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
On
I don't know if this would help you but here is a hint for you.
$\left \{ \begin{aligned} x, y, z \ge 0\\ x + y^2 + z^3 = 1 \end{aligned} \right. \implies 0 \le x, y, z \le 1 \implies \left\{ \begin{aligned} x = x\\ y \ge y^2\\ z \ge z^3 \end{aligned} \right. \implies x + y + z \ge x + y^2 + z^3 = 2$
Now, you can use the Contradiction method.
Indeed, since $$xy+xz+yz\geq x^2y+y^2z+z^2x,$$ it's enough to prove that $$\sum_{cyc}(2x^2+3xy)\geq2(x+y+z).$$ Let $\sum\limits_{cyc}(2x^2+3xy)<2(x+y+z),$ $x=ka$, $y=kb$ and $z=kc$ such that $k>0$ and $$\sum_{cyc}(2a^2+3ab)=2(a+b+c).$$ Thus, $$k\sum_{cyc}(2a^2+3ab)<2(a+b+c)=\sum_{cyc}(2a^2+3ab),$$ which gives $0<k<1$ and $$1=x+y^2+z^3=ka+k^2b^2+k^3c^3<a+b^2+c^3,$$ which is a contradiction because we'll prove now that $$a+b^2+c^3\leq1.$$ Indeed, since $$\frac{\sum\limits_{cyc}(2a^2+3ab)}{2(a+b+c)}=1,$$ we need to prove that $$a\left(\frac{\sum\limits_{cyc}(2a^2+3ab)}{2(a+b+c)}\right)^2+b^2\left(\frac{\sum\limits_{cyc}(2a^2+3ab)}{2(a+b+c)}\right)+c^3\leq\left(\frac{\sum\limits_{cyc}(2a^2+3ab)}{2(a+b+c)}\right)^3$$ or $$\left(\sum_{cyc}(2a^2+3ab)\right)^2\geq$$ $$\geq8c^3(a+b+c)^3+4b^2(a+b+c)^2\sum_{cyc}(2a^2+3ab)+2a(a+b+c)\left(\sum_{cyc}(2a^2+3ab)\right)^2$$ or $$4(a^5b+a^5c+2b^5c+c^5a+3c^5b)+12a^4b^2+20a^4c^2+6b^4a^2+38b^4c^2+22c^4a^2+46c^4b^2+$$ $$+13a^3b^3+33a^3c^3+63b^3c^3+abc(36a^3+36b^3+64c^3+79a^2b+107a^2c+77b^2a+127b^2c+129c^2a+151c^2b+178abc)\geq0,$$ which is obvious.