Prove inequality: $\frac x{2+xy+yz}+\frac y{2+yz+zx}+\frac z{2+zx+xy}\le \frac{x+y+z}{x+y+z+xyz}$ 
Numbers $x,y,z$ satisfy $x\in(0,1], y\in(0,1], z\in(0,1]$. Prove inequality:
  $$\frac x{2+xy+yz}+\frac y{2+yz+zx}+\frac z{2+zx+xy}\le \frac{x+y+z}{x+y+z+xyz}$$

My work so far:
$\frac x{2+xy+yz}\le \frac x{x^2+xz+xy+yz}=\frac x{(x+z)(x+y)}$
Then
$$\frac x{2+xy+yz}+\frac y{2+yz+zx}+\frac z{2+zx+xy}\le$$
$$\le \frac x{(x+z)(x+y)}+\frac y{(y+z)(x+y)}+\frac z{(x+z)(z+y)}=$$
$$=\frac{2(xy+yz+zx)}{(x+y)(y+z)(z+x)}$$

Hint:
$$(1-x)(1-y)(1-z)+(1-zx)\ge 0 \implies 2+ xy+yz \ge x+y+z+xyz$$
