# Problem 299 Van Khea

គេអោយបីចំនួនពិតវិជ្ជមាន $a, b, c$ ផ្ទៀងផ្ទាត់ $a^4+b^4+c^4=3$ ។ ស្រាយបញ្ជាក់ថាៈ
$\displaystyle \frac{a^3}{\sqrt{b^4+c^4}}+\frac{b^3}{\sqrt{c^4+a^4}}+\frac{c^3}{\sqrt{a^4+b^4}}\geq \frac{3}{\sqrt{2}}$
សំរាយបញ្ជាក់
តាង $a=\sqrt{x}; b=\sqrt{y}; c=\sqrt{z}\Rightarrow x^2+y^2+z^2=3$
ជំនួសតំលៃ $a, b, c$ ចូលវិសមភាពខាងលើយើងបានៈ
$\displaystyle \frac{x^{\frac{3}{2}}}{\sqrt{y^2+z^2}}+\frac{y^{\frac{3}{2}}}{\sqrt{z^2+x^2}}+\frac{z^{\frac{3}{2}}}{\sqrt{x^2+y^2}}\geq \frac{3}{\sqrt{2}}$
តាមវិសមភាព $Cauchy-Schwarz$ យើងមានៈ
$\displaystyle (x^2+y^2+z^2)(x^2+y^2+z^2)\biggl(\frac{x^{\frac{3}{2}}}{\sqrt{y^2+z^2}}+\frac{y^{\frac{3}{2}}}{\sqrt{z^2+x^2}}+\frac{z^{\frac{3}{2}}}{\sqrt{x^2+y^2}}\biggl)$$\displaystyle \geq \biggl(\frac{x^{\frac{11}{6}}}{\sqrt[6]{y^2+z^2}}+\frac{y^{\frac{11}{6}}}{\sqrt[6]{z^2+x^2}}+\frac{z^{\frac{11}{6}}}{\sqrt[6]{x^2+y^2}}\biggl)^3$
ម្យ៉ាងទៀតយើងមានៈ $\displaystyle \frac{x^{\frac{11}{6}}}{\sqrt[6]{y^2+z^2}}=\frac{(x^2)^{\frac{7}{6}}}{\sqrt[6]{x^3y^2+z^2x^3}}$
ធ្វើដូចគ្នាយើងបានៈ
$\displaystyle \frac{x^{\frac{11}{6}}}{\sqrt[6]{y^2+z^2}}+\frac{y^{\frac{11}{6}}}{\sqrt[6]{z^2+x^2}}+\frac{z^{\frac{11}{6}}}{\sqrt[6]{x^2+y^2}}$$\displaystyle =\frac{(x^2)^{\frac{7}{6}}}{\sqrt[6]{x^3y^2+z^2x^3}}+\frac{(y^2)^{\frac{7}{6}}}{\sqrt[6]{y^3z^2+x^2y^3}}+\frac{(z^2)^{\frac{7}{6}}}{\sqrt[6]{z^3x^2+y^2z^3}}$
ដោយ $\displaystyle \frac{7}{6}-\frac{1}{6}=1$ នោះតាម $Problem$ $VanKhea$ inequality យើងបានៈ
$\displaystyle \frac{(x^2)^{\frac{7}{6}}}{\sqrt[6]{x^3y^2+z^2x^3}}+\frac{(y^2)^{\frac{7}{6}}}{\sqrt[6]{y^3z^2+x^2y^3}}+\frac{(z^2)^{\frac{7}{6}}}{\sqrt[6]{z^3x^2+y^2z^3}}$$\displaystyle \geq \frac{(x^2+y^2+z^2)^{\frac{7}{6}}}{\sqrt[6]{x^3y^2+y^3z^2+z^3x^2+x^2y^3+y^2z^3+z^2x^3}}$
ម្យ៉ាងទៀតចំពោះ $x^2+y^2+z^2=3\Rightarrow x^3y^2+y^3z^2+z^3x^2+x^2y^3+y^2z^3+z^2x^3\leq 6$
$\displaystyle \Rightarrow \frac{x^{\frac{11}{6}}}{\sqrt[6]{y^2+z^2}}+\frac{y^{\frac{11}{6}}}{\sqrt[6]{z^2+x^2}}+\frac{z^{\frac{11}{6}}}{\sqrt[6]{x^2+y^2}}$$\displaystyle \geq \frac{3}{2^{\frac{1}{6}}}$
$\displaystyle \Rightarrow 9\biggl(\frac{x^{\frac{3}{2}}}{\sqrt{y^2+z^2}}+\frac{y^{\frac{3}{2}}}{\sqrt{z^2+x^2}}+\frac{z^{\frac{3}{2}}}{\sqrt{x^2+y^2}}\biggl)\geq \frac{3^3}{\sqrt{2}}$
$\displaystyle \Rightarrow \frac{x^{\frac{3}{2}}}{\sqrt{y^2+z^2}}+\frac{y^{\frac{3}{2}}}{\sqrt{z^2+x^2}}+\frac{z^{\frac{3}{2}}}{\sqrt{x^2+y^2}}\geq \frac{3}{\sqrt{2}}$
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។ សមភាពកើតមានពេល $x=y=z=1\Leftrightarrow a=b=c=1$

## ទាញយកចំលើយ problem 299 នៅទីនេះ

សូមមើលផងដែរ $Problem$ $Van Khea$ inequality