Quadratic inequality puzzle: Prove$ |cx^2 + bx + a| ≤ 2$ given $|ax^2+bx+c| ≤ 1$

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.

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*}