Example 6 For all $a, b, c \in \mathbf{R}_{+}$, and $abc=1$, find the minimum value of $S=\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}$.
【Analysis】Since the condition $abc=1$ is not linear, we can use the method of taking logarithms to convert it into a linear form. Meanwhile, $S=\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}$ can be transformed into the form $S=\frac{1}{\mathrm{e}^{x}+1}+\frac{1}{\mathrm{e}^{y}+1}+\frac{1}{\mathrm{e}^{x}+1}$, and then use the derivative method to solve it.
Solution Let $x=\ln 2a, y=\ln 2b, z=\ln 2c$.
Then the problem is equivalent to:
For all $x, y, z \in \mathbf{R}$, and $x+y+z=3 \ln 2$, find the minimum value of $S=\frac{1}{\mathrm{e}^{x}+1}+\frac{1}{\mathrm{e}^{y}+1}+\frac{1}{\mathrm{e}^{z}+1}$.
Let $f(x)=\frac{1}{\mathrm{e}^{x}+1}(x \in \mathbf{R})$. Then $f^{\prime}(x)=-\frac{\mathrm{e}^{x}}{\left(\mathrm{e}^{x}+1\right)^{2}}<0$ when $x>0$, and $f^{\prime \prime}(x)>0$, i.e., the function is convex; when $x<0, f(-x)+f(x)=1$ and $f(0)=\frac{1}{2}$.
(1) When $x, y, z \geqslant 0$, we have
$$
\begin{array}{l}
f^{\prime}(\ln 2)=-\frac{2}{9}, f(\ln 2)=\frac{1}{3} \\
\Rightarrow f(x) \geqslant-\frac{2}{9}(x-\ln 2)+\frac{1}{3} \\
\Rightarrow S=\frac{1}{\mathrm{e}^{x}+1}+\frac{1}{\mathrm{e}^{y}+1}+\frac{1}{\mathrm{e}^{z}+1} \\
\geqslant-\frac{2}{9}((x+y+z)-3 \ln 2)+3 \times \frac{1}{3}=1 \\
\Rightarrow S=\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1} \text { has a minimum value of } 1.
\end{array}
$$

(2) When two of $x, y, z$ are not greater than 0, without loss of generality, assume $x, y \leqslant 0$, then
$$
\begin{array}{l}
f(x) \geqslant f(0)=\frac{1}{2}, f(y) \geqslant f(0)=\frac{1}{2} \\
\Rightarrow S=f(x)+f(y)+f(z) \geqslant 1+f(z)>1 . \\
\text { Also, } z=3 \ln 2-x-y>-x>0, \text { so } \\
f(z)+f(x)>1.
\end{array}
$$
In summary, the minimum value of $S$ is 1.