TITLE: Quadratic inequality puzzle: Prove$ |cx^2 + bx + a| ≤ 2$ given $|ax^2+bx+c| ≤ 1$
QUESTION [17 upvotes]: I came across this problem as part of a recreational mathematics challenge on university:
Suppose $a, b, c$ are real numbers where for all $ -1 \le x \le 1 $ we have $|ax^2 + bx + c| \le 1$. 
Prove that for all $-1 \le x \le 1 ,$

$$    |cx^2 + bx + a| \le 2$$

Was interested to know how you can approach this as I have not so far been successful, especially frustrating as it does not appear to be that hard of a problem? 
The inequality to prove is presented accurately by professors and I am sure it holds. 
edit: I have tried with a graph, as I only have to worry about $x=1, -1$ and at the maxima/minima of where the quadratic will be at most 2 but I don't know if a graph can constitute a 'prove' and in any case it hasn't worked.
I have tried a lot of algebraic manipulation like completing the square but again, nothing. 
I have noticed the symettrey in the inequalities, specifically, if we add them then we get 
$$|(a+c)x^2 + 2bx + (a+c)| = |dx^2 + 2b + d|$$

REPLY [9 votes]: Denote $$f(x)=ax^2+bx+c.$$ Hence $$f(0)=c,~~~f(1)=a+b+c,~~~f(-1)=a-b+c.$$
Further $$a=\frac{f(1)+f(-1)}{2}-f(0),~~~b=\frac{f(1)-f(-1)}{2},~~~c=f(0).$$
Thus \begin{align*}|cx^2+bx+a|&=\big|f(0)\cdot x^2+\frac{f(1)-f(-1)}{2}\cdot x+\frac{f(1)+f(-1)}{2}-f(0)\big|\\&=\big|f(0)\cdot (x^2-1)+f(1)\cdot\frac{x+1}{2}+f(-1)\cdot\frac{1-x}{2}\big|\\
&\leq(1-x^2)\cdot|f(0)|+\frac{x+1}{2}\cdot|f(1)|+\frac{1-x}{2}\cdot|f(-1)|\\
&\leq(1-x^2)+\frac{x+1}{2}+\frac{1-x}{2}\\
&=2-x^2\\
&\leq 2.
\end{align*}