Có $\frac{a^3}{b+1}+\frac{b+1}{4}+\frac{1}{2}\geq \frac{3a}{2}\Rightarrow \frac{a^3}{b+1}\geq \frac{3a}{2}-\frac{b}{4}-\frac{3}{4}$Cmtt: $\frac{b^3}{a+1}\geq \frac{3b}{2}-\frac{a}{4}-\frac{3}{4}$
$\Rightarrow A\geq \frac{5}{4}(a+b)-\frac{3}{2}\geq \frac{5}{2}\sqrt{ab}-\frac{3}{2}=1$
Dấu $=$ có khi $a=b=1$