Ketaksamaan Schur

4. Ketaksamaan Schur

(Bagian Pertama)

Misal $a,b,c$ adalah bilangan real positif dan $r$ adalah bilangan positif, maka pertidaksamaan berikut berlaku

$\begin{array}{r}{a}^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b)\ge 0\end{array}$.

Atau dapat pula dituliskan dengan lebih sederhana

${\displaystyle \sum _{\text{siklik}}^{.}{a}^r(a-b)(a-c)\ge 0}$.

Kesamaan terjadi ketika $a=b=c$ atau jika dua di antara $a,b,c$ bernilai sama dan nilai yang lain sama dengan nol.

Bukti:

$\begin{array}{rl}& \text{Asumsikan dengan tanpa mengurangi keumuman}\\ & \text{yaitu}a\ge b\ge c,\text{maka kita akan mendapatkan}\\ & (a-b)(a^r(a-c)-b^r(b-c))+c^r(c-a)(c-b)\ge 0◼\end{array}$.

Cukup jelas bahwa ruas kiri bernilai tak negatif.

Misalkan untuk $r=1$, didapatkan hasil berikut,:

$\begin{array}{rl}& a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\\ & =a^3+b^3+c^3-(a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b)+3abc\\ & ={\displaystyle \sum _{\text{sym}}^{.}(a^3-2a^{2}b+abc)\ge 0}\end{array}$.

$\begin{array}{|l|}\hline \begin{array}{rl}& \text{Sebagai pengingat bahwa}\\ & \begin{array}{rl}1.& \sum _{sym}^{.}{a}^3=2a^3+2b^3+2c^3\\ 2.& \sum _{sym}^{.}{a}^{2}b=a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2\\ 3.& \sum _{sym}^{.}abc=6abc\end{array}\end{array}\\ \hline\end{array}$.

Selanjutnya saat  $r=1$, kita bisa mendaptkan

$\begin{array}{rl}1.& a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ca(c+a)\\ 2.& abc\ge (a+b-c)(b+c-a)(c+a-b)\\ 3.& (a+b+c{)}^3+9abc\ge 4(a+b+c)(ab+bc+ca)\end{array}$.

$\begin{array}{rl}& \text{Selanjutnya, masih saat}r=1\\ & \bullet \text{untuk}c=0,\text{maka}(a-b)(a^{k+1}-b^{k+1})\ge 0\\ & \bullet \text{untuk}b=c=0,\text{maka}{a}^{k+2}\ge 0\\ & \bullet \text{untuk}b=c,\text{maka}{a}^k(a-c{)}^2\ge 0\end{array}$.

Dan saat  $r=2$, kita bisa mendaptkan

$\begin{array}{rl}{a}^4+& b^4+c^4+abc(a+b+c)\ge ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)\end{array}$.

$\text{CONTOH SOAL}$.

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

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

$\begin{array}{l}\\ 3.& \mathbf{\text{(IMO 2000)}}\\ & \text{Jika}a,b,c\text{bilangan real positif dengan}\\ & abc=1,\text{tunjukkan bahwa}\\ & (a+1-{\displaystyle \frac{1}{b}})(b+1-{\displaystyle \frac{1}{c}})(c+1-{\displaystyle \frac{1}{a}})\le 1\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Misalkan}a={\displaystyle \frac{x}{y},b=\frac{y}{z},\text{dan}c={\displaystyle \frac{z}{x},\text{maka}}}\\ & \left({\displaystyle \frac{x}{y}+1-\frac{z}{y}}\right)\left({\displaystyle \frac{y}{z}+1-\frac{x}{z}}\right)\left({\displaystyle \frac{z}{x}+1-\frac{y}{x}}\right)\le 1\\ & \iff \left({\displaystyle \frac{x+y-z}{y}}\right)\left({\displaystyle \frac{y+z-x}{z}}\right)\left({\displaystyle \frac{z+x-y}{x}}\right)\le 1\\ & \iff (x+y-z)(y+z-x)(z+x-y)\le xyz,\text{atau}\\ & \iff xyz\ge (x+y-z)(y+z-x)(z+x-y)\\ & \text{Bentuk terakhir memenuhi bentuk kedua dari}\\ & \mathbf{\text{Ketaksamaan Schur}}\text{saat}r=1.\\ & \text{Jadi},\\ & (a+1-{\displaystyle \frac{1}{b}})(b+1-{\displaystyle \frac{1}{c}})(c+1-{\displaystyle \frac{1}{a}})\le 1◼\end{array}$.

$\begin{array}{l}\\ 4.& \mathbf{\text{(IMO 1983)}}\\ & \text{Jika}a,b,c\text{adalah panjang sisi-sisi segitiga}\\ & \text{tunjukkan bahwa}\\ & a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\ge 0\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Pada sebuah segitiga dengan sisi}a,b,c\\ & \text{berlaku}\{\begin{array}{ll}a+b>c& \Rightarrow a>c-b\\ a+c>b& \Rightarrow c>b-c\\ b+c>a& \Rightarrow b>a-c\end{array}\\ & \text{Sehingga untuk ketaksamaan pada soal}\\ & a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(x-a)\\ & \ge a^2(a-c)(a-b)+b^2(b-c)(b-c)+c^2(c-b)(c-a)\ge 0\\ & \text{Bentuk terakhir memenuhi bentuk dari}\\ & \mathbf{\text{Ketaksamaan Schur}}\text{saat}r=2.\\ & \text{Jadi},\\ & a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\ge 0◼\end{array}$.

$\begin{array}{l}\\ 5.& \text{Misalkan}a,b,c\text{bilangan real positif dengan}\\ & a+b+c=2,\text{tunjukkan bahwa}\\ & a^4+b^4+c^4+abc\ge a^3+b^3+c^3\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Schur}}\text{saat}r=2\\ & \text{kita memiliki}\\ & a^2(a-b)(a-c)+b^2(b-a)(b-c)+c^2(c-a)(c-b)\ge 0\\ & \iff a^4+b^4+c^4+abc(a+b+c)-a^3(b+c)-b^3(a+c)-c^3(a+b)\ge 0\\ & \iff a^4+b^4+c^4+abc(a+b+c)\ge a^3(b+c)+b^3(a+c)+c^3(a+b)\\ & \iff a^4+b^4+c^4+abc(a+b+c)\ge (a^3+b^3+c^3)(a+b+c)-(a^4+b^4+c^4)\\ & \iff 2(a^4+b^4+c^4)+abc(a+b+c)\ge (a^3+b^3+c^3)(a+b+c)\\ & \iff 2(a^4+b^4+c^4)+abc(2)\ge (a^3+b^3+c^3)(2)\\ & \iff a^4+b^4+c^4+abc\ge a^3+b^3+c^3◼\\ \\ & \text{Bentuk di atas kadang dituliskan dengan bentuk}\\ & \text{berikut}:\\ & \begin{array}{rl}& \text{Dengan}\mathbf{\text{Ketaksamaan Schur}}\text{saat}r=2\\ & \text{kita memiliki}\\ & {\displaystyle \sum _{\text{siklik}}^{.}{a}^2(a-b)(a-c)\ge 0}\\ & \iff {\displaystyle \sum _{\text{siklik}}^{.}{a}^4+abc{\displaystyle \sum _{\text{siklik}}^{.}a-{\displaystyle \sum _{\text{siklik}}^{.}{a}^3(b+c)\ge 0}}}\\ & \iff {\displaystyle \sum _{\text{siklik}}^{.}{a}^4+abc{\displaystyle \sum _{\text{siklik}}^{.}a\ge {\displaystyle \sum _{\text{siklik}}^{.}{a}^3(b+c)}}}\\ & \iff {\displaystyle \sum _{\text{siklik}}^{.}{a}^4+abc{\displaystyle \sum _{\text{siklik}}^{.}a\ge \left({\displaystyle \sum _{\text{siklik}}^{.}{a}^3}\right)\left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)-\left({\displaystyle \sum _{\text{siklik}}^{.}{a}^4}\right)}}\\ & \iff 2{\displaystyle \sum _{\text{siklik}}^{.}{a}^4+abc{\displaystyle \sum _{\text{siklik}}^{.}a\ge \left({\displaystyle \sum _{\text{siklik}}^{.}{a}^3}\right)\left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)}}\\ & \iff {\displaystyle \sum _{\text{siklik}}^{.}{a}^4+abc\ge \left({\displaystyle \sum _{\text{siklik}}^{.}{a}^3}\right)◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& (\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}}\\ & \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}$.

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. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.