3. In $\triangle A B C$, $B C=a, C A=b, A B=c$. If $2 a^{2}+b^{2}+c^{2}=4$, then the maximum value of the area of $\triangle A B C$ is $\qquad$ (Provided by Li Hongchang)
3. $\frac{\sqrt{5}}{5}$.

Let the midpoint of $B C$ be $M$. By the median length formula, we get $M N=\frac{1}{2} \sqrt{2\left(b^{2}+c^{2}\right)-a^{2}}$. Then the area of $\triangle A B C$ is
$$
\begin{aligned}
S_{\triangle A B C} & =\frac{1}{2} \cdot a \cdot A M \cdot \sin \angle A M C \\
& \leqslant \frac{1}{2} a \sqrt{2\left(b^{2}+c^{2}\right)-a^{2}}=\frac{1}{4} a \sqrt{8-5 a^{2}} \\
& \leqslant \frac{1}{4 \sqrt{5}} \cdot \frac{5 a^{2}+\left(8-5 a^{2}\right)}{2}=\frac{\sqrt{5}}{5} .
\end{aligned}
$$

Equality holds if and only if $5 a^{2}=8-5 a^{2}$ and $A M \perp B C$, i.e., $a=\frac{2}{\sqrt{5}}, b=c=\sqrt{\frac{6}{5}}$. Therefore, the maximum area of $\triangle A B C$ is $\frac{\sqrt{5}}{5}$.