3. In $\triangle A B C$, if $B C=a, C A=b, A B=c$. If $2 a^{2}+b^{2}+c^{2}=4$, then the maximum value of the area of $\triangle A B C$ is
3. $\frac{\sqrt{5}}{5}$

Analysis: By the cosine theorem, $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{4-3 a^{2}}{2 b c}$, so $\sin A=\sqrt{1-\left(\frac{4-3 a^{2}}{2 b c}\right)^{2}}$, thus $4 S^{2}=b^{2} c^{2} \sin ^{2} A=b^{2} c^{2}-\left(2-\frac{3}{2} a^{2}\right)^{2} \leq\left(\frac{b^{2}+c^{2}}{2}\right)^{2}-\left(2-\frac{3}{2} a^{2}\right)^{2}=\left(2-a^{2}\right)^{2}-\left(2-\frac{3}{2} a^{2}\right)^{2}$ $=-\frac{5}{4}\left(a^{2}-\frac{4}{5}\right)^{2}+\frac{4}{5} \leq \frac{4}{5}$, hence $S \leq \frac{\sqrt{5}}{5}$, equality holds when $a=\frac{2 \sqrt{5}}{5}, b=c=\frac{\sqrt{30}}{5}$, so the maximum area is $\frac{\sqrt{5}}{5}$.