If $|ax^2+bx+c|\le \frac12$ for all $|x|\le1$, then $|ax^2+bx+c|\le x^2-\frac12$ for all $|x|\ge1$

Question

Prove that if $|ax^2+bx+c|\le \frac12$ for all $|x|\le1$ then $|ax^2+bx+c|\le x^2-\frac12$ for all $|x|\ge1$.

My attempts:

Let $f(x)=ax^2+bx+c$

I know that

1) if $f(a)<0$ and $f(b)>0$ then exist $x_0\in[a;b]$ then $f(x_0)=0$

2) $|ax^2+bx+c|\le \frac12 \Leftrightarrow -\frac12<ax^2+bx+c\le \frac12$

Answer 1

Since $$f(-1)=a-b+c,\quad f(0)=c,\quad f(1)=a+b+c$$ we can write $$a=\frac{f(1)+f(-1)-2f(0)}{2},\quad b=\frac{f(1)-f(-1)}{2},\quad c=f(0)$$

Suppose here that there exists a real number $p$ such that $$|p|\ge 1\qquad\text{and}\qquad |ap^2+bp+c|\gt p^2-\frac 12$$

Then, $$\begin{align}p^2-\frac 12&\lt\left|\frac{f(1)+f(-1)-2f(0)}{2}p^2+\frac{f(1)-f(-1)}{2}p+f(0)\right|\\\\&=\left|f(1)\cdot\frac{p^2+p}{2}+f(-1)\cdot\frac{p^2-p}{2}+f(0)(-p^2+1)\right|\\\\&\le \left|f(1)\right|\left|\frac{p^2+p}{2}\right|+\left|f(-1)\right|\left|\frac{p^2-p}{2}\right|+\left|f(0)\right||p^2-1|\\\\&\le \frac 12\left|\frac{p^2+p}{2}\right|+\frac 12\left|\frac{p^2-p}{2}\right|+\frac 12|p^2-1|\\\\&=\frac 14\left|p(p+1)\right|+\frac 14\left|p(p-1)\right|+\frac 12|(p-1)(p+1)|\\\\&=\frac 14p(p+1)+\frac 14p(p-1)+\frac 12(p-1)(p+1)\\\\&=p^2-\frac 12\end{align}$$ which is impossible.