Example 1 Let $x, y, z > 0$, and $x+y+z=1$. Find
$$
f(x, y, z)=\sum \frac{x(2 y-z)}{1+x+3 y}
$$

the maximum value.
Notice,
$$
\begin{array}{l}
f(x, y, z)=\sum \frac{x(2 y-z)}{1+x+3 y} \\
=\sum \frac{x(2 y-z)}{2+(2 y-z)} .
\end{array}
$$

Consider the function $f(x)=\frac{x}{x+2}$.
It is easy to see that, $f^{\prime}(x)=\frac{2}{(x+2)^{2}}>0$,
$$
f^{\prime \prime}(x)=\frac{-4}{(x+2)^{3}}<0 \text {. }
$$

Therefore, $f(x)$ is a convex function.
By the weighted Jensen's inequality, we get
$$
\begin{array}{l}
\sum x f(2 y-z) \leqslant f\left(\sum x(2 y-z)\right) \\
=f\left(\sum x y\right) .
\end{array}
$$

By the AM-GM inequality, we know
$$
x y+y z+z x \leqslant \frac{1}{3}(x+y+z)^{2}=\frac{1}{3} \text {. }
$$

Since $f(x)$ is an increasing function, we have
$$
f\left(\sum x y\right)=\frac{\sum x y}{2+\sum x y} \leqslant \frac{1}{7} \text {. }
$$

Therefore, $f(x, y, z) \leqslant \frac{1}{7}$, with equality holding if and only if $x=y=z=\frac{1}{3}$.