<s>[INST] 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.
 [/INST] 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...
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