Contoh Soal 16 (Segitiga dan Ketaksamaan)

 $\begin{array}{l}\\ 76.& \mathbf{\text{(IMO 1983)}}\\ & \text{Jika}a,b,c\text{adalah panjang sisi-sisi segitiga}\\ & \text{tunjukkan bahwa}\\ & a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\ge 0\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Pada sebuah segitiga dengan sisi}a,b,c\\ & \text{berlaku}\{\begin{array}{ll}a+b>c& \Rightarrow a>c-b\\ a+c>b& \Rightarrow c>b-c\\ b+c>a& \Rightarrow b>a-c\end{array}\\ & \text{Sehingga untuk ketaksamaan pada soal}\\ & a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(x-a)\\ & \ge a^2(a-c)(a-b)+b^2(b-c)(b-c)+c^2(c-b)(c-a)\ge 0\\ & \text{Bentuk terakhir memenuhi bentuk dari}\\ & \mathbf{\text{Ketaksamaan Schur}}\text{saat}r=2.\\ & \text{Jadi},\\ & a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\ge 0◼\end{array}$.

$\begin{array}{l}\\ 77.& \text{Misalkan}a,b,c\text{bilangan real positif dengan}\\ & a+b+c=2,\text{tunjukkan bahwa}\\ & a^4+b^4+c^4+abc\ge a^3+b^3+c^3\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Schur}}\text{saat}r=2\\ & \text{kita memiliki}\\ & a^2(a-b)(a-c)+b^2(b-a)(b-c)+c^2(c-a)(c-b)\ge 0\\ & \iff a^4+b^4+c^4+abc(a+b+c)-a^3(b+c)-b^3(a+c)-c^3(a+b)\ge 0\\ & \iff a^4+b^4+c^4+abc(a+b+c)\ge a^3(b+c)+b^3(a+c)+c^3(a+b)\\ & \iff a^4+b^4+c^4+abc(a+b+c)\ge (a^3+b^3+c^3)(a+b+c)-(a^4+b^4+c^4)\\ & \iff 2(a^4+b^4+c^4)+abc(a+b+c)\ge (a^3+b^3+c^3)(a+b+c)\\ & \iff 2(a^4+b^4+c^4)+abc(2)\ge (a^3+b^3+c^3)(2)\\ & \iff a^4+b^4+c^4+abc\ge a^3+b^3+c^3◼\\ \\ & \text{Bentuk di atas kadang dituliskan dengan bentuk}\\ & \text{berikut}:\\ & \begin{array}{rl}& \text{Dengan}\mathbf{\text{Ketaksamaan Schur}}\text{saat}r=2\\ & \text{kita memiliki}\\ & {\displaystyle \sum _{\text{siklik}}^{.}{a}^2(a-b)(a-c)\ge 0}\\ & \iff {\displaystyle \sum _{\text{siklik}}^{.}{a}^4+abc{\displaystyle \sum _{\text{siklik}}^{.}a-{\displaystyle \sum _{\text{siklik}}^{.}{a}^3(b+c)\ge 0}}}\\ & \iff {\displaystyle \sum _{\text{siklik}}^{.}{a}^4+abc{\displaystyle \sum _{\text{siklik}}^{.}a\ge {\displaystyle \sum _{\text{siklik}}^{.}{a}^3(b+c)}}}\\ & \iff {\displaystyle \sum _{\text{siklik}}^{.}{a}^4+abc{\displaystyle \sum _{\text{siklik}}^{.}a\ge \left({\displaystyle \sum _{\text{siklik}}^{.}{a}^3}\right)\left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)-\left({\displaystyle \sum _{\text{siklik}}^{.}{a}^4}\right)}}\\ & \iff 2{\displaystyle \sum _{\text{siklik}}^{.}{a}^4+abc{\displaystyle \sum _{\text{siklik}}^{.}a\ge \left({\displaystyle \sum _{\text{siklik}}^{.}{a}^3}\right)\left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)}}\\ & \iff {\displaystyle \sum _{\text{siklik}}^{.}{a}^4+abc\ge \left({\displaystyle \sum _{\text{siklik}}^{.}{a}^3}\right)◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 78.& \text{Misalkan}x,y,x\text{bilangan real positif dengan}\\ & \text{tunjukkan bahwa}\\ & a^2+b^2+c^2+2abc+1\ge 2(ab+acb+c)\\ & (\mathbf{\text{Darij Grinberg}})\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Dengan}\mathbf{\text{ketaksamaan AM-GM}}\text{dan}\\ & \text{dilanjutkan dengan}\mathbf{\text{ketaksamaan Schur}}\\ & \text{serta menggesernya ke ruas kiri, maka}\\ & a^2+b^2+c^2+2abc+1-2(ab+ac+bc)\\ & \ge a^2+b^2+c^2+3(abc{)}^{{}^{\frac{2}{3}}}+1\ge 2(ab+ac+bc)\\ & \ge {((a{)}^{{}^{\frac{2}{3}}})}^3+{((b{)}^{{}^{\frac{2}{3}}})}^3+{((c{)}^{{}^{\frac{2}{3}}})}^3+3(abc{)}^{{}^{\frac{2}{3}}}-2(ab+ac+bc)\\ & \ge a^{{.}^{\frac{2}{3}}}{b}^{{.}^{\frac{2}{3}}}(a^{{.}^{\frac{2}{3}}}+b^{{.}^{\frac{2}{3}}})+a^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}(a^{{.}^{\frac{2}{3}}}+c^{{.}^{\frac{2}{3}}})\\ & +b^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}(b^{{.}^{\frac{2}{3}}}+c^{{.}^{\frac{2}{3}}})-2(ab+ac+bc)\\ & \ge a^{{.}^{\frac{2}{3}}}{b}^{{.}^{\frac{2}{3}}}(a^{{.}^{\frac{2}{3}}}+b^{{.}^{\frac{2}{3}}})+a^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}(a^{{.}^{\frac{2}{3}}}+c^{{.}^{\frac{2}{3}}})\\ & +b^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}(b^{{.}^{\frac{2}{3}}}+c^{{.}^{\frac{2}{3}}})-2(ab+ac+bc)\\ & =a^{{.}^{\frac{2}{3}}}{b}^{{.}^{\frac{2}{3}}}{(a^{{.}^{\frac{2}{3}}}-b^{{.}^{\frac{2}{3}}})}^2+a^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}{(a^{{.}^{\frac{2}{3}}}-c^{{.}^{\frac{2}{3}}})}^2\\ & +b^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}{(b^{{.}^{\frac{2}{3}}}-c^{{.}^{\frac{2}{3}}})}^2\ge 0◼\end{array}$.

$\begin{array}{l}\\ 79.& (\mathbf{\text{APMO 2004}})\\ & \text{Misalkan}x,y,x\text{bilangan real positif dengan}\\ & \text{tunjukkan bahwa}\\ & (x^2+2)(y^2+2)(z^2+2)\ge 9(xy+yz+zx)\\ \\ & \mathbf{\text{Bukti}}\\ & \mathbf{\text{Alternatif 1}}\\ & \text{Dengan menjabarkan akan didapatkan}\\ & x^2{y}^2{z}^2+2{\displaystyle \sum _{\text{siklik}}^{.}{x}^2{y}^2+4{\displaystyle \sum _{\text{siklik}}^{.}{x}^2+8\ge 9{\displaystyle \sum _{\text{siklik}}^{.}xy}}}\\ & \text{Perhatikan bahwa}\\ & \bullet 2{\displaystyle \sum _{\text{siklik}}^{.}{x}^2{y}^2-4{\displaystyle \sum _{\text{siklik}}^{.}xy+6=2{\displaystyle \sum _{\text{siklik}}^{.}(xy-1{)}^2\ge 0}}}\\ & \bullet {\displaystyle \sum _{\text{siklik}}^{.}{x}^2\ge {\displaystyle \sum _{\text{siklik}}^{.}xy\text{atau}{x}^2+y^2+z^2\ge xy+xz+yz}}\\ & \text{Kita cukup membuktikan bahwa}\\ & x^2{y}^2{z}^2+{\displaystyle \sum _{\text{siklik}}^{.}{x}^2+2\ge 2{\displaystyle \sum _{\text{siklik}}^{.}xy}}\\ & \iff x^2{y}^2{z}^2+2\ge {\displaystyle \sum _{\text{siklik}}^{.}xy}\\ & \begin{array}{rl}& \text{Untuk}a,b,c\text{bilangan real positif},\\ & \mathbf{\text{Ketaksamaan Schur}}\text{saat}r=1\\ & \text{memberikan}\\ & {\displaystyle \sum _{\text{siklik}}^{.}{a}^3+3abc\ge {\displaystyle \sum _{\text{siklik}}^{.}{a}^{2}b+{\displaystyle \sum _{\text{siklik}}^{.}a{b}^2}}}\\ & =ab(a+b)+bc(b+c)+ca(c+a)\\ & \text{Dengan}\mathbf{\text{Ketaksamaan AM-GM}}\text{didapakan}\\ & {\displaystyle \sum _{\text{siklik}}^{.}{a}^3+3abc\ge 2{\displaystyle \sum _{\text{siklik}}^{.}(ab{)}^{\frac{3}{2}}}}\\ & \text{Pilih}a=x^{\frac{2}{3}},b=y^{\frac{2}{3}},c=z^{\frac{2}{3}},\text{maka didapatkan}\\ & (x^{\frac{2}{3}}{)}^3+(y^{\frac{2}{3}}{)}^3+(z^{\frac{2}{3}}{)}^3+3(xyz{)}^{\frac{2}{3}}\ge 2(xy+yz+zx)\\ & \text{Selanjutnya kita selesaikan ini},x^2{y}^2{z}^2+2\ge {\displaystyle \sum _{\text{siklik}}^{.}xy}\\ & \iff x^2{y}^2{z}^2+2\ge 3(xyz{)}^{\frac{2}{3}}\\ & \text{Misalkan}(xyz{)}^{\frac{2}{3}}=t,\text{maka}\\ & t^3+2\ge 3t\iff t^3-3t+2\ge 0\\ & (t-1{)}^2(t+2)\ge 0\text{adalah hal benar}\end{array}\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Alternatif 2}}\\ & x^2{y}^2{z}^2+2{\displaystyle \sum _{\text{siklik}}^{.}{x}^2{y}^2+4{\displaystyle \sum _{\text{siklik}}^{.}{x}^2+8\ge 9{\displaystyle \sum _{\text{siklik}}^{.}xy}}}\\ & \text{atau dalam bentuk utuhnya, yaitu}\\ & x^2{y}^2{z}^2+2(x^2{y}^2+x^2{z}^2+y^2{z}^2)+4(x^2+y^2+z^2)+8\ge 9(xy+xz+yz)\\ & \text{Sekarang kira uraikan satu persatu bagian}\\ & \bullet x^2{y}^2{z}^2+1+1\ge 3\sqrt[3]{(xyz{)}^2}\ge {\displaystyle \frac{9abc}{a+b+c}=\frac{9r}{p}}\\ & \text{ingat bahwa}\mathbf{\text{jika ada}}{\displaystyle \frac{9r}{p}\ge 4q-p^2}\\ & =4(xy+xz+yz)-(x+y+z{)}^2\\ & \text{adalah}\mathbf{\text{ketaksamaan Schur saat}}r=1\\ & \bullet x^2{y}^2+1+x^2{z}^2+1+y^2{z}^2+1\ge 2(xy+xz+yz)\\ & \bullet x^2+y^2+z^2\ge xy+xz+yz\\ & \text{keduanya didapat dengan ketaksamaan}\mathbf{\text{AM-GM}}\\ & x^2{y}^2{z}^2+2(x^2{y}^2+x^2{z}^2+y^2{z}^2)+4(x^2+y^2+z^2)+8\\ & =x^2{y}^2{z}^2+2+2(x^2{y}^2+x^2{z}^2+y^2{z}^2+3)+4(x^2+y^2+z^2)\\ & \ge 4(xy+xz+yz)-(x+y+z{)}^2+4(xy+xz+yz)+4(xy+xz+yz)\\ & \ge 12(xy+xz+yz)-(x+y+z{)}^2\\ & \ge 12(xy+xz+yz)-3(xy+xz+yz)\\ & =9(xy+xz+yz)◼\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Alternatif 3}}\\ & (x^2+2)(y^2+2)(z^2+2)-9(xy+yz+zx)\\ & \text{Dengan}\mathbf{\text{ketaksamaan AM-GM}}\\ & =x^2{y}^2{z}^2+2(x^2{y}^2+x^2{z}^2+y^2{z}^2)\\ & +4(x^2+y^2+z^2)+8-9(xy+xz+yz)\\ & =4(x^2+y^2+z^2)+2((x^2{y}^2+1)+(x^2{z}^2+1)+(y^2{z}^2+1))\\ & +(x^2{y}^2{z}^2+1)+1-9(xy+xz+yz)\\ & \ge 4(x^2+y^2+z^2)+4(xy+xz+yz)\\ & +2xyz+1-9(xy+xz+yz)\\ & =(x^2+y^2+z^2)+3(x^2+y^2+z^2)\\ & +2xyz+1-5(x^2+y^2+z^2)\\ & \ge x^2+y^2+z^2+3(xy+xz+yz)\\ & +2xyz-5(xy+xz+yz)\\ & =x^2+y^2+z^2+2xyz+1-2(xy+xz+yz)\ge 0\\ & \text{adalah benar dengan bukti ada pada}\\ & \text{nomor soal sebelumnya}\end{array}$.

$\begin{array}{l}\\ 80.& \text{Misalkan}x,y,x\text{bilangan real positif dengan}\\ & \text{tunjukkan bahwa}\\ & 2(a^2+b^2+c^2)+abc+8\ge 5(a+b+c)\\ & (\mathbf{\text{Tran Nam Dung}})\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Dengan}\mathbf{\text{ketaksamaan AM-GM}}\text{dan}\\ & \text{menggeser ke ruas kiri dan masing-masing}\\ & \text{serta mengalikan semunya dengan 6, maka}\\ & 12(a^2+b^2+c^2)+6abc+48-30(a+b+c)\\ & =12(a^2+b^2+c^2)+3(2abc+1)+45-5.2.3(a+b+c)\\ & \ge 2(a^2+b^2+c^2)+9\sqrt[3]{(abc{)}^2}+45-5((a+b+c{)}^2+9)\\ & =12(a^2+b^2+c^2)+{\displaystyle \frac{9abc}{\sqrt[3]{abc}}-5((a^2+b^2+c^2)+2(ab+ac+bc))}\\ & =7(a^2+b^2+c^2)+{\displaystyle \frac{9abc}{\sqrt[3]{abc}}-10(ab+ac+bc)}\\ & \ge 7(a^2+b^2+c^2)+{\displaystyle \frac{27abc}{a+b+c}-10(ab+ac+bc)}\\ & \begin{array}{rl}& \text{dengan}\mathbf{\text{ketaksamaan Schur}},\text{yaitu}:\\ & p^3+9r\ge 4pq\iff {\displaystyle \frac{9r}{p}\ge 4q-p^2}\\ & \text{maka ketaksamaan akan menjadi}\\ & \ge 7(a^2+b^2+c^2)+3(4q-p^2)-10q\\ & \ge 7(a^2+b^2+c^2)+2q-3p^2\\ & =7(a^2+b^2+c^2)+2(ab+ac+bc)-3(a+b+c{)}^2\\ & =7(a^2+b^2+c^2)+2q-3((a^2+b^2+c^2)+2q)\\ & =4(a^2+b^2+c^2)+2q-6q\\ & =4(a^2+b^2+c^2)-4q\\ & =4(a^2+b^2+c^2)-4(ab+ac+bc)\\ & =4(a^2+b^2+c^2-ab-ac-bc)\ge 0◼\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Tung, K.Y. 2013. Ayo Raih Medali Emas Olimpiade Matematika SMA. Yogyakarta: ANDI.
2. Venkatachala, B.J. 2009. Inequalities An Approach Through Problems (2nd). India: SPRINGER.
3. Vo Tranh Van..... Bat Dang Thuc Schur Va Phuong Phap Doi Bien P, Q, R.
4. Vo Quoc Ba Can. 2007. Bai Viet Ve Bat Dang Thuc Schur Va Vornicu Schur.
5. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.

Contoh Soal 15 (Segitiga dan Ketaksamaan)

$\begin{array}{l}\\ 71.& (\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 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 \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}$.

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

$\begin{array}{l}\\ 72.& \text{Jika}a,b,\text{dan}c\text{bilangan real positif}\\ & \text{dengan}a+b+c=1\text{tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}\ge {\displaystyle \frac{81}{4}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Dengan Ketaksamaan Holder pilih}\\ & {\lambda }_1={\lambda }_2={\lambda }_3={\lambda }_4={\displaystyle \frac{1}{4}\left({\displaystyle \sum _{i=1}^4{\lambda }_i=1}\right)\text{akan diperoleh}}\\ & (a+b+c)(b+c+a+c+a+b)(b+c+a+c+a+b)\left({\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}}\right)\\ & \ge (1^{{.}^{\frac{1}{4}}}+1^{{.}^{\frac{1}{4}}}+1^{{.}^{\frac{1}{4}}}{)}^4=3^4=81\\ & \iff (1)(2)(2)\left({\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}}\right)\ge 81\\ & \iff {\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}\ge {\displaystyle \frac{81}{4}◼}}\end{array}\end{array}$.

$\begin{array}{l}\\ 73.& \text{Buktikan bahwa setiap bilangan real}\\ & \text{positif}a,b\text{dan}c\text{berlaku}\\ & a^2+b^2+c^2\ge ab+ac+bc\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & \text{Perhatikan bahwa}(a-b{)}^2\ge 0\\ & (a-c{)}^2\ge 0,\text{dan}(b-c{)}^2\ge 0\\ & \text{adalah benar, maka}\\ & (a-b{)}^2=a^2-2ab+b^2\ge 0\\ & \iff a^2+b^2\ge 2ab.....(1)\\ & \text{Dengan cara yang kurang lebih sama}\\ & \text{akan didapatkan}\\ & \bullet a^2+c^2\ge 2ac.....(2)\\ & \bullet b^2+c^2\ge 2bc.....(1)\\ & \text{Jika ketaksamaan}(1),(2),\mathrm{\&}(3)\text{dijumlahkan}\\ & \text{akan didapatkan bentuk}\\ & 2a^2+2b^2+2c^2\ge 2ab+2ac+2bc\\ & \iff a^2+b^2+c^2\ge ab+ac+bc◼\\ & \text{Alternatif 2}\\ & \text{Dengan ketaksamaan}\mathbf{\text{Cauchy-Schwarz}}\\ & (a^2+b^2+c^2)(b^2+c^2+a^2)\ge (ab+bc+ca{)}^2\\ & \iff a^2+b^2+c^2\ge ab+ac+bc◼\\ & \text{Alternatif 3}\\ & \text{Untuk barisan}(a,b,c),\text{asumsikan}a\ge b\ge c\\ & \text{maka dengan}\mathbf{\text{Ketaksamaan Renata}}\text{diperoleh}\\ & a.a+b.b+c.c\ge ab+bc+ca\\ & a^2+b^2+c^2\ge ab+ac+bc◼\\ & \text{Alternatif 4}\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Schur}}\text{saat}r=0,\\ & \text{yaitu}\\ & a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b)\ge 0\\ & \iff a^0(a-b)(a-c)+b^0(b-a)(b-c)+c^0(c-a)(c-b)\ge 0\\ & \iff (a-b)(a-c)+(b-a)(b-c)+(c-a)(c-b)\ge 0\\ & \iff a^2+b^2+c^2-ab-ac-bc\ge 0\\ & \iff a^2+b^2+c^2\ge ab+ac+bc◼\end{array}\end{array}$.

$\begin{array}{l}\\ 74.& \mathbf{\text{(IMO 1964)}}\\ & \text{Jika}a,b,c\text{adalah panjang sisi-sisi segitiga}\\ & \text{tunjukkan bahwa}\\ & a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)\le 3abc\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Schur}}\text{saat}r=1.\\ & a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ca(c+a)\\ & \iff 3abc\ge ab(a+b)-a^3+bc(b+c)-b^3+ca(c+a)-c^3\\ & \iff 3abc\ge a^{2}b+a{b}^2-c^3+b^{2}c+b{c}^2-b^3+c^{2}a+c{a}^2-c^3\\ & \iff 3abc\ge a^{2}b+c{a}^2-a^3+a{b}^2+b^{2}c-b^3+c^{2}a+b{c}^2-c^3\\ & \iff 3abc\ge a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)\\ & \text{atau}\\ & \iff a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)\le 3abc◼\end{array}$ .

$\begin{array}{l}\\ 75.& \mathbf{\text{(IMO 2000)}}\\ & \text{Jika}a,b,c\text{bilangan real positif dengan}\\ & abc=1,\text{tunjukkan bahwa}\\ & (a+1-{\displaystyle \frac{1}{b}})(b+1-{\displaystyle \frac{1}{c}})(c+1-{\displaystyle \frac{1}{a}})\le 1\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Misalkan}a={\displaystyle \frac{x}{y},b=\frac{y}{z},\text{dan}c={\displaystyle \frac{z}{x},\text{maka}}}\\ & \left({\displaystyle \frac{x}{y}+1-\frac{z}{y}}\right)\left({\displaystyle \frac{y}{z}+1-\frac{x}{z}}\right)\left({\displaystyle \frac{z}{x}+1-\frac{y}{x}}\right)\le 1\\ & \iff \left({\displaystyle \frac{x+y-z}{y}}\right)\left({\displaystyle \frac{y+z-x}{z}}\right)\left({\displaystyle \frac{z+x-y}{x}}\right)\le 1\\ & \iff (x+y-z)(y+z-x)(z+x-y)\le xyz,\text{atau}\\ & \iff xyz\ge (x+y-z)(y+z-x)(z+x-y)\\ & \text{Bentuk terakhir memenuhi bentuk kedua dari}\\ & \mathbf{\text{Ketaksamaan Schur}}\text{saat}r=1.\\ & \text{Jadi},\\ & (a+1-{\displaystyle \frac{1}{b}})(b+1-{\displaystyle \frac{1}{c}})(c+1-{\displaystyle \frac{1}{a}})\le 1◼\end{array}$.

Contoh Soal 14 (Segitiga dan Ketaksamaan)

$\begin{array}{l}\\ 66.& \text{Jika}x,y\text{bilangan positif, tunjukkan}\\ & x^2+y^2\ge {\displaystyle \frac{(x+y{)}^2}{2}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Alternatif 1}\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & (x-y{)}^2\ge 0\iff x^2-2xy+y^2\ge 0\\ & \iff x^2+y^2\ge 2xy\\ & \iff 2x^2+2y^2\ge x^2+y^2+2xy\\ & \iff 2(x^2+y^2)\ge (x+y{)}^2\\ & \iff x^2+y^2\ge {\displaystyle \frac{(x+y{)}^2}{2}◼}\end{array}\\ & \text{Alternatif 2}\\ & \text{Dengan Ketaksamaan Holder pilih}\\ & {\lambda }_1={\lambda }_2={\displaystyle \frac{1}{2}\left({\displaystyle \sum _{i=1}^2{\lambda }_i=1}\right)\text{akan diperoleh}}\\ & \begin{array}{rl}& {\left({\displaystyle \frac{x^2}{1}+\frac{y^2}{1}}\right)}^{\frac{1}{2}}(1+1{)}^{\frac{1}{2}}\ge x+y\\ & \iff (x^2+y^2)(1+1)\ge (x+y{)}^2\\ & \iff (x^2+y^2)\ge {\displaystyle \frac{(x+y{)}^2}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 67.& \text{Untuk}x,y\text{bilangan positif, tunjukkan}\\ & \text{kebenaran ketaksamaan Cauchy-Schwarz}\\ & {\displaystyle \frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\frac{x_3^2}{y_3}+\cdots +\frac{x_n^2}{y_n}\ge {\displaystyle \frac{(x_1+x_2+\cdots +x_n{)}^2}{y_1+y_2+\cdots +y_n}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Dengan Ketaksamaan Holder pilih}\\ & {\lambda }_1={\lambda }_2={\displaystyle \frac{1}{2}\left({\displaystyle \sum _{i=1}^2{\lambda }_i=1}\right)\text{akan diperoleh}}\\ & {\left({\displaystyle \frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\cdots +\frac{x_n^2}{y_n}}\right)}^{\frac{1}{2}}(y_1+y_2+\cdots +y_n{)}^{\frac{1}{2}}\ge (x_1+x_2+\cdots +x_n)\\ & \left({\displaystyle \frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\cdots +\frac{x_n^2}{y_n}}\right)(y_1+y_2+\cdots +y_n)\ge (\sqrt{x_1^2}+\sqrt{x_2^2}+\cdots +\sqrt{x_n^2}{)}^2\\ & \left({\displaystyle \frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\cdots +\frac{x_n^2}{y_n}}\right)(y_1+y_2+\cdots +y_n)\ge (x_1+x_2+\cdots +x_n{)}^2\\ & \left({\displaystyle \frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\cdots +\frac{x_n^2}{y_n}}\right)\ge {\displaystyle \frac{(x_1+x_2+\cdots +x_n{)}^2}{(y_1+y_2+\cdots +y_n)}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 68.& (\mathbf{\text{National Mathematical Contest, Belarus-2000}})\\ & \text{Jika}a,b,c,x,y\text{dan}z\text{bilangan real positif, tunjukkan}\\ & {\displaystyle \frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}\ge {\displaystyle \frac{(a+b+c{)}^3}{3(x+y+z)}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Dengan Ketaksamaan Holder pilih}\\ & {\lambda }_1={\lambda }_2={\lambda }_3={\displaystyle \frac{1}{3}\left({\displaystyle \sum _{i=1}^3{\lambda }_i=1}\right)\text{akan diperoleh}}\\ & {\left({\displaystyle \frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}}\right)}^{\frac{1}{3}}(1+1+1{)}^{\frac{1}{3}}(x+y+z{)}^{\frac{1}{3}}\ge (a+b+c)\\ & \iff \left({\displaystyle \frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}}\right)(3)(x+y+z)\ge (a+b+c{)}^3\\ & \iff \left({\displaystyle \frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}}\right)\ge {\displaystyle \frac{(a+b+c{)}^3}{3(x+y+z)}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 69.& \text{Jika}a,b,c,x,y\text{dan}z\text{bilangan real positif}\\ & \text{dengan}a+b+c=x+y+z\text{tunjukkan bahwa}\\ & {\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}\ge a+b+c}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Dengan Ketaksamaan Holder pilih}\\ & {\lambda }_1={\lambda }_2={\lambda }_3={\displaystyle \frac{1}{3}\left({\displaystyle \sum _{i=1}^3{\lambda }_i=1}\right)\text{akan diperoleh}}\\ & {\left({\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}}\right)}^{\frac{1}{3}}(x+y+z{)}^{\frac{1}{3}}(x+y+z{)}^{\frac{1}{3}}\ge (a+b+c)\\ & \iff \left({\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}}\right)(x+y+z{)}^2\ge (a+b+c{)}^3\\ & \iff \left({\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}}\right)(a+b+c{)}^2\ge (a+b+c{)}^3\\ & \iff \left({\displaystyle \frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}}\right)\ge {\displaystyle \frac{(a+b+c{)}^3}{(a+b+c{)}^2}}\\ & \iff {\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}\ge a+b+c◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 70.& \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}$.

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.

Contoh Soal 13 (Segitiga dan Ketaksamaan)

 $.\begin{array}{rl}& \text{Mengenal penulisan pola}\mathbf{\text{Siklik dan Simetri}}\\ & \text{Misal untuk}n=3,\text{pada penulisan unsur}\\ & x,y,\text{dan}z,\text{maka}\\ & \begin{array}{|ll|}\hline \mathbf{\text{Pola Siklik}}& \mathbf{\text{Pola Simetri}}\\ \begin{array}{r}{\displaystyle \sum _{\text{siklik}}^{.}{x}^2=x^2+y^2+z^2}\\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}{\displaystyle \sum _{\text{sym}}^{.}{x}^2}& =x^2+x^2\\ & +y^2+y^2\\ \\ & +z^2+z^2\\ \\ & =2(x^2+y^2+z^2)\end{array}\\ \begin{array}{r}{\displaystyle \sum _{\text{siklik}}^{.}{x}^3=x^3+y^3+z^3}\end{array}& \begin{array}{r}{\displaystyle \sum _{\text{sym}}^{.}{x}^3=2(x^3+y^3+z^3)}\end{array}\\ \begin{array}{r}{\displaystyle \sum _{\text{siklik}}^{.}{x}^{2}y=x^{2}y+y^{2}z+z^{2}x}\\ & \\ & \\ & \end{array}& \begin{array}{rl}{\displaystyle \sum _{\text{sym}}^{.}{x}^{2}y}& =x^{2}y+x^{2}z\\ & +y^{2}x+y^{2}z\\ \\ & +z^{2}x+z^{2}y\end{array}\\ \begin{array}{rl}{\displaystyle \sum _{\text{siklik}}^{.}xyz}& =xyz+yzx+zxy\\ & =3xyz\end{array}& \begin{array}{rl}{\displaystyle \sum _{\text{sym}}^{.}xyz}& =xyz+xzy+\cdots \\ & =6xyz\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 61.& (\mathbf{\text{IMO 1995}})\\ & \text{Jika}a,b,c\text{bilangan-bilangan real positif}\\ & \text{dengan}abc=1,\text{maka tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\ge {\displaystyle \frac{3}{2}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Misalkan}x={\displaystyle \frac{1}{a},y={\displaystyle \frac{1}{b},\text{dan}z={\displaystyle \frac{1}{c},}}}\\ & \text{maka}\\ & {\displaystyle \frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}}\\ & ={\displaystyle \frac{x^{3}yz}{y+z}+\frac{y^{3}xz}{x+z}+\frac{z^{3}xy}{x+y},\text{karena}xyz=1}\\ & ={\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\\ & \text{Dengan ketaksamaan}\mathbf{\text{Cauchy-Schwarz}}\\ & \left(2{\displaystyle \sum _{\text{siklik}}^{.}y+z}\right)\left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge (x+y+z{)}^2\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{(x+y+z{)}^2}{2(x+y+z)}}\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{(a+b+c)}{2}}\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{3\sqrt[3]{xyz}}{2}}\\ & \iff \left({\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}}\right)\ge {\displaystyle \frac{3}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 62.& \text{Diketahui}a,b\text{bilangan real positif}\\ & \text{Tunjukkan bahwa}{\displaystyle \frac{a^2}{b}+\frac{b^2}{a}\ge a+b}\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Asumsikan bahwa}a\ge b,\text{maka}{a}^2\ge b^2\\ & \text{dan}{\displaystyle \frac{1}{b}\ge \frac{1}{a}.}\\ & \text{Perhatikan bahwa baik}(a^2,b^2)\text{dan}\left({\displaystyle \frac{1}{b},\frac{1}{a}}\right)\\ & \text{adalah kumpulan dua barisan yang monoton}\\ & \text{sama yaitu sama-sama naik. Sehingga}\\ & \text{dengan}\mathbf{\text{ketaksamaan Renata}}\text{diperoleh}\\ & a^2.{\displaystyle \frac{1}{b}+b^2.{\displaystyle \frac{1}{a}\ge a^2.{\displaystyle \frac{1}{a}+b^2.{\displaystyle \frac{1}{b}}}}}\\ & \iff {\displaystyle \frac{a^2}{b}+\frac{b^2}{a}\ge a+b◼}\end{array}$.

$\begin{array}{l}\\ 63.& \text{Diberikan}a,b,c>0,\text{tunjukkan bahwa}\\ & {\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Alternatif 1}\\ & \text{Dengan AM-GM diperoleh}\\ & \bullet {\displaystyle \frac{a}{c}+\frac{c}{a}\ge 2\iff {\displaystyle \frac{c}{a}\ge 2-{\displaystyle \frac{a}{c}}}}\\ & \bullet {\displaystyle \frac{b}{c}+\frac{c}{b}\ge 2\iff {\displaystyle \frac{b}{c}\ge 2-{\displaystyle \frac{c}{b}}}}\\ & \bullet {\displaystyle \frac{a}{b}+\frac{b}{a}\ge 2\iff {\displaystyle \frac{a}{b}\ge 2-{\displaystyle \frac{b}{a}}}}\\ & \text{Sehingga}\\ & \begin{array}{rl}{\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}}& \ge a(2-{\displaystyle \frac{c}{b}})+b(2-{\displaystyle \frac{a}{c}})+c(2-{\displaystyle \frac{b}{a}})\\ & =2a-{\displaystyle \frac{ac}{b}+2b-{\displaystyle \frac{ab}{c}+2c-{\displaystyle \frac{bc}{a}}}}\\ {\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}}& \ge 2(a+b+c)-\left({\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}}\right)\\ 2& \left({\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}}\right)\ge 2(a+b+c)\\ {\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}}& \ge a+b+c◼\end{array}\\ & \text{Alternatif 2}\\ & \text{Dengan ketaksamaan Renata}\\ & \begin{array}{rl}& \text{Asumsikan}a\ge b\ge c,\text{maka}ab\ge ca\ge bc\\ & \text{dan}{\displaystyle \frac{1}{c}\ge \frac{1}{b}\ge \frac{1}{a}.}\\ & \text{Perhatikan bahwa}\\ & (ab\ge ca\ge bc)\text{dan}\left({\displaystyle \frac{1}{c}\ge \frac{1}{b}\ge \frac{1}{a}}\right)\\ & \text{memiliki kemonotonan yang sama}\\ & \text{maka dengan}\mathbf{\text{ketaksamaan Renata}}\\ & \text{dapat diperoleh bentuk}\\ & ab.{\displaystyle \frac{1}{c}+ac.{\displaystyle \frac{1}{b}+bc.{\displaystyle \frac{1}{a}\ge ab.{\displaystyle \frac{1}{b}+ac.{\displaystyle \frac{1}{a}+bc.{\displaystyle \frac{1}{c}}}}}}}\\ & \iff {\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+c+b}\\ & \iff {\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 64.& \text{Diberikan}a,b,c>0,\text{tunjukkan kebenaran}\\ & \mathbf{\text{ketaksamaan Nesbitt}}\text{berikut}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Alternatif 1}\\ & \text{Dengan AM-GM diperoleh}\\ & {\displaystyle \frac{(a+b)+(b+c)+(c+a)}{3}\ge {\displaystyle \frac{3}{{\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}}}}\\ & \iff ((a+b)+(b+c)+(c+a))\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge 9\\ & \iff 2\left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)+6\ge 9\\ & \iff 2\left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge 3\\ & \iff \left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge {\displaystyle \frac{3}{2}◼}\\ & \text{Alternatif 2}\\ & \text{Dengan ketaksamaan Renata}\\ & \begin{array}{rl}& \text{Asumsikan}a\le b\le c,\text{maka}a+b\le a+c\le b+c\\ & \text{dan}{\displaystyle \frac{1}{b+c}\le \frac{1}{a+c}\le \frac{1}{a+b}.}\\ & \text{Perhatikan bahwa}\\ & (a\le b\le c)\text{dan}{\displaystyle \frac{1}{b+c}\le \frac{1}{a+c}\le \frac{1}{a+b}}\\ & \text{memiliki kemonotonan yang sama}\\ & \text{maka dengan}\mathbf{\text{ketaksamaan Renata}}\\ & \text{dapat diperoleh bentuk}\\ & a.{\displaystyle \frac{1}{b+c}+b.{\displaystyle \frac{1}{a+c}+c.{\displaystyle \frac{1}{a+b}\ge b.{\displaystyle \frac{1}{b+c}+c.{\displaystyle \frac{1}{a+c}+a.{\displaystyle \frac{1}{a+b}}}}}}}\\ & \text{dan}\\ & a.{\displaystyle \frac{1}{b+c}+b.{\displaystyle \frac{1}{a+c}+c.{\displaystyle \frac{1}{a+b}\ge c.{\displaystyle \frac{1}{b+c}+a.{\displaystyle \frac{1}{a+c}+b.{\displaystyle \frac{1}{a+b}}}}}}}\\ & \text{Jika dijumlahkan keduanya, maka}\\ & 2\left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge 3\\ & \iff \left({\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\right)\ge {\displaystyle \frac{3}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 65.& (\mathbf{\text{OSN 2015}})\\ & \text{Diberikan}a,b,c>0,\text{Buktikan bahwa}\\ & \sqrt{{\displaystyle \frac{a}{b+c}+\frac{b}{c+a}}}+\sqrt{{\displaystyle \frac{b}{c+a}+\frac{c}{a+b}}}+\sqrt{{\displaystyle \frac{c}{a+b}+\frac{a}{b+c}}}\ge 3\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Perhatikan bukti soal no. 4 di atas}\\ & \text{Dengan}\mathbf{\text{keksamaan Renata}}\text{dapat diperoleh}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{a+c}\ge \frac{b}{b+c}+\frac{a}{a+c}}\\ & \text{Misalkan}\\ & x=b+c,y=c+a,y=a+b,\text{maka}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{a+c}\ge \frac{b}{b+c}+\frac{a}{a+c}}\\ & \iff {\displaystyle \frac{a}{b+c}+\frac{b}{a+c}\ge \frac{y+z-x}{2x}+\frac{x+z-y}{2y}}\\ & \begin{array}{rl}& \iff \sqrt{{\displaystyle \frac{a}{b+c}+\frac{b}{a+c}}}\ge \sqrt{\frac{y+z-x}{2x}+\frac{x+z-y}{2y}}\\ & \iff =\sqrt{{\displaystyle \frac{1}{2}}}\sqrt{{\displaystyle \frac{y}{x}+\frac{z}{x}-1+\frac{x}{y}+\frac{z}{y}-1}}\\ & \iff \text{dengan AM-GM}\\ & \iff \ge {\displaystyle \frac{1}{\sqrt{2}}\sqrt{{\displaystyle \frac{z}{x}+\frac{z}{y}+2\sqrt{\frac{y}{x}.\frac{x}{y}}-2}}}\\ & \iff ={\displaystyle \frac{1}{\sqrt{2}}\sqrt{{\displaystyle \frac{z}{x}+\frac{z}{y}+2-2}}}\\ & \iff \ge {\displaystyle \frac{1}{\sqrt{2}}\sqrt{{\displaystyle \frac{z}{x}+\frac{z}{y}}}}\\ & \iff \ge {\displaystyle \frac{\sqrt{2}}{\sqrt{2}}\sqrt{{\displaystyle \frac{z^2}{xy}}}={\displaystyle \sqrt{{\displaystyle \frac{z^2}{xy}}}}}\end{array}\\ & \begin{array}{rl}& \bullet \sqrt{{\displaystyle \frac{a}{b+c}+\frac{b}{a+c}}}\ge {\displaystyle \sqrt{{\displaystyle \frac{z^2}{xy}}}}\\ & \bullet \sqrt{{\displaystyle \frac{b}{c+a}+\frac{c}{a+b}}}\ge \sqrt{{\displaystyle \frac{x^2}{yz}}}\\ & \bullet \sqrt{{\displaystyle \frac{c}{a+b}+\frac{a}{b+c}}}\ge \sqrt{{\displaystyle \frac{y^2}{xz}}}\end{array}\\ & \begin{array}{rl}& \text{Selanjutnya}\\ & \sqrt{{\displaystyle \frac{a}{b+c}+\frac{b}{c+a}}}+\sqrt{{\displaystyle \frac{b}{c+a}+\frac{c}{a+b}}}+\sqrt{{\displaystyle \frac{c}{a+b}+\frac{a}{b+c}}}\\ & \ge {\displaystyle \sqrt{{\displaystyle \frac{z^2}{xy}}}+\sqrt{{\displaystyle \frac{x^2}{yz}}}+\sqrt{{\displaystyle \frac{y^2}{xz}}}}\\ & \text{Dengan AM-GM lagi}\\ & \ge 3\sqrt[3]{\sqrt{{\displaystyle \frac{z^2}{xy}}}\times \sqrt{{\displaystyle \frac{x^2}{yz}}}\times \sqrt{{\displaystyle \frac{y^2}{xz}}}}\\ & \ge 3\sqrt[3]{\sqrt{{\displaystyle \frac{(xyz{)}^2}{(xyz{)}^2}}}}\\ & \ge 3◼\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.

WEBSITE

1. https://holdenlee.github.io/high_school/omc/23-rearrange.pdf diakses 18 Januari 2022.
2. https://www.gotohaggstrom.com/Advanced%20inequality%20manipulations.pdf  diakses 20 Januari 2022

Contoh Soal 12 (Segitiga dan Ketaksamaan)

$\begin{array}{l}\\ 56.& (\mathbf{\text{OSK 2018}})\\ & \text{Diketahui bilangan real}x\text{dan}y\\ & \text{yang memenuhi}{\displaystyle \frac{1}{2}<\frac{x}{y}<2}\\ & \text{Nilai minimum}{\displaystyle \frac{x}{2y-x}+\frac{2y}{2x-y}\text{adalah}....}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & \text{Misal}t={\displaystyle \frac{x}{y}}\\ & \text{Misalkan juga}f(t)={\displaystyle \frac{x}{2y-x}+{\displaystyle \frac{2y}{2x-y}}}\\ & \text{maka}f(t)={\displaystyle \frac{2t^2-3t+4}{-2t^2+5t-2}}\\ & \bullet \text{Agar minimum, maka}{f}^{\prime }(t)=0\\ & \text{Sehingga}\\ & f^{\prime }(t)=4t^2+8t-14=0\\ & \iff t_{1,2}=-1\pm {\displaystyle \frac{3}{2}\sqrt{2}}\\ & \text{Pilih yang positif, yaitu}t=-1+{\displaystyle \frac{3}{2}\sqrt{2}}\\ & \bullet \text{Dengan proses substistusi harga}t\\ & \text{di atas, maka akan didapatkan }\\ & \text{nilai}f(t)=1+{\displaystyle \frac{4}{3}\sqrt{2}}\end{array}\\ & \begin{array}{rl}& \text{Alternatif 2}\\ & \text{Menurut bentuk}{\displaystyle \frac{1}{2}<\frac{x}{y}<2}\\ & \text{jelas bahwa baik}2y-x\text{dan}2x-y\\ & \text{keduanya}\mathbf{\text{positif}}\\ & \text{Lihat tabel berikut}\\ & \begin{array}{|ccc|}\hline \text{Bentuk}& \text{Pengecekan 1}& \text{Pengecekan 2}\\ \begin{array}{rl}& {\displaystyle \frac{1}{2}<\frac{x}{y}<2}\\ & \text{Jelas bahwa}\\ & x,y\ne 0\\ & \\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}& \text{Saat}(\times y)\\ & \text{Yaitu}:\\ & {\displaystyle \frac{1}{2}(y)<\frac{x}{y}(y)<2(y)}\\ & \iff {\displaystyle \frac{1}{2}y<x<2y}\\ & \text{Jelas bahwa}\\ & 2y-x>0\\ & \\ & \\ & \end{array}& \begin{array}{rl}& \text{Saat dibali posisinya}\\ & {\displaystyle \frac{1}{2}<\frac{y}{x}<2}\\ & \text{Saat}(\times x)\\ & \text{Yaitu}:\\ & {\displaystyle \frac{1}{2}(x)<\frac{y}{x}(x)<2(x)}\\ & \iff {\displaystyle \frac{1}{2}x<y<2x}\\ & \text{Jelas bahwa}\\ & 2x-y>0\end{array}\\ \hline\end{array}\\ & \text{Saat masing-masing}\\ & \bullet {\displaystyle \frac{x}{2y-x}={\displaystyle \frac{1}{3}+\frac{2}{3}\left({\displaystyle \frac{2x-y}{2y-x}}\right)\text{dan}}}\\ & \bullet {\displaystyle \frac{2y}{2x-y}={\displaystyle \frac{2}{3}+\frac{4}{3}\left({\displaystyle \frac{2y-x}{2x-y}}\right)}}\\ & \text{Dengan ketaksamaan AM-GM diperoleh}\\ & \begin{array}{rl}{\displaystyle \frac{x}{2y-x}+{\displaystyle \frac{2y}{2x-y}}}& =1+\frac{2}{3}\left({\displaystyle \frac{2x-y}{2y-x}}\right)+\frac{4}{3}\left({\displaystyle \frac{2y-x}{2x-y}}\right)\\ & \ge 1+2\sqrt{\frac{2}{3}\left({\displaystyle \frac{2x-y}{2y-x}}\right)\frac{4}{3}\left({\displaystyle \frac{2y-x}{2x-y}}\right)}\\ & =1+2\sqrt{{\displaystyle \frac{8}{9}}}\\ & =1+2\left({\displaystyle \frac{2}{3}}\right)\sqrt{2}\\ & =1+{\displaystyle \frac{4}{3}\sqrt{2}}\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 57.& \text{Diketahui}a,b\text{bilangan real positif}\\ & \text{dan}a+b=1.\text{Tunjukkan bahwa}\\ & {(a+{\displaystyle \frac{1}{a}})}^2+{(b+{\displaystyle \frac{1}{b}})}^2\ge {\displaystyle \frac{25}{2}}\\ \\ & \mathbf{\text{Bukti}}\end{array}$.

$\begin{array}{rl}& \text{Dengan AM-GM kita akan mendapatkan}\\ & 1=a+b\ge 2\sqrt{ab}\iff {\displaystyle \frac{1}{2}\ge \sqrt{ab}\iff 2\le {\displaystyle \frac{1}{\sqrt{ab}}}}\\ & \iff {\displaystyle \frac{1}{\sqrt{ab}}\ge 2\iff {\displaystyle \frac{1}{ab}\ge 4}}\\ & \text{Perhatikan soal, dengan CS-Engel}\text{akan}\\ & \text{didapatkan}\\ & \begin{array}{rl}& (1+1)\left({\displaystyle \frac{{(a+\frac{1}{a})}^2}{1}+{\displaystyle \frac{{(b+\frac{1}{b})}^2}{1}}}\right)\ge {(a+\frac{1}{a}+b+\frac{1}{b})}^2\\ & \iff 2({\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2)\ge {(a+b+\frac{1}{a}+\frac{1}{b})}^2\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{a+b}{ab}})}^2}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{1}{ab}})}^2}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}(1+4{)}^2}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}\times 25}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{25}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 58.& \text{Jika}a,b\text{bilangan positif, buktikan}\\ & \text{a}.2(a^2+b^2)\ge (a+b{)}^2\\ & \text{b}.4(a^3+b^3)\ge (a+b{)}^3\\ & \text{c}.8(a^4+b^4)\ge (a+b{)}^4\\ & \text{d}.16(a^5+b^5)\ge (a+b{)}^5\\ & \text{e}.32(a^6+b^6)\ge (a+b{)}^6\\ & \text{f}.64(a^7+b^7)\ge (a+b{)}^7\\ & \text{g}.128(a^8+b^8)\ge (a+b{)}^8\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Akan ditunjukkan bukti poin 6.c saja}\\ & \text{untuk poin yang lain, silahkan pembaca}\\ & \text{sekalian untuk dibuktikan sendiri sebagai}\\ & \text{bahan latihan mandiri}.\\ & \text{Adapun bukti poin 6.c adalah sebagaimana}\\ & \text{berikut ini}\\ & \begin{array}{rl}& \text{Dengan ketaksamaan CS-Engel}\text{akan}\\ & \text{didapatkan}\\ & \bullet (1+1)(a^4+b^4)\ge (a^2+b^2{)}^2\\ & \iff 2(a^4+b^4)\ge (a^2+b^2{)}^2..........(1)\\ & \bullet (1+1)(a^2+b^2)\ge (a+b{)}^2\\ & \iff 2(a^2+b^2)\ge {\displaystyle (a+b{)}^2,(\text{kuadratkan})}\\ & \iff 4{\displaystyle {(a^2+b^2)}^2\ge (a+b{)}^4}\\ & \iff {(a^2+b^2)}^2\ge {\displaystyle \frac{(a+b{)}^4}{4}...........(2)}\\ & \text{Dari (1) dan (2) didapatkan hubungan}\\ & 2(a^4+b^4)\ge {(a^2+b^2)}^2\ge {\displaystyle \frac{(a+b{)}^4}{4}}\\ & \iff 8(a^4+b^4)\ge 4{(a^2+b^2)}^2\ge (a+b{)}^4\\ & \iff 8(a^4+b^4)\ge (a+b{)}^4◼\end{array}\end{array}$.

$\begin{array}{l}\\ 59.& \text{Jika}x,y,z\text{adalah bilangan real positif}\\ & \text{dengan}{x}^2+y^2+z^2=27.\text{Tunjukkan}\\ & \text{bahwa}{x}^3+y^3+z^3\ge 81\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Pada contoh soal no.2 terdapat}\\ & (ax+by+cz{)}^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)\\ & \text{ganti mengganti}a=b=c=1,\text{maka menjadi}\\ & (x+y+z{)}^2\le (1^2+1^2+1^2)(x^2+y^2+z^2)\\ & \iff (x+y+z{)}^2\le (3)(x^2+y^2+z^2)........(1)\\ & \text{Selanjutnya dengan mengganti dengan}\\ & x^{{.}^{\frac{3}{2}}},y^{{.}^{\frac{3}{2}}},z^{{.}^{\frac{3}{2}}}\text{dan}{x}^{{.}^{\frac{1}{2}}},y^{{.}^{\frac{1}{2}}},z^{{.}^{\frac{1}{2}}},\text{pada}\\ & (ax+by+cz{)}^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)\\ & \text{Kita akan dapatkan}\\ & (x^2+y^2+z^2)\le (x^3+y^3+z^3)(x+y+z)........(2)\\ & \text{Jika masing-masing ruas dikuadratkan, maka}\\ & (x^2+y^2+z^2{)}^4\le (x^3+y^3+z^3{)}^2(x+y+z{)}^2\\ & \iff (x^2+y^2+z^2{)}^4\le 3(x^3+y^3+z^3{)}^2(x^2+y^2+z^2)\\ & \iff (x^2+y^2+z^2{)}^3\le 3(x^3+y^3+z^3{)}^2\\ & \iff (27{)}^3\le 3(x^3+y^3+z^3{)}^2\\ & \iff 3^9\le 3(x^3+y^3+z^3{)}^2\\ & \iff 3^8\le (x^3+y^3+z^3{)}^2\\ & \iff (x^3+y^3+z^3{)}^2\ge 3^8\\ & \iff (x^3+y^3+z^3)\ge 3^{{.}^{\frac{8}{2}}}\\ & \iff (x^3+y^3+z^3)\ge 3^4=81◼\end{array}$.

$\begin{array}{l}\\ 60.& \text{Untuk}a,b,c,d\text{adalah bilangan real }\\ & \text{positif, tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\ge {\displaystyle \frac{64}{a+b+c+d}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Dengan ketaksamaan CS-Engel}\\ & (a+b+c+d)\left({\displaystyle \frac{1^2}{a}+{\displaystyle \frac{1^2}{b}+{\displaystyle \frac{2^2}{c}+{\displaystyle \frac{4^2}{d}}}}}\right)\ge (1+1+2+4{)}^2\\ & \iff {\displaystyle \frac{1^2}{a}+{\displaystyle \frac{1^2}{b}+{\displaystyle \frac{2^2}{c}+{\displaystyle \frac{4^2}{d}\ge {\displaystyle \frac{(1+1+2+4{)}^2}{a+b+c+d}}}}}}\\ & \iff {\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\ge {\displaystyle \frac{64}{a+b+c+d}◼}}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Muslim, M.S. 2020. Kumpulan Soal dan Pembahasan Olimpiade Matematika SMA Tahun 2007-2019 Tingkat Kota/Kabupaten. Bandung: YRAMA WIDYA.
2. Widodo, T. 2018. Booklet OSN SMA 2018: Soal dan Solusi OSK, OSP, OSN SMA Bidang Matematika.
3. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.

Contoh Soal 11 (Segitiga dan Ketaksamaan)

$\begin{array}{l}\\ 51.& \text{Diketahui rata-rata bilangan positif}\\ & m_1,m_2,...,m_k\text{adalah}A\\ & \text{Buktikan bahwa}\\ & {(m_1+{\displaystyle \frac{1}{m_1}})}^2+{(m_2+{\displaystyle \frac{1}{m_2}})}^2+...+{(m_k+{\displaystyle \frac{1}{m_k}})}^2\ge k{(A+{\displaystyle \frac{1}{A}})}^2\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & {\displaystyle \frac{m_1+m_2+m_3+...+m_k}{k}=A}\\ & \text{Dengan ketaksamaan QM-AM akan diperoleh}\\ & \sqrt{{\displaystyle \frac{(m_1{)}^2+(m_2{)}^2+...+(m_k{)}^2}{k}}}\ge {\displaystyle \frac{m_1+m_2+...+m_k}{k}}\\ & \iff \sqrt{{\displaystyle \frac{(m_1{)}^2+(m_2{)}^2+...+(m_k{)}^2}{k}}}\ge A\\ & \iff {\displaystyle \frac{(m_1{)}^2+(m_2{)}^2+...+(m_k{)}^2}{k}\ge A^2}\\ & \iff (m_1{)}^2+(m_2{)}^2+...+(m_k{)}^2\ge k{A}^2\\ & \iff {\displaystyle \sum _{i=1}^k(m_i{)}^2\ge k{A}^2........(1)}\end{array}\\ & \begin{array}{rl}& \text{Dengan ketaksamaan QM-HM diperoleh juga}\\ & \sqrt{{\displaystyle \frac{{\left(\frac{1}{m_1}\right)}^2+{\left(\frac{1}{m_2}\right)}^2+...+{\left(\frac{1}{m_k}\right)}^2}{k}}}\ge {\displaystyle \frac{k}{m_1+m_2+...+m_k}}\\ & \iff \sqrt{{\displaystyle \frac{{\left(\frac{1}{m_1}\right)}^2+{\left(\frac{1}{m_2}\right)}^2+...+{\left(\frac{1}{m_k}\right)}^2}{k}}}\ge {\displaystyle \frac{1}{A}}\\ & \iff {\displaystyle \frac{{\left(\frac{1}{m_1}\right)}^2+{\left(\frac{1}{m_2}\right)}^2+...+{\left(\frac{1}{m_k}\right)}^2}{k}\ge {\displaystyle \frac{1}{A^2}}}\\ & \iff {\left(\frac{1}{m_1}\right)}^2+{\left(\frac{1}{m_2}\right)}^2+...+{\left(\frac{1}{m_k}\right)}^2\ge {\displaystyle \frac{k}{A^2}}\\ & \iff \sum _{i=1}^k{\left({\displaystyle \frac{1}{m_i}}\right)}^2\ge {\displaystyle \frac{k}{A^2}........(2)}\end{array}\\ & \begin{array}{rl}& \text{Dengan (1) dan (2) akan diperoleh}\\ & {\displaystyle \sum _{i=1}^k{(m_i+{\displaystyle \frac{1}{m_i}})}^2={\displaystyle \sum _{i=1}^k((m_i{)}^2+2+{\left({\displaystyle \frac{1}{m_i}}\right)}^2)}}\\ & ={\displaystyle \sum _{i=1}^k(m_i{)}^2+{\displaystyle \sum _{i=1}^{k}2+{\displaystyle \sum _{i=1}^k{\left(\frac{1}{m_i}\right)}^2}}}\\ & \ge k{A}^2+2k+{\displaystyle \frac{k}{A^2}}\\ & =k(A^2+2+{\displaystyle \frac{1}{A^2}})\\ & =k{(A+{\displaystyle \frac{1}{A}})}^2◼\end{array}\end{array}$.

$\begin{array}{l}\\ 52.& \text{Buktikan bahwa untuk bilangan asli}n>1\\ & \text{berlaku}\sqrt[n]{1+{\displaystyle \frac{\sqrt[n]{n}}{n}}}+\sqrt[n]{1-{\displaystyle \frac{\sqrt[n]{n}}{n}}}<2\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa untuk}\sqrt[n]{1-{\displaystyle \frac{\sqrt[n]{n}}{n}}}\\ & \text{dengan}n>1,\text{maka}0<{\displaystyle \frac{\sqrt[n]{n}}{n}<1}\\ & \text{Dengan ketaksamaan AM-GM dapat diperoleh}\\ & \begin{array}{rl}{\displaystyle \frac{(1+{\displaystyle \frac{\sqrt[n]{n}}{n}})+\underset{\text{sebanyak}(n-1)}{\underbrace{1+1+1+...+1}}}{n}}& >\sqrt[n]{(1+{\displaystyle \frac{\sqrt[n]{n}}{n}})111...1}\\ & =\sqrt[n]{(1+{\displaystyle \frac{\sqrt[n]{n}}{n}})}......(1)\\ {\displaystyle \frac{(1-{\displaystyle \frac{\sqrt[n]{n}}{n}})+\underset{\text{sebanyak}(n-1)}{\underbrace{1+1+1+...+1}}}{n}}& >\sqrt[n]{(1-{\displaystyle \frac{\sqrt[n]{n}}{n}})111...1}\\ & =\sqrt[n]{(1-{\displaystyle \frac{\sqrt[n]{n}}{n}})}......(2)\end{array}\\ & \text{Jika ketaksamaan (1) dan (2) dijumlahkna, maka}\\ & \begin{array}{rl}{\displaystyle \sqrt[n]{1+{\displaystyle \frac{\sqrt[n]{n}}{n}}}+{\displaystyle \sqrt[n]{1-{\displaystyle \frac{\sqrt[n]{n}}{n}}}}}& <{\displaystyle \frac{\underset{\text{sebanyak}(2n-2)}{\underbrace{1+1+1+...+1}}+(1+{\displaystyle \frac{\sqrt[n]{n}}{n}})+(1-{\displaystyle \frac{\sqrt[n]{n}}{n}})}{n}}\\ & ={\displaystyle \frac{\underset{\text{sebanyak}(2n-2)}{\underbrace{1+1+1+...+1}}+1+1}{n}}\\ & =2◼\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 53.& \text{Jika bilangan real positif}x,y,z\text{dengan}\\ & xyz=1.\text{tentukan nilai minimum dari}\\ & 2x^3+12y^2+24z\\ \\ & \mathbf{\text{Jawab}}\\ & \text{Misal}\\ & A=2x^3+12y^2+24z\\ & \iff A=x^3+x^3+4y^2+4y^2+4y^2+\underset{\text{sebanyak}6\text{kali}}{\underbrace{4z+4z+...+4z}}\\ & \begin{array}{rl}& \text{Dengan ketaksamaan AM-GM dapat diperoleh}\\ & {\displaystyle \frac{A}{11}\ge \sqrt[11]{(x^3)(x^3)(4y^2)(4y^2)(4y^2)\underset{\text{sebanyak}6}{\underbrace{(4z)...(4z)}}}}\\ & \iff {\displaystyle \frac{A}{11}\ge \sqrt[11]{(xyz{)}^6(4{)}^9}}\\ & \iff A\ge 11\sqrt[11]{4^9}\end{array}\end{array}$.

$\begin{array}{l}\\ 54.& (\mathbf{\text{OSK 2019}})\\ & \text{Tentukan bilangan real terbesar}M,\\ & \text{sehingga untuk setiap}x\text{positif berlaku}\\ & (x+1)(x+3)(x+5)(x+11)\ge Mx\\ \\ & \mathbf{\text{Jawab}}\\ & \text{Diketahui}\\ & (x+1)(x+3)(x+5)(x+11)\ge Mx\\ & \iff x^4+20x^3+122x^2+268x+165\ge Mx\\ & \text{Dengan AM-GM kita akan mendapatkan}\\ & {\displaystyle \frac{x^4+\stackrel{\text{sebanyak}20}{\overbrace{x^3+x^3+...+x^3}}+\stackrel{\text{sebanyak}122}{\overbrace{x^2+x^2+...+x^2}}+\stackrel{\text{sebanyak}268}{\overbrace{x+x+...+x}}+\stackrel{\text{sebanyak}165}{\overbrace{1+1+...+1}}}{1+20+122+268+165}}\\ & \ge \sqrt[576]{x^4\left(\underset{\text{sebanyak}20}{\underbrace{x^3{x}^3...x^3}}\right)\left(\underset{\text{sebanyak}268}{\underbrace{x^2{x}^2...x^2}}\right)\left(\underset{\text{sebanyak}165}{\underbrace{11...1}}\right)}\\ & \iff {\displaystyle \frac{x^4+20x^3+122x^2+268x+165}{576}\ge \sqrt[576]{x^{576}}}\\ & \iff x^4+20x^3+122x^2+268x+165\ge 576x\\ & \text{Jadi, nilai}M=576\end{array}$.

$\begin{array}{l}\\ 55.& (\mathbf{\text{OMITS 2012}})\\ & \text{Jika diketahui}{x}_1,x_2,x_3,...,x_{2012}\in \left({\displaystyle \frac{1}{4},1}\right),\\ & \text{maka nilai minimum dari bentuk}\\ & {}^{{.}^{x_1}}\log (x_2-{\displaystyle \frac{1}{4}}){+}^{{.}^{x_2}}\log (x_3-{\displaystyle \frac{1}{4}})+...{+}^{{.}^{x_{2012}}}\log (x_1-{\displaystyle \frac{1}{4}})\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & \bullet \text{untuk}:x\ge 1,\text{berlaku}{(x-{\displaystyle \frac{1}{2}})}^2\ge 0\\ & \text{sehingga berlaku juga}(x-{\displaystyle \frac{1}{4}})\le x^2\\ & \bullet \text{untuk}:x_1,x_2,x_3,...,x_{2012}\in \left({\displaystyle \frac{1}{4},1}\right)\\ & \text{kita akan memperoleh fakta bahwa}\\ & {}^{{.}^{x_i}}\log (x_{i+1}-{\displaystyle \frac{1}{4}}){\ge }^{{.}^{x_1}}\log x_{x_{i+1}}^2={2.}^{{.}^{x_1}}\log x_{x_{i+1}}\\ & \text{Selanjutnya}\\ & {}^{{.}^{x_1}}\log (x_2-{\displaystyle \frac{1}{4}}){+}^{{.}^{x_2}}\log (x_3-{\displaystyle \frac{1}{4}})+...{+}^{{.}^{x_n}}\log (x_1-{\displaystyle \frac{1}{4}})\\ & ={\displaystyle \sum _{i=1}^n{.}^{{.}^{x_i}}\log (x_{i+1}-{\displaystyle \frac{1}{4}})\ge 2{\displaystyle \sum _{i=1}^n{.}^{{.}^{x_i}}\log (x_{i+1})=2{\displaystyle \sum _{i=1}^n{\displaystyle \frac{\log x_{i+1}}{\log x_i}}}}}\\ & \text{Dengan AM-GM kita mendapatkan}\\ & {\displaystyle \sum _{i=1}^n{.}^{{.}^{x_i}}\log (x_{i+1}-{\displaystyle \frac{1}{4}})}\\ & \ge 2{\displaystyle \sum _{i=1}^n{\displaystyle \frac{\log x_{i+1}}{\log x_i}\ge 2n.\sqrt[n]{\prod _{i=1}^n{\displaystyle \frac{\log x_{i+1}}{\log x_i}}}}}\\ & =2n.\sqrt[n]{{\displaystyle \frac{\log x_2}{\log x_1}.\frac{\log x_3}{\log x_2}.\frac{\log x_4}{\log x_3}...\frac{\log x_n}{\log x_{n-1}}.\frac{\log x_1}{\log x_n}}}\\ & =2n.1=2n\\ & \text{Sehingga nilai minimum dari}\\ & {}^{{.}^{x_1}}\log (x_2-{\displaystyle \frac{1}{4}}){+}^{{.}^{x_2}}\log (x_3-{\displaystyle \frac{1}{4}})+...{+}^{{.}^{x_{2012}}}\log (x_1-{\displaystyle \frac{1}{4}})\\ & =2n=2(2012)=4024\end{array}\end{array}$. 

DAFTAR PUSTAKA

1. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
2. Muslim, M.S. 2020. Kumpulan Soal dan Pembahasan Olimpiade Matematika SMA Tahun 2007-2019 Tingkat Kota/Kabupaten. Bandung: YRAMA WIDYA.
3. Sidi, A.B. 2010. Aljabar: Alhaqibiyyah Attadribiyyah lita'hil Attullab litasfiyat Oulimbiyat Arriyadliyyat bi Hazakhistan. Saudi Arabia.

Contoh Soal 10 (Segitiga dan Ketaksamaan)

 $\begin{array}{l}\\ 46.& \text{Misalkan}a,b,c\text{bilangan real non negatif}\\ & \text{dengan}a+b+c=1,\text{Tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge {\displaystyle \frac{9}{2}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui}a+b+c=1\\ & \text{Dengan AM-GM kita memiliki}\\ & 2(a+b+c)\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\\ & =((a+b)+(b+c)+(c+a))\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\\ & \ge 3\sqrt[3]{(a+b)(b+c)(c+a)}\times 3\sqrt[3]{{\displaystyle \frac{1}{(a+b)(b+c)(c+a)}}}\\ & =9\sqrt[3]{{\displaystyle \frac{(a+b)(b+c)(c+a)}{(a+b)(b+c)(c+a)}}}=9\sqrt[3]{1}=9\\ & \text{Sehingga}\\ & 2\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge 9\\ & \iff \left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge {\displaystyle \frac{9}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 47.& (\mathbf{\text{OSN 2013}})\\ & \text{Tentukan semua bilangan real}M\\ & \text{sedemikian sehingga untuk sebarang}\\ & \text{bilangan real}a,b,c\text{paling sedikit}\\ & \text{satu di antara tiga bilangan berikut}\\ & a+{\displaystyle \frac{M}{ab},b+{\displaystyle \frac{M}{bc},c+{\displaystyle \frac{M}{ca}}}}\\ & \text{bernilai lebih dari atau sama dengan}\\ & 1+M\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \text{min}\{a+{\displaystyle \frac{M}{ab},b+{\displaystyle \frac{M}{bc},c+{\displaystyle \frac{M}{ca}}}}\}\ge 1+M\\ & \text{Perhatikan bahwa}\\ & a+{\displaystyle \frac{M}{ab}+b+{\displaystyle \frac{M}{bc}+c+{\displaystyle \frac{M}{ca}}}}\\ & =a+b+c+{\displaystyle \frac{M}{ab}+{\displaystyle \frac{M}{bc}+{\displaystyle \frac{M}{ca}}}}\\ & ={\displaystyle \frac{1}{2}(2(a+b+c)+{\displaystyle \frac{2M}{ab}+{\displaystyle \frac{2M}{bc}+{\displaystyle \frac{2M}{ca}}}})}\\ & ={\displaystyle \frac{1}{2}(a+b+{\displaystyle \frac{2M}{ab}+b+c+{\displaystyle \frac{2M}{bc}+c+a+{\displaystyle \frac{2M}{ca}}}})}\\ & \text{Dengan AM-GM akan diperoleh bentuk}\\ & \ge {\displaystyle \frac{1}{2}(3\sqrt[3]{2M}+3\sqrt[3]{2M}+3\sqrt[3]{2M})}\\ & ={\displaystyle \frac{9}{2}\sqrt[3]{2M}}\\ & \text{Pilih nilai minimum}\\ & \{a+{\displaystyle \frac{M}{ab},b+{\displaystyle \frac{M}{bc},c+{\displaystyle \frac{M}{ca}}}}\}=1+M={\displaystyle \frac{3\sqrt[3]{2M}}{2}}\\ & \text{Selanjutnya}\\ & 1+M={\displaystyle \frac{3\sqrt[3]{2M}}{2}}\\ & \iff M={\displaystyle \frac{1}{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 48.& (\mathbf{\text{OSK 2017}})\\ & \text{Diketahui bilangan real positif}a,b,c\\ & \text{yang memenuhi}a+b+c=1.\text{Nilai}\\ & \text{minimum dari}{\displaystyle \frac{a+b}{abc}\text{adalah}....}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}a,b,c\in R^{+}\\ & \text{dengan}a+b+c=1.\text{kita dapat peroleh}\\ & a+b=1-c\\ & \text{Selanjutnya}\\ & {\displaystyle \frac{a+b}{abc}={\displaystyle \frac{a}{abc}+\frac{b}{abc}={\displaystyle \frac{1}{bc}+\frac{1}{ac}}}}\\ & \text{Sebelumnya ingat ketaksamaan AM-HM}\\ & {\displaystyle \frac{m+n}{2}\ge {\displaystyle \frac{2}{{\displaystyle \frac{1}{m}+\frac{1}{n}}}\iff {\displaystyle \frac{1}{m}+\frac{1}{n}\ge {\displaystyle \frac{4}{m+n}}}}}\\ & \text{Sehingga}\\ & {\displaystyle \frac{a+b}{abc}={\displaystyle \frac{1}{bc}+\frac{1}{ac}\ge {\displaystyle \frac{4}{ac+bc}={\displaystyle \frac{4}{c(a+b)}}}}}\\ & \ge {\displaystyle \frac{4}{c(1-c)}={\displaystyle \frac{4}{c-c^2}={\displaystyle \frac{4}{{\displaystyle \frac{1}{4}-{\displaystyle \frac{1}{4}+c-c^2}}}}}}\\ & ={\displaystyle \frac{4}{{\displaystyle \frac{1}{4}-{(c-{\displaystyle \frac{1}{2}})}^2}}}\\ & \text{Saat}c-{\displaystyle \frac{1}{2}=0,\text{maka akan diperoleh}}\\ & \text{nilai minimum yaitu}:\\ & {\left({\displaystyle \frac{a+b}{abc}}\right)}_{\text{minimum}}={\displaystyle \frac{4}{{\displaystyle \frac{1}{4}-0}}=16}\end{array}\end{array}$.

$\begin{array}{l}\\ 49.& \text{Jika pada soal 48 diubah dengan}\\ & a,b,c\text{adalah sisi segitiga}ABC\\ & \text{yang memenuhi}a+b+c=1.\\ & \text{Tentukan nilai dari}{\displaystyle \frac{a+b}{abc}}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}a,b,c\text{sisi}△ABC\\ & \text{dengan}a+b+c=1.\text{kita dapat peroleh}\\ & a+b=a+b\\ & \text{Ingat bahwa dalam}△ABC,\text{berlaku}\\ & \{\begin{array}{ll}\bullet & a+b>c\\ \bullet & a+c>b\\ \bullet & b+c>a\end{array}\\ & \text{maka}a+b+a+b>a+b+c\\ & \iff 2(a+b)>1\iff a+b>{\displaystyle \frac{1}{2}}\\ & \text{Dan dengan AM-GM diperoleh}\\ & {\displaystyle \frac{a+b+c}{3}\ge \sqrt[3]{abc}\iff {\displaystyle \frac{1}{3}\ge \sqrt[3]{abc}}}\\ & \iff {\displaystyle \frac{1}{27}\ge abc}\\ & \text{Selanjutnya}\\ & {\displaystyle \frac{a+b}{abc}>{\displaystyle \frac{\frac{1}{2}}{\frac{1}{27}}={\displaystyle \frac{27}{2}=13,5}}}\\ & \text{Jadi, nilai}{\displaystyle \frac{a+b}{abc}>13,5}\end{array}\end{array}$.

$\begin{array}{l}\\ 50.& \text{Diketahui bilangan real positif}{x}_1,x_2,...,x_k\\ & \text{memenuhi persamaan}{x}_1+x_2+...+x_k=1\\ & \text{Buktikan bahwa}\\ & {\displaystyle \frac{1}{(x_1{)}^{2013}}+\frac{1}{(x_2{)}^{2013}}+...+\frac{1}{(x_k{)}^{2013}}\ge k^{2014}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Dengan ketaksamaan AM-GM kita}\\ & \text{memiliki}\\ & {\displaystyle \frac{x_1+x_2+...+x_k}{k}\ge \sqrt[k]{x_1.x_2...x_k}}\\ & \iff {\displaystyle \frac{1}{k}\ge \sqrt[k]{x_1.x_2...x_k}}\\ & \iff \sqrt[k]{x_1.x_2...x_k}\le {\displaystyle \frac{1}{k}}\\ & \iff x_1.x_2...x_k\le {\left({\displaystyle \frac{1}{k}}\right)}^k\\ & \text{Selanjutnya}\\ & {\displaystyle \frac{{\displaystyle \frac{1}{(x_1{)}^{2013}}+\frac{1}{(x_2{)}^{2013}}+...+\frac{1}{(x_k{)}^{2013}}}}{k}}\\ & \ge \sqrt[k]{{\displaystyle \frac{1}{(x_1{)}^{2013}.(x_2{)}^{2013}...(x_k{)}^{2013}}}}\\ & {\displaystyle \frac{1}{(x_1{)}^{2013}}+\frac{1}{(x_2{)}^{2013}}+...+\frac{1}{(x_k{)}^{2013}}}\\ & \ge k\sqrt[k]{{\displaystyle \frac{1}{(x_1{)}^{2013}.(x_2{)}^{2013}...(x_k{)}^{2013}}}}\\ & \ge k\sqrt[k]{{\displaystyle \frac{1}{(x_1.x_2...x_k{)}^{2013}}}}\ge k.\sqrt[k]{{\displaystyle \frac{1}{{\left({\left({\displaystyle \frac{1}{k}}\right)}^k\right)}^{2013}}}}\\ & \ge k.\sqrt[k]{k^{2013k}}\\ & \ge k.k^{2013}\\ & \ge k^{2014}◼\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
2. Muslim, M.S. 2020. Kumpulan Soal dan Pembahasan Olimpiade Matematika SMA Tahun 2007-2019 Tingkat Kota/Kabupaten. Bandung: YRAMA WIDYA.
3. Widodo, T. 2013. Pembahasan OSN Matematika SMA Tahun 2013 Seleksi Tingkat Nasional. 

Contoh Soal 9 (Segitiga dan Ketaksamaan)

$\begin{array}{l}\\ 41.& \text{Jika}a,b,c>0,\text{tunjukkan bahwa}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge {\displaystyle \frac{3}{2}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& ((a+b)+(b+c)+(c+a))\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge 9\\ & \iff 2(a+b+c)\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge 9\\ & \iff (a+b+c)\left({\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\right)\ge {\displaystyle \frac{9}{2}}\\ & \iff {\displaystyle \frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\ge {\displaystyle \frac{9}{2}}}\\ & \iff 1+{\displaystyle \frac{c}{a+b}+1+{\displaystyle \frac{a}{b+c}+1+{\displaystyle \frac{b}{c+a}\ge {\displaystyle \frac{9}{2}}}}}\\ & \iff 3+{\displaystyle \frac{c}{a+b}+{\displaystyle \frac{a}{b+c}+{\displaystyle \frac{b}{c+a}\ge {\displaystyle \frac{9}{2}}}}}\\ & \iff {\displaystyle \frac{c}{a+b}+{\displaystyle \frac{a}{b+c}+{\displaystyle \frac{b}{c+a}\ge {\displaystyle \frac{9}{2}-3}}}}\\ & \iff {\displaystyle \frac{c}{a+b}+{\displaystyle \frac{a}{b+c}+{\displaystyle \frac{b}{c+a}\ge {\displaystyle \frac{3}{2}◼}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 42.& \text{Jika}a,b,c\text{adalah sisi-sisi segitiga}\\ & \text{ABC, tunjukkan bahwa}\\ & {\displaystyle \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\ge 3}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa dengan AM-HM diperoleh}\\ & (a+b+c)\left({\displaystyle \frac{1}{b+c-a}+\frac{1}{c+a-b}+\frac{1}{a+b-c}}\right)\ge 9\\ & \iff ((b+c-a)+(c+a-b)+(a+b-c))\\ & \times \left({\displaystyle \frac{1}{b+c-a}+\frac{1}{c+a-b}+\frac{1}{a+b-c}}\right)\ge 9\\ & \iff {\displaystyle \frac{a+b+c}{b+c-a}+\frac{a+b+c}{c+a-b}+\frac{a+b+c}{a+b-c}\ge 9}\\ & \iff 1+{\displaystyle \frac{2a}{b+c-a}+1+{\displaystyle \frac{2b}{c+a-b}+1+{\displaystyle \frac{2c}{a+b-c}\ge 9}}}\\ & \iff 3+{\displaystyle \frac{2a}{b+c-a}+{\displaystyle \frac{2b}{c+a-b}+{\displaystyle \frac{2c}{a+b-c}\ge 9}}}\\ & \iff {\displaystyle \frac{2a}{b+c-a}+{\displaystyle \frac{2b}{c+a-b}+{\displaystyle \frac{2c}{a+b-c}\ge 6}}}\\ & \iff {\displaystyle \frac{a}{b+c-a}+{\displaystyle \frac{b}{c+a-b}+{\displaystyle \frac{c}{a+b-c}\ge 3◼}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 43.& \text{Diketahui bilangan real positif}a,b,c\\ & \text{dengan}abc=1.\text{Buktikan bahwa}\\ & {\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\ge 3}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}abc=1\\ & \text{Perhatikan bahwa}\\ & \{\begin{array}{ll}\bullet & {\displaystyle \frac{1+ab}{1+a}+{\displaystyle \frac{abc+ab}{1+a}=ab\left({\displaystyle \frac{1+c}{1+a}}\right)}}\\ \bullet & {\displaystyle \frac{1+bc}{1+b}+{\displaystyle \frac{abc+bc}{1+b}=bc\left({\displaystyle \frac{1+a}{1+b}}\right)}}\\ \bullet & {\displaystyle \frac{1+ca}{1+c}+{\displaystyle \frac{abc+ca}{1+c}=ac\left({\displaystyle \frac{1+b}{1+c}}\right)}}\end{array}\\ & \text{Jika hasilnya dijumlahkan},\text{maka}\\ & \text{kita akan mendapatkan hasil}\\ & ab\left({\displaystyle \frac{1+c}{1+a}}\right)+bc\left({\displaystyle \frac{1+a}{1+b}}\right)+ac\left({\displaystyle \frac{1+b}{1+c}}\right)\\ & \text{Selanjutnya dengan AM-GM akan diperoleh}\\ & {\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}}\\ & =ab\left({\displaystyle \frac{1+c}{1+a}}\right)+bc\left({\displaystyle \frac{1+a}{1+b}}\right)+ac\left({\displaystyle \frac{1+b}{1+c}}\right)\ge 3\sqrt[3]{(abc{)}^2}\\ & \iff {\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\ge 3\sqrt[3]{(abc{)}^2}}\\ & \iff {\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\ge 3\sqrt[3]{1}}\\ & \iff {\displaystyle \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\ge 3◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 44.& (\mathbf{\text{SEAMO III-Malaysia}})\\ & \text{Misalkan}a,b,c\text{bilangan real positif}\\ & \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge {\displaystyle \frac{3}{1+abc}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & {\displaystyle \frac{1}{a(1+b)}+\frac{1}{1+abc}=\frac{1}{1+abc}\left({\displaystyle \frac{1+a}{a(1+b)}+\frac{b(1+c)}{(1+b)}}\right)}\\ & \text{Sehingga}\\ & {\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc}}\\ & ={\displaystyle \frac{1}{1+abc}\left({\displaystyle \frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a}}\right)}\\ & \text{Secara AM-GM kita mendapatkan}\\ & \left({\displaystyle \frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a}}\right)\ge 6\\ & \text{Kembali ke soal}\\ & {\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc}\ge {\displaystyle \frac{1}{1+abc}(6)}}\\ & \iff {\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge {\displaystyle \frac{3}{1+abc}◼}}\end{array}\end{array}$.

$\begin{array}{l}\\ 45.& \text{Misalkan}a,b,c\text{bilangan real non negatif}\\ & \text{dengan}a+b+c=1,\text{Tunjukkan bahwa}\\ & {\displaystyle \frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}\ge {\displaystyle \frac{9}{10}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui}a+b+c=1\\ & \text{Dengan AM-GM kita memiliki}\\ & {\displaystyle \frac{a+b+c}{3}\ge \sqrt[3]{abc}\iff {\displaystyle \frac{1}{3}\ge \sqrt[3]{abc}\iff {\displaystyle \frac{1}{27}\ge abc}}}\\ & \text{Sehingga}\\ & {\displaystyle \frac{1}{1+3abc}\ge {\displaystyle \frac{1}{1+3\left({\displaystyle \frac{1}{27}}\right)}\ge {\displaystyle \frac{1}{1+{\displaystyle \frac{1}{9}}}\ge {\displaystyle \frac{9}{10}}}}}\\ & \text{Selanjutnya dengan perluasan AM-HM}\\ & \text{Sebagaimana bentuk berikut}\\ & \left({\displaystyle \frac{p_1{a}_1+p_2{a}_2+p_3{a}_3}{p_1+p_2+p_3}}\right)\ge \left({\displaystyle \frac{p_1+p_2+p_3}{p_1{a}_1^{-1}+p_2{a}_2^{-1}+p_3{a}_3^{-1}}}\right)\\ & \text{maka}\\ & {\displaystyle \frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}\ge {\displaystyle \frac{1}{a(1+bc)+b(1+ac)+c(1+ab)}}}\\ & \ge {\displaystyle \frac{1}{a+b+c+3abc}}\\ & \ge {\displaystyle \frac{1}{1+3abc}}\\ & \ge {\displaystyle \frac{9}{10}◼}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Bambang, S. 2012. Materi, Soal dan Penyelesaian Olimpiade Matematika Tingkat SMA/MA. Jakarta: BINA PRESTASI INSANI.
2. Bintari, N., Gunarto, D. 2007. Panduan Menguasai Soal-Soal Olimpiade Matematika Nasional dan Internasional. Yogyakarta: INDONESIA CERDAS.
3. Tung, K.Y. 2013. Ayo Raih Medali Emas Olimpiade Matematika SMA. Yogyakarta: ANDI.

Contoh Soal 8 (Segitiga dan Ketaksamaan)

 $\begin{array}{l}\\ 36.& \text{Jika}a,b,c>0,\text{dengan}a+b+c=1\\ & \text{tunjukkan bahwa}\\ & (1-a)(1-b)(1-c)\ge 8abc\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & \{\begin{array}{ll}1-a& =b+c\\ 1-b& =a+c\\ 1-c& =a+b\end{array}\\ & \text{Kurang lebih seperti pembuktian}\\ & \text{pada no.35 di atas dengan tetap}\\ & \text{menggunakan AM-GM akan}\\ & \text{didapatkan}\\ & \bullet a+b\ge 2\sqrt{ab}\\ & \bullet b+c\ge 2\sqrt{bc},\text{dan}\\ & \bullet c+a\ge 2\sqrt{ca}\\ & \text{Selanjutnya}\\ & (1-a)(1-b)(1-c)=(b+c)(a+c)(a+b)\\ & \ge 2\sqrt{bc}\times 2\sqrt{ac}\times 2\sqrt{ab}\\ & \ge 8\sqrt{a^2{b}^2{c}^2}\\ & \ge 8abc◼\end{array}\end{array}$.

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

$\begin{array}{l}\\ 38.& \text{Diketahui}a,b,c\text{bilangan real positif}\\ & \text{dan}(a+1)(b+1)(c+1)=8.\\ & \text{tunjukkan bahwa}abc<1\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & (a+1)(b+1)(c+1)=8\\ & abc+(ab+bc+ca)+(a+b+c)+1=8\\ & \text{Dengan AM-GM kita akan mendapatkan}\\ & abc+3(abc{)}^{{.}^{\frac{2}{3}}}+3(abc{)}^{{.}^{\frac{1}{3}}}+1<8\\ & \iff {((abc{)}^{{.}^{\frac{1}{3}}}+1)}^3<2^3\\ & \iff (abc{)}^{{.}^{\frac{1}{3}}}+1<2\\ & \iff (abc{)}^{{.}^{\frac{1}{3}}}<1◼\end{array}\end{array}$.

$\begin{array}{l}\\ 39.& \text{Jika}a,b,c>0\\ & \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{a}{a^2+1}+{\displaystyle \frac{b}{b^2+1}+{\displaystyle \frac{c}{c^2+1}\le {\displaystyle \frac{3}{2}}}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & (a-1{)}^2\ge 0\\ & \iff a^2-2a+1\ge 0\\ & \iff a^2+1\ge 2a\\ & {\displaystyle \frac{1}{2}\ge {\displaystyle \frac{a}{a^2+1}}}\\ & \text{atau}\\ & {\displaystyle \frac{a}{a^2+1}\le {\displaystyle \frac{1}{2}..................(1)}}\\ & \text{dengan cara yang sama akan}\\ & \text{pula jiika}b\text{dan}c\text{dikondisikan}\\ & \text{akan didapatkan ketaksamaan}\\ & {\displaystyle \frac{b}{b^2+1}\le {\displaystyle \frac{1}{2}..................(2)}}\\ & {\displaystyle \frac{c}{c^2+1}\le {\displaystyle \frac{1}{2}..................(3)}}\\ & \text{Jika ketaksamaan}(1),(2),\mathrm{\&}(3)\\ & \text{dijumlahkan akan menghasilkan}\\ & {\displaystyle \frac{a}{a^2+1}+{\displaystyle \frac{b}{b^2+1}+{\displaystyle \frac{c}{c^2+1}\le {\displaystyle \frac{1}{2}+{\displaystyle \frac{1}{2}+{\displaystyle \frac{1}{2}}}}}}}\\ & \le 3\left({\displaystyle \frac{1}{2}}\right)\\ & \le {\displaystyle \frac{3}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 40.& \text{Jika}a,b,c\text{adalah sisi-sisi segitiga}\\ & \text{ABC, tunjukkan bahwa}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa pada segitiga}\\ & \text{ABC berlaku}\\ & \{\begin{array}{ll}a+b& >c\\ a+c& >b\\ b+c& >a\end{array}\\ & \text{Misalkan}2s=a+b+c,\text{maka}\\ & a+b=a+b\\ & \iff a+b+a+b>a+b+c\\ & \iff 2(a+b)>2s\\ & \iff (a+b)>s\\ & \text{Demikian juga akan berlaku}\\ & \{\begin{array}{ll}b+c& >s\\ c+a& >s\end{array}\\ & \text{Sehingga}\\ & \{\begin{array}{ll}{\displaystyle \frac{a}{b+c}}& <{\displaystyle \frac{a}{s}}\\ {\displaystyle \frac{b}{c+a}}& <{\displaystyle \frac{b}{s}}\\ {\displaystyle \frac{c}{a+b}}& <{\displaystyle \frac{c}{s}}\end{array}\\ & \text{Selanjutnya kita kembali ke soal}\\ & {\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<{\displaystyle \frac{a+b+c}{s}}}\\ & <{\displaystyle \frac{2s}{s}}\\ & <2◼\end{array}\end{array}$.

Contoh Soal 7 (Segitiga dan Ketaksamaan)

$\begin{array}{l}\\ 31.& (\mathbf{\text{Russia 1992}})\\ & \text{Jika}a,b>1,\text{buktikan bahwa}\\ & {\displaystyle \frac{x^2}{y-1}+{\displaystyle \frac{y^2}{x-1}\ge 8}}\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Alternatif 1}\\ & \begin{array}{rl}& \text{Gunakan AM-GM untuk mendapatkan}\\ & {\displaystyle \frac{x^2}{y-1}+{\displaystyle \frac{y^2}{x-1}\ge 2\sqrt{{\displaystyle \frac{x^2{y}^2}{\sqrt{(x-1)(y-1)}}}}}}\\ & \ge {\displaystyle \frac{2xy}{\sqrt{(x-1)(y-1)}}}\\ & \text{Sebelum kita lanjutkan, ingat bahwa}\\ & (x-2{)}^2\ge 0\iff x^2-4x+4\ge 0\\ & \iff x^2\ge 4x-4\iff x^2\ge 4(x-1)\\ & \iff {\displaystyle \frac{x^2}{x-1}\ge 4\iff {\displaystyle \frac{x}{\sqrt{x-1}}\ge 2}}\\ & \text{Demikian juga}{\displaystyle \frac{y}{\sqrt{y-1}}\ge 2}\\ & \text{Selanjutnya kembali ke semula yaitu}\\ & \ge {\displaystyle \frac{2xy}{\sqrt{(x-1)(y-1)}}}\\ & \ge {\displaystyle 2.2.2}\\ & {\displaystyle \frac{x^2}{y-1}+{\displaystyle \frac{y^2}{x-1}\ge 8◼}}\end{array}\\ & \text{Alternatif 2}\\ & \begin{array}{rl}& \text{Misalkan saja}\\ & \{\begin{array}{ll}a& =x-1\Rightarrow x=a+1\\ b& =y-1\Rightarrow y=b+1\end{array}\\ & \text{Maka}\\ & {\displaystyle \frac{x^2}{y-1}+{\displaystyle \frac{y^2}{x-1}={\displaystyle \frac{(a+1{)}^2}{b}+{\displaystyle \frac{(b+1{)}^2}{a}}}}}\\ & \text{Dengan AM-GM akan diperoleh}\\ & {\displaystyle \frac{(a+1{)}^2}{b}+{\displaystyle \frac{(b+1{)}^2}{a}\ge 4\left({\displaystyle \frac{a}{b}+{\displaystyle \frac{b}{a}}}\right)}}\\ & \text{Bentuk di atas didapatkan dari}\\ & (a-1{)}^2\ge 0\iff a^2-2a+1\ge 0\\ & \iff a^2+2a-4a+1\ge 0\\ & \iff a^2+2a+1\ge 4a\iff (a+1{)}^2\ge 4a\\ & \text{Demikian juga yabf satunya, yaitu}\\ & (b+1{)}^2\ge 4b.\text{Serta bentuk}\\ & \left({\displaystyle \frac{a}{b}+{\displaystyle \frac{b}{a}}}\right)\ge 2\sqrt{{\displaystyle \frac{a}{b}.\frac{b}{s}}}=2\\ & \text{Selanjutnya kembali ke soal, yaitu}:\\ & {\displaystyle \frac{(a+1{)}^2}{b}+{\displaystyle \frac{(b+1{)}^2}{a}\ge 4\left({\displaystyle \frac{a}{b}+{\displaystyle \frac{b}{a}}}\right)}}\\ & \ge 4(2)\\ & \ge 8◼\end{array}\end{array}$.

$\begin{array}{l}\\ 32.& \text{Untuk}a>0,\text{tentukan nilai }\\ & \text{minimum dari bentuk}{\displaystyle \frac{4a^2+8a+13}{6+6a}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui}a>0,\text{maka bentuk}\\ & {\displaystyle \frac{4a^2+8a+13}{6+6a}={\displaystyle \frac{4(a+1{)}^2+9}{6(a+1)}}}\\ & ={\displaystyle \frac{2(a+1)}{3}+\frac{3}{2(a+1)}}\\ & \text{Dengan AM-GM akan diperoleh}\\ & {\displaystyle \frac{2(a+1)}{3}+\frac{3}{2(a+1)}}\\ & \ge 2\sqrt{{\displaystyle \frac{2}{3}\times \frac{3}{2}\times \frac{(a+1)}{(a+1)}}}=2\\ & \text{Jadi},{\displaystyle \frac{2(a+1)}{3}+\frac{3}{2(a+1)}\ge 2}\end{array}\end{array}$.

$\begin{array}{l}\\ 33.& \text{Untuk}0\le a<6,\text{tentukan nilai }\\ & \text{minimum dari bentuk}:a{(6-a)}^2\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui}0\le a<6,\text{maka bentuk}\\ & a{(6-a)}^2={\displaystyle \frac{1}{2}(2a)(6-a)(6-a)}\\ & \text{Dengan GM-AM akan kita peroleh}\\ & {\displaystyle \frac{1}{2}(2a)(6-a)(6-a)\le {\displaystyle \frac{1}{2}{\left({\displaystyle \frac{2a+6-a+6-a}{3}}\right)}^3}}\\ & \le {\displaystyle \frac{1}{2}{\left({\displaystyle \frac{12}{3}}\right)}^3}\\ & \le {\displaystyle \frac{1}{2}{\left(4\right)}^3}\\ & \le {\displaystyle \frac{1}{2}\left(64\right)}\\ & \le 32\end{array}\end{array}$.

$\begin{array}{l}\\ 34.& \text{Jika}a,b>0,\text{tunjukkan bahwa}\\ & (a+1)(b+1)(ab+1)\ge 8ab\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa dengan AM-GM}\\ & \bullet a+1\ge 2\sqrt{a}\\ & \bullet b+1\ge 2\sqrt{b},\text{dan}\\ & \bullet ab+1\ge 2\sqrt{ab}\\ & \text{Selanjutnya}\\ & (a+1)(b+1)(ab+1)\ge 2\sqrt{a}\times 2\sqrt{b}\times 2\sqrt{ab}\\ & \ge 8\sqrt{a^2{b}^2}\\ & \ge 8ab◼\end{array}\end{array}$.

$\begin{array}{l}\\ 35.& \text{Jika}a,b,c>0,\text{tunjukkan bahwa}\\ & (a+b)(b+c)(c+a)\ge 8abc\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Kurang lebih seperti pembuktian}\\ & \text{pada no.34 di atas dengan tetap}\\ & \text{menggunakan AM-GM akan}\\ & \text{didapatkan}\\ & \bullet a+b\ge 2\sqrt{ab}\\ & \bullet b+c\ge 2\sqrt{bc},\text{dan}\\ & \bullet c+a\ge 2\sqrt{ca}\\ & \text{Selanjutnya}\\ & (a+b)(b+c)(c+a)\ge 2\sqrt{ab}\times 2\sqrt{bc}\times 2\sqrt{ca}\\ & \ge 8\sqrt{a^2{b}^2{c}^2}\\ & \ge 8abc◼\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Bambang, S. 2012. Materi, Soal dan Penyelesaian Olimpiade Matematika Tingkat SMA/MA. Jakarta: BINA PRESTASI INSANI.
2. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
3. Manfrino, R.B., dkk. 2009. Inequalities A Mathematical Olympiad Approach. Basel: Birkhauser Verlag AG.