$x,y,z > 0$ and $x+y+z=3$, prove $$(x^2y+y^2z+z^2x)\cdot \left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2} +\frac{1}{(z+x)^2}\right) \geqslant \frac94.$$
My immediate thought is that this inequality is similar to the famous Iran inequality
$$(xy+yz+zx)\cdot \left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2} +\frac{1}{(z+x)^2}\right) \geqslant \frac94$$
then if I can prove that
$$x^2y+y^2z+z^2x \geqslant xy+yz+zx$$ for positive $x,y,z$ satistifes $x+y+z=3$ then the problem is solved.
However, it turns out that $x^2y+y^2z+z^2x$ is neither always greater or lesser than $xy+yz+zx$, so I get stuck here.
I don't like solution involved computer or numerical methods. I will down vote all of answers that showing these methods