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Consider $a,b,c\in\mathbb{R}$ such that $|ax^2+bx+c|\leq 1\;\forall x\in \left[0,1\right]$. Prove that $|a|\leq 8\;\;,|b| \leq 8$ and $|c| \leq 1$.

My Attempt:

Set $x = 0$ in $|ax^2+bx+c|\leq 1$ to get $|c| \leq 1$. Similarly set $x = 1$ in $|ax^2+bx+c| \leq 1$ to get $|a+b+c|\leq 1$.

From here, how can I calculate bounds for $a$ and $b$?

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  • 2
    $\begingroup$ It may help the intuition to note that the parabola $y=8x^3-8x+1$ is $1$ at the endpoints, and $-1$ at its minimum at $x=1/2$. $\endgroup$ – André Nicolas Feb 26 '13 at 8:20
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Let $f(x)=ax^2+bx+c$.

Hence, $f(0)=c$, $f(1)=a+b+c$ and $f\left(\frac{1}{2}\right)=\frac{a}{4}+\frac{b}{2}+c$.

After solving of this system we obtain:

$|a|=|2f(1)-4f\left(\frac{1}{2}\right)+2f(0)|\leq2|f(1)|+4|f\left(\frac{1}{2}\right)|+2|f(0)|\leq8$,

$|b|=|-f(1)+4f\left(\frac{1}{2}\right)-3f(0)|\leq|f(1)|+4|f\left(\frac{1}{2}\right)|+3|f(0)|\leq8$,

$|c|=|f(0)|\leq1$.

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