Ta có: \(x;y;z\ge0\)\(\left(x+y-z\right)^2+x^2y^2\ge0\Leftrightarrow x^2+y^2+z^2+2xy-2yz-2xz+x^2y^2\ge0\)
\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)\le x^2y^2+4xy+4\)\(\Leftrightarrow\left(x+y+z\right)^2\le\left(xy+2\right)^2\)\(\Leftrightarrow x+y+z\le xy+2\)
Từ đó \(P=\dfrac{x}{2+yz}+\dfrac{y}{2+xz}+\dfrac{z}{2+xy}\)
Ta có: \(x+y+z\le xy+2\Rightarrow\dfrac{z}{x+y+z}\ge\dfrac{z}{xy+2}\)
\(\Rightarrow P\le\sum\dfrac{z}{x+y+z}=1\)\(\Rightarrow MaxB=1\)
Đẳng thức xảy ra chẳng hạn x=y=1; z=0
Mà:\(x\left(2+yz\right)\le\left(\dfrac{x^2+2}{2\sqrt{2}}\right).\left(2+\dfrac{y^2+z^2}{2}\right)\) (cauchy)
\(=\dfrac{\left(x^2+2\right)\left(4+y^2+z^2\right)}{4\sqrt{2}}\le\dfrac{\left(x^2+2+4+y^2+z^2\right)^2}{16\sqrt{2}}\)(cauchy) = \(\dfrac{\left(2+2+4\right)^2}{16\sqrt{2}}=2\sqrt{2}\)\(\Rightarrow\dfrac{x}{2+yz}\ge\dfrac{x^2}{2\sqrt{2}}\)
Từ đó \(P\ge\sum\dfrac{x^2}{2\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\) (vì \(x^2+y^2+z^2=2\))
Đẳng thức xảy ra chẳng hạn x=y=0; z=\(\sqrt{2}\)
