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$x,y,z > 0$ 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

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  • $\begingroup$ Have you looked at the proof of the Iran inequality? Does a similar proof work here? $\endgroup$
    – Servaes
    May 20, 2016 at 20:17
  • $\begingroup$ If you say about brute force proof, I would say this is a lot of work for this inequality $\endgroup$
    – HN_NH
    May 20, 2016 at 20:18
  • $\begingroup$ I'm not sure what you mean by a brute force proof here. Either way, Chebyshev's sum inequality proves this without too much force. Maybe this also works for your inequality? $\endgroup$
    – Servaes
    May 20, 2016 at 20:25
  • $\begingroup$ by brute force, I means we clear fraction and use Muihead to prove. I have no idea what you means by Chebyshev's sum inequality. $\endgroup$
    – HN_NH
    May 20, 2016 at 20:27
  • $\begingroup$ See en.wikipedia.org/wiki/Chebyshev's_sum_inequality $\endgroup$
    – Servaes
    May 20, 2016 at 20:27

1 Answer 1

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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!

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