If $|ax^2+bx+c|\le100$ for all $|x|\le 1$, What is the maxima for $|a|+|b|+|c|$ Here is a similar problem posted before. I try to use this method to solve this problem.
$$f(-1)=a-b+c, f(0)=c, f(1)=a+b+c$$
So we have $|c|=|f(0)|\le 100$
$$|2a|=|f(-1)-2f(0)+f(1)|\le|f(-1)|+2|f(0)|+|f(1)|\le400$$
So we have $|a|\le200$
But how to find an upper bound for $b$?
Due to the symmetry, I make a guess for the maxima of $|a|+|b|+|c|$ occurring when $b=0$, then we have
$$y=200x^2-100~~~\text{or}~~~y=-200x^2+100$$
But is there a rigorous way to prove it?

Let $|ax^2+bx+c|\le100$ for all $|x|\le 1$
Firstly if $x=0, |c|\le 100(1)$
And if $x = \dfrac{1}{2}, then \left| \dfrac{a}{4}+\dfrac{b}{2}+c \right| < 100\Longrightarrow| a + 2b + 4c | < 400$
$| a | = | 2a + 2b + 2c - (a + 2b + 4c) + 2c | < 2 | a + b + c |   + | a + 2b + 4c | + 2 | c | < 200+ 400 + 200< 800$ (if $x=1$ then $|a+b+c|\le 100$)\ then $|a|\le 800(2)$
$ | b | = | a + 2b + 4c - (a + b + c) - 3c |  <  | a + 2b + 4c |  + | a + b + c | + 3 | c | < 400 + 100 + 300< 800$\ then $| b | < 800 (3)$
Finally $| a | +  | b | + | c | \le 100+800+800\le 1700$ i.e $| a | +  | b | + | c | \le1700$
