Given $f(x)=a x^{2}+b x+c(a, b, c$ are real numbers) and its absolute value on $[-1,1]$ is $\leqslant 1$, find the maximum value of $|a|+|b|+|c|$.
- $c=f(0)$.
$$
\begin{array}{l}
a+b+c=f(1) \\
a-b+c=f(-1) .
\end{array}
$$

If $a$ and $b$ have the same sign, then
$$
\begin{aligned}
|a|+|b|+|c| & =|a+b|+|c|=|f(1)-c|+|c| \\
& \leqslant|f(1)|+|c|+|c| \\
& \leqslant|f(1)|+2|f(0)| \leqslant 3 .
\end{aligned}
$$

If $a$ and $b$ have opposite signs, then
$$
|a|+|b|+|c|=|a-b|+|c|=|f(-1)-c|+|c| \leqslant 3 .
$$

When $f(x)=2 x^{2}-1$, $|a|+|b|+|c|=3$. Therefore, the maximum value of $|a|+|b|+|c|$ is 3.