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.
Solution: 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 $x \geqslant \frac{1}{3} \geqslant z$. Fix $y$, and let $f(x)=\sum \frac{\sqrt{x}}{4 x+1}$, which is strictly decreasing $\Leftrightarrow^{\prime}(x)0, x \in\left[\frac{1}{4}, \frac{1}{2}\right]$
$\Leftrightarrow x(4 x+1)-(4 x+1)(4 x-1)-$
$16 x(4 x-1)>0, x \in\left[\frac{1}{4}, \frac{1}{2}\right]$
$\Leftrightarrow+24 x-48 x^{2}>0, x \in\left[\frac{1}{4}, \frac{1}{2}\right]$
$\Leftrightarrow+6 t-3 t^{2}>0, t \in[1,2]$.
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 $x+z$ unchanged), and $f(x)$ will increase. Continue this until one of $x$ or $z$ becomes $\frac{1}{3}$, let's say $z=\frac{1}{3}, x \geqslant y$. 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 $\frac{3 \sqrt{3}}{7}$.

Example 6 uses the same function as Example 5, but the condition $x+y+z=\frac{1}{2}$ is changed to $x+y+z=1$, making the adjustment slightly more complex than in Example 5.

It is worth noting that many people prefer to use Jensen's inequality to handle such problems. However, using Jensen's inequality requires that the function be convex (concave). To prove that a function is convex (concave), one needs to find the second derivative and show that it is always negative (positive) (except for a few known convex (concave) functions, such as $y=\lg x, y=a x^{2}+b x+c, y=\sin x(x \in(0, \pi)$).