Lanjutan Ketaksamaan Schur

(Bagian Kedua)

Bagian Pertama silahkan klik di sini

1. Penyederhanaan dengan pola siklik dan simetri

$.\quad\begin{aligned}&\color{red}\textrm{Mengenal penulisan pola}\: \textbf{Siklik dan Simetri}\\ &\textrm{Misal untuk}\: \: n=3,\: \: \textrm{pada penulisan unsur}\\ &x,y,\: \: \textrm{dan}\: \: z,\: \textrm{maka}\\ &\begin{array}{|l|l|}\hline \textbf{Pola Siklik}&\textbf{Pola Simetri}\\\hline\begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}=x^{2}+y^{2}+z^{2}\\ &\\ &\\ &\\ &\\ & \end{aligned} &\begin{aligned}\displaystyle\sum_{\textrm{sym}}^{.}x^{2}&=x^{2}+x^{2}\\ &+y^{2}+y^{2}\\ \\ &+z^{2}+z^{2}\\ \\ &=2\left (x^{2}+y^{2}+z^{2}  \right ) \end{aligned}\\ \begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}x^{3}=x^{3}+y^{3}+z^{3}\end{aligned}&\begin{aligned}\displaystyle \sum_{\textrm{sym}}^{.}x^{3}=2\left (x^{3}+y^{3}+z^{3}  \right ) \end{aligned}\\ \begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y=x^{2}y+y^{2}z+z^{2}x\\ &\\ &\\ & \end{aligned}&\begin{aligned}\displaystyle \sum_{\textrm{sym}}^{.}x^{2}y&=x^{2}y+x^{2}z\\ &+y^{2}x+y^{2}z\\\\ &+z^{2}x+z^{2}y \end{aligned}\\ \begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}xyz&=xyz+yzx+zxy\\ &=3xyz \end{aligned}&\begin{aligned}\displaystyle \sum_{\textrm{sym}}^{.}xyz&=xyz+xzy+\cdots \\ &=6xyz \end{aligned}  \\\hline \end{array} \end{aligned}$

2. Bentuk ketaksamaan berdasar nilai r

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

$\begin{aligned}&a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\\ &=a^{3}+b^{3}+c^{3}-\left( a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b \right)+3abc\\ &=\displaystyle \sum_{\textrm{sym}}^{.}\left( a^3-2a^{2}b+abc \right)\ge 0 \end{aligned}$.

Selanjutnya saat  $\color{red}r=1$, kita bisa mendaptkan

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

Dan saat  $\color{red}r=2$, kita bisa mendaptkan

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

3. Beberapa formulasi bantu

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

4. Penyederhanaan ketaksamaan metode pqr

$\begin{array}{|l|l|}\hline .\qquad\qquad\textbf{Kesamaan}&\quad \qquad\textbf{Ketaksamaan}\\\hline\begin{aligned}1.\quad&\displaystyle \sum_{sik}^{.}x^{2}=p^{2}-2q\\ 2.\quad&\displaystyle \sum_{sik}^{.}x^{3}=p(p^{2}-3q)+3r\\ 3.\quad&\sum_{sik}^{.}x^{2}y^{2}=q^{2}-2pr\\ 4.\quad&\prod (x+y)=pq-r\\ 5.\quad&\sum_{sik}^{.}xy(x+y)=pq-3r\\ 6.\quad&\sum_{sik}^{.}x^{2}(y+z)=pq-3r\\ 7.\quad&\prod (1+x)=1+p+q+r\\ &\\ &\\ &\\ &\\ & \end{aligned} &\begin{aligned}1.\quad&pq\geq 9r\\ 2.\quad&p^{2}\geq 3q\\ 3.\quad&q^{2}\geq 3pr\\ 4.\quad&p^{3}\geq 27r\\ 5.\quad&q^{3}\geq 27r^{2}\\ 6.\quad&p^{3}r\geq q^{3}\\ 7.\quad&p^{3}+9r\geq 4pq\\ 8.\quad&2p^{3}+9r\geq 7pq\\ 9.\quad&2p^{3}+27r\geq 9pq\\ 10.\quad&2p^{3}+9r^{2}\geq 7pqr\\ 11.\quad&q^{3}+9r^{2}\geq 4pqr\\ 12.\quad&2q^{3}+27r^{2}\geq 9pqr\\ 13.\quad&p^{4}+3q^{2}\geq 4p^{2}q\\ 14.\quad&p^{4}+4q^{2}+6pr\geq 5p^{2}q\\ 15.\quad&p^{2}q+3pr\geq 4q^{2}\\ 16.\quad&pq^{2}\geq 2p^{2}r+3qr\\ 17.\quad&p^{2}q^{2}+12r^{2}\geq 4p^{3}r+pqr \end{aligned}  \\\hline \end{array}$.

5. Ketaksamaan Schur bentuk pqr

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

$\begin{aligned}&(a+b+c)^{3}+9abc\geq 4(a+b+c)(ab+bc+ca)\\ &\Leftrightarrow \color{red}p^{3}+9r\geq 4pq \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1&\textrm{Untuk}\: \: a,b,c\: \: \textrm{tak negatif, tunjukkan bahwa}\\ &(a+b+c)(ab+ac+bc)\geq 9abc\\\\ &\textbf{Bukti}\\ &\textrm{Misalkan}\\ &p=a+b+c,\: q=ab+ac+bc,\\ &\textrm{dan}\: \: r=abc\\ &\textrm{Untuk pembuktian pernyataan di atas}\\ &\textrm{dengan AM-GM kita memiliki}\\ &\bullet\quad a+b+c\geq 3\sqrt[3]{abc}\\ &\bullet\quad ab+ac+bc\geq 3\sqrt[3]{(abc)^{2}}\\ &\textrm{maka hasil dari}\\ &(a+b+c)(ab+ac+bc)\geq 3\sqrt{abc}.3\sqrt[3]{(abc)^{2}}\\ &\Leftrightarrow pq\geq 3\sqrt[3]{r}.3\sqrt[3]{r^{2}}\\ &\Leftrightarrow pq\geq 9\sqrt[3]{r^{3}}\\ &\Leftrightarrow \color{red}pq\geq 9r\qquad \color{black}\blacksquare   \end{array}$.

$\begin{array}{ll}\\ 2&\textrm{Untuk}\: \: a,b,c\: \: \textrm{tak negatif, tunjukkan bahwa}\\ &(a+b+c)^{2}\geq 3(ab+ac+bc)\\\\ &\textbf{Bukti}\\ &\color{blue}\textbf{Alternatif 1}\\ &\textrm{Misalkan}\\ &p=a+b+c,\: q=ab+ac+bc,\\ &\textrm{dan}\: \: r=abc\\ &\textrm{Untuk pembuktian pernyataan di atas}\\ &\textrm{dengan AM-GM kita memiliki}\\ &\bullet\quad a^{2}+b^{2}+c^{2}\geq ab+ac+bc\\ &\textrm{Dan juga sebuah kesamaan}\\ &\bullet \quad a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+ac+bc)\\ &\textrm{maka dari kedua bentuk di atas kita akan}\\ &\textrm{dapatkan bentuk}\\ &a^{2}+b^{2}+c^{2}\geq ab+ac+bc\\ &\Leftrightarrow (a+b+c)^{2}-2(ab+ac+bc)\geq ab+ac+bc\\ &\Leftrightarrow (a+b+c)^{2}\geq 3(ab+ac+bc)\\ &\Leftrightarrow  \color{red}p^{2}\geq 3q\qquad \color{black}\blacksquare\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &\textrm{Telah kita ketahui bahwa}\\ &(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+ac+bc)\\ &\textrm{Dengan ketaksamaan}\: \textbf{renata}\: \textrm{kita akan}\\ &\textrm{dapatkan bentuk}\\ &(a+b+c)^{2}\geq ab+bc+ca+2(ab+ac+bc)\\ &\Leftrightarrow (a+b+c)^{2}\geq 3(ab+ac+bc)\\ &\Leftrightarrow  \color{red}p^{2}\geq 3q\qquad \color{black}\blacksquare  \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Untuk}\: \: a,b,c\: \: \textrm{tak negatif, tunjukkan}\\ &\textrm{bahwa}\: \: (ab+ac+bc)^{2}\geq 3abc(a+b+c)\\\\ &\textbf{Bukti}\\ &\textrm{Misalkan}\\ &p=a+b+c,\: q=ab+ac+bc,\\ &\textrm{dan}\: \: r=abc\\ &\textrm{Perhatikan kesamaan berikut}\\ &\begin{aligned}&(ab+ac+bc)^{2}=a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}+2abc(a+b+c)\\  &\textrm{Dengan AM-GM dan GM-AM kita dapatkan}\\ &\Leftrightarrow (ab+ac+bc)^{2}\geq 3\sqrt[3]{a^{4}b^{4}c^{4}}+2abc(a+b+c)\\ &\Leftrightarrow (ab+ac+bc)^{2}\geq 3abc\sqrt[3]{abc}+2abc(a+b+c)\\ &\Leftrightarrow (ab+ac+bc)^{2}\geq abc(a+b+c)+2abc(a+b+c)\\ &\Leftrightarrow (ab+ac+bc)^{2}\geq 3abc(a+b+c)\\ &\Leftrightarrow \color{red}q^{2}\geq 3pr\qquad \color{black}\blacksquare  \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Untuk}\: \: a,b,c\: \: \textrm{tak negatif, tunjukkan bahwa}\\ &(a+b+c)^{3}\geq 27abc\\\\ &\textbf{Bukti}\\ &\textrm{Misalkan}\\ &p=a+b+c,\: q=ab+ac+bc,\\ &\textrm{dan}\: \: r=abc\\ &\textrm{Untuk pembuktian pernyataan di atas}\\ &\textrm{dengan AM-GM kita memiliki}\\ &\bullet\quad a+b+c\geq 3\sqrt[3]{abc}\\ &\textrm{Masing-masing ruas pangkatkan 3, maka}\\ &(a+b+c)^{3}\geq 27abc\Leftrightarrow \color{red}p^{3}\geq 27r\qquad \color{black}\blacksquare  \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Untuk}\: \: a,b,c\: \: \textrm{tak negatif, tunjukkan bahwa}\\ &(a+b+c)^{3}\geq 27abc\\\\ &\textbf{Bukti}\\ &\textrm{Misalkan}\\ &p=a+b+c,\: q=ab+ac+bc,\\ &\textrm{dan}\: \: r=abc\\ &\textrm{Untuk pembuktian pernyataan di atas}\\ &\textrm{dengan AM-GM kita memiliki}\\ &\bullet\quad ab+ac+bc\geq 3\sqrt[3]{(abc)^{2}}\\ &\textrm{Masing-masing ruas pangkatkan 3, maka}\\ &(ab+ac+bc)^{3}\geq 27(abc)^{2}\Leftrightarrow \color{red}q^{3}\geq 27r^{2}\qquad \color{black}\blacksquare  \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad a^{2}+b^{2}+c^{2}+2abc+1\geq 2(ab+acb+c)\\&\qquad\qquad\qquad\qquad\qquad (\textbf{Darij Grinberg})\\\\ &\textbf{Bukti}\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\: \: \textrm{dan}\\ &\textrm{dilanjutkan dengan}\: \: \textbf{ketaksamaan Schur}\\ &\textrm{serta menggesernya ke ruas kiri, maka}\\ &a^{2}+b^{2}+c^{2}+2abc+1- 2(ab+ac+bc)\\ &\geq a^{2}+b^{2}+c^{2}+3(abc)^{^{\frac{2}{3}}}+1\geq 2(ab+ac+bc)\\ &\geq \left ((a)^{^{\frac{2}{3}}}  \right )^{3}+\left ((b)^{^{\frac{2}{3}}}  \right )^{3}+\left ((c)^{^{\frac{2}{3}}}  \right )^{3}+3(abc)^{^{\frac{2}{3}}}-2(ab+ac+bc)\\ &\geq a^{.^{\frac{2}{3}}}b^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +b^{.^{\frac{2}{3}}} \right )+a^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )\\ &\quad +b^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (b^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )-2(ab+ac+bc)\\ &\geq a^{.^{\frac{2}{3}}}b^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +b^{.^{\frac{2}{3}}} \right )+a^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )\\ &\quad +b^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (b^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )-2(ab+ac+bc)\\ &= a^{.^{\frac{2}{3}}}b^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} -b^{.^{\frac{2}{3}}} \right )^{2}+a^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} -c^{.^{\frac{2}{3}}} \right )^{2}\\ &\quad +b^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (b^{.^{\frac{2}{3}}} -c^{.^{\frac{2}{3}}} \right )^{2}\geq 0\qquad \blacksquare \end{array}$.

$\begin{array}{ll}\\ 7.&(\textbf{APMO 2004})\\ &\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad (x^{2}+2)(y^{2}+2)(z^{2}+2)\geq 9(xy+yz+zx)\\\\ &\textbf{Bukti}\\ &\color{blue}\textbf{Alternatif 1}\\ &\textrm{Dengan menjabarkan akan didapatkan}\\ &x^{2}y^{2}z^{2}+2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}+4\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+8\geq 9\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\textrm{Perhatikan bahwa}\\ &\bullet \quad\color{red}2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}-4\displaystyle \sum_{\textrm{siklik}}^{.}xy+6=2\displaystyle \sum_{\textrm{siklik}}^{.}(xy-1)^{2}\geq 0\\ &\bullet  \quad \color{red}\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}\geq \displaystyle \sum_{\textrm{siklik}}^{.}xy\: \: \color{black}\textrm{atau}\: \: \color{red}x^{2}+y^{2}+z^{2}\geq xy+xz+yz\\ &\textrm{Kita cukup membuktikan bahwa}\\ & x^{2}y^{2}z^{2}+\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+2\geq 2\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\Leftrightarrow  x^{2}y^{2}z^{2}+2\geq \displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\begin{aligned}&\textrm{Untuk}\: \: a,b,c\: \: \textrm{bilangan real positif},\\ &\textbf{Ketaksamaan Schur}\: \textrm{saat}\: \: \color{red}r=1\\ &\textrm{memberikan}\\ &\displaystyle \sum_{\textrm{siklik}}^{.}a^{3}+3abc\geq \displaystyle \sum_{\textrm{siklik}}^{.}a^{2}b+\displaystyle \sum_{\textrm{siklik}}^{.}ab^{2}\\ &\qquad\qquad\qquad =ab(a+b)+bc(b+c)+ca(c+a)\\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan AM-GM}\: \: \textrm{didapakan}\\ &\displaystyle \sum_{\textrm{siklik}}^{.}a^{3}+3abc\geq 2\displaystyle \sum_{\textrm{siklik}}^{.}(ab)^{\frac{3}{2}}\\ &\textrm{Pilih}\: \: a=x^{\frac{2}{3}},\: b=y^{\frac{2}{3}},\: c=z^{\frac{2}{3}},\: \textrm{maka didapatkan}\\ &(x^{\frac{2}{3}})^{3}+(y^{\frac{2}{3}})^{3}+(z^{\frac{2}{3}})^{3}+3(xyz)^{\frac{2}{3}}\geq 2(xy+yz+zx)\\ &\textrm{Selanjutnya kita selesaikan ini}\: ,\: x^{2}y^{2}z^{2}+2\geq \displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\Leftrightarrow  x^{2}y^{2}z^{2}+2 \geq 3(xyz)^{\frac{2}{3}}\\ &\textrm{Misalkan}\: \: (xyz)^{\frac{2}{3}}=t,\: \textrm{maka}\\ &t^{3}+2\geq 3t\Leftrightarrow t^{3}-3t+2\geq 0\\ &(t-1)^{2}(t+2)\geq 0\: \: \textrm{adalah hal benar} \end{aligned}   \end{array}$.

$.\: \quad\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &x^{2}y^{2}z^{2}+2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}+4\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+8\geq 9\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\textrm{atau dalam bentuk utuhnya, yaitu}\\ &x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+4(x^{2}+y^{2}+z^{2})+8\geq 9(xy+xz+yz)\\ &\textrm{Sekarang kira uraikan satu persatu bagian}\\ &\bullet\quad x^{2}y^{2}z^{2}+1+1\geq 3\sqrt[3]{(xyz)^{2}}\geq \displaystyle \frac{9abc}{a+b+c}=\frac{9r}{p}\\ &\qquad \textrm{ingat bahwa}\: \: \textbf{jika ada}\: \: \displaystyle \frac{9r}{p}\geq 4q-p^{2}\\ &\qquad =4(xy+xz+yz)-(x+y+z)^{2}\\ &\qquad \textrm{adalah}\: \: \textbf{ketaksamaan Schur saat}\: \: \color{red}r=1\\ &\bullet \quad x^{2}y^{2}+1+x^{2}z^{2}+1+y^{2}z^{2}+1\geq 2(xy+xz+yz)\\ &\bullet\quad  x^{2}+y^{2}+z^{2}\geq xy+xz+yz\\ &\qquad \textrm{keduanya didapat dengan ketaksamaan}\: \: \textbf{AM-GM}\\&x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+4(x^{2}+y^{2}+z^{2})+8\\ &=x^{2}y^{2}z^{2}+2+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}+3)+4(x^{2}+y^{2}+z^{2})\\ &\geq 4(xy+xz+yz)-(x+y+z)^{2}+4(xy+xz+yz)+4(xy+xz+yz)\\ &\geq 12(xy+xz+yz)-(x+y+z)^{2}\\ &\geq 12(xy+xz+yz)-3(xy+xz+yz)\\ &=9(xy+xz+yz)\qquad \blacksquare    \end{aligned}$.

$.\: \quad\begin{aligned}&\color{blue}\textbf{Alternatif 3}\\ &(x^{2}+2)(y^{2}+2)(z^{2}+2)- 9(xy+yz+zx)\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\\ &=x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})\\ &\quad +4(x^{2}+y^{2}+z^{2})+8- 9(xy+xz+yz)\\ &=4(x^{2}+y^{2}+z^{2})+2\left ((x^{2}y^{2}+1) +(x^{2}z^{2}+1)+(y^{2}z^{2}+1) \right )\\ &\quad +(x^{2}y^{2}z^{2}+1)+1-9(xy+xz+yz)\\ &\geq 4(x^{2}+y^{2}+z^{2})+4(xy+xz+yz)\\ &\quad +2xyz+1-9(xy+xz+yz)\\ &=(x^{2}+y^{2}+z^{2})+3(x^{2}+y^{2}+z^{2})\\ &\quad +2xyz+1-5(x^{2}+y^{2}+z^{2})\\ &\geq x^{2}+y^{2}+z^{2}+3(xy+xz+yz)\\ &+2xyz-5(xy+xz+yz)\\ &=x^{2}+y^{2}+z^{2}+2xyz+1-2(xy+xz+yz)\geq 0\\ &\textrm{adalah benar dengan bukti ada pada}\\ &\textrm{nomor soal sebelumnya}  \end{aligned}$.

$\begin{array}{ll}\\ 8.&\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad 2(a^{2}+b^{2}+c^{2})+abc+8\geq 5(a+b+c)\\&\qquad\qquad\qquad\qquad\qquad (\textbf{Tran Nam Dung})\\\\ &\textbf{Bukti}\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\: \: \textrm{dan}\\ &\textrm{menggeser ke ruas kiri dan masing-masing}\\ &\textrm{serta mengalikan semunya dengan 6, maka}\\ &12(a^{2}+b^{2}+c^{2})+6abc+48- 30(a+b+c)\\ &= 12(a^{2}+b^{2}+c^{2})+3(2abc+1)+45- 5.2.3(a+b+c)\\ &\geq 2(a^{2}+b^{2}+c^{2})+9\sqrt[3]{(abc)^{2}}+45- 5\left ((a+b+c)^{2}+9  \right )\\ &=12(a^{2}+b^{2}+c^{2})+\displaystyle \frac{9abc}{\sqrt[3]{abc}}-5\left ((a^{2}+b^{2}+c^{2})+2(ab+ac+bc)  \right )\\ &=7(a^{2}+b^{2}+c^{2})+ \displaystyle \frac{9abc}{\sqrt[3]{abc}}-10(ab+ac+bc)\\ &\geq  7(a^{2}+b^{2}+c^{2})+\displaystyle \frac{27abc}{a+b+c}-10(ab+ac+bc)\\ &\begin{aligned}&\textrm{dengan}\: \: \textbf{ketaksamaan Schur},\: \: \textrm{yaitu}:\\ &p^{3}+9r\geq 4pq\Leftrightarrow \displaystyle \color{red}\frac{9r}{p}\geq 4q-p^{2}\\ &\textrm{maka ketaksamaan akan menjadi}\\ &\geq 7(a^{2}+b^{2}+c^{2})+\color{blue}3(4q-p^{2})-10q\\ &\geq 7(a^{2}+b^{2}+c^{2})+2q-3p^{2}\\ &=7(a^{2}+b^{2}+c^{2})+2(ab+ac+bc)-3(a+b+c)^{2}\\ &=7(a^{2}+b^{2}+c^{2})+2q-3\left ((a^{2}+b^{2}+c^{2})+2q  \right )\\ &=4(a^{2}+b^{2}+c^{2})+2q-6q\\ &=4(a^{2}+b^{2}+c^{2})-4q\\ &=4(a^{2}+b^{2}+c^{2})-4(ab+ac+bc)\\ &=4(a^{2}+b^{2}+c^{2}-ab-ac-bc)\geq 0\quad \blacksquare  \end{aligned} \end{array}$.

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

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

Atau dapat pula dituliskan dengan lebih sederhana

$\displaystyle \sum_{\textrm{siklik}}^{.}a^{r}(a-b)(a-c)\geq 0$.

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

Bukti:

$\begin{aligned}&\textrm{Asumsikan dengan tanpa mengurangi keumuman}\\ &\textrm{yaitu}\: \: \color{red}a\ge b\ge c,\:\: \color{black}\textrm{maka kita akan mendapatkan}\\ &(a-b)\left( a^{r}(a-c)-b^{r}(b-c) \right)+c^{r}(c-a)(c-b)\ge 0\qquad \blacksquare \end{aligned}$.

Cukup jelas bahwa ruas kiri bernilai tak negatif.

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

$\begin{aligned}&a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\\ &=a^{3}+b^{3}+c^{3}-\left( a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b \right)+3abc\\ &=\displaystyle \sum_{\textrm{sym}}^{.}\left( a^3-2a^{2}b+abc \right)\ge 0 \end{aligned}$.

$\begin{array}{|l|}\hline \begin{aligned}&\color{red}\textrm{Sebagai pengingat bahwa}\\ &\begin{aligned}1.\quad&\sum_{sym}^{.}a^{3}=2a^{3}+2b^{3}+2c^{3}\\ 2.\quad &\sum_{sym}^{.}a^{2}b=a^{2}b+ab^{2}+a^{2}c+ac^{2}+b^{2}c+bc^{2}\\ 3.\quad &\sum_{sym}^{.}abc=6abc \end{aligned} \end{aligned}\\\hline \end{array}$.

Selanjutnya saat  $\color{red}r=1$, kita bisa mendaptkan

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

$\begin{aligned}&\textrm{Selanjutnya, masih saat}\: \: \color{red}r=1\\ &\bullet \quad \textrm{untuk}\: \: c=0,\: \textrm{maka}\: \: (a-b)(a^{k+1}-b^{k+1})\geq 0\\ &\bullet \quad \textrm{untuk}\: \: b=c=0,\: \textrm{maka}\: \: a^{k+2}\geq 0\\ &\bullet \quad \textrm{untuk}\: \: b=c,\: \textrm{maka}\: \: a^{k}(a-c)^{2}\geq 0 \end{aligned}$.

Dan saat  $\color{red}r=2$, kita bisa mendaptkan

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

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Buktikan bahwa setiap bilangan real}\\ &\textrm{positif}\: \: a,\: b\: \: \textrm{dan}\: \: c\: \: \textrm{berlaku}\\ & a^{2}+b^{2}+c^{2}\geq ab+ac+bc\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\color{blue}\textrm{Alternatif 1}\\ &\textrm{Perhatikan bahwa}\: \: (a-b)^{2}\geq 0\\  &(a-c)^{2}\geq 0,\: \: \textrm{dan}\: \: (b-c)^{2}\geq 0\\ &\textrm{adalah benar, maka}\\ &(a-b)^{2}=a^{2}-2ab+b^{2}\geq 0\\ &\Leftrightarrow a^{2}+b^{2}\geq 2ab\: .....(1)\\ &\textrm{Dengan cara yang kurang lebih sama}\\ &\textrm{akan didapatkan}\\ &\bullet \quad a^{2}+c^{2}\geq 2ac\: .....(2)\\ &\bullet \quad b^{2}+c^{2}\geq 2bc\: .....(1)\\ &\textrm{Jika ketaksamaan}\quad (1),(2), \& \: (3)\: \: \textrm{dijumlahkan}\\ &\textrm{akan didapatkan bentuk}\\ &2a^{2}+2b^{2}+2c^{2}\geq 2ab+2ac+2bc\\ &\Leftrightarrow \: a^{2}+b^{2}+c^{2}\geq ab+ac+bc\quad \blacksquare\\ &\color{blue}\textrm{Alternatif 2}\\ &\textrm{Dengan ketaksamaan}\: \: \color{red}\textbf{Cauchy-Schwarz}\\ &(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+a^{2})\geq (ab+bc+ca)^{2}\\ &\Leftrightarrow a^{2}+b^{2}+c^{2}\geq ab+ac+bc\qquad \blacksquare\\ &\color{blue}\textrm{Alternatif 3}\\  &\textrm{Untuk barisan}\: \: (a,b,c),\: \textrm{asumsikan}\: a\geq b\geq c\\ &\textrm{maka dengan}\: \: \color{red}\textbf{Ketaksamaan Renata}\: \: \color{black}\textrm{diperoleh}\\ &a.a+b.b+c.c\geq ab+bc+ca\\ &a^{2}+b^{2}+c^{2}\geq ab+ac+bc\qquad \blacksquare \\  &\color{blue}\textrm{Alternatif 4}\\ &\textrm{Dengan}\: \: \color{red}\textbf{Ketaksamaan Schur}\: \: \color{black}\textrm{saat}\: \: \color{red}r=0\color{black},\\ &\textrm{yaitu}\\ &a^{r}(a-b)(a-c)+b^{r}(b-a)(b-c)+c^{r}(c-a)(c-b)\geq 0\\ &\Leftrightarrow a^{0}(a-b)(a-c)+b^{0}(b-a)(b-c)+c^{0}(c-a)(c-b)\geq 0\\ &\Leftrightarrow (a-b)(a-c)+(b-a)(b-c)+(c-a)(c-b)\geq 0\\ &\Leftrightarrow a^{2}+b^{2}+c^{2}-ab-ac-bc\geq 0\\ &\Leftrightarrow a^{2}+b^{2}+c^{2}\geq ab+ac+bc\qquad \blacksquare    \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textbf{(IMO 1964)}\\ &\textrm{Jika}\: \: a,b,c\: \: \textrm{adalah panjang sisi-sisi segitiga}\\ &\textrm{tunjukkan bahwa}\\ &\quad a^{2}(b+c-a)+b^{2}(a+c-b)+c^{2}(a+b-c)\leq 3abc\\\\ &\textbf{Bukti}:\\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan Schur}\: \: \textrm{saat}\: \: \color{red}r=1.\\ &a^{3}+b^{3}+c^{3}+3abc\geq ab(a+b)+bc(b+c)+ca(c+a)\\ &\Leftrightarrow 3abc\geq ab(a+b)-a^{3}+bc(b+c)-b^{3}+ca(c+a)-c^{3}\\ &\Leftrightarrow 3abc\geq a^{2}b+ab^{2}-c^{3}+b^{2}c+bc^{2}-b^{3}+c^{2}a+ca^{2}-c^{3}\\ &\Leftrightarrow 3abc\geq a^{2}b+ca^{2}-a^{3}+ab^{2}+b^{2}c-b^{3}+c^{2}a+bc^{2}-c^{3}\\ &\Leftrightarrow  3abc\geq a^{2}(b+c-a)+b^{2}(a+c-b)+c^{2}(a+b-c)\\ &\textrm{atau}\\ &\Leftrightarrow a^{2}(b+c-a)+b^{2}(a+c-b)+c^{2}(a+b-c)\leq 3abc\quad \blacksquare   \end{array}$ .

$\begin{array}{ll}\\ 3.&\textbf{(IMO 2000)}\\ &\textrm{Jika}\: \: a,b,c\: \: \textrm{bilangan real positif dengan}\\ &abc=1,\: \: \textrm{tunjukkan bahwa}\\ &\quad \left (a+1- \displaystyle \frac{1}{b} \right )\left (b+1- \displaystyle \frac{1}{c} \right )\left (c+1- \displaystyle \frac{1}{a} \right )\leq 1\\\\ &\textbf{Bukti}:\\ &\textrm{Misalkan}\: \: a=\displaystyle \frac{x}{y},\: b=\frac{y}{z},\: \: \textrm{dan}\: \: c=\displaystyle \frac{z}{x},\: \: \textrm{maka}\\ &\left ( \displaystyle \frac{x}{y}+1-\frac{z}{y} \right )\left ( \displaystyle \frac{y}{z}+1-\frac{x}{z} \right )\left ( \displaystyle \frac{z}{x}+1-\frac{y}{x} \right ) \leq 1\\ &\Leftrightarrow \left ( \displaystyle \frac{x+y-z}{y} \right )\left ( \displaystyle \frac{y+z-x}{z} \right )\left ( \displaystyle \frac{z+x-y}{x} \right )\leq 1\\ &\Leftrightarrow (x+y-z)(y+z-x)(z+x-y)\leq xyz,\: \: \textrm{atau}\\ &\Leftrightarrow \color{red}xyz\geq (x+y-z)(y+z-x)(z+x-y)\\ &\textrm{Bentuk terakhir memenuhi bentuk kedua dari}\\ &\textbf{Ketaksamaan Schur}\: \: \textrm{saat}\: \: \color{red}r=1.\\ &\textrm{Jadi},\\ &\: \: \left (a+1- \displaystyle \frac{1}{b} \right )\left (b+1- \displaystyle \frac{1}{c} \right )\left (c+1- \displaystyle \frac{1}{a} \right )\leq 1\quad \blacksquare   \end{array}$.

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

$\begin{array}{ll}\\ 5.&\textrm{Misalkan}\: \: a,b,c\: \: \textrm{bilangan real positif dengan}\\ &a+b+c=2\: ,\: \textrm{tunjukkan bahwa}\\ &\quad a^{4}+b^{4}+c^{4}+abc\geq a^{3}+b^{3}+c^{3}\\\\ &\textbf{Bukti}\\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan Schur}\: \textrm{saat}\: \: \color{red}r=2\\ &\textrm{kita memiliki}\\ &a^{2}(a-b)(a-c)+b^{2}(b-a)(b-c)+c^{2}(c-a)(c-b)\geq 0\\ &\Leftrightarrow a^{4}+b^{4}+c^{4}+abc(a+b+c)-a^{3}(b+c)-b^{3}(a+c)-c^{3}(a+b)\geq 0\\ &\Leftrightarrow a^{4}+b^{4}+c^{4}+abc(a+b+c)\geq a^{3}(b+c)+b^{3}(a+c)+c^{3}(a+b)\\ &\Leftrightarrow a^{4}+b^{4}+c^{4}+abc(a+b+c)\geq (a^{3}+b^{3}+c^{3})(a+b+c)-(a^{4}+b^{4}+c^{4})\\ &\Leftrightarrow 2(a^{4}+b^{4}+c^{4})+abc(a+b+c)\geq (a^{3}+b^{3}+c^{3})(a+b+c)\\ &\Leftrightarrow 2(a^{4}+b^{4}+c^{4})+abc(2)\geq (a^{3}+b^{3}+c^{3})(2)\\ &\Leftrightarrow a^{4}+b^{4}+c^{4}+abc\geq a^{3}+b^{3}+c^{3}\qquad \blacksquare \\\\ &\color{blue}\textrm{Bentuk di atas kadang dituliskan dengan bentuk}\\ &\color{blue}\textrm{berikut}:\\ &\begin{aligned}&\textrm{Dengan}\: \: \textbf{Ketaksamaan Schur}\: \textrm{saat}\: \: \color{red}r=2\\ &\textrm{kita memiliki}\\ &\displaystyle \sum_{\textrm{siklik}}^{.}a^{2}(a-b)(a-c)\geq 0\\ &\Leftrightarrow \displaystyle \sum_{\textrm{siklik}}^{.}a^{4}+abc\displaystyle \sum_{\textrm{siklik}}^{.}a-\displaystyle \sum_{\textrm{siklik}}^{.}a^{3}(b+c)\geq 0\\ &\Leftrightarrow \displaystyle \sum_{\textrm{siklik}}^{.}a^{4}+abc\displaystyle \sum_{\textrm{siklik}}^{.}a\geq \displaystyle \sum_{\textrm{siklik}}^{.}a^{3}(b+c)\\ &\Leftrightarrow \displaystyle \sum_{\textrm{siklik}}^{.}a^{4}+abc\displaystyle \sum_{\textrm{siklik}}^{.}a\geq \left ( \displaystyle \sum_{\textrm{siklik}}^{.}a^{3} \right )\left ( \displaystyle \sum_{\textrm{siklik}}^{.}a \right )-\left ( \displaystyle \sum_{\textrm{siklik}}^{.}a^{4} \right )\\ &\Leftrightarrow 2\displaystyle \sum_{\textrm{siklik}}^{.}a^{4}+abc\displaystyle \sum_{\textrm{siklik}}^{.}a\geq \left ( \displaystyle \sum_{\textrm{siklik}}^{.}a^{3} \right )\left ( \displaystyle \sum_{\textrm{siklik}}^{.}a \right )\\ &\Leftrightarrow \displaystyle \sum_{\textrm{siklik}}^{.}a^{4}+abc\geq \left ( \displaystyle \sum_{\textrm{siklik}}^{.}a^{3} \right )\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&(\textbf{APMO 2004})\\ &\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad (x^{2}+2)(y^{2}+2)(z^{2}+2)\geq 9(xy+yz+zx)\\\\ &\textbf{Bukti}\\ &\textrm{Dengan menjabarkan akan didapatkan}\\ &x^{2}y^{2}z^{2}+2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}+4\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+8\geq 9\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\textrm{Perhatikan bahwa}\\ &\bullet \quad\color{red}2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}-4\displaystyle \sum_{\textrm{siklik}}^{.}xy+6=2\displaystyle \sum_{\textrm{siklik}}^{.}(xy-1)^{2}\geq 0\\ &\bullet  \quad \color{red}\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}\geq \displaystyle \sum_{\textrm{siklik}}^{.}xy\: \: \color{black}\textrm{atau}\: \: \color{red}x^{2}+y^{2}+z^{2}\geq xy+xz+yz\\ &\textrm{Kita cukup membuktikan bahwa}\\ & x^{2}y^{2}z^{2}+\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+2\geq 2\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\Leftrightarrow  x^{2}y^{2}z^{2}+2\geq \displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\begin{aligned}&\textrm{Untuk}\: \: a,b,c\: \: \textrm{bilangan real positif},\\ &\textbf{Ketaksamaan Schur}\: \textrm{saat}\: \: \color{red}r=1\\ &\textrm{memberikan}\\ &\displaystyle \sum_{\textrm{siklik}}^{.}a^{3}+3abc\geq \displaystyle \sum_{\textrm{siklik}}^{.}a^{2}b+\displaystyle \sum_{\textrm{siklik}}^{.}ab^{2}\\ &\qquad\qquad\qquad =ab(a+b)+bc(b+c)+ca(c+a)\\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan AM-GM}\: \: \textrm{didapakan}\\ &\displaystyle \sum_{\textrm{siklik}}^{.}a^{3}+3abc\geq 2\displaystyle \sum_{\textrm{siklik}}^{.}(ab)^{\frac{3}{2}}\\ &\textrm{Pilih}\: \: a=x^{\frac{2}{3}},\: b=y^{\frac{2}{3}},\: c=z^{\frac{2}{3}},\: \textrm{maka didapatkan}\\ &(x^{\frac{2}{3}})^{3}+(y^{\frac{2}{3}})^{3}+(z^{\frac{2}{3}})^{3}+3(xyz)^{\frac{2}{3}}\geq 2(xy+yz+zx)\\ &\textrm{Selanjutnya kita selesaikan ini}\: ,\: x^{2}y^{2}z^{2}+2\geq \displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\Leftrightarrow  x^{2}y^{2}z^{2}+2 \geq 3(xyz)^{\frac{2}{3}}\\ &\textrm{Misalkan}\: \: (xyz)^{\frac{2}{3}}=t,\: \textrm{maka}\\ &t^{3}+2\geq 3t\Leftrightarrow t^{3}-3t+2\geq 0\\ &(t-1)^{2}(t+2)\geq 0\: \: \textrm{adalah hal benar} \end{aligned}   \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.