ta có $\frac{a^{3}}{2\sqrt{b^{2}+3}}+\frac{a^{3}}{2\sqrt{b^{2}+3}} +\frac{b^{2}+3}{16}\geq \frac{3a^{2}}{4} (cosi cho 3 số)$ $\Rightarrow \frac{a^{3}}{\sqrt{b^{2}+3}}+\frac{b^{2}+3}{16}\geq \frac{3a^{2}}{4}$
TT $\Rightarrow P +\frac{a^{2} +b^{2} +c^{2}+9}{16}\geq \frac{3}{4}(a^{2}+b^{2}+c^{2})$
$\Rightarrow P\geq \frac{3}{2}$
dấu "=" $\Leftrightarrow a=b=c=1$