An Olympiad inequality Iran96

One day, Xiaoke asks me an Olympiad inequality, which recalls me of a lot of beautiful memories.

For any $a,b,c \in \mathbb{R}$, try to prove $$(ab + bc + ca)(\frac{1}{(a+b)^2} + \frac{1}{(b+c)^2} + \frac{1}{(c+a)^2}) \geq \frac{9}{4}$$

This is a very hard inequality named "Iran96". In the WeChat post, it is said that this one is so hard that many students who participates in competitions cannot solve  it. I believe that it is a little exaggerated. Even for a Olympiad question, this is not the most difficult one.

Let us do some transform.

$$LHS = 4\sum [a^5b + b^5 a + 2a^4 b^2 + 2 a^2 b^4 + 5  a^4 bc + 3 b^3 c^3 + 13 a^3  b^2 c + 13  a^3 b c^2 + 8  a^2b^2c^2]$$

$$RHS = 90a^2b^2c^2 + 9\sum[a^4b^2 + a^2b^4 + 2a^3b^3 + 6a^3b^2c + a^2b^3c + 2 a^4bc]$$

We do some simplification and finally we have to prove this inequality $$\sum[4 a^5b + 4ab^5 + 2a^4bc + 2a^2b^2c^2] \geq \sum[a^4b^2 + a^2b^4 + 6b^3c^3 + 2a^3b^2c + 2a^3bc^2]$$

It reduces to three inequalities $$\begin{eqnarray*}3\sum(a^5b + ab^5) &\geq& 3\sum a^3b^3\\\sum(a^5b + ab^5) &\geq& \sum(a^4b^2 + a^2 b^4)\\2abc(\sum a^3 + 3abc) &\geq& 2abc\sum(a^2b + ab^2)\end{eqnarray*}$$

The last one is a very famous inequality called inequality of Schur.