ta có
$\sum_{}^{}\frac{a^2(b+1)}{ab+a+b}=\sum_{}^{}\frac{a^2}{a+\frac{b}{b+1}}\geq \frac{(a+b+c)^2}{a+b+c+\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}}=\frac{9}{3+\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}}$
vậy chỉ cần chứng minh $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\leq \frac{3}{2}$
ta có $\frac{a}{a+1}=\frac{a+1-1}{a+1}=1-\frac{1}{a+1}$
$\Rightarrow\sum_{}^{}\frac{a}{a+1}=3-\left ( \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} \right )\leq 3-\frac{9}{a+b+c+3}=\frac{3}{2} $
Vậy ta đã có ĐPCM :D
dấu đẳng thức xẩy ra $ \Leftrightarrow a=b=c=1$