Maximum value of the sum of absolute values of cubic polynomial coefficients $a,b,c,d$ 
If $p(x) = ax^3+bx^2+cx+d$ and $|p(x)|\leq 1\forall |x|\leq 1$,  what is the $\max$ value of $|a|+|b|+|c|+|d|$?

My try:


*

*Put $x=0$, we get $p(0)=d$,

*Similarly put $x=1$, we get $p(1)=a+b+c+d$,

*similarly put $x=-1$, we get $p(-1)=-a+b-c+d$,

*similarly put $\displaystyle x=\frac{1}{2}$, we get $\displaystyle p\left(\frac{1}{2}\right)=\frac{a}{8}+\frac{b}{4}+\frac{c}{2}+d$


So, given that $|p(x)|\leq 1\forall |x|\leq 1$, we get $|d|\leq 1$.
Similarly $$\displaystyle |b|=\left|\frac{p(1)+p(-1)}{2}-p(0)\right|\leq \left|\frac{p(1)}{2}\right|+\left|\frac{p(1)}{2}\right|+|p(0)|\leq 2$$
Now I do  not understand how can I calculate the  $\max$ of $|a|$ and $|c|$.

Let $A=\max(|a|,|c|),C=\min(|a|,|c|),B=\max(|b|,|d|),D=\min(|b|,|d|).$ Then $$|A|+|B|+|C|+|D|=|a|+|b|+|c|+|d|.$$
For $|x|\le1$, $|Ax^2-C|\le|ax^2+c|$ and $|Bx^2-D|\le|bx^2+d|$. Then
$$|(A+B)x^3-(C+D)x|\le |Ax^3-Cx|+|Bx^2-D|\le |ax^3+cx|+|bx^2+d|\le |p(x)|\text{ or } |p(-x)|.$$
Therefore we only need consider $p(x)=ax^3-cx$ with $a\ge c\ge0$. 
At $x=1$ we need $a\le c+1$ and at $x=\frac{1}{2}$ we need $|a-4c|\le8$. The required maximum is attained for $a=4,c=3$.
