Example 2 If the quadratic function $f(x)=a x^{2}+b x+c$ has values whose absolute values do not exceed 1 on $[0,1]$, what is the maximum value of $|a|+$ $|b|+|c|$?
Solution 1: For $x \in [0,1]$, we have $|f(x)| \leqslant 1$, thus
$$
\begin{array}{l}
|f(0)| \leqslant 1, \left|f\left(\frac{1}{2}\right)\right| \leqslant 1, |f(1)| \leqslant 1. \\
\text { Also } f(0)=c, \\
f\left(\frac{1}{2}\right)=\frac{1}{4} a+\frac{1}{2} b+c, \\
f(1)=a+b+c,
\end{array}
$$

Solving simultaneously, we get
$$
\begin{array}{l}
a=2 f(0)-4 f\left(\frac{1}{2}\right)+2 f(1), \\
b=-3 f(0)+4 f\left(\frac{1}{2}\right)-f(1), \\
c=f(0) .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence }|a|=\left|2 f(0)-4 f\left(\frac{1}{2}\right)+2 f(1)\right| \\
\leqslant 2|f(0)|+4\left|f\left(\frac{1}{2}\right)\right|+2|f(1)| \\
\leqslant 2+4+2=8 .
\end{array}
$$

Similarly, $|b| \leqslant 3+4+1=8, |c| \leqslant 1$.
Therefore, $|a|+|b|+|c| \leqslant 17$.
Moreover, when $f(0)=f(1)=-f\left(\frac{1}{2}\right)=1$, i.e.,
$$
a=8, b=-8, c=1
$$

or $f(0)=f(1)=-f\left(\frac{1}{2}\right)=-1$,
i.e., $a=-8, b=8, c=-1$, the equality holds.
Thus, the maximum value of $|a|+|b|+|c|$ is 17.
Solution 2: For $x \in (0,1)$, we have
$$
\left\{\begin{array}{l}
f(0)=c, \\
f(1)=a+b+c, \\
f(x)=a x^{2}+b x+c .
\end{array}\right.
$$

Substituting (1) into (2) and (3), we get
$$
\begin{array}{l}
a=\frac{1}{x(1-x)}[-f(x)+f(1) x+f(0)(1-x)], \\
b=\frac{1}{x(1-x)}\left[f(x)-f(1) x^{2}-f(0)\left(1-x^{2}\right)\right] .
\end{array}
$$

From (4), for all $x \in (0,1)$, we have
$$
\begin{array}{l}
|a| \leqslant \frac{1}{x(1-x)}[|f(x)|+|f(1)| x+|f(0)|(1-x)] \\
\leqslant \frac{1}{x(1-x)}[1+x+(1-x)] \\
=\frac{2}{x(1-x)} .
\end{array}
$$

But $x(1-x) \leqslant\left[\frac{x+(1-x)}{2}\right]^{2}=\frac{1}{4}$, with equality holding only when $x=\frac{1}{2}$, hence
$$
|a| \leqslant \min _{x \in(0,1)} \frac{2}{x(1-x)}=8,
$$
with equality holding only when $x=\frac{1}{2}$.
Substituting into (6), the conditions for equality are
$$
\left\{\begin{array}{l}
x=\frac{1}{2}, \\
f(1)=1, \\
f(0)=1, \\
f\left(\frac{1}{2}\right)=-1
\end{array} \text { or } \left\{\begin{array}{l}
x=\frac{1}{2}, \\
f(1)=-1, \\
f(0)=-1, \\
f\left(\frac{1}{2}\right)=1 .
\end{array}\right.\right.
$$

This gives $\left\{\begin{array}{l}a=8, \\ c=1\end{array}\right.$ or $\left\{\begin{array}{l}a=-8, \\ c=-1 .\end{array}\right.$
Similarly, from (5) we have
$$
\begin{array}{l}
|b| \leqslant \frac{1}{x(1-x)}\left[1+x^{2}+\left(1-x^{2}\right)\right]=\frac{2}{x(1-x)}, \\
|b| \leqslant \min _{x \in(0,1)} \frac{2}{x(1-x)}=8 .
\end{array}
$$

The conditions for equality are
$$
\left\{\begin{array}{l}
x=\frac{1}{2}, \\
f(1)=1, \\
f(0)=1, \\
f\left(\frac{1}{2}\right)=-1
\end{array} \text { or } \left\{\begin{array}{l}
x=\frac{1}{2}, \\
f(1)=-1, \\
f(0)=-1, \\
f\left(\frac{1}{2}\right)=1 .
\end{array}\right.\right.
$$

This gives $\left\{\begin{array}{l}b=-8, \\ c=1\end{array}\right.$ or $\left\{\begin{array}{l}b=8, \\ c=-1 .\end{array}\right.$
From (1), (7), and (8), we get
$|a|+|b|+|c| \leqslant 8+8+1=17$.
Furthermore, when $a=8, b=-8, c=1$, for $x \in [0,1]$, we have
$$
\begin{array}{l}
f(x)=8 x^{2}-8 x+1=-1+2(2 x-1)^{2} \geqslant-1, \\
f(x)=8 x^{2}-8 x+1=1-8 x(1-x) \leqslant 1 .
\end{array}
$$

Therefore, the maximum value of $|a|+|b|+|c|$ is 17.