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 $p(1)=u,$ $p(-1)=v$, $p\left(\frac{1}{2}\right)=w$ and $p\left(-\frac{1}{2}\right)=t$.
Thus, we have the following system:
$$a+b+c+d=u,$$
$$-a+b-c+d=v,$$
$$\frac{a}{8}+\frac{b}{4}+\frac{c}{2}+d=w$$ and
$$-\frac{a}{8}+\frac{b}{4}-\frac{c}{2}+d=t,$$ which gives
$$a=\frac{2u-2v-4w+4t}{3},$$
$$b=\frac{2u+2v-2w-2t}{3},$$
$$c=\frac{-u+v+8w-8t}{6}$$ and
$$d=\frac{-u-v+4w+4t}{6}.$$
Now, $$a+b+c+d=u\leq1,$$
$$a+b+c-d=\frac{4u+v-4w-4t}{3}\leq\frac{13}{3},$$
$$a+b-c+d=\frac{4u-v-8w+8t}{3}\leq7,$$
$$a+b-c-d=\frac{5u-12w+4t}{3}\leq7,$$
$$a-b+c+d=\frac{-u-4v+4w+4t}{3}\leq\frac{13}{3},$$
$$a-b+c-d=-v\leq1,$$
$$a-b-c+d=\frac{-5v-4w+12t}{3}\leq7$$ and
$$a-b-c-d=\frac{u-4v-8w+8t}{3}\leq7,$$
which gives
$$|a|+|b|+|c|+|d|\leq7.$$
But for $p(x)=4x^3-3x$ the equality occurs, which says that $7$ is a maximal value.
