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.
 A: 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.


*

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

*$y=z=1$, which gives $x(x-1)^2\geq0$. Done!
A: 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.
A: Each of the two expressions that are multiplying are symmetric with respect to the three variables x, y, z.
Making a change of varibales:
$$a = x + y\\
b = y + z\\
c = z + x$$
It causes the second expression to become another symmetric expression:
$$ \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$$
And that the first expression also becomes another symmetrical expression:
$$ \frac{(a+b+c)^2-2(a^2+b^2+c^2)}{4}$$
Multiplying and eliminating the 4 of the denominator of the inequality:
$$ ((a+b+c)^2-2(a^2+b^2+c^2))(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) \geq 9 $$
Renaming $M_n = \sum_{k} a_k^n$, then:
$$ (M_1^2-2M_2)M_{-2} \geq 9 $$
$$ M_1^2-2M_2 \geq 9M_{-2}^{-1} $$
All this is related to statistics, arithmetic mean, variance, moments, etc.
$$\frac{M_1}{3} = \mu; \frac{M_2}{3}=\alpha_2; \frac{M_{-2}}{3}=\alpha_{-2} $$
$$ 9\mu^2-6\alpha_2 \geq 3\alpha_{-2}^{-1} $$
$$ 3\mu^2 \geq \alpha_{-2}^{-1}+2\alpha_2 $$
Continue...
