TITLE: Prove $(x^2y+y^2z+z^2x)\cdot \left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2} +\frac{1}{(z+x)^2}\right) \geqslant \frac94$
QUESTION [5 upvotes]: $x,y,z > 0$ 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

REPLY [2 votes]: A full expanding gives 
$$\sum\limits_{cyc}(4x^6y+5x^5y^2+x^5z^2+3x^4y^3-x^4z^3+$$
$$+2x^5yz+9x^4y^2z-7x^4z^2y-2x^3y^3z-14x^3y^2z^2)\geq0$$
and since by Muirhead and AM-GM 
$$\sum\limits_{cyc}(x^5y^2+x^5z^2)\geq\sum\limits_{cyc}(x^4y^3+x^4z^3)$$
$$6x^4y^3+2y^4z^3+5z^4x^3\geq13x^3y^2z^2$$
$$19x^6y+2y^6z+10z^6x\geq31x^4z^2y$$
$$9x^5y^2+4y^5z^2+6z^5x^2\geq19x^3y^2z^2$$
which gives $4\sum\limits_{cyc}x^6y\geq4\sum\limits_{cyc}x^4z^2y$, $4\sum\limits_{cyc}x^4y^3\geq4\sum\limits_{cyc}x^3y^2z^2$ and $4\sum\limits_{cyc}x^5y^2\geq4\sum\limits_{cyc}x^3y^2z^2$,
it remains to prove that $$\sum\limits_{cyc}(2x^5yz+9x^4y^2z-3x^4z^2y-2z^3y^3z-6x^3y^2z^2)\geq0$$ or
$$\sum\limits_{cyc}(x^4+9x^3y-3x^3z-2x^2y^2-6x^2yz)\geq0$$ or
$$\sum\limits_{cyc}(2x^4+3x^3y+3x^3z-2x^2y^2-6x^2yz)\geq6\sum\limits_{cyc}(x^3z-x^3y)$$ or
$$\sum\limits_{cyc}((x^2-y^2)^2+3z(x+y)(x-y)^2)\geq6(x+y+z)(x-y)(y-z)(z-x)$$ or
$$\sum\limits_{cyc}(x-y)^2(x+y)(x+y+3z)\geq6(x+y+z)(x-y)(y-z)(z-x)$$
Since for $x\geq y\geq z$ we get $\prod\limits_{cyc}(x-y)\leq0$, we can assume $x\geq z\geq y$ and since 
$(x+y+z)(x-y)(y-z)(z-x)=(x+y+z)(x-y)(x-z)(z-y)\leq(x+z)(x-z)xz$, 
it remains to prove 
$$\sum\limits_{cyc}(x-y)^2(x+y)(x+y+3z)\geq6(x+y+z)(x-y)(y-z)(z-x)$$
for $y\rightarrow0^+$, which gives 
$$x^4-3x^3z-2x^2z^2+9xz^3+z^4\geq0$$
which is obviously true.
Done!