4. Determine the smallest real number $m$, for which there exist real numbers $a$ and $b$ such that the inequality

$$
\left|x^{2}+a x+b\right| \leqq m\left(x^{2}+1\right)
$$

holds for every $x \in\langle-1,1\rangle$.

The regional round of category A takes place

## on Tuesday, January 12, 2016

so that it starts in the morning and the contestants have 4 hours of pure time to solve the problems. Allowed aids are writing and drawing supplies and school MF tables. Calculators, laptops, and any other electronic aids are not allowed. Each problem can earn the contestant 6 points; the score threshold $\mathrm{k}$ for determining successful solvers will be centrally set after evaluating the statistical point results from all regions. This information will be communicated to the students before the competition begins.
4. Suppose that the numbers $a, b, m$ satisfy the condition

$$
\forall x \in [-1,1]: |f(x)| \leq m(x^2 + 1), \quad \text{where} \quad f(x) = x^2 + ax + b
$$

First, we will show that at least one of the differences $f(1) - f(0)$ and $f(-1) - f(0)$ is greater than or equal to one: for any function $f(x) = x^2 + ax + b$, we have

$$
f(0) = b, \quad f(1) = 1 + a + b, \quad f(-1) = 1 - a + b,
$$

so

$$
\max(f(1) - f(0), f(-1) - f(0)) = \max(1 + a, 1 - a) = 1 + |a| \geq 1
$$

The assumption from the beginning of the solution means that $|f(1)| \leq 2m$, $|f(-1)| \leq 2m$, and $|f(0)| \leq m$. Therefore, either

$$
1 \leq 1 + |a| = f(1) - f(0) \leq |f(1)| + |f(0)| \leq 2m + m = 3m
$$

or

$$
1 \leq 1 + |a| = f(-1) - f(0) \leq |f(-1)| + |f(0)| \leq 2m + m = 3m.
$$

In both cases, we get the estimate $m \geq \frac{1}{3}$.

We will try to show that $m = \frac{1}{3}$ satisfies the requirements of the problem. For such $m$, however, equality holds everywhere in (1) or (2), so it must be that $a = 0$, $-f(0) = |f(0)|$, and $|f(0)| = m = \frac{1}{3}$, hence $b = f(0) = -\frac{1}{3}$. Note that for the found values of $m, a, b$, we have not yet considered the validity of the inequality from the problem statement for values $x \in [-1,1]$ different from the numbers $-1, 0$, and $1$, which we have not considered so far.

Let's verify that the found function $f(x) = x^2 - \frac{1}{3}$ for $m = \frac{1}{3}$ satisfies the conditions of the problem: The inequality $\left|x^2 - \frac{1}{3}\right| \leq \frac{1}{3}(x^2 + 1)$ is equivalent to the inequalities

$$
-\frac{1}{3}(x^2 + 1) \leq x^2 - \frac{1}{3} \leq \frac{1}{3}(x^2 + 1) \quad \text{or} \quad -x^2 - 1 \leq 3x^2 - 1 \leq x^2 + 1
$$

and these are equivalent to the inequalities $0 \leq x^2 \leq 1$, which are clearly satisfied on the interval $[-1,1]$.

Answer: The smallest $m$ is equal to the fraction $\frac{1}{3}$.

For a complete solution, award 6 points. For finding the values of the parameters $a, b$ for $m = 1/3$, award one point, and for directly verifying the relation from the problem statement, award another (second) point. For deriving the necessary condition $m \geq 1/3$, award four points. If the solver uses a geometric argument similar to the algebraic use of the triangle inequality, do not deduct points. Conversely, for considerations about the shape of the graph of the quadratic function (for example, silently using its convexity), do not award points unless these considerations are accompanied by additional arguments.

[^0]: The distinction is motivated by the intersection of the specified sets $\{2,4,6,8,12\}$ and $\{2,4,8,10\}$.