Chia 2 vế PT cho z^2 được :
\(xy^2+\frac{x^2}{z}+\frac{y}{z^2}=3\)
ta có: \(\left(x^2y^2+y^2\right)+\left(x^2+\frac{x^2}{z^2}\right)+\left(\frac{y^2}{z^2}+\frac{1}{z^2}\right)\ge2\left(xy^2+\frac{x^2}{z}+\frac{y}{z^2}\right)=6\)
Đặt \(a=\frac{1}{z^2},b=y^2,c=x^2\).Ta có
P=\(\frac{1}{\frac{1}{z^4}+x^4+y^4}=\frac{1}{a^2+b^2+c^2}\)
Có: \(3\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca+a+b+c\right)-3=2\left(\frac{x^2}{z^2}+x^2y^2+\frac{y^2}{z^2}+\frac{1}{z^2}+x^2+y^2\right)\ge9\Rightarrow a^2+b^2+c^2\ge3\)
\(\Rightarrow P\le\frac{1}{3}\Leftrightarrow x=y=z=1\)
