Prove that $|a|+|b|+|c|\le17$ if $p(x)=ax^2+bx+c$ is a real polynomial with $|p|\le1$ for $0\le x\le1$ 
Let $ax^2+bx+c$ be a quadratic polynomial with real coefficients such that $$|ax^2+bx+c| \leq 1,$$ for $ 0\leq x\leq 1$. Prove that $$|a|+|b|+|c|\leq 17$$

How to proceed in this particular question. Sorry I can't show any work because I really not getting how to initiate.
 A: Hint: Let $P$ your polynomial. Put $P(0)=u$, $P(1/2)=v$ and $P(1)=w$. Then $|u|, |v|, |w|$ are $\leq 1$. Find $a,b,c$ in function of $u,v,w$, and bound them. 
A: Let $f(x) = ax^2 + bx + c$ We know that
$$
\left|\frac{a}2\right| = |f(0) + f(1) - 2f(0.5)| = |[f(0) - f(0.5)] - [f(0.5) - f(1)]|\\
\leq |f(0) - f(0.5)| + |f(0.5) - f(1)| \leq 2+2 = 4
$$
so $|a| \leq 8$. Clearly, $|c| = |f(0)| \leq 1$. That leaves $b$. We get
$$
|b| = |4f(0.5) - f(1) - 3f(0)| \leq  3|f(0.5)-f(0)| + |f(0.5) - f(1)| \leq 3\cdot 2 + 2 = 8
$$
which is what we need.
It's also worth noting that
$$
f(x) = 8x^2 - 8x + 1
$$
demonstrates that $17$ is a strict bound, so we cannot do any better.
A: There are a few additional posts like this quora post link, which try to solve this problem via double difference, clever manipulation of triangle inequality and translating the inequalities at $x=0,\frac 12, 1$ to arrive at $a \leq 8$. 
Unfortunately I have been finding it difficult to defend the approach against the following arguments:


*

*Why can't we use double difference on sample points like $x=0, 1/4, 1/2$ as a viable approach

*Why does the method of triangle inequality give us different results when we use points like $x=\frac 14$


Following is an alternate approach to solve the problem. Would appreciate if this can be verified/challenged.
The problem can be translated to finding a bounding condition for the set of quadratics which can pass through the bounding box defined by $y=\pm 1, x=0,1$, without intersecting top and bottom edges ($y=\pm 1$). This takes care of the $|\ |$ condition too. 

To find the bounding condition, we can simplify the ask by translating the bounding box to $y=0,-2; x=0,1$. We can see that quadratics whose difference in roots is less than 1 and $V_y \lt -2$ won't qualify. 
Hence, we can conclude that all quadratics which have distance between roots > 1 and $V_y \leq 2$ can be translated to fit the bounding box.

Therefore, the limiting quadratic is $8x^2 - 8x$, which when translated to the original bounding box $(y=\pm 1; x=0,1)$ becomes $8x^2 - 8x + 1$ and $-8x^2 + 8x -1$.
Thus $|a| \leq 8$, $|b| \leq 8$ and $|c| \leq 1$. Hence, proving the ask: 
$|a| + |b| + |c| \leq 17$
