solved here

This can also be done as follows. First, \[ \sum \frac{1}{a+1} \ge \frac{9}{a+b+c+3} = \frac32, \]by AM-HM. Equipped with this, we apply Cauchy-Schwarz to find that \[ \left(\sum \frac{a^2(b+1)}{ab+a+b}\right)\left(\sum \frac{ab+a+b}{b+1}\right)\ge (a+b+c)^2 = 9. \]Moreover, \[ \sum \frac{ab+a+b}{b+1} = \sum \frac{(a+1)(b+1)-1}{b+1} = 6 - \sum \frac{1}{a+1}\le \frac92. \]Combining the last two displays, we establish the desired inequality.