Contoh Soal 14 (Segitiga dan Ketaksamaan)

 $\begin{array}{l}\\ 66.& \text{Diberikan}a,b,c>0,\text{tunjukkan bahwa}\\ & {\displaystyle \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\ge 6}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Alternatif 1}\\ & \text{Dengan mengaplikasikan AM-GM-HM}\\ & \text{pada}{\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\text{kita dapat menemukan}}\\ & {\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge {\displaystyle \frac{3}{(abc{)}^{{.}^{\frac{1}{3}}}}\ge {\displaystyle \frac{9}{a+b+c}}}}\\ & \text{Jika kedua ruas dikalikan dengan}a+b+c,\\ & \text{maka}\\ & 3+{\displaystyle \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\ge {\displaystyle \frac{9(a+b+c)}{a+b+c}}}\\ & \iff {\displaystyle \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\ge 6◼}\\ & \begin{array}{rl}& \text{Alternatif 2}\\ & \text{Asumsikan}a\le b\le c,\text{maka}a+b\le a+c\le b+c\\ & \text{dan}{\displaystyle \frac{1}{c}\le \frac{1}{b}\le \frac{1}{a}.}\\ & \text{Perhatikan bahwa}\\ & (a+b\le a+c\le b+c)\text{dan}{\displaystyle \frac{1}{c}\le \frac{1}{b}\le \frac{1}{a}}\\ & \text{memiliki kemonotonan yang sama}\\ & \text{maka dengan}\mathbf{\text{ketaksamaan Renata}}\\ & \text{dapat diperoleh bentuk}\\ & (b+c).{\displaystyle \frac{1}{a}+(c+a).{\displaystyle \frac{1}{b}+(a+b).{\displaystyle \frac{1}{c}\ge (b+c).{\displaystyle \frac{1}{b}+(c+a).{\displaystyle \frac{1}{c}+(a+b).{\displaystyle \frac{1}{a}}}}}}}\\ & \doteq 1+{\displaystyle \frac{c}{b}+1+\frac{a}{c}+1+\frac{b}{a}}\\ & \doteq 3+\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\\ & \doteq 3+3{\left({\displaystyle \frac{abc}{abc}}\right)}^{{.}^{\frac{1}{3}}}\\ & \doteq 3+3\\ & \doteq 6◼\end{array}\end{array}$.

$\begin{array}{l}\\ 67.& \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}$ 

$\begin{array}{l}\\ 68.& \text{Jika}a,b,c>0,\text{dengan}a\ne b\ne c\\ & \text{tunjukkan bahwa}\\ & (a^3+b^3+c^3)>{\displaystyle \frac{(a+b+c{)}^3}{9}>3abc}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{ketaksamaan Chebyshev}}\\ & \text{untuk}a\ge b\ge c,\text{dapat diperoleh bentuk}\\ & {\displaystyle \frac{(a^3+b^3+c^3)}{3}>\left({\displaystyle \frac{a+b+c}{3}}\right)\left({\displaystyle \frac{a+b+c}{3}}\right)\left({\displaystyle \frac{a+b+c}{3}}\right)}\\ & \iff {\displaystyle \frac{(a^3+b^3+c^3)}{3}>{\left({\displaystyle \frac{a+b+c}{3}}\right)}^3>{\left({\displaystyle \frac{3\sqrt[3]{abc}}{3}}\right)}^3}\\ & \iff {\displaystyle \frac{(a^3+b^3+c^3)}{3}>{\displaystyle \frac{(a+b+c{)}^3}{27}>abc}}\\ & \iff (a^3+b^3+c^3)>{\displaystyle \frac{(a+b+c{)}^3}{9}>3abc◼}\end{array}$.

$\begin{array}{l}\\ 69.& \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}\\ 70.& (\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}}^{.}}\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.