Example 10 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Find
$$
\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}
$$

the minimum value.
Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}(x, y, z \in \mathbf{R}_{+})$, then
$$
\begin{array}{l}
\frac{1}{2 a+1}+\frac{1}{2 b+1}+\frac{1}{2 c+1} \\
=\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z} .
\end{array}
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{[y(y+2 x)+z(z+2 y)+x(x+2 z)] \cdot} \\
\quad\left(\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z}\right) \\
\geqslant(x+y+z)^{2} .
\end{array}
$$

Thus, $\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z}$
$$
\begin{array}{l}
\geqslant \frac{(x+y+z)^{2}}{[y(y+2 x)+z(z+2 y)+x(x+2 z)]} \\
=1,
\end{array}
$$

which means $\frac{1}{2 a+1}+\frac{1}{2 b+1}+\frac{1}{2 c+1} \geqslant 1$.
The equality holds if and only if $a=b=c=1$. Therefore, the minimum value is 1.