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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}$.
$.\: \quad\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &x^{2}y^{2}z^{2}+2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}+4\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+8\geq 9\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\textrm{atau dalam bentuk utuhnya, yaitu}\\ &x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+4(x^{2}+y^{2}+z^{2})+8\geq 9(xy+xz+yz)\\ &\textrm{Sekarang kira uraikan satu persatu bagian}\\ &\bullet\quad x^{2}y^{2}z^{2}+1+1\geq 3\sqrt[3]{(xyz)^{2}}\geq \displaystyle \frac{9abc}{a+b+c}=\frac{9r}{p}\\ &\qquad \textrm{ingat bahwa}\: \: \textbf{jika ada}\: \: \displaystyle \frac{9r}{p}\geq 4q-p^{2}\\ &\qquad =4(xy+xz+yz)-(x+y+z)^{2}\\ &\qquad \textrm{adalah}\: \: \textbf{ketaksamaan Schur saat}\: \: \color{red}r=1\\ &\bullet \quad x^{2}y^{2}+1+x^{2}z^{2}+1+y^{2}z^{2}+1\geq 2(xy+xz+yz)\\ &\bullet\quad x^{2}+y^{2}+z^{2}\geq xy+xz+yz\\ &\qquad \textrm{keduanya didapat dengan ketaksamaan}\: \: \textbf{AM-GM}\\&x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+4(x^{2}+y^{2}+z^{2})+8\\ &=x^{2}y^{2}z^{2}+2+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}+3)+4(x^{2}+y^{2}+z^{2})\\ &\geq 4(xy+xz+yz)-(x+y+z)^{2}+4(xy+xz+yz)+4(xy+xz+yz)\\ &\geq 12(xy+xz+yz)-(x+y+z)^{2}\\ &\geq 12(xy+xz+yz)-3(xy+xz+yz)\\ &=9(xy+xz+yz)\qquad \blacksquare \end{aligned}$.
$.\: \quad\begin{aligned}&\color{blue}\textbf{Alternatif 3}\\ &(x^{2}+2)(y^{2}+2)(z^{2}+2)- 9(xy+yz+zx)\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\\ &=x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})\\ &\quad +4(x^{2}+y^{2}+z^{2})+8- 9(xy+xz+yz)\\ &=4(x^{2}+y^{2}+z^{2})+2\left ((x^{2}y^{2}+1) +(x^{2}z^{2}+1)+(y^{2}z^{2}+1) \right )\\ &\quad +(x^{2}y^{2}z^{2}+1)+1-9(xy+xz+yz)\\ &\geq 4(x^{2}+y^{2}+z^{2})+4(xy+xz+yz)\\ &\quad +2xyz+1-9(xy+xz+yz)\\ &=(x^{2}+y^{2}+z^{2})+3(x^{2}+y^{2}+z^{2})\\ &\quad +2xyz+1-5(x^{2}+y^{2}+z^{2})\\ &\geq x^{2}+y^{2}+z^{2}+3(xy+xz+yz)\\ &+2xyz-5(xy+xz+yz)\\ &=x^{2}+y^{2}+z^{2}+2xyz+1-2(xy+xz+yz)\geq 0\\ &\textrm{adalah benar dengan bukti ada pada}\\ &\textrm{nomor soal sebelumnya} \end{aligned}$.
$\begin{array}{ll}\\ 8.&\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad 2(a^{2}+b^{2}+c^{2})+abc+8\geq 5(a+b+c)\\&\qquad\qquad\qquad\qquad\qquad (\textbf{Tran Nam Dung})\\\\ &\textbf{Bukti}\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\: \: \textrm{dan}\\ &\textrm{menggeser ke ruas kiri dan masing-masing}\\ &\textrm{serta mengalikan semunya dengan 6, maka}\\ &12(a^{2}+b^{2}+c^{2})+6abc+48- 30(a+b+c)\\ &= 12(a^{2}+b^{2}+c^{2})+3(2abc+1)+45- 5.2.3(a+b+c)\\ &\geq 2(a^{2}+b^{2}+c^{2})+9\sqrt[3]{(abc)^{2}}+45- 5\left ((a+b+c)^{2}+9 \right )\\ &=12(a^{2}+b^{2}+c^{2})+\displaystyle \frac{9abc}{\sqrt[3]{abc}}-5\left ((a^{2}+b^{2}+c^{2})+2(ab+ac+bc) \right )\\ &=7(a^{2}+b^{2}+c^{2})+ \displaystyle \frac{9abc}{\sqrt[3]{abc}}-10(ab+ac+bc)\\ &\geq 7(a^{2}+b^{2}+c^{2})+\displaystyle \frac{27abc}{a+b+c}-10(ab+ac+bc)\\ &\begin{aligned}&\textrm{dengan}\: \: \textbf{ketaksamaan Schur},\: \: \textrm{yaitu}:\\ &p^{3}+9r\geq 4pq\Leftrightarrow \displaystyle \color{red}\frac{9r}{p}\geq 4q-p^{2}\\ &\textrm{maka ketaksamaan akan menjadi}\\ &\geq 7(a^{2}+b^{2}+c^{2})+\color{blue}3(4q-p^{2})-10q\\ &\geq 7(a^{2}+b^{2}+c^{2})+2q-3p^{2}\\ &=7(a^{2}+b^{2}+c^{2})+2(ab+ac+bc)-3(a+b+c)^{2}\\ &=7(a^{2}+b^{2}+c^{2})+2q-3\left ((a^{2}+b^{2}+c^{2})+2q \right )\\ &=4(a^{2}+b^{2}+c^{2})+2q-6q\\ &=4(a^{2}+b^{2}+c^{2})-4q\\ &=4(a^{2}+b^{2}+c^{2})-4(ab+ac+bc)\\ &=4(a^{2}+b^{2}+c^{2}-ab-ac-bc)\geq 0\quad \blacksquare \end{aligned} \end{array}$.
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