ÁP dụng bất đẳng thức AM-GM ta có:
\(P=\dfrac{x^2}{x^2+2yz}+\dfrac{y^2}{y^2+2xz}+\dfrac{z^2}{z^2+2xy}\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2\left(xy+yz+xz\right)}\)\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)
Dấu "=" xảy ra\(\Leftrightarrow x=y=z>0\)
Vậy \(MinP=1\Leftrightarrow x=y=z>0\)