Example 2 Let non-negative real numbers $a, b, c$ satisfy $a+b+c=1$. Find the
$$
S=\sqrt{4 a+1}+\sqrt{4 b+1}+\sqrt{4 c+1}
$$

extreme values.
Let $a \leqslant b \leqslant c$, for any given $a_{0}\left(0 \leqslant a_{0} \leqslant \frac{1}{3}\right)$.
Then $b+c=1-a_{0}$.
Let $x=c-b$. Then
$$
c=\frac{1-a_{0}+x}{2}, b=\frac{1-a_{0}-x}{2} .
$$

Substitute into $S$ to get
$$
S=\sqrt{4 a_{0}+1}+\sqrt{3-2 a_{0}-2 x}+\sqrt{3-2 a_{0}+2 x} \text {. }
$$

Differentiate with respect to $x$ to get
$$
\begin{array}{l}
S^{\prime}(x)=\frac{1}{2} \cdot \frac{-2}{\sqrt{3-2 a_{0}-2 x}}+\frac{1}{2} \cdot \frac{2}{\sqrt{3-2 a_{0}+2 x}} \\
=\frac{\sqrt{3-2 a_{0}-2 x}-\sqrt{3-2 a_{0}+2 x}}{\sqrt{3-2 a_{0}-2 x} \cdot \sqrt{3-2 a_{0}+2 x}} \leqslant 0 .
\end{array}
$$

This indicates that, for a given $a_{0}$, $S$ is a decreasing function of $x$.
Therefore, when $x=0(b=c)$, $S\left(a_{0}\right)$ is maximized;
when $x$ is maximized, $S\left(a_{0}\right)$ is minimized.
First, find the maximum value (at $x=0$).
From equation (1), we have
$$
S\left(a_{0}\right)=\sqrt{4 a_{0}+1}+2 \sqrt{3-2 a_{0}} \text {. }
$$

Differentiate with respect to $a_{0}$ to get
$$
\begin{array}{l}
S^{\prime}\left(a_{0}\right)=\frac{1}{2} \cdot \frac{4}{\sqrt{4 a_{0}+1}}+2 \cdot \frac{1}{2} \cdot \frac{-2}{\sqrt{3-2 a_{0}}} \\
=2 \cdot \frac{\sqrt{3-2 a_{0}}-\sqrt{4 a_{0}+1}}{\sqrt{4 a_{0}+1} \cdot \sqrt{3-2 a_{0}}} \\
\geqslant 0\left(3-2 a_{0} \geqslant 4 a_{0}+1 \Leftrightarrow a_{0} \leqslant \frac{1}{3}\right) .
\end{array}
$$

This indicates that $S$ is an increasing function of $a_{0}$.
When $a_{0}=\frac{1}{3}$ (then $b=c=\frac{1}{3}$),
$$
S_{\text {max }}=\sqrt{4 \times \frac{1}{3}+1} \times 3=\sqrt{21} \text {. }
$$

Next, find the minimum value of $S$.
It is easy to see that the minimum value of $S$ is obtained when $x=c-b$ is maximized.

Since $b \geqslant a_{0}$, when $b$ is minimized to $a_{0}$ (then $c=1-2 a_{0}$), the difference $x=c-b$ is maximized. At this time,
$$
\begin{array}{l}
S=\sqrt{4 a_{0}+1}+\sqrt{4 a_{0}+1}+\sqrt{4\left(1-2 a_{0}\right)+1} \\
=2 \sqrt{4 a_{0}+1}+\sqrt{5-8 a_{0}} .
\end{array}
$$

Differentiate with respect to $a_{0}$ to get
$$
\begin{array}{l}
S^{\prime}\left(a_{0}\right)=2 \cdot \frac{1}{2} \cdot \frac{4}{\sqrt{4 a_{0}+1}}+\frac{1}{2} \cdot \frac{-8}{\sqrt{5-8 a_{0}}} \\
=4\left[\frac{1}{\sqrt{4 a_{0}+1}}-\frac{1}{\sqrt{5-8 a_{0}}}\right] \\
\geqslant 0\left(4 a_{0}+1 \leqslant 5-8 a_{0} \Leftrightarrow a_{0} \leqslant \frac{1}{3}\right) .
\end{array}
$$

Thus, $S$ is an increasing function of $a_{0}$.
When $a_{0}$ is minimized to 0 (then $b=0, c=1$),
$$
\begin{array}{l}
S_{\min }=\sqrt{4 \times 0+1}+\sqrt{4 \times 0+1}+\sqrt{4 \times 1+1} \\
=2+\sqrt{5} .
\end{array}
$$

In summary, $2+\sqrt{5} \leqslant S \leqslant \sqrt{21}$.