
4. Find the least real $m$, for which there exists real $a, b$ such, that

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

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


Solution. Let us assume that $a, b, m$ satisfy the condition:

$$
\forall x \in\langle-1,1\rangle:|f(x)| \leqslant m\left(x^{2}+1\right), \quad \text { kde } \quad f(x)=x^{2}+a x+b .
$$

Firstly we prove that at least one of the $f(1)-f(0) \geqslant 0$ and $f(-1)-f(0) \geqslant 0$ 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

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

Our assumption means $|f(1)| \leqslant 2 m,|f(-1)| \leqslant 2 m$ a $|f(0)| \leqslant m$. Consequently

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

or

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

In both cases we get $m \geqslant \frac{1}{3}$.

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 $f(x)=x^{2}-\frac{1}{3}$ pro $m=\frac{1}{3}$ satisfies the conditions of the problem: The inequality $\left|x^{2}-\frac{1}{3}\right| \leqslant \frac{1}{3}\left(x^{2}+1\right)$ is equivalent with the inequalities

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

which are equivalent to $0 \leqslant x^{2} \leqslant 1$, which is evidently fulfilled on $\langle-1,1\rangle$.

Answer. The thought $m$ is $\frac{1}{3}$.

Final Round of the 65th Czech and Slovak

Mathematical Olympiad

(April 4-5, 2016)

$\mathbb{N} / /(2)$
