Analysis We use homogeneity to simplify the inequality first.
Solution By homogeneity, without loss of generality, assume $x+y+z=1$,
$$
\text { LHS }=\frac{x}{\sqrt{1-x}}+\frac{y}{\sqrt{1-y}}+\frac{z}{\sqrt{1-z}}=\frac{x^{\frac{5}{4}}}{\sqrt[4]{x(1-x)^{2}}}+\frac{y^{\frac{5}{4}}}{\sqrt[4]{y(1-y)^{2}}}+\frac{z^{\frac{5}{4}}}{\sqrt[4]{z(1-z)^{2}}} \text {, }
$$

RHS $=k$, and $x(1-x)^{2}=\frac{1}{2} \cdot 2 x(1-x)(1-x) \leqslant \frac{1}{2} \cdot\left(\frac{2}{3}\right)^{3}=\frac{4}{27}$,
so LHS $\geqslant \frac{x^{\frac{5}{4}}}{\sqrt[4]{\frac{4}{27}}}+\frac{y^{\frac{5}{4}}}{\sqrt[4]{\frac{4}{27}}}+\frac{z^{\frac{5}{4}}}{\sqrt[4]{\frac{4}{27}}}=\sqrt[4]{\frac{27}{4}}\left(x^{\frac{5}{4}}+y^{\frac{5}{4}}+z^{\frac{5}{4}}\right)$. Since $y=x^{\frac{5}{4}}$ is a concave function in the interval $(0,+\infty)$,
by Jensen's inequality: LHS $\geqslant \sqrt[4]{\frac{27}{4}} \cdot 3\left(\frac{x+y+z}{3}\right)^{\frac{5}{4}}=\frac{\sqrt{6}}{2}$, equality holds if and only if $x=y=z=$ $\frac{1}{3}$, so the maximum value of $k$ is $\frac{\sqrt{6}}{2}$.
!Comment Here, the properties of power functions are cleverly used to solve the problem.
Example 4 Find the largest constant $k$, such that for $x, y, z \in \mathbf{R}^{+}$, we have
$$
\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\frac{z}{\sqrt{x+y}} \leqslant k \sqrt{x+y+z} .
$$