5 Let $u$, $v$, $w$ be positive real numbers, satisfying the condition
$$
u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v} \geqslant 1,
$$

Find the minimum value of $u+v+w$.
(Supplied by Chen Yonggao)
By the AM-GM inequality and the conditions given in the problem, we know that
$$
u \frac{v+w}{2}+v \frac{w+u}{2}+w \frac{u+v}{2} \geqslant u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v} \geqslant 1,
$$

which implies
$$
u v+v w+w u \geqslant 1 .
$$

Therefore,
$$
\begin{aligned}
(u+v+w)^{2} & =u^{2}+v^{2}+w^{2}+2 u v+2 v w+2 w u \\
& =\frac{u^{2}+v^{2}}{2}+\frac{v^{2}+w^{2}}{2}+\frac{w^{2}+u^{2}}{2}+2 u v+2 v w+2 w u \\
& \geqslant 3 u v+3 v w+3 w u \geqslant 3,
\end{aligned}
$$

which means
$$
u+v+w \geqslant \sqrt{3} .
$$

On the other hand, it is clear that $u=v=w=\frac{\sqrt{3}}{3}$ satisfies the conditions given in the problem, and in this case, $u+v+w=\sqrt{3}$.
Combining these two aspects, we conclude that the minimum value of $u+v+w$ is $\sqrt{3}$.