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Consider $a,b,c\in \mathbb{R}$ such that $|a{x}^{2}+bx+c|\le 1\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall }x\in \left[0,1\right]$. Prove that $|a|\le 8\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}},|b|\le 8$ and $|c|\le 1$.
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Ryan Newman
Let $f\left(x\right)=a{x}^{2}+bx+c$.
Hence, $f\left(0\right)=c$, $f\left(1\right)=a+b+c$ and $f\left(\frac{1}{2}\right)=\frac{a}{4}+\frac{b}{2}+c$
After solving of this system we obtain:
$|a|=|2f\left(1\right)-4f\left(\frac{1}{2}\right)+2f\left(0\right)|\le 2|f\left(1\right)|+4|f\left(\frac{1}{2}\right)|+2|f\left(0\right)|\le 8$
$|b|=|-f\left(1\right)+4f\left(\frac{1}{2}\right)-3f\left(0\right)|\le |f\left(1\right)|+4|f\left(\frac{1}{2}\right)|+3|f\left(0\right)|\le 8$
$|c|=|f\left(0\right)|\le 1$
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