Prove the following inequality $a,b,c\in R_{+}^{*}$ and $2(a+b+c)=3$

Question

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 ?

Answer 1

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 $c=\frac32$.