Let $F=\max _{1<x<3}\left|x^{3}-a x^{2}-b x-c\right|$, when $a$, $b$, $c$ take all real numbers, find the minimum value of $F$.
Below is another simplified solution:
Let $f(x)=x^{3}-a x^{2}-b x-c$, and take 4 points symmetrically in the interval $[1,3]$: $x=1, \frac{3}{2}, \frac{5}{2}, 3$. It is easy to verify that
$$
-\frac{2}{3} f(1)+\frac{4}{3} f\left(\frac{3}{2}\right)-\frac{4}{3} f\left(\frac{5}{2}\right)+\frac{2}{3} f(3)=1
$$
holds.
$$
\begin{array}{l}
\text { Since } F \geqslant|-f(1)|, F \geqslant\left|f\left(\frac{3}{2}\right)\right|, \\
F \geqslant\left|-f\left(\frac{5}{2}\right)\right|, F \geqslant|f(3)|, \text { then } \\
4 F=\frac{2}{3} F+\frac{4}{3} F+\frac{4}{3} F+\frac{2}{3} F \\
\geqslant\left|-\frac{2}{3} f(1)\right|+\left|\frac{4}{3} f\left(\frac{3}{2}\right)\right|+ \\
\geqslant\left|-\frac{4}{3} f\left(\frac{5}{2}\right)\right|+\left|\frac{2}{3} f(3)\right| \\
\geqslant \left\lvert\,-\frac{2}{3} f(1)+\frac{4}{3} f\left(\frac{3}{2}\right)-\right. \\
\left.\quad \frac{4}{3} f\left(\frac{5}{2}\right)+\frac{2}{3} f(3) \right\rvert\, \\
=1 .
\end{array}
$$

Therefore, $F \geqslant \frac{1}{4}$.
Equality holds if and only if
$$
\begin{array}{l}
\left.1-f(1)|=| f\left(\frac{3}{2}\right)|=|-f\left(\frac{5}{2}\right) \right\rvert\, \\
=|f(3)|=\frac{1}{4},
\end{array}
$$

and $-f(1), f\left(\frac{3}{2}\right), -f\left(\frac{5}{2}\right), f(3)$ have the same sign. From equation (1), it is only possible that
$$
f(1)=f\left(\frac{5}{2}\right)=-\frac{1}{4}, f\left(\frac{3}{2}\right)=f(3)=\frac{1}{4} .
$$

Solving yields $a=6, b=-\frac{45}{4}, c=\frac{13}{2}$.
Next, we prove that when $f(x)=x^{3}-6 x^{2}+\frac{45}{4} x-\frac{13}{2}$, $F=\max _{1 \leq x<3}|f(x)|=\frac{1}{4}$.
It is easy to verify that $f(x)$ can be rewritten as the following two expressions:
$$
\begin{array}{l}
f(x)=\left(x-\frac{3}{2}\right)^{2}(x-3)+\frac{1}{4}, \\
f(x)=\left(x-\frac{5}{2}\right)^{2}(x-1)-\frac{1}{4},
\end{array}
$$

Thus, when $1 \leqslant x \leqslant 3$,
$$
-\frac{1}{4} \leqslant f(x) \leqslant \frac{1}{4},|f(x)| \leqslant \frac{1}{4} \text {. }
$$

Therefore, when $a=6, b=-\frac{45}{4}, c=\frac{13}{2}, x=1$, $\frac{3}{2}, \frac{5}{2}, 3$, $F$ attains its minimum value $\frac{1}{4}$.
Reference:
[1] Yang Riwu. A Simplified Solution to a Competition Problem [J]. Middle School Mathematics, 2002(3).