Let $x,y,z$ are non-negative numbers such that $x^2+y^2+z^2=12$.Prove that $$\frac{x+y}{4+yz}+\frac{y+z}{4+xz}+\frac{x+z}{4+xy}\ge \frac{3}{2}$$
$$\frac{x+y}{4+yz}\ge \frac{x+y}{4+\frac{\left(y+z\right)^2}{4}}=\frac{4a}{16+b^2}=4a-\frac{4ab^2}{b^2+16}\ge 4a-\frac{ab}{2}$$
But $x^2+y^2+z^2=12 \Rightarrow a+b+c\le 12$ so $4a-\frac{ab}{2}\le....$ (wrong)
I also use C-S but wrong too
