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?

The solution by Vo Quoc Ba Can
Multiplying both sides by $a^2+b^2+c^2+ab+bc+ca,$ the inequality is equivalent to
$$\sum {\frac{a^2+b^2+c^2+ab+bc+ca}{a^2+ab+b^2}} \geqslant \frac{9(a^2+b^2+c^2+ab+bc+ca)}{(a+b+c)^2},$$
equivalent to
$$3+(a+b+c)\sum {\frac{c}{a^2+ab+b^2}} \geqslant 9-\frac{9(ab+bc+ca)}{(a+b+c)^2},$$
or
$$(a+b+c)\sum {\frac{c}{a^2+ab+b^2}}+\frac{9(ab+bc+ca)}{(a+b+c)^2}\geq6.$$
By theCauchy-Schwartz inequality, we have
$$\sum {\frac{c}{a^2+ab+b^2}}=\sum {\frac{c^2}{c(a^2+ab+b^2)}}\geq\frac{(a+b+c)^2}{\sum{c(a^2+ab+b^2)}}=\frac{a+b+c}{ab+bc+ca}.$$
Hence, it suffices to prove that
$$\frac{(a+b+c)^2}{ab+bc+ca}+\frac{9(ab+bc+ca)}{(a+b+c)^2}\geq6.$$
Which is just AM-GM.
