text{tunjukkan bahwa}\\ & (a^3+b^3+c^3)>{\displaystyle \frac{(a+b+c{)}^3}{9}>3abc}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{untuk}a\ge b\ge c,\text{dapat diperoleh bentuk}\\ & {\displaystyle \frac{(a^3+b^3+c^3)}{3}>\left({\displaystyle \frac{a+b+c}{3}}\right)\left({\displaystyle \frac{a+b+c}{3}}\right)\left({\displaystyle \frac{a+b+c}{3}}\right)}\\ & \iff {\displaystyle \frac{(a^3+b^3+c^3)}{3}>{\left({\displaystyle \frac{a+b+c}{3}}\right)}^3>{\left({\displaystyle \frac{3\sqrt[3]{abc}}{3}}\right)}^3}\\ & \iff {\displaystyle \frac{(a^3+b^3+c^3)}{3}>{\displaystyle \frac{(a+b+c{)}^3}{27}>abc}}\\ & \iff (a^3+b^3+c^3)>{\displaystyle \frac{(a+b+c{)}^3}{9}>3abc◼}\end{array}$.

$\begin{array}{l}\\ 8.& \text{tunjukkan bahwa untuk}n\text{bilangan asli}\\ & \text{berlaku}\\ & 1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n}\le n\sqrt{{\displaystyle \frac{n+1}{2}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{dapat diperoleh bentuk berikut}\\ & {\displaystyle \frac{1+2+3+\cdots +n}{n}\ge {\displaystyle \frac{(1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n})}{n}\times {\displaystyle \frac{(1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n})}{n}}}}\\ & \iff {\displaystyle \frac{1+2+3+\cdots +n}{n}\ge {\displaystyle \frac{{(1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n})}^2}{n^2}}}\\ & \iff {\displaystyle \frac{{(1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n})}^2}{n^2}\le {\displaystyle \frac{1+2+3+\cdots +n}{n}}}\\ & \iff {(1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n})}^2\le n(1+2+3+\cdots +n)\\ & \iff {(1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n})}^2\le n\left({\displaystyle \frac{n(n+1)}{2}}\right)\\ & \iff (1+\sqrt{2}+\sqrt{3}+\cdots +\sqrt{n})\le n\sqrt{{\displaystyle \frac{n+1}{2}}}◼\end{array}$.

$\begin{array}{l}\\ 9.& \text{tunjukkan bahwa untuk}n\text{bilangan asli}\\ & \text{berlaku}\\ & {\displaystyle \frac{1}{\sqrt{n}}(1+\sqrt{{\displaystyle \frac{1}{2}}}+\sqrt{{\displaystyle \frac{1}{3}}}+\cdots +\sqrt{{\displaystyle \frac{1}{n}}})\le (2n-1{)}^{{.}^{\frac{1}{4}}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{untuk}:(1\ge {\displaystyle \frac{1}{2}\ge \frac{1}{3}\ge \cdots \ge \frac{1}{n}})\\ & \text{dapat diperoleh bentuk berikut}\\ & {(1+{\displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}})}^2\le n(1+{\displaystyle \frac{1}{2.2}+\frac{1}{3.3}+\cdots +\frac{1}{n.n}})\\ & \iff {(1+{\displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}})}^2\le n(1+{\displaystyle \frac{1}{1.2}+\frac{1}{2.3}+\cdots +\frac{1}{(n-1).n}})\\ & \iff {(1+{\displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}})}^2\le n(1+(1-{\displaystyle \frac{1}{2}})+\left({\displaystyle \frac{1}{2}-\frac{1}{3}}\right)+\cdots +\left({\displaystyle \frac{1}{(n-1)}-\frac{1}{n}}\right))\\ & \iff {(1+{\displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}})}^2\le n(1+1-{\displaystyle \frac{1}{n}})=n(2-{\displaystyle \frac{1}{n}})\\ & \iff (1+{\displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}})\le \sqrt{2n-1}..........(1)\\ & \text{Gunakan lagi}\mathbf{\text{ketaksamaan Chebyshev}}\\ & {(1+\sqrt{{\displaystyle \frac{1}{2}}}+\sqrt{{\displaystyle \frac{1}{3}}}+\cdots +\sqrt{{\displaystyle \frac{1}{n}}})}^2\le n(1+{\displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}})\\ & \iff {\displaystyle \frac{1}{n}{(1+\sqrt{{\displaystyle \frac{1}{2}}}+\sqrt{{\displaystyle \frac{1}{3}}}+\cdots +\sqrt{{\displaystyle \frac{1}{n}}})}^2\le (1+{\displaystyle \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}})...(2)}\\ & \text{Dari ketaksamaan (1) dan (2), dapat diperoleh}\\ & {\displaystyle \frac{1}{n}{(1+\sqrt{{\displaystyle \frac{1}{2}}}+\sqrt{{\displaystyle \frac{1}{3}}}+\cdots +\sqrt{{\displaystyle \frac{1}{n}}})}^2\le \sqrt{2n-1}=(2n-1{)}^{{.}^{\frac{1}{2}}}}\\ & \iff {\displaystyle \frac{1}{\sqrt{n}}(1+\sqrt{{\displaystyle \frac{1}{2}}}+\sqrt{{\displaystyle \frac{1}{3}}}+\cdots +\sqrt{{\displaystyle \frac{1}{n}}})\le (2n-1{)}^{{.}^{\frac{1}{4}}}◼}\end{array}$.

$\begin{array}{l}\\ 10.& (\mathbf{\text{OSN 2011}})\\ & \text{Jika}a,b,c>0,\text{dengan}abc=1\\ & \text{Jika diketahui}\\ & a^{2011}+b^{2011}+c^{2011}<{\displaystyle \frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}}\\ & \text{tunjukkan bahwa}\\ & (a+b+c)>{\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Asumsikan}\\ & \{\begin{array}{ll}& a\ge b\ge c\\ & {\displaystyle \frac{1}{c}\ge \frac{1}{b}\ge \frac{1}{a}}\end{array}\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{Perhatikan}\\ & \begin{array}{rl}& {\displaystyle \frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}\ge {\displaystyle \frac{1}{3}\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)\left({\displaystyle \frac{1}{a^{2010}}+\frac{1}{b^{2010}}+\frac{1}{c^{2010}}}\right)}}\\ & \iff {\displaystyle \frac{1}{a^{2010}}+\frac{1}{b^{2010}}+\frac{1}{c^{2010}}\ge {\displaystyle \frac{1}{3}\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)\left({\displaystyle \frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}}\right)}}\\ & \iff {\displaystyle \frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}\ge {\displaystyle \frac{1}{3}\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)\left({\displaystyle \frac{1}{a^{2008}}+\frac{1}{b^{2008}}+\frac{1}{c^{2008}}}\right)}}\\ & ⋮\\ & \iff {\displaystyle \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\ge {\displaystyle \frac{1}{3}\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)\left({\displaystyle \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\right)}}\\ & \iff {\displaystyle \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge {\displaystyle \frac{1}{3}\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)}}\\ & \text{Sehingga}\\ & {\displaystyle \frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}\ge {\displaystyle \frac{1}{3^{2010}}{\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)}^{2011}.....(1)}}\end{array}\\ & \text{dan}\\ & \begin{array}{rl}& a^{2011}+b^{2011}+c^{2011}\ge {\displaystyle \frac{1}{3}(a+b+c)(a^{2010}+b^{2010}+c^{2010})}\\ & \iff a^{2010}+b^{2010}+c^{2010}\ge {\displaystyle \frac{1}{3}(a+b+c)(a^{2009}+b^{2009}+c^{2009})}\\ & \iff a^{2009}+b^{2009}+c^{2009}\ge {\displaystyle \frac{1}{3}(a+b+c)(a^{2008}+b^{2008}+c^{2008})}\\ & ⋮\\ & \iff a^3+b^3+c^3\ge {\displaystyle \frac{1}{3}(a+b+c)(a^2+b^2+c^2)}\\ & \iff a^2+b^2+c^2\ge {\displaystyle \frac{1}{3}(a+b+c)(a+b+c)}\\ & \text{Sehingga}\\ & \iff a^{2011}+b^{2011}+c^{2011}\ge {\displaystyle \frac{1}{3^{2010}}(a+b+c{)}^{2011}........(2)}\end{array}\\ & \begin{array}{rl}& \text{Dari ketaksamaan (1) dan (2) didapatkan}\\ & \frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}\ge {\displaystyle \frac{1}{3^{2010}}{\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)}^{2011}}\\ & >a^{2011}+b^{2011}+c^{2011}\ge {\displaystyle \frac{1}{3^{2010}}(a+b+c{)}^{2011}}\\ & \text{atau}\\ & {\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{1}{a^{2011}}\ge {\displaystyle \frac{1}{3^{2010}}{\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{1}{a}}}\right)}^{2011}>{\displaystyle \sum _{\text{siklik}}^{.}{a}^{2011}\ge {\displaystyle \frac{1}{3^{2010}}{\left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)}^{2011}}}}}}\\ & \iff {\displaystyle \frac{1}{3^{2010}}{\left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)}^{2011}>{\displaystyle \frac{1}{3^{2010}}(a+b+c{)}^{2011}}}\\ & \iff \left({\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)>(a+b+c)\\ & \iff a+b+c<{\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}◼}\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.