Czech and Slovak Mathematical Olympiad

Let us assume that $a$, $b$, $m$ satisfy the condition:
$$\forall x \in (-1, 1): |f(x)| \leq m(x^2 + 1), \quad \text{where} \quad f(x) = x^2 + a x + b.$$
Firstly we prove that at least one of the $f(1) - f(0) \geq 0$ and $f(-1) - f(0) \geq 0$ holds:
for arbitrary $f(x) = x^2 + a x + b$ there is
$$f(0) = b, \quad f(1) = 1 + a + b, \quad f(-1) = 1 - a + b,$$
and
$$\max(f(1) - f(0), f(-1) - f(0)) = \max(1 + a, 1 - a) = 1 + |a| \geq 1.$$
Our assumption means $|f(1)| \le 2m$, $|f(-1)| \le 2m$ and $|f(0)| \le m$. Consequently
$$1 \le 1 + |a| = f(1) - f(0) \le |f(1)| + |f(0)| \le 2m + m = 3m, \quad (1)$$
or
$$1 \le 1 + |a| = f(-1) - f(0) \le |f(-1)| + |f(0)| \le 2m + m = 3m. \quad (2)$$
In both cases we get $m \ge \frac{1}{3}$.
We show, that $m = \frac{1}{3}$ fulfills the problem conditions. For this $m$ either (1) or (2) is equality, that is $a = 0$, $-f(0) = |f(0)|$ and $|f(0)| = m = \frac{1}{3}$, thus $b = f(0) = -\frac{1}{3}$.
We will verify, that the function $f(x) = x^2 - \frac{1}{3}$ for $m = \frac{1}{3}$ satisfies the conditions of the problem: The inequality $|x^2 - \frac{1}{3}| \le \frac{1}{3}(x^2 + 1)$ is equivalent with the inequalities
$$-\frac{1}{3}(x^2 + 1) \le x^2 - \frac{1}{3} \le \frac{1}{3}(x^2 + 1) \quad \text{or} \quad -x^2 - 1 \le 3x^2 - 1 \le x^2 + 1$$
which are equivalent to $0 \le x^2 \le 1$, which is evidently fulfilled on $(-1, 1]$.

The sought $m$ is $\frac{1}{3}$.