Contoh Soal 15 (Segitiga dan Ketaksamaan)

$\begin{array}{ll}\\ 71.&(\textbf{IMO 2001})\\ &\textrm{Diberikan}\: \: a,b\: \: \textrm{dan}\: \: c\: \: \textrm{bilangan real positif, tunjukkan}\\  &\displaystyle \frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\geq 1\\\\ &\textbf{Bukti}:\\   &\begin{aligned}&\textrm{Dengan Ketaksamaan Holder pilih}\\ &\color{red}\lambda _{1}=\lambda _{2}=\lambda _{3}=\displaystyle \frac{1}{3}\: \left ( \displaystyle \sum_{i=1}^{3}\lambda _{i}=1 \right )\: \: \color{black}\textrm{akan diperoleh}\\ &\left ( \displaystyle \sum_{\textrm{siklik}}^{.} a(a^{2}+8bc)\right )^{\frac{1}{3}}\left ( \displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{a}{\sqrt{a^{2}+8bc}} \right )^{\frac{1}{3}}\left ( \displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{a}{\sqrt{a^{2}+8bc}} \right )^{\frac{1}{3}}\geq (\displaystyle \sum_{\textrm{siklik}}^{.}a)\\ &\textrm{dapat juga dituliskan lebih sederhana}\\ &\left ( \displaystyle \sum_{\textrm{siklik}}^{.} a(a^{2}+8bc)\right )^{\frac{1}{3}}\left ( \displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{a}{\sqrt{a^{2}+8bc}} \right )^{\frac{2}{3}}\geq \left ( \displaystyle \sum_{\textrm{siklik}}^{.}a \right )\\ &\Leftrightarrow \left ( \displaystyle \sum_{\textrm{siklik}}^{.} a(a^{2}+8bc)\right )\left ( \displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{a}{\sqrt{a^{2}+8bc}} \right )^{2}\geq \left ( \displaystyle \sum_{\textrm{siklik}}^{.}a \right )^{3}\\ &\Leftrightarrow \left ( \displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{a}{\sqrt{a^{2}+8bc}} \right )^{2}\geq \displaystyle \frac{\left ( \displaystyle \sum_{\textrm{siklik}}^{.}a \right )^{3}}{\left ( \displaystyle \sum_{\textrm{siklik}}^{.} a(a^{2}+8bc)\right )}\\ &\color{red}\textrm{Perhatikan bahwa}\\&\bullet \: \left ( \displaystyle \sum_{\textrm{siklik}}^{.}a \right )^{3}=(a+b+c)^{3}\geq \color{blue}a^{3}+b^{3}+c^{3}+24abc\\ &\bullet \: \left ( \displaystyle \sum_{\textrm{siklik}}^{.} a(a^{2}+8bc)\right )=\color{blue}a^{3}+b^{3}+c^{3}+24abc\\ &\textrm{maka}\\ &\left ( \displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{a}{\sqrt{a^{2}+8bc}} \right )^{2}\geq \displaystyle \frac{a^{3}+b^{3}+c^{3}+24abc}{a^{3}+b^{3}+c^{3}+24abc}=1\\ &\displaystyle \sum_{\textrm{siklik}}^{.}\displaystyle \frac{a}{\sqrt{a^{2}+8bc}}\geq 1\qquad \blacksquare     \end{aligned} \end{array}$.

$.\qquad\begin{aligned}&\color{blue}\textbf{Sebagai catatan}\\&(a+b+c)^{3}=\color{red}a^{3}+b^{3}+c^{3}+3a^{2}b+3ab^{2}\\&\qquad \color{red}+3b^{2}c+3bc^{2}+3a^{2}c+3ac^{2}+6abc\\ &(a+b+c)^{3}-(a^{3}+b^{3}+c^{3})=\color{red}3a^{2}b+3ab^{2}\\ &\qquad \color{red}+3b^{2}c+3bc^{2}+3a^{2}c+3ac^{2}+6abc\\&\textrm{Dengan AM-GM akan diperoleh bentuk}\\&(a+b+c)^{3}-(a^{3}+b^{3}+c^{3})\\ &\geq 3\left ( 3\sqrt[3]{(abc)^{3}}+3\sqrt[3]{(abc)^{3}} \right )+6abc\\&\geq 3\left ( 3abc+3abc \right )+6abc=24abc\\ &\textrm{Sehingga}\\ &(a+b+c)^{3}\geq a^{3}+b^{3}+c^{3}+24abc     \end{aligned}$.

$\begin{array}{ll}\\ 72.&\textrm{Jika}\: \: a,b,\: \: \textrm{dan}\: \: c\: \: \textrm{bilangan real positif}\\ &\textrm{dengan}\: \: a+b+c=1\: \: \textrm{tunjukkan bahwa}\\ &\displaystyle \frac{1}{a(b+c)^{2}}+\frac{1}{b(a+c)^{2}}+\frac{1}{c(a+b)^{2}}\geq \displaystyle \frac{81}{4}\\\\ &\textbf{Bukti}:\\  &\begin{aligned}&\textrm{Dengan Ketaksamaan Holder pilih}\\ &\color{red}\lambda _{1}=\lambda _{2}=\lambda _{3}=\lambda _{4}=\displaystyle \frac{1}{4}\: \left ( \displaystyle \sum_{i=1}^{4}\lambda _{i}=1 \right )\: \: \color{black}\textrm{akan diperoleh}\\ &(a+b+c)(b+c+a+c+a+b)(b+c+a+c+a+b)\left (\displaystyle \frac{1}{a(b+c)^{2}}+\frac{1}{b(a+c)^{2}}+\frac{1}{c(a+b)^{2}}  \right )\\  &\geq (1^{.^{\frac{1}{4}}}+1^{.^{\frac{1}{4}}}+1^{.^{\frac{1}{4}}})^{4}=3^{4}=81\\ &\Leftrightarrow (1)(2)(2)\left (\displaystyle \frac{1}{a(b+c)^{2}}+\frac{1}{b(a+c)^{2}}+\frac{1}{c(a+b)^{2}}  \right )\geq 81\\ &\Leftrightarrow \displaystyle \frac{1}{a(b+c)^{2}}+\frac{1}{b(a+c)^{2}}+\frac{1}{c(a+b)^{2}}\geq \displaystyle \frac{81}{4}\qquad \blacksquare   \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 73.&\textrm{Buktikan bahwa setiap bilangan real}\\ &\textrm{positif}\: \: a,\: b\: \: \textrm{dan}\: \: c\: \: \textrm{berlaku}\\ & a^{2}+b^{2}+c^{2}\geq ab+ac+bc\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\color{blue}\textrm{Alternatif 1}\\ &\textrm{Perhatikan bahwa}\: \: (a-b)^{2}\geq 0\\  &(a-c)^{2}\geq 0,\: \: \textrm{dan}\: \: (b-c)^{2}\geq 0\\ &\textrm{adalah benar, maka}\\ &(a-b)^{2}=a^{2}-2ab+b^{2}\geq 0\\ &\Leftrightarrow a^{2}+b^{2}\geq 2ab\: .....(1)\\ &\textrm{Dengan cara yang kurang lebih sama}\\ &\textrm{akan didapatkan}\\ &\bullet \quad a^{2}+c^{2}\geq 2ac\: .....(2)\\ &\bullet \quad b^{2}+c^{2}\geq 2bc\: .....(1)\\ &\textrm{Jika ketaksamaan}\quad (1),(2), \& \: (3)\: \: \textrm{dijumlahkan}\\ &\textrm{akan didapatkan bentuk}\\ &2a^{2}+2b^{2}+2c^{2}\geq 2ab+2ac+2bc\\ &\Leftrightarrow \: a^{2}+b^{2}+c^{2}\geq ab+ac+bc\quad \blacksquare\\ &\color{blue}\textrm{Alternatif 2}\\ &\textrm{Dengan ketaksamaan}\: \: \color{red}\textbf{Cauchy-Schwarz}\\ &(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+a^{2})\geq (ab+bc+ca)^{2}\\ &\Leftrightarrow a^{2}+b^{2}+c^{2}\geq ab+ac+bc\qquad \blacksquare\\ &\color{blue}\textrm{Alternatif 3}\\  &\textrm{Untuk barisan}\: \: (a,b,c),\: \textrm{asumsikan}\: a\geq b\geq c\\ &\textrm{maka dengan}\: \: \color{red}\textbf{Ketaksamaan Renata}\: \: \color{black}\textrm{diperoleh}\\ &a.a+b.b+c.c\geq ab+bc+ca\\ &a^{2}+b^{2}+c^{2}\geq ab+ac+bc\qquad \blacksquare \\  &\color{blue}\textrm{Alternatif 4}\\ &\textrm{Dengan}\: \: \color{red}\textbf{Ketaksamaan Schur}\: \: \color{black}\textrm{saat}\: \: \color{red}r=0\color{black},\\ &\textrm{yaitu}\\ &a^{r}(a-b)(a-c)+b^{r}(b-a)(b-c)+c^{r}(c-a)(c-b)\geq 0\\ &\Leftrightarrow a^{0}(a-b)(a-c)+b^{0}(b-a)(b-c)+c^{0}(c-a)(c-b)\geq 0\\ &\Leftrightarrow (a-b)(a-c)+(b-a)(b-c)+(c-a)(c-b)\geq 0\\ &\Leftrightarrow a^{2}+b^{2}+c^{2}-ab-ac-bc\geq 0\\ &\Leftrightarrow a^{2}+b^{2}+c^{2}\geq ab+ac+bc\qquad \blacksquare    \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 74.&\textbf{(IMO 1964)}\\ &\textrm{Jika}\: \: a,b,c\: \: \textrm{adalah panjang sisi-sisi segitiga}\\ &\textrm{tunjukkan bahwa}\\ &\quad a^{2}(b+c-a)+b^{2}(a+c-b)+c^{2}(a+b-c)\leq 3abc\\\\ &\textbf{Bukti}:\\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan Schur}\: \: \textrm{saat}\: \: \color{red}r=1.\\ &a^{3}+b^{3}+c^{3}+3abc\geq ab(a+b)+bc(b+c)+ca(c+a)\\ &\Leftrightarrow 3abc\geq ab(a+b)-a^{3}+bc(b+c)-b^{3}+ca(c+a)-c^{3}\\ &\Leftrightarrow 3abc\geq a^{2}b+ab^{2}-c^{3}+b^{2}c+bc^{2}-b^{3}+c^{2}a+ca^{2}-c^{3}\\ &\Leftrightarrow 3abc\geq a^{2}b+ca^{2}-a^{3}+ab^{2}+b^{2}c-b^{3}+c^{2}a+bc^{2}-c^{3}\\ &\Leftrightarrow  3abc\geq a^{2}(b+c-a)+b^{2}(a+c-b)+c^{2}(a+b-c)\\ &\textrm{atau}\\ &\Leftrightarrow a^{2}(b+c-a)+b^{2}(a+c-b)+c^{2}(a+b-c)\leq 3abc\quad \blacksquare   \end{array}$ .

$\begin{array}{ll}\\ 75.&\textbf{(IMO 2000)}\\ &\textrm{Jika}\: \: a,b,c\: \: \textrm{bilangan real positif dengan}\\ &abc=1,\: \: \textrm{tunjukkan bahwa}\\ &\quad \left (a+1- \displaystyle \frac{1}{b} \right )\left (b+1- \displaystyle \frac{1}{c} \right )\left (c+1- \displaystyle \frac{1}{a} \right )\leq 1\\\\ &\textbf{Bukti}:\\ &\textrm{Misalkan}\: \: a=\displaystyle \frac{x}{y},\: b=\frac{y}{z},\: \: \textrm{dan}\: \: c=\displaystyle \frac{z}{x},\: \: \textrm{maka}\\ &\left ( \displaystyle \frac{x}{y}+1-\frac{z}{y} \right )\left ( \displaystyle \frac{y}{z}+1-\frac{x}{z} \right )\left ( \displaystyle \frac{z}{x}+1-\frac{y}{x} \right ) \leq 1\\ &\Leftrightarrow \left ( \displaystyle \frac{x+y-z}{y} \right )\left ( \displaystyle \frac{y+z-x}{z} \right )\left ( \displaystyle \frac{z+x-y}{x} \right )\leq 1\\ &\Leftrightarrow (x+y-z)(y+z-x)(z+x-y)\leq xyz,\: \: \textrm{atau}\\ &\Leftrightarrow \color{red}xyz\geq (x+y-z)(y+z-x)(z+x-y)\\ &\textrm{Bentuk terakhir memenuhi bentuk kedua dari}\\ &\textbf{Ketaksamaan Schur}\: \: \textrm{saat}\: \: \color{red}r=1.\\ &\textrm{Jadi},\\ &\: \: \left (a+1- \displaystyle \frac{1}{b} \right )\left (b+1- \displaystyle \frac{1}{c} \right )\left (c+1- \displaystyle \frac{1}{a} \right )\leq 1\quad \blacksquare   \end{array}$.