3. Let the function be
$$
y(x)=(\sqrt{1+x}+\sqrt{1-x}+2)\left(\sqrt{1-x^{2}}+1\right) \text {, }
$$

where, $x \in[0,1]$. Then the minimum value of $y(x)$ is
$$
\begin{array}{l}
\text { 3. } 2+\sqrt{2} \text {. } \\
\text { Let } u=\sqrt{1+x}+\sqrt{1-x}(x \in[0,1]) \text {. }
\end{array}
$$

Given that $u(x)$ is monotonically decreasing, we know $u \in[\sqrt{2}, 2]$. Then $y=\frac{u^{2}(u+2)}{2}$ is monotonically increasing. Therefore, when $u=\sqrt{2}$, the original function achieves its minimum value $2+\sqrt{2}$.