\(b^2+c^2\le a^2\Rightarrow\frac{a^2}{b^2+c^2}\ge1\)
\(A\ge\frac{4a^2}{b^2+c^2}+\frac{b^2+c^2}{a^2}=\frac{a^2}{b^2+c^2}+\frac{b^2+c^2}{a^2}+\frac{3a^2}{b^2+c^2}\)
\(A\ge2\sqrt{\frac{a^2\left(b^2+c^2\right)}{a^2\left(b^2+c^2\right)}}+3.1=5\)
\(A_{min}=5\) khi \(b=c=\frac{a}{\sqrt{2}}\)