Xét $a-\frac{a^3+ab^2}{a^2+b+b^2}=\frac{ab}{a^2+b^2+b} \Rightarrow \frac{a^3+ab^2}{a^2+b+b^2}=a-\frac{ab}{a^2+b^2+b}$$\Rightarrow \frac{a^3+ab^2}{a^2+b^2+b}\geq a-\frac{ab}{2ab+b}=a-\frac{a}{2a+1}$
$\frac{a^3+ab^2}{a^2+b+b^2}\geq \frac{2a^2+a-a}{2a+1}\Rightarrow \frac{a^3+ab^2}{a^2+b^2+b}\geq \frac{2a^2}{2a+1}$
Ta có : $VT=\sum\frac{a^3+ab^2}{a^2+b+b^2}\geq \sum\frac{2a^2}{2a+1}\geq \frac{2(a+b+c)^2}{2.(a+b+c)+3}=2$
(Theo bất đẳng thức $Cauchy-Swart$ )
Vậy ta có $VT\geq 2$. Dấu bằng xảy ra khi a=b=c=1