Example 1 Let non-negative real numbers $a, b, c$ satisfy $a+b+c=1$. Find the extremum of $S=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$.
By symmetry, we can assume
$$
a \leqslant b \leqslant c\left(0 \leqslant a \leqslant \frac{1}{3} \leqslant c \leqslant 1\right) \text {. }
$$

First, arbitrarily set \( a = a_{0} \). Then \( c + b = 1 - a_{0} \).
Let \( x = c - b \geqslant 0 \). Then
$$
c = \frac{1 - a_{0} + x}{2}, \quad b = \frac{1 - a_{0} - x}{2} .
$$

Substituting into \( S \) gives
$$
\begin{array}{l} 
S = \frac{1}{1 + a_{0}} + \frac{1}{1 + \frac{1 - a_{0} + x}{2}} + \frac{1}{1 + \frac{1 - a_{0} - x}{2}} \\
= \frac{1}{1 + a_{0}} + \frac{4(3 - a_{0})}{(3 - a_{0})^2 - x^2} .
\end{array}
$$

Considering \( a_{0} \) as a constant, \( S \) is a function of \( x \).
From equation (1), \( S \) increases as \( x \) increases. Hence, \( S \) is an increasing function of \( x \).

Since \( x = c - b \geqslant 0 \), \( S \) is minimized when \( x = 0 \) (i.e., \( b = c \)), and maximized when \( x \) is maximized.
First, we find the minimum value.
For any \( a_{0} \), when \( x = 0 \), \( S(a_{0}) \) is minimized.
From equation (1), we get
$$
S(a_{0}) = \frac{1}{1 + a_{0}} + \frac{4}{3 - a_{0}} = \frac{3a_{0} + 7}{-a_{0}^2 + 2a_{0} + 3} \text {. }
$$

Next, we use the discriminant method to find the minimum value.
Equation (2) can be rearranged as
$$
\begin{array}{l}
S a_{0}^2 + (3 - 2S) a_{0} + 7 - 3S = 0 \\
\Rightarrow \Delta = (3 - 2S)^2 - 4S(7 - 3S) \geqslant 0 \\
\Rightarrow S \geqslant \frac{9}{4} \left(S \leqslant \frac{1}{4}, \text{ discard} \right) \\
\Rightarrow S_{\min} = \frac{9}{4} \text{ (when } a = \frac{1}{3}, b = c = \frac{1}{3} \text{)}.
\end{array}
$$

Next, we find the maximum value of \( S \).
It is easy to see that the maximum value of \( S \) is achieved when \( x \) is maximized.
For a given \( a_{0} \), \( b \) is minimized at \( a_{0} \) (since \( b \geqslant a \)), and \( c \) is maximized at \( 1 - 2a_{0} \), at which point \( x = c - b = 1 - 3a_{0} \) is maximized.
Substituting into \( S \) gives
$$
\begin{array}{c}
S = \frac{1}{1 + a_{0}} + \frac{1}{1 + a_{0}} + \frac{1}{1 + 1 - 2a_{0}} \\
= \frac{2}{1 + a_{0}} + \frac{1}{2 - 2a_{0}} .
\end{array}
$$

Taking the derivative with respect to \( a_{0} \) gives
$$
\begin{array}{l}
S'(a_{0}) = \frac{-2 \times 1}{(1 + a_{0})^2} + \frac{(-1)(-2)}{4(1 - a_{0})^2} \\
= \frac{-3a_{0}^2 + 10a_{0} - 3}{2(1 + a_{0})^2(1 - a_{0})^2} \\
= \frac{(1 - 3a_{0})(a_{0} - 3)}{2(1 + a_{0})^2(1 - a_{0})^2} \leqslant 0,
\end{array}
$$

indicating that \( S \) is a decreasing function of \( a_{0} \).
Since \( 0 \leqslant a_{0} \leqslant \frac{1}{3} \), \( S(a_{0}) \) is maximized when \( a_{0} = 0 \) (at which point, \( b = 0 \), \( c = 1 \), and \( x = c - b \) is maximized). Then \( S_{\text{max}} = \frac{1}{1 + 0} + \frac{1}{1 + 0} + \frac{1}{1 + 1} = \frac{5}{2} \).
In summary, \(\frac{9}{4} \leqslant S \leqslant \frac{5}{2}\).