Quadratic inequality puzzle: Prove$ |cx^2 + bx + a| ≤ 2$ given $|ax^2+bx+c| ≤ 1$ 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 
$$|(a+c)x^2 + 2bx + (a+c)| = |dx^2 + 2b + d|$$
 A: 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*}
A: Let $f(x):=ax^2+bx+c$ and $g(x):=cx^2+bx+a$ for all $x\in\mathbb{R}$.  Note that
$$g(x)=\frac{1+x}{2}\,f(+1)-\left(1-x^2\right)\,f(0)+\frac{1-x}{2}\,f(-1)\text{ for all }x\in\mathbb{R}\,.$$
By the Triangle Inequality, we see that
$$\big|g(x)\big|\leq \frac{1+x}{2}\,\big|f(+1)\big|+\left(1-x^2\right)\,\big|f(0)\big|+\frac{1-x}{2}\,\big|f(-1)\big|$$
for all $x\in[-1,+1]$.  Since $\big|f(t)\big|\leq 1$ for all $t\in[-1,+1]$, we obtain
$$\big|g(x)\big|\leq \frac{1+x}{2}+\left(1-x^2\right)+\frac{1-x}{2}=2-x^2\leq 2\text{ for each }x\in[-1,+1]\,.$$
The inequality becomes an equality iff $(a,b,c)=\pm(2,0,-1)$ and $x=0$.

Let $\mathbb{D}$ be the open unit disc $\big\{z\in\mathbb{C}\,\big|\,|z|< 1\big\}$ in the complex plane, and write $\mathbb{U}$ for the boundary $\partial\mathbb{D}=\big\{z\in\mathbb{C}\,\big|\,|z|=1\big\}$ of $\mathbb{D}$ (that is, $\mathbb{U}$ is the unit circle centered at $0$).  Then, $\bar{\mathbb{D}}$ denote the topological closure $\mathbb{D}\cup\partial\mathbb{D}=\big\{z\in\mathbb{C}\,\big|\,|z|\leq 1\big\}$ of $\mathbb{D}$.
Now, suppose that $a_0$, $a_1$, $a_2$, $\ldots$, $a_n$ are complex numbers such that the entire function $f$ defined by $$f(z):=a_0+a_1z+a_2z^2+\ldots+a_{n-1}z^{n-1}+a_nz^n\text{ for all }z\in\mathbb{C}$$ satisfies the inequality $\big|f(z)\big|\leq M$ for all $z\in\bar{\mathbb{D}}$, where $M$ is a fixed positive real number.  I shall prove that, if the entire function $g$ is given by $$g(z):=a_n+a_{n-1}z+a_{n-2}z^2+\ldots+a_1z^{n-1}+a_0z^n\text{ for every }z\in\mathbb{C}\,,$$ then $\big|g(z)\big|\leq M$ for all $z\in\bar{\mathbb{D}}$ as well.
To prove this, we note by the Maximum Modulus Principle that the maximum of $\big|g(z)\big|$ for $z\in\bar{\mathbb{D}}$ is attained on $\mathbb{U}$.  Since $g(z)=z^n\,f(\bar{z})$ for every $z\in\mathbb{U}$, we conclude that
$$\big|g(z)\big|=\big|f(\bar{z})\big|\leq M\text{ for every }z\in\mathbb{U}\,,$$
and the claim follows immediately.
