inequality with roots of unity Do you know proofs or references for the following inequality:
There exists a positive constant $C>0$ such that for any complex numbers $a_1,\ldots,a_n$
$$
|a_1|+\cdots+|a_n| \leq C\sup_{z_1^3=1,\ldots,z_n^3=1 } |a_1z_1+\cdots + a_n z_n|
$$
where the supremum is taken over the complex numbers $z_1,\ldots,z_n$ such that $z_1^3=1,\ldots,z_n^3=1$?
 A: The strategy is to rotate the $a$ values so that they are roughly pointing in the same  direction. More formally, for each complex $a$ choose a third root of unity $z$  so
that the argument $\theta $ of $az$ lies between plus and minus $\pi/3$.
Then 
\begin{eqnarray*}
|a_1z_1+\cdots +a_nz_n|&\geq&|\mbox{Re}(a_1z_1+\cdots +a_nz_n)| \cr
&=&|a_1z_1|\cos(\theta_1)+\cdots +|a_nz_n|\cos(\theta_n) \cr
&\geq &(|a_1|+\cdots +|a_n|)\cos(\pi/3)\cr
&=&{1\over 2}\ (|a_1|+\cdots +|a_n|)\cr
\end{eqnarray*} 
This gives your result with $C=2$.
A: Notice that the mapping $\|\cdot\|_* \colon \mathbb{C}^n \to \mathbb{R}$ given by
$$\|(a_1,\dots,a_n)\|_* = \sup_{z_1^3=z_2^3=\dots=1} |a_1z_1 + \dots + a_n z_n|$$
defines a norm. The result then follows by the equivalence of norms in finite dimensional vector spaces.
The triangle inequality is easily seen to be true: Just use the ordinary triangle inequality. To prove that if $\|(a_1,\dots,a_n)\|_* = 0$ then $(a_1,\dots,a_n) = 0$ notice that if one of the coordinates (let's say $a_1$) wasn't $0$, then there would be three different numbers: $a_1 z_1 + a_2 + \dots + a_n$, $a_1 z_2 + a_2 + \dots + a_n$ and $a_1 z_3 + a_2 + \dots + a_n$ one of which is non-zero and thus the sup we get would be non-zero also.
