<s>[INST] Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$. Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$.  
I made quite a few attempts but could not explicitly derive the required condition. I can understand that it happens, but cannot derive that. Please help.
 [/INST] WLOG, let us suppose $f(x)=a(x-u)^2-v$ for $a>0$ and $0< u\le 1$, $v\ge 0$.
Then the premise says $1\ge au^2-v\ge -1$, $1\ge a(1-u)^2-v\ge -1$, $1 \ge a(1+u)^2-v\ge -1$. We need to show that $v\le \frac{3}{2}$.
The above inequalities implies that $1\ge au$ and $2\ge a(1+2u)$. 
If $u\in (0,1/2]$, then
$$
v\le 1+au^2\le 1+2u^2/(1+2u)\le\frac{5}{4}.
$$
If $u\in [1/2, 1]$, then
$$
v\le 1+a(1-u)^2\le 1+(1-u)^2/u\le \frac{5}{4}.
$$
Let $u=\frac{1}{2}$, $a=2$, $v=\frac{5}{4}$, then $f=2(x-\frac{1}{2})^2-\frac{5}{4}$ attains the extreme value.
I suppose there should be more effective proof.
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