Example 2 Find the maximum constant $k$, such that for any $x, y, z \in R^{+}$, we have
$$
\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\frac{z}{\sqrt{x+y}} \geqslant k \sqrt{x+y+z} .
$$
Solution: First, let $x=y=z=1$, we get $k \leqslant \sqrt{\frac{3}{2}}$.
Next, we prove: $\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\frac{z}{\sqrt{x+y}} \geqslant \sqrt{\frac{3}{2}(x+y+z)}$.
By the symmetry of the inequality, we may assume $x \geqslant y \geqslant z>0$, then
$$
\sqrt{x+y} \geqslant \sqrt{x+z} \geqslant \sqrt{y+z} \text {. }
$$

By Chebyshev's inequality and Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
x \sqrt{y+z}+y \sqrt{z+x}+z \sqrt{x+y} \\
\leqslant \frac{1}{3}(x+y+z)(\sqrt{y+z}+\sqrt{z+x}+\sqrt{x+y}) \\
\leqslant \frac{1}{3}(x+y+z) \cdot \sqrt{3[(y+z)+(z+x)+(x+y)]} \\
=\frac{\sqrt{6}}{3}(x+y+z)^{\frac{3}{2}} \text {. } \\
\end{array}
$$

And $[x \sqrt{y+z}+y \sqrt{z+x}+z \sqrt{x+y}]\left[\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\right.$
$$
\left.\frac{z}{\sqrt{x+y}}\right] \geqslant(x+y+z)^{2} \text {, }
$$

Thus, $\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\frac{z}{\sqrt{x+y}}$
$$
\geqslant \frac{(x+y+z)^{2}}{\frac{\sqrt{6}}{3}(x+y+z)^{\frac{3}{2}}}=\sqrt{\frac{3}{2}(x+y+z)} \text {. }
$$

Therefore, the maximum value of $k$ is $\sqrt{\frac{3}{2}}$.