# Problem 220 (van khea)

គេអោយ $a, b, c$ ជាបីចំនួនពិតវិជ្ជមាន។ ស្រាយបញ្ជាក់ថាៈ
$\displaystyle (a^6+b^6+c^6)^7\geq 81\biggl[(ab)^7+(bc)^7+(ca)^7\biggl]^3$
សំរាយបញ្ជាក់
តាង $x=a^6; y=b^6; z=c^6$ ដូចនេះយើងត្រូវស្រាយថាៈ
$\displaystyle (x+y+z)^7\geq 81\biggl[(xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}}\biggl]^3$
យើងមានៈ $\displaystyle (xy)^{\frac{7}{6}}=(x^3)^{\frac{1}{18}}(y^3)^{\frac{1}{18}}(xy^2)^{\frac{1}{9}}(x^2y)^{\frac{1}{9}}(xy)^{\frac{2}{3}}$
$\displaystyle \Rightarrow (xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}}$
$\displaystyle =(x^3)^{\frac{1}{18}}(y^3)^{\frac{1}{18}}(xy^2)^{\frac{1}{9}}(x^2y)^{\frac{1}{9}}(xy)^{\frac{2}{3}}+(y^3)^{\frac{1}{18}}(z^3)^{\frac{1}{18}}(yz^2)^{\frac{1}{9}}(y^2z)^{\frac{1}{9}}(yz)^{\frac{2}{3}}$$\displaystyle +(z^3)^{\frac{1}{18}}(x^3)^{\frac{1}{18}}(zx^2)^{\frac{1}{9}}(z^2x)^{\frac{1}{9}}(zx)^{\frac{2}{3}}$
ដោយ $\displaystyle \frac{1}{18}+\frac{1}{18}+\frac{1}{9}+\frac{1}{9}+\frac{2}{3}=1$ តាមវិសមភាព $Holder's$ យើងបានៈ
$\displaystyle (xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}}$$\displaystyle \leq (x^3+y^3+z^3)^{\frac{2}{18}}(x^2y+y^2z+z^2x)^{\frac{1}{9}}(xy^2+yz^2+zx^2)^{\frac{1}{9}}(xy+yz+zx)^{\frac{2}{3}}$
$\displaystyle \Leftrightarrow (xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}}$$\displaystyle \leq ((x^3+y^3+z^3)(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2))^{\frac{1}{9}}(xy+yz+zx)^{\frac{2}{3}}$
តាមវិសមភាព $AM-GM$ inequality យើងមានៈ
$\displaystyle (x^3+y^3+z^3)(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\leq \biggl(\frac{x^3+y^3+z^3+x^2y+y^2z+z^2x+xy^2+yz^2+zx^2}{3}\biggl)^3$
$\displaystyle \Leftrightarrow (x^3+y^3+z^3)(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\leq \biggl(\frac{(x+y+z)(x^2+y^2+z^2)}{3}\biggl)^3$
$\displaystyle \Rightarrow (xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}}\leq \biggl(\frac{(x+y+z)(x^2+y^2+z^2)}{3}\biggl)^{\frac{1}{3}}$$\displaystyle (xy+yz+zx)^{\frac{2}{3}}$
$\displaystyle \Leftrightarrow (xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}}\leq \frac{(x+y+z)^{\frac{1}{3}}((x^2+y^2+z^2)(xy+yz+zx)^2)^{\frac{1}{3}}}{\sqrt[3]{3}}$
តាមវិសមភាព $AM-GM$ យើងមានៈ
$\displaystyle (x^2+y^2+z^2)(xy+yz+zx)^2\leq \biggl(\frac{x^2+y^2+z^2+2(xy+yz+zx)}{3}\biggl)^3$
$\displaystyle \Leftrightarrow (x^2+y^2+z^2)(xy+yz+zx)^2\leq \frac{(x+y+z)^6}{27}$
$\displaystyle \Rightarrow (xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}}\leq \frac{(x+y+z)^{\frac{1}{3}}(x+y+z)^2}{3\sqrt[3]{3}}$$\displaystyle =\frac{(x+y+z)^{\frac{7}{3}}}{3\sqrt[3]{3}}$
$\displaystyle \Rightarrow (x+y+z)^7\geq 81((xy)^{\frac{7}{6}}+(yz)^{\frac{7}{6}}+(zx)^{\frac{7}{6}})^3$
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់ ។ សមភាពកើមានពេល $x=y=z\Leftrightarrow a=b=c$