$\begin{array}{ll}\\ 76.&\textbf{(IMO 1983)}\\ &\textrm{Jika}\: \: a,b,c\: \: \textrm{adalah panjang sisi-sisi segitiga}\\ &\textrm{tunjukkan bahwa}\\ &\quad a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\geq 0\\\\ &\textbf{Bukti}:\\ &\textrm{Pada sebuah segitiga dengan sisi}\: \: a,b,c\\ &\textrm{berlaku}\: \: \begin{cases} a+b>c & \Rightarrow a>c-b\\ a+c>b & \Rightarrow c>b-c\\ b+c>a & \Rightarrow b>a-c \end{cases}\\ &\textrm{Sehingga untuk ketaksamaan pada soal}\\ &\color{red}a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(x-a)\\ &\geq a^{2}(a-c)(a-b)+b^{2}(b-c)(b-c)+c^{2}(c-b)(c-a)\color{red}\geq 0\\ &\textrm{Bentuk terakhir memenuhi bentuk dari}\\ &\textbf{Ketaksamaan Schur}\: \: \textrm{saat}\: \: \color{red}r=2.\\ &\textrm{Jadi},\\ &\quad a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\geq 0\quad \blacksquare \end{array}$.
$\begin{array}{ll}\\ 77.&\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}$.
$\begin{array}{ll}\\ 78.&\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad a^{2}+b^{2}+c^{2}+2abc+1\geq 2(ab+acb+c)\\&\qquad\qquad\qquad\qquad\qquad (\textbf{Darij Grinberg})\\\\ &\textbf{Bukti}\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\: \: \textrm{dan}\\ &\textrm{dilanjutkan dengan}\: \: \textbf{ketaksamaan Schur}\\ &\textrm{serta menggesernya ke ruas kiri, maka}\\ &a^{2}+b^{2}+c^{2}+2abc+1- 2(ab+ac+bc)\\ &\geq a^{2}+b^{2}+c^{2}+3(abc)^{^{\frac{2}{3}}}+1\geq 2(ab+ac+bc)\\ &\geq \left ((a)^{^{\frac{2}{3}}} \right )^{3}+\left ((b)^{^{\frac{2}{3}}} \right )^{3}+\left ((c)^{^{\frac{2}{3}}} \right )^{3}+3(abc)^{^{\frac{2}{3}}}-2(ab+ac+bc)\\ &\geq a^{.^{\frac{2}{3}}}b^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +b^{.^{\frac{2}{3}}} \right )+a^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )\\ &\quad +b^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (b^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )-2(ab+ac+bc)\\ &\geq a^{.^{\frac{2}{3}}}b^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +b^{.^{\frac{2}{3}}} \right )+a^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )\\ &\quad +b^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (b^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )-2(ab+ac+bc)\\ &= a^{.^{\frac{2}{3}}}b^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} -b^{.^{\frac{2}{3}}} \right )^{2}+a^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} -c^{.^{\frac{2}{3}}} \right )^{2}\\ &\quad +b^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (b^{.^{\frac{2}{3}}} -c^{.^{\frac{2}{3}}} \right )^{2}\geq 0\qquad \blacksquare \end{array}$.
$\begin{array}{ll}\\ 79.&(\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}$.
$.\: \quad\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &x^{2}y^{2}z^{2}+2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}+4\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+8\geq 9\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\textrm{atau dalam bentuk utuhnya, yaitu}\\ &x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+4(x^{2}+y^{2}+z^{2})+8\geq 9(xy+xz+yz)\\ &\textrm{Sekarang kira uraikan satu persatu bagian}\\ &\bullet\quad x^{2}y^{2}z^{2}+1+1\geq 3\sqrt[3]{(xyz)^{2}}\geq \displaystyle \frac{9abc}{a+b+c}=\frac{9r}{p}\\ &\qquad \textrm{ingat bahwa}\: \: \textbf{jika ada}\: \: \displaystyle \frac{9r}{p}\geq 4q-p^{2}\\ &\qquad =4(xy+xz+yz)-(x+y+z)^{2}\\ &\qquad \textrm{adalah}\: \: \textbf{ketaksamaan Schur saat}\: \: \color{red}r=1\\ &\bullet \quad x^{2}y^{2}+1+x^{2}z^{2}+1+y^{2}z^{2}+1\geq 2(xy+xz+yz)\\ &\bullet\quad x^{2}+y^{2}+z^{2}\geq xy+xz+yz\\ &\qquad \textrm{keduanya didapat dengan ketaksamaan}\: \: \textbf{AM-GM}\\&x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+4(x^{2}+y^{2}+z^{2})+8\\ &=x^{2}y^{2}z^{2}+2+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}+3)+4(x^{2}+y^{2}+z^{2})\\ &\geq 4(xy+xz+yz)-(x+y+z)^{2}+4(xy+xz+yz)+4(xy+xz+yz)\\ &\geq 12(xy+xz+yz)-(x+y+z)^{2}\\ &\geq 12(xy+xz+yz)-3(xy+xz+yz)\\ &=9(xy+xz+yz)\qquad \blacksquare \end{aligned}$.
$.\: \quad\begin{aligned}&\color{blue}\textbf{Alternatif 3}\\ &(x^{2}+2)(y^{2}+2)(z^{2}+2)- 9(xy+yz+zx)\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\\ &=x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})\\ &\quad +4(x^{2}+y^{2}+z^{2})+8- 9(xy+xz+yz)\\ &=4(x^{2}+y^{2}+z^{2})+2\left ((x^{2}y^{2}+1) +(x^{2}z^{2}+1)+(y^{2}z^{2}+1) \right )\\ &\quad +(x^{2}y^{2}z^{2}+1)+1-9(xy+xz+yz)\\ &\geq 4(x^{2}+y^{2}+z^{2})+4(xy+xz+yz)\\ &\quad +2xyz+1-9(xy+xz+yz)\\ &=(x^{2}+y^{2}+z^{2})+3(x^{2}+y^{2}+z^{2})\\ &\quad +2xyz+1-5(x^{2}+y^{2}+z^{2})\\ &\geq x^{2}+y^{2}+z^{2}+3(xy+xz+yz)\\ &+2xyz-5(xy+xz+yz)\\ &=x^{2}+y^{2}+z^{2}+2xyz+1-2(xy+xz+yz)\geq 0\\ &\textrm{adalah benar dengan bukti ada pada}\\ &\textrm{nomor soal sebelumnya} \end{aligned}$.
$\begin{array}{ll}\\ 80.&\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad 2(a^{2}+b^{2}+c^{2})+abc+8\geq 5(a+b+c)\\&\qquad\qquad\qquad\qquad\qquad (\textbf{Tran Nam Dung})\\\\ &\textbf{Bukti}\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\: \: \textrm{dan}\\ &\textrm{menggeser ke ruas kiri dan masing-masing}\\ &\textrm{serta mengalikan semunya dengan 6, maka}\\ &12(a^{2}+b^{2}+c^{2})+6abc+48- 30(a+b+c)\\ &= 12(a^{2}+b^{2}+c^{2})+3(2abc+1)+45- 5.2.3(a+b+c)\\ &\geq 2(a^{2}+b^{2}+c^{2})+9\sqrt[3]{(abc)^{2}}+45- 5\left ((a+b+c)^{2}+9 \right )\\ &=12(a^{2}+b^{2}+c^{2})+\displaystyle \frac{9abc}{\sqrt[3]{abc}}-5\left ((a^{2}+b^{2}+c^{2})+2(ab+ac+bc) \right )\\ &=7(a^{2}+b^{2}+c^{2})+ \displaystyle \frac{9abc}{\sqrt[3]{abc}}-10(ab+ac+bc)\\ &\geq 7(a^{2}+b^{2}+c^{2})+\displaystyle \frac{27abc}{a+b+c}-10(ab+ac+bc)\\ &\begin{aligned}&\textrm{dengan}\: \: \textbf{ketaksamaan Schur},\: \: \textrm{yaitu}:\\ &p^{3}+9r\geq 4pq\Leftrightarrow \displaystyle \color{red}\frac{9r}{p}\geq 4q-p^{2}\\ &\textrm{maka ketaksamaan akan menjadi}\\ &\geq 7(a^{2}+b^{2}+c^{2})+\color{blue}3(4q-p^{2})-10q\\ &\geq 7(a^{2}+b^{2}+c^{2})+2q-3p^{2}\\ &=7(a^{2}+b^{2}+c^{2})+2(ab+ac+bc)-3(a+b+c)^{2}\\ &=7(a^{2}+b^{2}+c^{2})+2q-3\left ((a^{2}+b^{2}+c^{2})+2q \right )\\ &=4(a^{2}+b^{2}+c^{2})+2q-6q\\ &=4(a^{2}+b^{2}+c^{2})-4q\\ &=4(a^{2}+b^{2}+c^{2})-4(ab+ac+bc)\\ &=4(a^{2}+b^{2}+c^{2}-ab-ac-bc)\geq 0\quad \blacksquare \end{aligned} \end{array}$.
DAFTAR PUSTAKA
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