<s>[INST] If $|ax^2 + bx + c| \le 1$ for all $x$ in $[0,1]$ then evaluate $|a|$, $|b|$ and $|c|$ 
Solve the following:
If $|ax^2 + bx + c| \le 1$ for all $x$ in $[0,1]$, then
i) $|a| \le 8$
ii) $|b| \ge 8$
iii) $|c| \le 1$
iv) $|a| + |b| + |c| \le 17$

Solution from my textbook:

Put $x = 0, 1 \text{ and } \frac{1} {2}$ to get:
$$|c| \le 1$$
$$|a + b + c| \le 1$$
$$|a + 2b + 4c| \le 4$$
From the above three equations, we get,
$|b| \le 8 \text{ and } |a| \le 8$.
Therefore, $|a| + |b| + |c| \le 17$

However, I don't understand: how, from the three equations, can you find the values of $|a|$ and $|b|$?
 [/INST] Let $f(x)=ax^2+bx+c$.
Thus, $$a+b+c=f(1),$$
$$\frac{1}{4}a+\frac{1}{2}b+c=f\left(\frac{1}{2}\right)$$ and $$c=f(0),$$ which gives
$$b=4f\left(\frac{1}{2}\right)-f(1)-3f(0),$$ $$a=2f(1)+2f(0)-4f\left(\frac{1}{2}\right)$$ and by the triangle inequality we obtain: $$|a|+|b|+|c|\leq2+2+4+4+1+3+1=17.$$
It's interesting that for $f(x)=8x^2-8x+1$ we have an equality.
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