Ta có:
$\dfrac{2x^2+y^2+z^2}{4-yz}$
$\ge\dfrac{4\sqrt[4]{x^4y^2z^2}}{4-yz}$
$=\dfrac{4xyz}{\sqrt{yz}(2-\sqrt{yz})(2+\sqrt{yz})}$
$\ge\dfrac{4xyz}{\dfrac{(\sqrt{yz}+2-\sqrt{yz})^2}{4}.\left(2+\dfrac{y+z}{2}\right)}$
$=\dfrac{8xyz}{y+z+4}$
Suy ra:
$\sum\dfrac{2x^2+y^2+z^2}{4-yz}\ge\sum\dfrac{8xyz}{x+y+4}\ge\dfrac{72xyz}{\sum(x+y+4)}=4xyz$