# Problem 219 (Van Khea)

គេអោយបីចំនួនពិតវិជ្ជមាន $x, y, z$ ផ្ទៀងផ្ទាត់ $x+y+z=3$ ។ ស្រាយបញ្ចាក់ថាៈ
$\displaystyle \frac{1}{x^2y+2}+\frac{1}{y^2z+2}+\frac{1}{z^2x+2}\geq 1$
សំរាយបញ្ជាក់
យើងមានៈ $\displaystyle \frac{1}{x^2y+2}=\frac{1}{2}(1-\frac{x^2y}{x^2y+2})\geq \frac{1}{2}(1-\frac{x^2y}{3\sqrt[3]{x^2y}})=\frac{1}{2}(1-\frac{1}{3}x^{\frac{4}{3}}y^{\frac{2}{3}})$
ដូចនេះយើងបានៈ
$\displaystyle \frac{1}{x^2y+2}+\frac{1}{y^2z+2}+\frac{1}{z^2x+2}\geq \frac{3}{2}-\frac{1}{6}(x^{\frac{4}{3}}y^{\frac{2}{3}}+y^{\frac{4}{3}}z^{\frac{2}{3}}+z^{\frac{4}{3}}x^{\frac{2}{3}})$
ឥឡូវយើងត្រូវស្រាយថាចំពោះ $x+y+z=3$ នាំអោយយើងបាន $\displaystyle x^{\frac{4}{3}}y^{\frac{2}{3}}+y^{\frac{4}{3}}z^{\frac{2}{3}}+z^{\frac{4}{3}}x^{\frac{2}{3}}\leq 3$
យើងមានៈ $\displaystyle x^{\frac{4}{3}}y^{\frac{2}{3}}=\sqrt[3]{x^2y}.\sqrt[3]{xy}.\sqrt[3]{x}$
ដូចនេះយើងបានៈ
$\displaystyle x^{\frac{4}{3}}y^{\frac{2}{3}}+y^{\frac{4}{3}}z^{\frac{2}{3}}+z^{\frac{4}{3}}x^{\frac{2}{3}}$$\displaystyle =\sqrt[3]{x^2y}.\sqrt[3]{xy}.\sqrt[3]{x}+\sqrt[3]{y^2z}.\sqrt[3]{yz}.\sqrt[3]{y}+\sqrt[3]{z^2x}.\sqrt[3]{zx}.\sqrt[3]{z}$
ម្យ៉ាងទៀតតាមវិសមភាព $Holder$ យើងមានៈ
$\displaystyle (\sqrt[3]{x^2y}.\sqrt[3]{xy}.\sqrt[3]{x}+\sqrt[3]{y^2z}.\sqrt[3]{yz}.\sqrt[3]{y}+\sqrt[3]{z^2x}.\sqrt[3]{zx}.\sqrt[3]{z})^3$$\displaystyle \leq (x^2y+y^2z+z^2x)(xy+yz+zx)(x+y+z)$
$\displaystyle \Leftrightarrow (x^{\frac{4}{3}}y^{\frac{2}{3}}+y^{\frac{4}{3}}z^{\frac{2}{3}}+z^{\frac{4}{3}}x^{\frac{2}{3}})^3$$\leq (x^2y+y^2z+z^2x)(xy+yz+zx)(x+y+z)$
យើងនឹងស្រាយថាៈ $\displaystyle (xy+yz+zx)(x^2y+y^2z+z^2x)\leq \frac{(x+y+z)^5}{27}$
មិនបាត់បង់លក្ខណៈទូទៅនៃលំហាត់ទេ យើងឧបមាថា $x\leq y\leq z$ នោះយើងបានៈ
$(y-x)(z-y)\geq 0\Leftrightarrow y^2+zx\leq y(z+x)$
$\Rightarrow x^2y+z(y^2+zx)\leq x^2y+yz(z+x)$
$\Leftrightarrow x^2y+y^2z+z^2x\leq y(z^2+zx+x^2)$
$\displaystyle \Rightarrow (xy+yz+zx)(x^2y+y^2z+z^2x)\leq y(xy+yz+zx)(z^2+zx+x^2)$
តាមវិសមភាព $AM - GM$ យើងមានៈ
$\displaystyle (xy+yz+zx)(z^2+zx+x^2)\leq \biggl(\frac{xy+yz+zx+z^2+zx+x^2}{2}\biggl)^2$$\displaystyle \leq \frac{(z+x)^2(x+y+z)^2}{4}$
$\displaystyle \Rightarrow y(xy+yz+zx)(z^2+zx+x^2)\leq \frac{y(z+x)^2(x+y+z)^2}{4}$
ម្យ៉ាងទៀតយើងមានៈ
$\displaystyle \frac{y(z+x)^2}{4}=y.\frac{z+x}{2}.\frac{z+x}{2}\leq \biggl(\frac{y+\frac{z+x}{2}+\frac{z+x}{2}}{3}\biggl)^3$
$\displaystyle \Leftrightarrow \frac{y(z+x)^2}{4}\leq \frac{(x+y+z)^3}{27}$
$\displaystyle \Rightarrow (xy+yz+zx)(x^2y+y^2z+z^2x)\leq \frac{(x+y+z)^5}{27}$ ពិត។
ដូចនេះយើងបានៈ
$\displaystyle \Rightarrow (xy+yz+zx)(x^2y+y^2z+z^2x)(x+y+z)\leq \frac{(x+y+z)^6}{27}$
$\displaystyle \Rightarrow (x^{\frac{4}{3}}y^{\frac{2}{3}}+y^{\frac{4}{3}}z^{\frac{2}{3}}+z^{\frac{4}{3}}x^{\frac{2}{3}})^3\leq \frac{(x+y+z)^6}{27}=3^3$
$\displaystyle \Rightarrow x^{\frac{4}{3}}y^{\frac{2}{3}}+y^{\frac{4}{3}}z^{\frac{2}{3}}+z^{\frac{4}{3}}x^{\frac{2}{3}}\leq 3$
$\displaystyle \Rightarrow \frac{1}{x^2y+2}+\frac{1}{y^2z+2}+\frac{1}{z^2x+2}\geq \frac{3}{2}-\frac{1}{6}.3=1$
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។ សមភាពកើតមានពេល $x=y=z=1$

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