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?

I give another idea. 

WLOG, we may assume that $a \ge b \ge c$. It's easy to see that the original problem follows immediately by adding two following inequalities:
\begin{align} & \frac{1}{a^2+ab+b^2}+\frac{1}{b^2+bc+c^2} \ge \frac{4}{ab+bc+2ca+2b^2} 
 \ \ \ \ \ \ \ \ (1) \\ &\frac{4}{ab+bc+2ca+2b^2} + \frac{1}{c^2+ca+a^2} \ge \frac{9}{\left(a+b+c \right)^2} \ \ \ \ \ \ \ \ (2)\end{align}
Indeed, by reducing and factoring, the first inequality is equivalent with:
\begin{align}  \left(c-a \right)^2 \left(ab+bc+2ca-b^2 \right) \ge 0 \end{align}
which is true because $a \ge b \ge c $. 
For the second one, it can be proved by applying Cauchy-Schwarz as follows:
\begin{align} \frac{4}{ab+bc+2ca+2b^2} &+ \frac{1}{c^2+ca+a^2} \ge \frac{ \left(2+1 \right)^2}{a^2+2b^2+c^2+ab+bc+3ca}\\ &= \frac{9}{ \left(a+b+c \right)^2 -\left(a-b \right) \left(b-c \right)} \ge \frac{9}{ \left(a+b+c \right)^2}\end{align}
Hence, our proof is completed.
