Solution: It's easy to guess that when $x=y=z=\frac{1}{6}$, equation (1) takes the maximum value $\frac{3}{5} \sqrt{\frac{3}{2}}$ (at least this is a "candidate" for the maximum value. Another candidate is when $x=\frac{1}{2}, y=z=0$, the function value is $\frac{1}{3} \sqrt{\frac{1}{2}}$. These special values are worth noting).

Assume $x>\frac{1}{6}>z$. Fix $y$, then $z=\frac{1}{2}-y-x$.
Consider the function of $x$, $f(x)=\sum \frac{\sqrt{x}}{4 x+1}$, $f(x)$ is strictly decreasing $\Leftrightarrow^{\prime}(x)2$ under the constraint that $z=\frac{1}{6}$, and $x>y$. For
$$\phi(x)=\frac{\sqrt{x}}{4 x+1}+\frac{\sqrt{y}}{4 y+1}+\frac{\sqrt{\frac{1}{6}}}{4 \times \frac{1}{6}+1}$$
(where $y=\frac{1}{3}-x$), similarly, we find that when $x=y=z=\frac{1}{6}$, $\sum \frac{\sqrt{x}}{4 x+1}$ achieves the maximum value $\frac{3}{5} \sqrt{\frac{3}{2}}$.

The above solution also shows that when $y=z=0, x=\frac{1}{2}$, $\sum \frac{\sqrt{x}}{4 x+1}$ achieves the minimum value $\frac{1}{3} \sqrt{\frac{1}{2}}$.
Example 5 Given $x, y, z \in \mathbf{R}_{+} \cup\{0\}$, and $x+y+z=\frac{1}{2}$. Find
$$\frac{\sqrt{x}}{4 x+1}+\frac{\sqrt{y}}{4 y+1}+\frac{\sqrt{z}}{4 z+1}$$

the maximum value.