Example 6 Given $x, y, z \in \mathbf{R}_{+}$, and $x+y+z=1$. Find
$$
\frac{\sqrt{x}}{4 x+1}+\frac{\sqrt{y}}{4 y+1}+\frac{\sqrt{z}}{4 z+1}
$$

the maximum value.
Solving: The difference between this problem and Example 5 is that the condition $x + y + z = \frac{1}{2}$ is changed to $x + y + z = 1$.
Assume without loss of generality that $x \geqslant \frac{1}{3} \geqslant z$. Fix $y$, then
$$
f(x) = \sum \frac{\sqrt{x}}{4 x + 1} \text{ is strictly decreasing} \Leftrightarrow f'(x) < 0, x \in \left[\frac{1}{4}, \frac{1}{2}\right] \\
\Leftrightarrow 8 x(4 x + 1) - (4 x + 1)(4 x - 1) - \\
16 x(4 x - 1) > 0, x \in \left[\frac{1}{4}, \frac{1}{2}\right] \\
\Leftrightarrow 1 + 24 x - 48 x^2 > 0, x \in \left[\frac{1}{4}, \frac{1}{2}\right] \\
\Leftrightarrow 1 + 6 t - 3 t^2 > 0, t \in [1, 2].
$$

It is not hard to see that the final inequality holds.
Therefore, equation (1) holds.
Thus, $f(x)$ is decreasing. We can decrease $x$ and increase $z$ (keeping $x + z$ constant), and $f(x)$ will increase. Continue until one of $x$ or $z$ becomes $\frac{1}{3}$, without loss of generality, let $z = \frac{1}{3}, x \geqslant y$. For
$$
g(x) = \frac{\sqrt{x}}{4 x + 1} + \frac{\sqrt{y}}{4 y + 1} + \frac{\sqrt{\frac{1}{3}}}{4 \times \frac{1}{3} + 1}
$$

Similarly, we find that when $x = y = z = \frac{1}{3}$, $\sum \frac{\sqrt{x}}{4 x + 1}$ achieves its maximum value $\frac{3 \sqrt{3}}{7}$.