<s>[INST] If $|ax^2+bx+c|\leq 2\ \ \forall x\in[-1,1]$ then find the maximum value of $|cx^2+2bx+4a|\ \ \forall x\in [-2,2]$. 
If
$$\left|ax^2+bx+c\right|\leq 2\quad \forall x\in[-1,1]$$
then find the maximum value of
$$\left|cx^2+2bx+4a\right|\quad \forall x\in [-2,2].$$

My Attempt
Let $f(x)=ax^2+bx+c$, then
$$|cx^2+2bx+4a|=x^2|4\frac{a}{x^2}+2\frac{b}{x}+c|=x^2\left|f\left(\frac{2}{x}\right)\right|$$
But then I couldn't make any headway.
I also tried to express $a,b,c$ in terms of $f(-1),f(0),f(1)$ i.e $$c=f(0);b=\frac{f(1)-f(-1)}{2};a=\frac{f(-1)-2f(0)+f(1)}{2}$$ but again couldn't go further
 [/INST] A solution using elementary calculus.
First observe by changing variables $x\mapsto\frac{x}{2}$ that the question becomes finding the maximum value of $4|cx^2+bx+a|$ on $[-1,1]$. Let's set $f(x)=ax^2+bx+c$ and $g(x)=cx^2+bx+a$. Also note that by swapping between $f$ and $-f$ (and $g$ and $-g$ respectively), we can assume wlog that $c\geq 0$. Also wlog, since the domains are symmetric around the origin, we can assume $b\geq 0$.
Now there are four main ingredients to the elementary proof.
Step 1: As Maverick pointed out, one can get bounds on the coefficients. Indeed, $a=\frac{f(-1)-2f(0)+f(1)}{2}$, $b=\frac{f(1)-f(-1)}{2}$, and $c=f(0)$ implying with the above that $b,c\in[0,2]$ and $|a|\leq4$.
Step 2: The extremal values a parabola attains on a closed interval are either at the endpoints of the interval or at the global extremum of the parabola.
Therefore
\begin{equation*}
\max_{[-1,1]}|g(x)|\in\{|g(1)|,|g(-1)|,|g(-\frac{b}{2c})|\}=\{|c+b+a|,|c-b+a|,|a-\frac{b^2}{4c}|\}=\{|f(1)|,|f(-1)|,|a-\frac{b^2}{4c}|\}. 
\end{equation*}
So $\max_{[-1,1]}|g(x)|\leq \max(2,|a-\frac{b^2}{4c}|)$.
As a technicality, if $c=0$, $g$ is linear, so it will attain extremal values only at the endpoints of the interval, so we don't need to worry about $g(-\frac{b}{2c})$.
Step 3: Note that by taking $a=-4,b=0,c=2$, we have that $|g(0)|=4$, so we are only further interested in looking at the case when the extremal value of $|g|$ is attained at the global extremal value of $g$ and when this value is attained inside the interval $[-1,1]$.
This implies $c>0$ and $-1\leq -\frac{b}{2c}\leq 0$. Observe that if $g(-\frac{b}{2c})\geq 0$, since we assumed $c\geq0$, $g$ increases away from $-\frac{b}{2c}$ and therefore $g(1)$ and $g(-1)$ will both be bigger than $g(-\frac{b}{2c})$. Therefore we can restrict our analysis to the case when $g(-\frac{b}{2c})<0$.
Step 4: Arguing that $g$ is bounded below on $[-1,1]$ by $-4$.
Assume this is not the case. Then necessarily $g(-\frac{b}{2c})<-4$. We now apply the mean value theorem on the interval $[-1,-\frac{b}{2c}]\subset [-1,0]$ to find some $t\in[-1,0]$ such that
\begin{equation*}
g'(t)=\frac{g(-\frac{b}{2c})-g(-1)}{-\frac{b}{2c}-(-1)}\leq g(-\frac{b}{2c})-g(-1)<- 2,
\end{equation*}
because $0\geq-\frac{b}{2c}\geq -1, g(-1)=f(-1)\geq -2$ and because we assumed $g(-\frac{b}{2c})<-4$.
But this implies that $2ct+b<-2$, or, since $t\in[-1,0]$, that $c>\frac{b}{2|t|}+\frac{1}{|t|}\geq\frac{1}{|t|}\geq 1$.
Now since we assumed that $g(-\frac{b}{2c})<-4$, we have
\begin{equation*}
-4>a-\frac{b^2}{4c}=(a-b+c)+b-c-\frac{b^2}{4c}\geq -2 +b-c-\frac{b^2}{4c}.
\end{equation*}
Rearranging, this means $\frac{1}{4c}b^2-b+c-2>0$. Viewing this as a degree $2$ inequality in $b$, we can compute the discriminant $\Delta=\sqrt{\frac{2}{c}}$ and roots $b_{1,2}=\frac{1\pm\sqrt{\frac{2}{c}}}{\frac{1}{2c}}=2c\pm 2\sqrt{2c}$. For the inequality to hold, we must therefore have that either $b<2c-2\sqrt{2c}=2\sqrt{c}(\sqrt{c}-\sqrt{2})<0$ (since $c\in(1,2]$), impossible since $b\geq 0 $, or that $b>2c+2\sqrt{2c}>2c>2$ (since $c>1$), impossible as $b\leq 2$. So we have a contradiction, concluding Step 4.
This means that the required maximum is $16$ and it can be achieved by taking $a=-4,b=0,c=2$.
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