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.

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...
