Example 2 Non-negative real numbers $a$, $b$, $c$ satisfy
$$
S=\frac{1}{1+a b}+\frac{1}{1+b c}+\frac{1}{1+a c}=1 \text{. }
$$

Find the extremum of the elementary symmetric polynomial.
$$
\begin{array}{l}
\text { Given } \\
S(1+a b+1+b c+1+c a) \geqslant 3^{2} \\
\Rightarrow a b+b c+a c \geqslant 6 .
\end{array}
$$
$$
\begin{array}{l}
\text { Since } a^{2}+b^{2}+c^{2} \geqslant a b+b c+a c \\
\Rightarrow a^{2}+b^{2}+c^{2} \geqslant 6 .
\end{array}
$$
$$
\begin{array}{l}
\text { Since }(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c \\
\Rightarrow a+b+c \geqslant 3 \sqrt{2} .
\end{array}
$$

Given the condition $S=1$, we simplify to get
$$
\begin{array}{l}
(a b c)^{2}=2+(a b+b c+a c) \\
\Rightarrow a b c \geqslant 2 \sqrt{2} . \\
\text { Since } \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a b+b c+a c}{a b c} \\
=\frac{(a b c)^{2}-2}{a b c}=a b c-\frac{2}{a b c} \geqslant 2 \sqrt{2}-\frac{2}{2 \sqrt{2}} \\
\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant \frac{3 \sqrt{2}}{2} .
\end{array}
$$

Equations (1) and (5) both achieve equality when $a=b=c=\sqrt{2}$, and these are the minimum values.