Find the smallest real number m for which there exist real numbers $a;b$ such that $|x^2+ax+b|\le m(x^2+1)$ for $x\in [-1;1]$

Question

we can easily see that $m\ge 0$

then it is obvious that $m\neq 0$, because if $m=0$

the equation $x^2+ax+b=0$ will have infinitely many solutions

then I have proved that $m=\frac{1}{3}$ always works

but I don't if it's the smallest real number

Answer 1

Let $$f(x)=\frac{\left|x^2+a x +b\right|}{x^2+1}$$ where $x\in [-1,1]$

For $m=\frac{1}{3}$ and $a=0,b=-\frac{1}{3}$ we have $f(x)\leq m $ for all $x\in [-1,1]$

To show $\frac{1}{3}$ is minimal it suffices to prove that for $0<m<\frac{1}{3}$ and for all $a,b\in R$ there is an $x\in [-1,1]$ such that $f(x)>m$

Now, $$ 1=\frac{1+a+b}{2}+\frac{1-a+b}{2}-b$$

Using the triangle inequality we get $$1\leq \frac{|1+a+b|}{2}+\frac{|1-a+b|}{2}+|b|$$

$$1\leq f(1)+f(-1)+f(0)$$

this shows that one of $f(1),f(-1),f(0)$ must be $>m$ since otherwise we would have $1\leq 3m$ contradicting $m<\frac{1}{3}$