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?
 A: $$y=200x^2-100~~~\text{or}~~~y=-200x^2+100$$
are examples where $\vert a \vert + \vert b \vert + \vert c \vert = 300.\quad (*)$
If $\vert a \vert + \vert b \vert \leq 200,$ then $\vert a \vert + \vert b \vert + \vert c \vert \leq 300,\ $ so not an improvement over $(*).$
If $\vert a \vert + \vert b \vert > 200,\ $ then one of $\vert f(1)\vert$ or $\vert f(-1)\vert$ are equal to $ \vert a\vert +  \vert b \vert + c\ \geq \vert a\vert +  \vert b\vert - \vert c\vert > 100, $ which contradicts the condition "$\vert ax^2+bx+c\vert \le 100$ for all $\vert x\vert \le 1$" in the question.
A: Remark: I used the approach in my answer for
this question.
We have
\begin{align*}
 |a - b + c| &\le 100, \tag{1}\\
 |a + b + c| &\le 100, \tag{2}\\
 |c| &\le 100. \tag{3}
\end{align*}
Using (1) and (3), we have
$$|a - b| \le |a - b + c| + |-c| \le 200. \tag{4}$$
Using (2) and (3), we have
$$|a + b| \le |a + b + c| + |-c| \le 200. \tag{5}$$
Using (4) and (5), we have
$$|a| + |b| \le 200. \tag{6}$$
(Note: If $ab \ge 0$, then $|a| + |b| = |a + b| \le 200$.
If $ab < 0$, then $|a| + |b| = |a - b| \le 200$.)
Using (3) and (6), we have
$$|a| + |b| + |c| \le 300.$$
On the other hand, when $a = 200, b = 0, c = -100$ (so $|a| + |b| + |c| = 300$),
we have, for all $|x| \le 1$,
$$|200x^2 - 100|\le
\max(|200\cdot 1^2 - 100|,
|200\cdot (-1)^2 - 100|, |200\cdot 0^2 - 100|) = 100.$$
Thus, the maximum of $|a| + |b| + |c|$ is $300$.
A: Since $300$ is the optimum value for the linear programming problems
$$ \eqalign{\text{maximize}\ & a + b - c \cr
      \text{subject to}\ & a + b - c \le 100 \cr
                         & a + b - c \ge -100\cr
                        & a - b - c \le 100\cr
                        & a - b - c \ge -100\cr
                        & -c \le 100\cr
                        & -c \ge -100\cr}$$
and the same with objective $a - b - c$,
you certainly can't do better than that in your problem with $a \ge 0$ and $c \le 0$.
By symmetry the same holds for $a \le 0$ and $c \ge 0$.  Now consider the other cases.
A: 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 $| a | +  | b | + | c | \le1700$
