9. 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 $\qquad$
9. $[0,1]$.

Solution: Clearly, $u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2} \geqslant 0$, and when $u=2$, $v=0, w=0$, the equality holds. Therefore, the minimum value is 0. Without loss of generality, assume $u=\max \{u, v, w\}$, then $v w \leqslant u^{2}<2 u^{2}$,

Thus, $u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2}=u^{2}(v+w)^{2}+$ $v w\left(v w-2 u^{2}\right)$
$$
\leqslant[u(v+w)]^{2} \leqslant\left[\frac{(u+v+w)^{2}}{4}\right]^{2} \leqslant 1,
$$

When $u=v=1, w=0$, the equality holds in the above expression. Therefore, the maximum value is 1. Hence, the range of values is $[0,1]$.