[tex]\frac{a^3}{a^2+b^2}=a-\frac{ab^2}{a^2+b^2}\geq a-\frac{ab^2}{2ab}=a-\frac{b}{2}[/tex]
[tex]\Rightarrow P\geq \frac{a+b+c}{4}-\sqrt{\frac{a+b+c}{4}}+\frac{1}{4}+\frac{3}{4}=\left ( \sqrt{\frac{a+b+c}{4}}-\frac{1}{2} \right )^2+\frac{3}{4}\geq \frac{3}{4}[/tex]