$\sum \frac{a}{b^{3}+ab}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\sum \frac{b}{a+b^{2}}\geq \sum \frac{1}{a}-\frac{1}{2}\sum \frac{1}{\sqrt{a}}\geq \frac{(\sum \frac{1}{\sqrt{a}})^{2}}{3}-\frac{1}{2}\sum \frac{1}{\sqrt{a}}$
Đặt $t=\sum \frac{1}{^{\sqrt{a}}}\geq 3$
Ta chứng minh $\frac{t^{2}}{3}-\frac{t}{2}\geq \frac{3}{2}\Leftrightarrow (t-3)(2t+3)\geq 0$ (TRUE)
BĐT đc chứng minh