Đăt $x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}$
$\Rightarrow x^{2}+y^{2}+z^{2}\geq1$
A=$\frac{x^{3}}{y^{2}+z^{2}}+\frac{y^{3}}{x^{2}+z^{2}}+\frac{z^{3}}{x^{2}+y^{2}}$
ta có $x(y^{2}+z^{2})=\frac{1}{\sqrt{2}}\sqrt{2x^{2}(y^{2}+z^{2})(y^{2}+z^{2}})$
$ \frac{1}{\sqrt{2}}\sqrt{(\frac{2(x^{2}+y^{2}+z^{2})}{3})^{3}} \leq \frac{2\sqrt{3}}{9}(x^{2}+y^{2}+z^{2})\sqrt{x^{2}+y^{2}+z^{2}}$
$\Rightarrow A\geq \frac{(x^{2}+y^{2}+z^{2})^{2}}{ \sum x (y^{2}+z^{2})}$
$ \frac{(x^{2}+y^{2}+z^{2})^{2}}{3.\frac{2\sqrt{3}}{9}(x^{2}+y^{2}+z^{2})\sqrt{x^{2}+y^{2}+z^{2}}}$
= $\frac{\sqrt{3}}{2}\sqrt{x^{2}+y^{2}+z^{2}}=\frac{\sqrt{3}}{2}$
Dấu "=" $\Leftrightarrow a=b=c=\sqrt{3}$