For which values of the real parameters $a$ and $b$ will the maximum of the function $\left|x^{2}+a x+b\right|$ on the interval $[-1,1]$ be the smallest?
We will prove that

$$
\max _{x \in[-1,1]}\left|x^{2}+a x+b\right| \geq \frac{1}{2}
$$

and equality holds only for the polynomial $x^{2}-\frac{1}{2}$. Suppose for some $a, b$,

$$
\left|x^{2}+a x+b\right| \leq \frac{1}{2}
$$

holds for all $-1 \leq x \leq 1$. We will show that in this case $a=0, b=-\frac{1}{2}$.

The function $x^{2}+a x+b$ attains its maximum absolute value on the interval $[-1,1]$ either at one of the endpoints of the interval or at the vertex of the parabola $x^{2}+a x+b$, provided that the vertex lies within the interval $[-1,1]$. The transformation

$$
x^{2}+a x+b=\left(x+\frac{a}{2}\right)^{2}+b-\frac{a^{2}}{4}
$$

shows that the vertex of the parabola is at $x=-\frac{a}{2}$. It is therefore useful to write (1) at the points $-1, +1$, and $-\frac{a}{2}$ (if $\left|-\frac{a}{2}\right| \leq 1$). Let's first consider $+1$ and $-1$:

$$
-\frac{1}{2} \leq 1+a+b \leq \frac{1}{2},(2)-\frac{1}{2} \leq 1-a+b \leq \frac{1}{2} .(3)
$$

From (3),

$$
-\frac{1}{2} \leq-1+a-b \leq \frac{1}{2}
$$

Adding this to (2),

$$
-1 \leq 2 a \leq 1
$$

which implies

$$
|a| \leq \frac{1}{2}
$$

Accordingly, the vertex of the parabola lies within the interval $[-1,1]$, and we can also write (1) at this point:

$$
-\frac{1}{2} \leq b-\frac{a^{2}}{4} \leq \frac{1}{2}
$$

which gives

$$
\begin{gathered}
b \geq \frac{a^{2}}{4}-\frac{1}{2} \geq-\frac{1}{2} \\
-1 \leq 2+2 b \leq 1
\end{gathered}
$$

Adding (2) and (3) yields

from the right side of which

$$
b \leq-\frac{1}{2}
$$

Thus,

$$
-\frac{1}{2} \geq b \geq \frac{a^{2}}{4}-\frac{1}{2} \geq-\frac{1}{2}
$$

which can only hold if

$$
b=-\frac{1}{2}, \quad a=0
$$

In this case, for $x^{2}+a x+b$,

$$
\max _{x \in[-1,1]}\left|x^{2}+a x+b\right|=\max _{x \in[-1,1]}\left|x^{2}-\frac{1}{2}\right|=\frac{1}{2}
$$