show this inequality $\sum\frac{x}{2+xy}\ge\frac{1}{2}$ 
Let $x,y,z\ge 0$ such that $$x+y^2+z^3=1.$$
  Show that
  $$\dfrac{x}{2+xy}+\dfrac{y}{2+yz}+\dfrac{z}{2+zx}\ge\dfrac{1}{2}$$

I try do
$$\sum_{cyc}\dfrac{x}{2+xy}=\sum_{cyc}\dfrac{x^2}{2x+x^2y}\ge\dfrac{(x+y+z)^2}{(2x+2y+2z)+(x^2y+y^2z+z^2x)}$$
it have to prove 
$$2(x+y+z)^2\ge (2x+2y+2z)+(x^2y+y^2z+z^2x)$$
it seem this is hold,But I can't prove it 

I don't know if this would help you but here is a hint for you.
$\left \{ \begin{aligned} x, y, z \ge 0\\ x + y^2 + z^3 = 1 \end{aligned} \right. \implies 0 \le x, y, z \le 1 \implies \left\{ \begin{aligned} x = x\\ y \ge y^2\\ z \ge z^3 \end{aligned} \right. \implies x + y + z \ge x + y^2 + z^3 = 2$
