Hard inequality $ (xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4} $

I need to prove or disprove the following inequality: $$ (xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4} $$ For $x,y,z \in \mathbb R^+$. I found no counter examples, so I think it should be true. I tried Cauchy-Schwarz, but I didn't get anything useful. Is it possible to prove this inequality without using brute force methods like Bunching and Schur?

This inequality was in the Iran MO in 1996.

Solution 1:

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, our inequality is equivalent to

$\frac{3v^2\sum\limits_{cyc}(x^2+3v^2)^2}{(9uv^2-w^3)^2}\geq\frac{9}{4}$, which is $f(w^3)\geq0$, where $f$ is a concave function.

Thus, $f$ gets a minimal value for an extremal value of $w^3$, which happens in the following cases.

1. $z\rightarrow0^+$, $y=1$, which gives $(x-1)^2(4x^2+7x+4)\geq0$;

2. $y=z=1$, which gives $x(x-1)^2\geq0$. Done!

Solution 2:

Pursuing a link from a comment above, here's the Iran 1996 solution attributed to Ji Chen. Reproduced here to save click- & scrolling ...

The difference $$4(xy+yz+zx)\cdot\left[\sum_\text{cyc}{(x+y)^2(y+z)^2}\right]-9\prod_\text{cyc}{(x+y)^2}$$ which is equivalent to the given inequality, is presented as the sum of squares $$=\;\sum_{\text{cyc}}{xy(x-y)^2\left(4x^2 + 7xy + 4y^2\right)} \:+\:\frac{xyz}{x+y+z}\sum_{\text{cyc}}{(y-z)^2\left(2yz + (y+z-x)^2\right)}$$ whence non-negativity is clear.