65th Czech and Slovak Mathematical Olympiad

Let us assume that $a$, $b$, $m$ satisfy the condition: $$\forall x\in (-1,1):|f(x)|\le m(x^2+1),\text{where}f(x)=x^2+ax+b.$$ Firstly we prove that at least one of the $f(1)-f(0)\ge 0$ and $f(-1)-f(0)\ge 0$ holds: for arbitrary $f(x)=x^2+ax+b$ there is $$f(0)=b,f(1)=1+a+b,f(-1)=1-a+b,$$ and $$\max (f(1)-f(0),f(-1)-f(0))=\max (1+a,1-a)=1+|a|\ge 1.$$ Our assumption means $|f(1)|\le 2m$, $|f(-1)|\le 2m$ and $|f(0)|\le m$. Consequently $$1\le 1+|a|=f(1)-f(0)\le |f(1)|+|f(0)|\le 2m+m=3m,(1)$$ or $$1\le 1+|a|=f(-1)-f(0)\le |f(-1)|+|f(0)|\le 2m+m=3m.(2)$$ 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 $$-\frac{1}{3}(x^2+1)\le x^2-\frac{1}{3}\le \frac{1}{3}(x^2+1)\text{or}-x^2-1\le 3x^2-1\le x^2+1$$ which are equivalent to $0\le x^2\le 1$, which is evidently fulfilled on $(-1,1]$.

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