Three. (25 points) Let the quadratic function $f(x)=x^{2}+a x+b$, $F=\max _{|x| \leq 1} |f(x)|$. When $a, b$ traverse all real numbers, find the minimum value of $F$.


$$
\begin{array}{l}
1=\frac{f(-1)}{2}-f(0)+\frac{f(1)}{2} \\
\leqslant\left|\frac{f(-1)}{2}\right|+|-f(0)|+\left|\frac{f(1)}{2}\right| \\
\leqslant \frac{F}{2}+F+\frac{F}{2}=2 F .
\end{array}
$$

Thus, $F \geqslant \frac{1}{2}$, and $F$ achieves its minimum value $\frac{1}{2}$ if and only if $f(x)=x^{2}-\frac{1}{2}$, i.e., when $a=0, b=-\frac{1}{2}$.