<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] Here is a proof  following Momo answer.
$$\dfrac{1}{a^2+b^2+ab}+\dfrac{1}{b^2+c^2+bc}+\dfrac{1}{a^2+c^2+ac}\ge \dfrac{10.5}{a^2+b^2+c^2+2.5(ab+bc+ca)}$$ where all variables are nonnegative or positive. This is the inequality with $r=2.5$ also if the inequality in Momo answer is true for $r\ge 0$ it is also true for every $r_0\in[0,r]$ (not hard to prove), and $r=2.5$ is the  maximum possible trying $a=b=1$ and $c\to 0$.
Since the inequality is quite symmetric we can multiply out denominators and with Muirhead inequality the case $r=2$ is easy. For $r=2.5$ we obtain the numerator $$7(b^5a+a^5b+c^5b+b^5c+c^5a+a^5c-(b^4a^2+a^4b^2+c^4b^2+b^4c^2+c^4a^2+a^4c^2))+N   $$  where $N$ can be written as $$(a^3-c^3)(a-c)(a^2+c^2+ca-b^2)+(b^3-c^3)(b-c)(b^2+c^2+cb-a^2)+(a^3-b^3)(a-b)(a^2+b^2+ba-c^2).$$ Assuming $b\ge a\ge c$ we see that the only possible  negative term is $ a^2+c^2+ca-b^2$ but $(a^2+c^2+ca-b^2)+(b^2+c^2+bc-a^2)\ge 0$ and this completes the proof.
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