Prove the following inequality $a,b,c\in R_{+}^{*}$ and $2(a+b+c)=3$ Prove : 
$\sum_{cyc}\frac{4a}{b^2+2(b+1)}≤3$ 
Where : 
 $a,b,c\in R_{+}^{*}$ , $2(a+b+c)$=3$
I think we use Cauchy inequality : 
$\sum_{cyc}\frac{4a}{b^2+2(b+1)}≤4\sqrt{\sum_{cyc}\frac{a^2}{(b^2+2(b+1))^2}}$ 
Then I use : 
$a^{2}+b^{2}+c^{2}≤(a+b+c)^{2}=\frac{9}{4}$ 
But what about the rest ? 
I don't if my idea help me or no ? 

As $b^2+2b+2\geq 2$, we have $$\sum_{cyc}\frac{4a}{b^2+2b+2}\leq \sum_{cyc} \frac{4a}{2}=3$$
Equality holds at $a=b=0$ and $c=\frac32$.
