# Problem 222 (van khea)

គេអោយបីចំនួនពិតវិជ្ជមាន $a, b, c$ ផ្ទៀងផ្ទាត់ $a^3+b^3+c^3=3$ ។ ស្រាយបញ្ជាក់ថាៈ
$(ab)^4+(bc)^4+(ca)^4\leq a^6+b^6+c^6$
សំរាយបញ្ជាក់
តាង $x=a^3; y=b^3; z=c^3\Rightarrow x+y+z=3$ នោះយើងបានៈ $\displaystyle (xy)^{\frac{4}{3}}+(yz)^{\frac{4}{3}}+(zx)^{\frac{4}{3}}\leq x^2+y^2+z^2$
យើងមានៈ $\displaystyle (xy)^{\frac{4}{3}}=(x^2y)^{\frac{1}{3}}(xy^2)^{\frac{1}{3}}(xy)^{\frac{1}{3}}$
$\displaystyle \Rightarrow (xy)^{\frac{4}{3}}+(yz)^{\frac{4}{3}}+(zx)^{\frac{4}{3}}$$\displaystyle =(x^2y)^{\frac{1}{3}}(xy^2)^{\frac{1}{3}}(xy)^{\frac{1}{3}}+(y^2z)^{\frac{1}{3}}(yz^2)^{\frac{1}{3}}(yz)^{\frac{1}{3}}+(z^2x)^{\frac{1}{3}}(zx^2)^{\frac{1}{3}}(zx)^{\frac{1}{3}}$
តាមវិសមភាព $Holder$ យើងមានៈ
$\displaystyle ((x^2y)^{\frac{1}{3}}(xy^2)^{\frac{1}{3}}(xy)^{\frac{1}{3}}+(y^2z)^{\frac{1}{3}}(yz^2)^{\frac{1}{3}}(yz)^{\frac{1}{3}}+(z^2x)^{\frac{1}{3}}(zx^2)^{\frac{1}{3}}(zx)^{\frac{1}{3}})^3$
$\displaystyle \leq (x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)(xy+yz+zx)$
$\displaystyle \Leftrightarrow ((xy)^{\frac{4}{3}}+(yz)^{\frac{4}{3}}+(zx)^{\frac{4}{3}})^3$$\displaystyle \leq (x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)(xy+yz+zx)$
ម្យ៉ាងទៀតយើងមានៈ
$\displaystyle x^3+y^3+z^3\geq \frac{1}{3}(x+y+z)(x^2+y^2+z^2)$$\geq xy+yz+zx$
$\displaystyle \Rightarrow ((xy)^{\frac{4}{3}}+(yz)^{\frac{4}{3}}+(zx)^{\frac{4}{3}})^3$$\displaystyle \leq (x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)(x^3+y^3+z^3)$
តាមវិសមភាព $AM-GM$ យើងមានៈ
$\displaystyle (x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)(x^3+y^3+z^3)$$\displaystyle \leq \biggl(\frac{x^2y+y^2z+z^2x+xy^2+yz^2+zx^2+x^3+y^3+z^3}{3}\biggl)^3$$\displaystyle \leq \biggl(\frac{(x+y+z)(x^2+y^2+z^2)}{3}\biggl)^3=(x^2+y^2+z^2)^3$
$\displaystyle \Rightarrow ((xy)^{\frac{4}{3}}+(yz)^{\frac{4}{3}}+(zx)^{\frac{4}{3}})^3\leq (x^2+y^2+z^2)^3$
$\displaystyle \Rightarrow (xy)^{\frac{4}{3}}+(yz)^{\frac{4}{3}}+(zx)^{\frac{4}{3}}\leq x^2+y^2+z^2$
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។ សមភាពកើតមានពេល $x=y=z=1\Leftrightarrow a=b=c=1$