Prove $\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\ge \frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}$ where $x,y,z>0$ and $x+y+z=3$. 
Prove $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\ge \frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x},$$ where $x,y,z>0$ and $x+y+z=3$.

Maybe we can show $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\ge x^3+y^3+z^3,\tag1$$
then $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\ge \frac{1}{2}\left(\frac{x}{y^2}+x^3+\frac{y}{z^2}+y^3+\frac{z}{x^2}+z^3\right)\ge \frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}.$$
But it's also difficlut to show $(1)$.
 A: Let $z=\max\{x,y,z\}.$
Thus, $$\sum_{cyc}\left(\frac{x}{y^2}-\frac{x^2}{y}\right)=$$
$$=\left(\tfrac{x}{y^2}+\tfrac{y}{z^2}+\tfrac{z}{x^2}-\tfrac{1}{x}-\tfrac{1}{y}-\tfrac{1}{z}\right)-\left(\tfrac{x^2}{y}+\tfrac{y^2}{z}+\tfrac{z^2}{x}-x-y-z\right)+\sum_{cyc}\left(\tfrac{1}{x}-1\right)=$$
$$=\left(\tfrac{x}{y^2}+\tfrac{y}{x^2}-\tfrac{1}{x}-\tfrac{1}{y}+\tfrac{y}{z^2}+\tfrac{z}{x^2}-\tfrac{y}{x^2}-\tfrac{1}{z}\right)-\left(\tfrac{x^2}{y}+\tfrac{y^2}{x}-x-y+\tfrac{y^2}{z}+\tfrac{z^2}{x}-\tfrac{y^2}{x}-z\right)+$$
$$+\frac{1}{3}\left(\tfrac{x}{y}+\tfrac{y}{x}-2+\tfrac{y}{z}+\tfrac{z}{x}-\tfrac{y}{x}-1+\tfrac{x}{y}+\tfrac{y}{x}-2+\tfrac{x}{z}+\tfrac{z}{y}-\tfrac{x}{y}-1\right)=$$
$$=\left(\tfrac{(x-y)^2(x+y)}{x^2y^2}+\tfrac{(z-x)(z-y)(x+z)}{x^2z^2}\right)-\left(\tfrac{(x-y)^2(x+y)}{xy}+\tfrac{(z-x)(z-y)(y+z)}{xz}\right)+$$
$$+\frac{1}{3}\left(\frac{2(x-y)^2}{xy}+(z-x)(z-y)\left(\frac{1}{xz}+\frac{1}{yz}\right)\right)=$$
$$=(x-y)^2\left(\tfrac{x+y}{x^2y^2}-\tfrac{x+y}{xy}+\tfrac{2}{3xy}\right)+(z-x)(z-y)\left(\tfrac{x+z}{x^2z^2}-\tfrac{y+z}{xz}+\tfrac{x+y}{3xyz}\right).
$$
We'll prove that:$$\frac{x+y}{x^2y^2}-\frac{x+y}{xy}+\frac{2}{3xy}\geq0.$$
Indeed, by the River Li's beautiful point since $x+y\leq2$, by AM-GM we obtain:
$$\frac{x+y}{x^2y^2}-\frac{x+y}{xy}+\frac{2}{3xy}>\frac{x+y}{xy\left(\frac{x+y}{2}\right)^2}-\frac{x+y}{xy}\geq0.$$
Id est, it's enough to prove that:
$$\frac{x+z}{x^2z^2}-\frac{y+z}{xz}+\frac{x+y}{3xyz}\geq0$$
or
$$\frac{(x+y+z)(x+z)}{xz}+\frac{x+y}{y}+\frac{9x}{x+y+z}-3(x+y+z)\geq0$$ or
$$\frac{x^2+z^2+y(x+z)}{xz}+\frac{x}{y}+\frac{9x}{x+y+z}\geq6$$ or $f(z)\geq0,$ where $$f(z)=yz^3+(x^2-5xy+2y^2)z^2+(x^3+5x^2y-4xy^2+y^3)z+xy(x+y)^2.$$
But $$f''(z)=6yz+2(x^2-5xy+2y^2)\geq$$
$$\geq3y(x+y)+2(x^2-5xy+2y^2)=2x^2-7xy+7y^2>0,$$
which says $$f'(z)=3yz^2+2(x^2-5xy+2y^2)z+x^3+5x^2y-4xy^2+y^3\geq$$
$$\geq\frac{3y(x+y)^2}{4}+(x+y)(x^2-5xy+2y^2)+x^3+5x^2y-4xy^2+y^3=$$
$$=\frac{8x^3+7x^2y-22xy^2+15y^3}{4}>0,$$
which says $$f(z)\geq f\left(\frac{x+y}{2}\right)=\frac{3(x+y)(2x^3+7x^2y-4xy^2+3y^3)}{8}>0.$$
