[tex]\sum \frac{x^2}{x^4+yz}\leq \frac{1}{4}\sum \left ( \frac{1}{x^2}+\frac{x^2}{yz} \right )=\frac{1}{4}\left ( \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \right )+\frac{3}{4}[/tex]
Mặt khác [tex]\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{x^2y^2+y^2z^2+z^22x^2}{(xyz)^2}=\frac{9(x^2y^2+y^2z^2+z^2x^2)}{(x^2+y^2+z^2)^2}\leq \frac{3(x^2+y^2+z^2)^2}{(x^2+y^2+z^2)^2}=3[/tex]
[tex]\Rightarrow P\leq \frac{3}{4}+\frac{3}{4}=\frac{3}{2}[/tex]
Dấu "=" xảy ra khi [tex]x=y=z=1[/tex]