Example 9 Let $a, b, c \in \mathbf{R}^{+}$, and $a b c=1$, find the maximum value of $\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}$.
Let $a=x^{3}, b=y^{3}, c=z^{3}$, where $x, y, z \in \mathbf{R}^{+}$, then by the given conditions we have $x y z=1$. Using the comparison method, it is easy to prove: $x^{3}+y^{3} \geqslant x^{2} y+y^{2} x$, thus we have
$$
\begin{aligned}
\frac{1}{a+b+1} & =\frac{1}{x^{3}+y^{3}+1} \leqslant \frac{1}{x^{2} y+x y^{2}+1}=\frac{1}{x^{2} y+x y^{2}+x y z} \\
& =\frac{z}{x y z(x+y+z)}=\frac{z}{x+y+z}
\end{aligned}
$$

Similarly, we have $\frac{1}{b+c+1} \leqslant \frac{x}{x+y+z}, \frac{1}{c+a+1} \leqslant \frac{y}{x+y+z}$.
Adding the above three inequalities, we get $\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1} \leqslant 1$,
Therefore, when $a=b=c=1$, $\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}$ achieves its maximum value of 1.