Example 1 Find the largest constant $k$, such that for $x, y, z \in \mathbf{R}_{+}$, we have
$$
\sum \frac{x}{\sqrt{y+z}} \geqslant k \sqrt{\sum x},
$$

where, “$\sum$” denotes the cyclic sum.
【Analysis】By symmetry, let $x=y=z=1$, we get
$$
k \leqslant \sqrt{\frac{3}{2}} \text {. }
$$

Next, we prove: $\sum \frac{x}{\sqrt{y+z}} \geqslant \sqrt{\frac{3}{2}} \sqrt{\sum x}$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(\sum x \sqrt{y+z}\right)\left(\sum \frac{x}{\sqrt{y+z}}\right) \geqslant\left(\sum x\right)^{2} \\
\Leftrightarrow \sum \frac{x}{\sqrt{y+z}} \geqslant \frac{\left(\sum x\right)^{2}}{\sum x \sqrt{y+z}} .
\end{array}
$$

Without loss of generality, assume $x \geqslant y \geqslant z$. Then
$$
\sqrt{y+z} \leqslant \sqrt{z+x} \leqslant \sqrt{x+y} \text {. }
$$

By the Chebyshev's inequality, we get
$$
\begin{array}{l}
\sum x \sqrt{y+z} \leqslant \frac{1}{3}\left(\sum x\right)\left(\sum \sqrt{y+z}\right) \\
\leqslant \frac{1}{3}\left(\sum x\right) \sqrt{3 \sum(y+z)} \\
=\sqrt{\frac{2}{3}}\left(\sum x\right)^{\frac{3}{2}} .
\end{array}
$$

In conclusion, inequality (1) holds.
Thus, $k_{\max }=\sqrt{\frac{3}{2}}$.