Contoh Soal 8 (Segitiga dan Ketaksamaan)

 $\begin{array}{ll}\\ 36.&\textrm{Jika}\: \: a,b,c> 0\: ,\: \textrm{dengan}\: \: a+b+c=1\\ &\textrm{tunjukkan bahwa}\\ &\qquad  (1-a)(1-b)(1-c)\geq 8abc\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\textrm{Perhatikan bahwa}\\ &\begin{cases} 1-a &=b+c \\  1-b & =a+c \\  1-c & =a+b  \end{cases}\\ &\textrm{Kurang lebih seperti pembuktian}\\ &\textrm{pada no.35 di atas dengan tetap}\\ &\textrm{menggunakan AM-GM akan}\\ &\textrm{didapatkan}\\ &\bullet \: \: a+b\geq 2\sqrt{ab}\\ &\bullet \: \: b+c\geq 2\sqrt{bc},\: \: \textrm{dan}\\ &\bullet \: \: c+a\geq 2\sqrt{ca}\\ &\textrm{Selanjutnya}\\ &(1-a)(1-b)(1-c)= (b+c)(a+c)(a+b)\\ &\: \: \: \qquad\qquad\qquad\quad\quad \geq 2\sqrt{bc}\times 2\sqrt{ac}\times 2\sqrt{ab}\\ & \: \: \: \qquad\qquad\qquad\quad\quad \geq 8\sqrt{a^{2}b^{2}c^{2}}\\ &\: \: \:  \qquad\qquad\qquad\quad\quad \geq 8abc\qquad \blacksquare  \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 37.&\textrm{Misalkan}\: \: a,b,c\: \: \textrm{bilangan real positif}\\ &\textrm{dengan nilai}\: \: abc=1.\: \: \textrm{Nilai terkecil}\\  &\textrm{dari}\: \: (a+2b)(b+2c)(ac+1)\: \: \textrm{tercapai}\\ &\textrm{ketika}\: \: a+b+c\: \: \textrm{bernilai}\: ...\: .\\\\  &\textbf{Jawab}:\\   &\begin{aligned}&\textrm{Dengan AM-GM kita dapatkan}\\ &\bullet \: a+2b\geq 2\sqrt{2ab}\\ &\bullet \: b+2c\geq 2\sqrt{2bc}\\ &\bullet \: ac+1\geq 2\sqrt{ac}\\ &\textrm{Selanjutnya}\\ &(a+2b)(b+2c)(ac+1)\geq 2\sqrt{2ab}\times 2\sqrt{2bc}\times 2\sqrt{ac}\\ &\: \qquad\qquad\qquad\qquad\qquad \geq 8\sqrt{4a^{2}b^{2}c^{2}}\\ &\: \qquad\qquad\qquad\qquad\qquad \geq 8\times 2abc\\ &\: \qquad\qquad\qquad\qquad\qquad \geq 16\times 1\\ &\: \qquad\qquad\qquad\qquad\qquad \geq 16\\ &\textrm{Karena}\: \: abc=1,\: \textrm{dengan cara coba-coba}\\ &\textrm{dapat kita peroleh nilai}\\ &a=2,\: b=1,\: \: \textrm{dan}\: \: c=\displaystyle \frac{1}{2}\\ &\textrm{Kita cek ke}\: \: (a+2b)(b+2c)(ac+1)\geq 16\\ &(2+2)(1+1)(1+1)\geq 16\: \: \textrm{adalah}\: \textbf{benar}\\ &\textrm{Sehingga nilai}\: \: a+b+c=2+1+\displaystyle \frac{1}{2}=3\displaystyle \frac{1}{2}=\displaystyle \frac{7}{2}  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 38.&\textrm{Diketahui}\: \: a,b,c\: \: \textrm{bilangan real positif}\\ &\textrm{dan}\: \: (a+1)(b+1)(c+1)=8.\\ &\textrm{tunjukkan bahwa}\quad  abc< 1\\\\  &\textbf{Bukti}\\   &\begin{aligned}&\textrm{Diketahui bahwa}\\ &(a+1)(b+1)(c+1)=8\\ &abc+(ab+bc+ca)+(a+b+c)+1=8\\ &\color{blue}\textrm{Dengan AM-GM kita akan mendapatkan}\\ &abc+3(abc)^{.^{\frac{2}{3}}}+3(abc)^{.^{\frac{1}{3}}}+1< 8\\ &\Leftrightarrow \: \left ( (abc)^{.^{\frac{1}{3}}}+1 \right )^{3}< 2^{3}\\ &\Leftrightarrow \: (abc)^{.^{\frac{1}{3}}}+1< 2\\ &\Leftrightarrow \: (abc)^{.^{\frac{1}{3}}}< 1\qquad \blacksquare   \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 39.&\textrm{Jika}\: \: a,b,c> 0\\ &\textrm{Tunjukkan bahwa}\\ &\qquad  \displaystyle \frac{a}{a^{2}+1}+\displaystyle \frac{b}{b^{2}+1}+\displaystyle \frac{c}{c^{2}+1}\leq  \displaystyle \frac{3}{2}\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\textrm{Perhatikan bahwa}\\ &(a-1)^{2}\geq 0\\ &\Leftrightarrow a^{2}-2a+1\geq 0\\ &\Leftrightarrow  a^{2}+1\geq 2a\\ &\displaystyle \frac{1}{2}\geq \displaystyle \frac{a}{a^{2}+1}\\ &\textrm{atau}\\ &\displaystyle \frac{a}{a^{2}+1}\leq \displaystyle \frac{1}{2}\: \: ..................(1)\\ &\textrm{dengan cara yang sama akan} \\ &\textrm{pula jiika}\: \: b\: \: \textrm{dan}\: \: c\: \: \textrm{dikondisikan}\\ &\textrm{akan didapatkan ketaksamaan}\\ &\displaystyle \frac{b}{b^{2}+1}\leq \displaystyle \frac{1}{2}\: \: ..................(2)\\ &\displaystyle \frac{c}{c^{2}+1}\leq \displaystyle \frac{1}{2}\: \: ..................(3)\\ &\textrm{Jika ketaksamaan}\: \: (1),\: (2),\: \&\: \: (3)\\ &\textrm{dijumlahkan akan menghasilkan}\\ &\displaystyle \frac{a}{a^{2}+1}+\displaystyle \frac{b}{b^{2}+1}+\displaystyle \frac{c}{c^{2}+1}\leq \displaystyle \frac{1}{2}+\displaystyle \frac{1}{2}+\displaystyle \frac{1}{2}\\ &\qquad\qquad\qquad\qquad\qquad\quad \leq 3\left ( \displaystyle \frac{1}{2} \right )\\ &\qquad\qquad\qquad\qquad\qquad\quad \leq \displaystyle \frac{3}{2}\: \quad \blacksquare \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 40.&\textrm{Jika}\: \: a,b,c\: \: \textrm{adalah sisi-sisi segitiga}\\ &\textrm{ABC, tunjukkan bahwa}\\ & \qquad  \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< 2\\\\  &\textbf{Bukti}\\  &\begin{aligned}&\textrm{Perhatikan bahwa pada segitiga}\\ &\textrm{ABC berlaku}\\ &\begin{cases} \color{red}a+b & >\color{red}c \\  a+c & >b \\  b+c & >a  \end{cases}\\ &\textrm{Misalkan}\: \: 2s=a+b+c,\: \: \textrm{maka}\\ &a+b=a+b\\ &\Leftrightarrow \: a+b+\color{red}a+b\color{black}>a+b+\color{red}c\\ &\Leftrightarrow \: 2(a+b)> 2s\\ &\Leftrightarrow \: (a+b)>s\\ &\textrm{Demikian juga akan berlaku}\\ &\begin{cases} b+c &>s \\  c+a &>s  \end{cases}\\ &\textrm{Sehingga}\\ &\begin{cases} \displaystyle \frac{a}{b+c} & <\displaystyle \frac{a}{s} \\  \displaystyle \frac{b}{c+a} & <\displaystyle \frac{b}{s} \\  \displaystyle \frac{c}{a+b} & <\displaystyle \frac{c}{s}  \end{cases}\\ &\textrm{Selanjutnya kita kembali ke soal}\\ &\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< \displaystyle \frac{a+b+c}{s}\\ &\: \: \: \qquad\qquad\qquad\qquad\quad <\displaystyle \frac{2s}{s}\\ &\: \: \: \qquad\qquad\qquad\qquad\quad <2\qquad \blacksquare  \end{aligned}  \end{array}$.