5. Given non-negative real numbers $u, v, w$ satisfy $u+v+w=2$. Then the range of $u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2}$ is $\qquad$ -
5. $[0,1]$.

Obviously, $u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2} \geqslant 0$.
If we take $u=2, v=w=0$, then the equality holds. Therefore, the minimum value of $u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2}$ is 0.
Assume without loss of generality that $u=\max \{u, v, w\}$.
Thus, $v w \leqslant u^{2}<2 u^{2}$. Therefore,
$$
\begin{array}{l}
u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2}=u^{2}(v+w)^{2}+v w\left(v w-2 u^{2}\right) \\
\leqslant[u(v+w)]^{2} \leqslant\left[\frac{(u+v+w)^{2}}{4}\right]^{2}=1 .
\end{array}
$$

When $u=v=1, w=0$, the equality holds.
Hence, the maximum value of $u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2}$ is 1.
Therefore, the range of values for $u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2}$ is $[0,1]$.