ÁD BĐT AM-GM:
$2+2yz=x^{2}+y^{2}+z^{2}+2yz=x^{2}+(y+z)^{2}\geq2x(y+z)$
$\Rightarrow 1+yz\geq x(y+z)\Rightarrow x^{2}+x+yz+1\geq x(x+y+z+1)$
$\Rightarrow \frac{x^{2}}{x^{2}+x+yz+1}\leq \frac{x}{x+y+z+1}$
Ta sẽ CM:$x+y+z-xyz\leq 2$
Thật vậy:
ÁD BĐT C-S:
$x+y+z-xyz=x(1-yz)+(y+z)\leq \sqrt{(x^{2}+(y+z)^{2})((1-yz)^{2}+1)}$
$=\sqrt{2(1+yz)((yz)^{2}-2yz+2)}=\sqrt{y^{2}z^{2}(yz-1)+4}\leq2$
$\Rightarrow M \leq \frac{x}{x+y+z+1}+\frac{y+z}{x+y+z+1}+\frac{1}{x+y+z+1}=1$
Dấu''='' xra$\Leftrightarrow x=y=1;z=0$