We will show that all denominators not greater then $2(x+y+z)$ First, note that $x,y,z \in (0;1)$. $2+xy = y(1-y^2-z^3) + 2(x+y^2+z^3)<2(x+y+z)$ Last inequality holds because $2y^2 \le y+y^3$ and $2z^3 < 2z$ Obviously $z>1-x-y^2$, then $2(x+y+z)-yz = 2x+2y+z(2-y)>2x+2y + (1-x-y^2)(2-y) = y+2-2y^2+y^3+xy \ge 2+xy > 2$, so $yz+2<2(x+y+z)$ $2+xz = 2(x+y^2+z^3) + z(1-y^2-z^3) <2(x+y+z)$, which is obvious using that $2y^2 \le 2y$, $2z^3 \le 2z^{5/2} \le z+z^4$ So, $$\frac{x}{2 + xy} + \frac{y}{2 + yz} + \frac{z}{2 + zx} > \frac{x}{2(x+y+z)} + \frac{y}{2(x+y+z)} + \frac{z}{2(x+y+z)} =\frac{1}{2}$$