Example 3 Let $a, b, c$ be positive numbers, and $abc=1$, find the minimum value of $\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}$.
Solution: Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}\left(x, y, z \in \mathbf{R}^{+}\right)$, 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 the Cauchy-Schwarz inequality, we have
$$
\begin{array}{c}
{[y(y+2 x)+z(z+2 y)+x(x+2 z)] \cdot} \\
\left(\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z}\right) \geqslant(x+y+z)^{2} .
\end{array}
$$

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

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