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}$.

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

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

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

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