I have a great problem from Qvant magazine I can’t solve. Please help!
Suppose that for any $-1\leq x\leq 1$, $|ax^2+bx+c|\leq 1$. Find the maximum possible value of $|a|+|b|+|c|$.
My attempts:
It is suffices to find the maximum value of $|a|+|b|+|c|$ if for any $-1\leq x\leq 1$, $(ax^2+bx+c)^2\leq 1$. I inserted $x=-1, x=0, x=1$, and I got that $c\leq 1, a+c\leq 1, a+b+c\leq 1$. How to continue? Or maybe it is not ppssoble to solve the problem with my attempts? Any solution?
On
Hint: We have $$-1\le a+b+c\le 1$$ and $$|c|\le 1$$ and $$-1\le a-b+c\le 1$$ Multiplying $$-1\le a+b+c\le 1$$ by $(-1)$ we get
$$-1\le -a-b-c\le 1$$ adding this to $$-1\le a+b+c\le 1$$ we get
$$-2\le -2b\le 2$$ so $$|b|\le 1$$ The last $$|a|$$ is for you! Second solution: Plugging $$x=0,x=-1,x=1$$ we get $$|c|\le \epsilon,|a+b+c|\le \epsilon,|a-b+c|\le \epsilon$$ we get $$-2\epsilon \le -2c\le 2\epsilon$$ $$-\epsilon \le a+b+c\le \epsilon$$ $$-\epsilon\le a-b+c\le \epsilon$$ doing the same like above and using that $$|a|+|b|=|a-b|$$ we get $$|a|+|b|+|c|\le 4\epsilon$$.
Let $f(x)=ax^2+bx+c$.
Thus, $$f(1)=a+b+c$$, $$f\left(-1\right)=a-b+c$$ and $$f(0)=c,$$ which gives $$a=\frac{1}{2}f(1)+\frac{1}{2}f(-1)-f(0),$$ $$b=\frac{1}{2}f(1)-\frac{1}{2}f(-1).$$ We can assume that $a\geq0$ and since we can always change $x$ at $-x$, we can assume also that $b\geq0.$
Now, if $c\geq0$ we obtain: $$|a|+|b|+|c|=a+b+c=f(1)\leq1.$$
Let $c<0$.
Thus, $$|a|+|b|+|c|=a+b-c=f(1)-2f(0)\leq3.$$ The equality occurs for the Chebyshov's polynomial $f(x)=2x^2-1,$
which says that $3$ is a maximal value.