Example 8 Let $F=\max _{1 \leqslant x \leqslant 3}\left|x^{3}-a x^{2}-b x-c\right|$. Find the minimum value of $F$ when $a, b, c$ take all real numbers. (2001, IMO China National Training Team Selection Exam)
Let $f(x)=(x-2)^{3}-\frac{3}{4}(x-2)$, $x \in[1,3]$. Set $x-2=\cos \theta, \theta \in[0, \pi]$, then $|f(x)|=\left|\cos ^{3} \theta-\frac{3}{4} \cos \theta\right|=\frac{1}{4}|\cos 3 \theta| \leqslant \frac{1}{4}$. When $\theta=0, \frac{\pi}{3}, \frac{2 \pi}{3}, \pi$, i.e., $x=3, \frac{5}{2}, \frac{3}{2}, 1$, $|f(x)|=\frac{1}{4}$.
Also, $f(x)=(x-2)^{3}-\frac{3}{4}(x-2)$
$$
=x^{3}-6 x^{2}+\frac{45}{4} x-\frac{13}{2} \text {, }
$$

When $a=6, b=-\frac{45}{4}, c=\frac{13}{2}$, $f(x)_{\text {max }}=\frac{1}{4}$.
Now we prove: $F \geqslant \frac{1}{4}$.
Let $f^{\prime}(x)=x^{3}-a x^{2}-b x-c$. On the interval $[1,3]$, take values symmetrically about $x=2$, i.e., $x=1, \frac{3}{2}, \frac{5}{2}, 3$.
Since $f^{\prime}(1)=-a-b-c+1$,
$$
\begin{array}{l}
f^{\prime}\left(\frac{3}{2}\right)=-\frac{9}{4} a-\frac{3}{2} b-c+\frac{27}{8}, \\
f^{\prime}\left(\frac{5}{2}\right)=-\frac{25}{4} a-\frac{5}{2} b-c+\frac{125}{8}, \\
f^{\prime}(3)=-9 a-3 b-c+27,
\end{array}
$$

Thus, we have
$$
-\frac{2}{3} f^{\prime}(1)+\frac{4}{3} f^{\prime}\left(\frac{3}{2}\right)-\frac{4}{3} f^{\prime}\left(\frac{5}{2}\right)+\frac{2}{3} f^{\prime}(3)=1 .
$$

By $F \geqslant \max || f^{\prime}(1)|,| f^{\prime}\left(\frac{3}{2}\right)|,| f^{\prime}\left(\frac{5}{2}\right)|,| f^{\prime}(3)||$, we get $4 F \geqslant \frac{2}{3}\left|f^{\prime}(1)\right|+\frac{4}{3}\left|f^{\prime}\left(\frac{3}{2}\right)\right|+$
$$
\frac{4}{3}\left|f^{\prime}\left(\frac{5}{2}\right)\right|+\frac{2}{3}\left|f^{\prime}(3)\right| \geqslant 1 .
$$

Therefore, $F \geqslant \frac{1}{4}$.
Equality holds if and only if $-f^{\prime}(1)=f^{\prime}\left(\frac{3}{2}\right)=-f^{\prime}\left(\frac{5}{2}\right)$ $=f^{\prime}(3)=\frac{1}{4}$.
Hence, when $a=6, b=-\frac{45}{4}, c=\frac{13}{2}$, $F=\frac{1}{4}$.
Thus, $F_{\min }=\frac{1}{4}$.
Finally, it is noted that solving the extremum problem of multivariate functions often requires the comprehensive application of various methods. Due to the limitation of space, we will not provide more examples here.