76. Let $u, v, w$ be positive real numbers, satisfying the condition $u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v}=1$, find the minimum value of $u+v+$ $w$ ( (3rd China Girls Mathematical Olympiad problem)
76. By the AM-GM inequality and the conditions given in the problem, we have
$$u \cdot \frac{v+w}{2}+v \cdot \frac{w+u}{2}+w \cdot \frac{u+v}{2} \geqslant u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v}=1$$

Thus, \(uv + vw + wu \geqslant 1\). Therefore,
$$\begin{array}{l}
(u+v+w)^{2}=u^{2}+v^{2}+w^{2}+2uv+2vw+2wu= \\
\frac{u^{2}+v^{2}}{2}+2 \frac{v^{2}+w^{2}}{2}+\frac{w^{2}+u^{2}}{2}+2uv+2vw+2wu \geqslant \\
uv + vw + wu + 2uv + 2vw + 2wu = \\
3(uv + vw + wu) \geqslant 3
\end{array}$$

Thus, \(u + v + w \geqslant \sqrt{3}\). 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 \(\sqrt{3}\).