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.

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!

• I suspect the $x$ in the numerator should not be an $x$. – darij grinberg Jan 25 '16 at 16:49
• @darij grinberg because $(x+y)(x+z)=x^2+3v^2$ – Michael Rozenberg Jan 25 '16 at 17:20
• Ah! Forgot about that $\sum$. – darij grinberg Jan 25 '16 at 17:26

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.

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...