5. Let $x, y \geqslant 0$, and $x+y \leqslant \sqrt{\frac{2}{3}}$. Then
$$
\sqrt{2-3 x^{2}}+\sqrt{2-3 y^{2}}
$$

the minimum value is $\qquad$
5. $\sqrt{2}$.

Obviously, $0 \leqslant x, y \leqslant \sqrt{\frac{2}{3}}$.
Let $x=\sqrt{\frac{2}{3}} \cos \alpha, y=\sqrt{\frac{2}{3}} \cos \beta$, where $0^{\circ} \leqslant \alpha, \beta \leqslant 90^{\circ}$.
Then $\sqrt{2-3 x^{2}}+\sqrt{2-3 y^{2}}$
$$
=\sqrt{2}(\sin \alpha+\sin \beta),
$$

and $\cos \alpha+\cos \beta \leqslant 1$.
Let $t=\sin \alpha+\sin \beta$. Then
$$
t^{2}+1^{2} \geqslant 2+2 \cos (\alpha-\beta).
$$

It is easy to see that $0^{\circ} \leqslant|\alpha-\beta| \leqslant 90^{\circ}$.
Thus, $\cos (\alpha-\beta) \geqslant 0$.
Therefore, $t^{2}+1^{2} \geqslant 2 \Rightarrow t \geqslant 1$.
Then $\sqrt{2-3 x^{2}}+\sqrt{2-3 y^{2}}=\sqrt{2} t \geqslant \sqrt{2}$.
Hence, when one of $x, y$ is 0 and the other is $\sqrt{\frac{2}{3}}$,
the equality holds.