<s>[INST] Prove that $\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}$. 
Let $a,b,c$ be positive real numbers. Prove that $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}.$$

I want to prove the inequality with elementary inequalities ( Mean inequalities, Cauchy-Schwarz etc.). I tried to use Cauchy-Schwarz, but I got $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\frac{9}{2(a^2+b^2+c^2)+ab+bc+ca}.$$
Then we have to show that $$2(a^2+b^2+c^2)+ab+bc+ca\leq (a+b+c)^2 $$
which is not true.
Multiplying the numerators of the fractions in LHS by $a^2,b^2,c^2$ also makes the problem more difficult.
So, how to solve the problem with elementary inequalities?
 [/INST] Multiplying the numerators of the fractions  by $a^2$, $b^2$ and $c^2$ gives a wrong inequality:
$$\sum_{cyc}\frac{1}{b^2+c^2+bc}=\sum_{cyc}\frac{a^2}{a^2(b^2+c^2+bc)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2a^2b^2+a^2bc)}$$ and we need to prove that:
$$(a+b+c)^4\geq9\sum\limits_{cyc}(2a^2b^2+a^2bc),$$ which is wrong for $c\rightarrow0^+$ and $a=b=1$.
But the following C-S helps already:
$$\sum_{cyc}\frac{1}{b^2+c^2+bc}=\sum_{cyc}\frac{(3a+b+c)^2}{(3a+b+c)^2(b^2+c^2+bc)}\geq\frac{25(a+b+c)^2}{\sum\limits_{cyc}(3a+b+c)^2(b^2+c^2+bc)}$$ and it's enough to prove that:
$$25(a+b+c)^4\geq9\sum\limits_{cyc}(3a+b+c)^2(b^2+c^2+bc)$$ or
$$\sum_{cyc}(7a^4+19a^3b+19a^3c-48a^2b^2+3a^2bc)\geq0$$ or
$$3\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+\sum_{cyc}(4a^4+22a^3b+22a^3c-48a^2b^2)\geq0,$$ which is true by Schur and Muirhead.
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