Example 8 Let $a, b, c \in \mathbf{R}^{+}$, and $a b c=1$, find the minimum value of $\frac{1}{2 a+1}+\frac{1}{2 b+1}+\frac{1}{2 c+1}$.
Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, where $x, y, z \in \mathbf{R}^{+}$, then
$$
\frac{1}{2 a+1}+\frac{1}{2 b+1}+\frac{1}{2 c+1}=\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z}
$$

By Cauchy's inequality, we have
$$
[y(y+2 x)+z(z+2 y)+x(x+2 z)]\left(\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z}\right) \geqslant(x+y+z)^{2}
$$

Therefore, $\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z} \geqslant \frac{(x+y+z)^{2}}{[y(y+2 x)+z(z+2 y)+x(x+2 z)]}=1$
which means $\frac{1}{2 a+1}+\frac{1}{2 b+1}+\frac{1}{2 c+1} \geqslant 1$, with equality holding if and only if $a=b=c=1$.