3. Given $a, b, c, d \in[0, \sqrt[4]{2})$, satisfying: $a^{3}+b^{3}+c^{3}+d^{3}=2$.
Find the minimum value of $\frac{a}{\sqrt{2-a^{4}}}+\frac{b}{\sqrt{2-b^{4}}}+\frac{c}{\sqrt{2-c^{4}}}+\frac{d}{\sqrt{2-d^{4}}}$.

When $a>0$, $\frac{a}{\sqrt{2-a^{4}}}=\frac{a^{3}}{\sqrt{a^{4}\left(2-a^{4}\right)}} \geqslant \frac{a^{3}}{\frac{a^{4}+\left(2-a^{4}\right)}{2}}=a^{3}$, equality holds when $a=1$; it is easy to see that when $a=0$, $\frac{a}{\sqrt{2-a^{4}}}=a^{3}$ also holds.
Therefore, for $a \in[0, \sqrt[4]{2})$, we always have $\frac{a}{\sqrt{2-a^{4}}} \geqslant a^{3}$, equality holds if and only if $a=0$ or 1.
Thus, $\frac{a}{\sqrt{2-a^{4}}}+\frac{b}{\sqrt{2-b^{4}}}+\frac{c}{\sqrt{2-c^{4}}}+\frac{d}{\sqrt{2-d^{4}}} \geqslant a^{3}+b^{3}+c^{3}+d^{3}=2$,
equality holds when exactly two of $a, b, c, d$ are 1 and the other two are 0.
Therefore, the minimum value of $\frac{a}{\sqrt{2-a^{4}}}+\frac{b}{\sqrt{2-b^{4}}}+\frac{c}{\sqrt{2-c^{4}}}+\frac{d}{\sqrt{2-d^{4}}}$ is 2.