$$\sum_{cyc}\frac{9a}{b^2+2(b+1)}≥4$$
With $a,b,c>0$ and $2(a+b+c)=3$.
First I use AM-GM inequality
$$\sum_{cyc}\frac{9a}{b^2+2(b+1)}≥4\left(\frac{abc}{\prod_{cyc}(b^2+2(b+1))^2)}\right)^{1/3}$$ I don't if my idea help me or no and can complete my problem ?
$f(x)=\frac{1}{(x+1)^2+1}$ is convex for $x\geq 0$, so by Jensen: $$\sum_{cyc}\frac{a}{b^2+2(b+1)}≥(a+b+c)\frac{1}{\left(\frac{ab+bc+ca}{a+b+c}+1\right)^2+1}\geq (a+b+c) \frac{1}{\left(\frac{a+b+c}{3}+1\right)^2+1}=\frac{6}{13}$$
Last inequality is true because of $ab+bc+ca\leq \frac{(a+b+c)^2}{3}$