Lanjutan Ketaksamaan Schur

(Bagian Kedua)

Bagian Pertama silahkan klik di sini

1. Penyederhanaan dengan pola siklik dan simetri

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

2. Bentuk ketaksamaan berdasar nilai r

Masih ingat kita pada ketaksamaan Schur saat $r=1$, yaitu,:

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

Selanjutnya saat  $r=1$, kita bisa mendaptkan

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

Dan saat  $r=2$, kita bisa mendaptkan

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

3. Beberapa formulasi bantu

$\begin{array}{rl}1.& (a+b+c)(a^2+b^2+c^2)\\ & =a^3+b^3+c^3+a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)\\ 2.& (a+b+c)(a^2+b^2+c^2)\\ & =a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)\\ 3.& (a+b+c)(ab+ac+bc)\\ & =3abc+ab(a+b)+ac(a+c)+bc(b+c)\\ 4.& (a+b+c)(ab+ac+bc)\\ & =3abc+a^2(b+c)+b^2(a+c)+c^2(a+b)\\ 5.& (a+b+c{)}^3+3abc\\ & =a^3+b^3+c^3+3(a+b+c)(ab+ac+bc)\\ 6.& (a+b)(a+c)(b+c)\\ & =2abc++a^2(b+c)+b^2(a+c)+c^2(a+b)\\ 7.& (a+b-c)(a+c-b)(b+c-a)\\ & =-2abc-(a^3+b^3+c^3)+a^2(b+c)+b^2(a+c)+c^2(a+b)\\ 8.& (a-b)(b-c)(c-a)\\ & =a^2(c-b)+b^2(a-c)+c^2(b-a)\end{array}$.

4. Penyederhanaan ketaksamaan metode pqr

$\begin{array}{|ll|}\hline .\mathbf{\text{Kesamaan}}& \mathbf{\text{Ketaksamaan}}\\ \begin{array}{rl}1.& {\displaystyle \sum _{sik}^{.}{x}^2=p^2-2q}\\ 2.& {\displaystyle \sum _{sik}^{.}{x}^3=p(p^2-3q)+3r}\\ 3.& \sum _{sik}^{.}{x}^2{y}^2=q^2-2pr\\ 4.& \prod (x+y)=pq-r\\ 5.& \sum _{sik}^{.}xy(x+y)=pq-3r\\ 6.& \sum _{sik}^{.}{x}^2(y+z)=pq-3r\\ 7.& \prod (1+x)=1+p+q+r\\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}1.& pq\ge 9r\\ 2.& p^2\ge 3q\\ 3.& q^2\ge 3pr\\ 4.& p^3\ge 27r\\ 5.& q^3\ge 27r^2\\ 6.& p^{3}r\ge q^3\\ 7.& p^3+9r\ge 4pq\\ 8.& 2p^3+9r\ge 7pq\\ 9.& 2p^3+27r\ge 9pq\\ 10.& 2p^3+9r^2\ge 7pqr\\ 11.& q^3+9r^2\ge 4pqr\\ 12.& 2q^3+27r^2\ge 9pqr\\ 13.& p^4+3q^2\ge 4p^{2}q\\ 14.& p^4+4q^2+6pr\ge 5p^{2}q\\ 15.& p^{2}q+3pr\ge 4q^2\\ 16.& p{q}^2\ge 2p^{2}r+3qr\\ 17.& p^2{q}^2+12r^2\ge 4p^{3}r+pqr\end{array}\\ \hline\end{array}$.

5. Ketaksamaan Schur bentuk pqr

Perhatikan poin 2 di atas saat $r=1$, kitaa akan medapatkan bentuk berikut ini:

$\begin{array}{rl}& (a+b+c{)}^3+9abc\ge 4(a+b+c)(ab+bc+ca)\\ & \iff p^3+9r\ge 4pq\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1& \text{Untuk}a,b,c\text{tak negatif, tunjukkan bahwa}\\ & (a+b+c)(ab+ac+bc)\ge 9abc\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Misalkan}\\ & p=a+b+c,q=ab+ac+bc,\\ & \text{dan}r=abc\\ & \text{Untuk pembuktian pernyataan di atas}\\ & \text{dengan AM-GM kita memiliki}\\ & \bullet a+b+c\ge 3\sqrt[3]{abc}\\ & \bullet ab+ac+bc\ge 3\sqrt[3]{(abc{)}^2}\\ & \text{maka hasil dari}\\ & (a+b+c)(ab+ac+bc)\ge 3\sqrt{abc}.3\sqrt[3]{(abc{)}^2}\\ & \iff pq\ge 3\sqrt[3]{r}.3\sqrt[3]{r^2}\\ & \iff pq\ge 9\sqrt[3]{r^3}\\ & \iff pq\ge 9r◼\end{array}$.

$\begin{array}{l}\\ 2& \text{Untuk}a,b,c\text{tak negatif, tunjukkan bahwa}\\ & (a+b+c{)}^2\ge 3(ab+ac+bc)\\ \\ & \mathbf{\text{Bukti}}\\ & \mathbf{\text{Alternatif 1}}\\ & \text{Misalkan}\\ & p=a+b+c,q=ab+ac+bc,\\ & \text{dan}r=abc\\ & \text{Untuk pembuktian pernyataan di atas}\\ & \text{dengan AM-GM kita memiliki}\\ & \bullet a^2+b^2+c^2\ge ab+ac+bc\\ & \text{Dan juga sebuah kesamaan}\\ & \bullet a^2+b^2+c^2=(a+b+c{)}^2-2(ab+ac+bc)\\ & \text{maka dari kedua bentuk di atas kita akan}\\ & \text{dapatkan bentuk}\\ & a^2+b^2+c^2\ge ab+ac+bc\\ & \iff (a+b+c{)}^2-2(ab+ac+bc)\ge ab+ac+bc\\ & \iff (a+b+c{)}^2\ge 3(ab+ac+bc)\\ & \iff p^2\ge 3q◼\\ & \begin{array}{rl}& \mathbf{\text{Alternatif 2}}\\ & \text{Telah kita ketahui bahwa}\\ & (a+b+c{)}^2=a^2+b^2+c^2+2(ab+ac+bc)\\ & \text{Dengan ketaksamaan}\mathbf{\text{renata}}\text{kita akan}\\ & \text{dapatkan bentuk}\\ & (a+b+c{)}^2\ge ab+bc+ca+2(ab+ac+bc)\\ & \iff (a+b+c{)}^2\ge 3(ab+ac+bc)\\ & \iff p^2\ge 3q◼\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Untuk}a,b,c\text{tak negatif, tunjukkan}\\ & \text{bahwa}(ab+ac+bc{)}^2\ge 3abc(a+b+c)\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Misalkan}\\ & p=a+b+c,q=ab+ac+bc,\\ & \text{dan}r=abc\\ & \text{Perhatikan kesamaan berikut}\\ & \begin{array}{rl}& (ab+ac+bc{)}^2=a^2{b}^2+a^2{c}^2+b^2{c}^2+2abc(a+b+c)\\ & \text{Dengan AM-GM dan GM-AM kita dapatkan}\\ & \iff (ab+ac+bc{)}^2\ge 3\sqrt[3]{a^4{b}^4{c}^4}+2abc(a+b+c)\\ & \iff (ab+ac+bc{)}^2\ge 3abc\sqrt[3]{abc}+2abc(a+b+c)\\ & \iff (ab+ac+bc{)}^2\ge abc(a+b+c)+2abc(a+b+c)\\ & \iff (ab+ac+bc{)}^2\ge 3abc(a+b+c)\\ & \iff q^2\ge 3pr◼\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Untuk}a,b,c\text{tak negatif, tunjukkan bahwa}\\ & (a+b+c{)}^3\ge 27abc\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Misalkan}\\ & p=a+b+c,q=ab+ac+bc,\\ & \text{dan}r=abc\\ & \text{Untuk pembuktian pernyataan di atas}\\ & \text{dengan AM-GM kita memiliki}\\ & \bullet a+b+c\ge 3\sqrt[3]{abc}\\ & \text{Masing-masing ruas pangkatkan 3, maka}\\ & (a+b+c{)}^3\ge 27abc\iff p^3\ge 27r◼\end{array}$.

$\begin{array}{l}\\ 5.& \text{Untuk}a,b,c\text{tak negatif, tunjukkan bahwa}\\ & (a+b+c{)}^3\ge 27abc\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Misalkan}\\ & p=a+b+c,q=ab+ac+bc,\\ & \text{dan}r=abc\\ & \text{Untuk pembuktian pernyataan di atas}\\ & \text{dengan AM-GM kita memiliki}\\ & \bullet ab+ac+bc\ge 3\sqrt[3]{(abc{)}^2}\\ & \text{Masing-masing ruas pangkatkan 3, maka}\\ & (ab+ac+bc{)}^3\ge 27(abc{)}^2\iff q^3\ge 27r^2◼\end{array}$.

$\begin{array}{l}\\ 6.& \text{Misalkan}x,y,x\text{bilangan real positif dengan}\\ & \text{tunjukkan bahwa}\\ & a^2+b^2+c^2+2abc+1\ge 2(ab+acb+c)\\ & (\mathbf{\text{Darij Grinberg}})\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Dengan}\mathbf{\text{ketaksamaan AM-GM}}\text{dan}\\ & \text{dilanjutkan dengan}\mathbf{\text{ketaksamaan Schur}}\\ & \text{serta menggesernya ke ruas kiri, maka}\\ & a^2+b^2+c^2+2abc+1-2(ab+ac+bc)\\ & \ge a^2+b^2+c^2+3(abc{)}^{{}^{\frac{2}{3}}}+1\ge 2(ab+ac+bc)\\ & \ge {((a{)}^{{}^{\frac{2}{3}}})}^3+{((b{)}^{{}^{\frac{2}{3}}})}^3+{((c{)}^{{}^{\frac{2}{3}}})}^3+3(abc{)}^{{}^{\frac{2}{3}}}-2(ab+ac+bc)\\ & \ge a^{{.}^{\frac{2}{3}}}{b}^{{.}^{\frac{2}{3}}}(a^{{.}^{\frac{2}{3}}}+b^{{.}^{\frac{2}{3}}})+a^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}(a^{{.}^{\frac{2}{3}}}+c^{{.}^{\frac{2}{3}}})\\ & +b^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}(b^{{.}^{\frac{2}{3}}}+c^{{.}^{\frac{2}{3}}})-2(ab+ac+bc)\\ & \ge a^{{.}^{\frac{2}{3}}}{b}^{{.}^{\frac{2}{3}}}(a^{{.}^{\frac{2}{3}}}+b^{{.}^{\frac{2}{3}}})+a^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}(a^{{.}^{\frac{2}{3}}}+c^{{.}^{\frac{2}{3}}})\\ & +b^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}(b^{{.}^{\frac{2}{3}}}+c^{{.}^{\frac{2}{3}}})-2(ab+ac+bc)\\ & =a^{{.}^{\frac{2}{3}}}{b}^{{.}^{\frac{2}{3}}}{(a^{{.}^{\frac{2}{3}}}-b^{{.}^{\frac{2}{3}}})}^2+a^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}{(a^{{.}^{\frac{2}{3}}}-c^{{.}^{\frac{2}{3}}})}^2\\ & +b^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}{(b^{{.}^{\frac{2}{3}}}-c^{{.}^{\frac{2}{3}}})}^2\ge 0◼\end{array}$.

$\begin{array}{l}\\ 7.& (\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}$.

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

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

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

DAFTAR PUSTAKA

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

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.