Example 8. (1994, Communication Competition Question) Find the maximum constant $k$, such that for any $x, y, z \in R^{+}$, we have
$$
\sum \frac{x}{\sqrt{y+z}} \geqslant k \sqrt{\sum x} .
$$
Solution: First, let $x=y=z=1$, we get $k \leqslant \sqrt{\frac{3}{2}}$. Next, we prove $\sum \frac{x}{\sqrt{y+z}} \geqslant \sqrt{\frac{3}{2}} \sqrt{\sum x}$.
Assume without loss of generality that $x \geqslant y \geqslant z$, by Chebyshev's inequality, we have
$$
\begin{array}{l}
3 \sum_{x} \sqrt{y+z} \leqslant \sum x \sum \sqrt{y+z} \\
\leqslant \sum_{x} \sqrt{3 \sum(y+z)} \\
=\sqrt{6}\left(\sum_{x)^{\frac{3}{2}}}\right.
\end{array}
$$

By Cauchy's inequality, we have
$$
\begin{array}{l}
\sum x \sqrt{y+z} \sum \frac{x}{\sqrt{y+z}} \geqslant\left(\sum x\right)^{2}, \\
\therefore \sum \frac{x}{\sqrt{y+z}} \geqslant \frac{\left(\sum x\right)^{2}}{\sum x \sqrt{y+z}} \\
\quad \geqslant \sqrt{\frac{3}{2}} \sqrt{\sum x} . \quad \text { (to be continued on page 20) }
\end{array}
$$