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?

Another way.
Let $a\geq b\geq c$.
Thus, $$\sum_{cyc}\frac{1}{a^2+b^2+ab}-\frac{9}{(a+b+c)^2}=\sum_{cyc}\left(\frac{1}{a^2+b^2+ab}-\frac{3}{(a+b+c)^2}\right)=$$
$$=\frac{1}{(a+b+c)^2}\sum_{cyc}\frac{c^2-2a^2-2b^2-ab+2ac+2bc}{a^2+b^2+ab}=$$
$$=\frac{1}{2(a+b+c)^2}\sum_{cyc}\frac{(c-a)(c+b+4a)-(b-c)(a+c+4b)}{a^2+b^2+ab}=$$
$$=\frac{1}{2(a+b+c)^2}\sum_{cyc}(a-b)\left(\frac{a+c+4b}{b^2+c^2+bc}-\frac{c+b+4a}{a^2+c^2+ac}\right)=$$
$$=\frac{\sum\limits_{cyc}(a-b)^2(a^2+b^2-2c^2+5ab+2ac+2bc)(a^2+b^2+ab)}{2(a+b+c)^2\prod\limits_{cyc}(a^2+b^2+ab)}\geq$$
$$\geq\frac{\sum\limits_{cyc}(a-b)^2c(a+b-c)(a^2+b^2+ab)}{(a+b+c)^2\prod\limits_{cyc}(a^2+b^2+ab)}\geq\frac{(a-c)^2b(a+c-b)(a^2+c^2+ac)+(b-c)^2a(b+c-a)(b^2+c^2+bc)}{(a+b+c)^2\prod\limits_{cyc}(a^2+b^2+ab)}\geq$$
$$\geq\frac{(b-c)^2b(a-b)(a^2+c^2+ac)+(b-c)^2a(b-a)(b^2+c^2+bc)}{(a+b+c)^2\prod\limits_{cyc}(a^2+b^2+ab)}=$$
$$=\frac{(b-c)^2(a-b)^2(ab-c^2)}{(a+b+c)^2\prod\limits_{cyc}(a^2+b^2+ab)}\geq0.$$
