21. (2004 China Western Mathematical Olympiad) 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 $u+v+w$.
21. By the AM-GM inequality and the conditions given in the problem, we know
$$
u \frac{v+w}{2}+v \frac{w+u}{2}+w \frac{u+v}{2} \geqslant u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v} \geqslant 1 \text {, }
$$

which implies $u v+v w+r u \geqslant 1$.
$$
\begin{array}{l} 
\text { Hence } \\
(u+v+w)^{2}=u^{2}+v^{2}+w^{2}+2 u v+2 v w+2 w u \\
\quad \frac{u^{2}+v^{2}}{2}+\frac{v^{2}+w^{2}}{1}+\frac{w^{2}+u^{2}}{2}+2 u v+2 v w+2 w u \\
\quad \geqslant 3 u v+3 v w+3 w u \geqslant 3 .
\end{array}
$$

Therefore, $u+v+w \geqslant \sqrt{3}$.
On the other hand, $u=v=w=\frac{\sqrt{3}}{3}$ clearly satisfies the conditions given in the problem, and in this case, $u+v+w=\sqrt{3}$.
In conclusion, the minimum value of $u+v+w$ is $\sqrt{3}$.