TITLE: An inequality for two independent identically distributed random vectors in a normed space
QUESTION [16 upvotes]: Suppose that $X$ and $Y$ are independent identically distributed random vectors in a separable Banach space $B$. Does it always follow that $E\|X-Y\|\le E\|X+Y\|$? 
Some background information on this question can be found at the end of the note posted on arXiv at additive decomposition of norms. In particular, the inequality in question holds if $B$ is two-dimensional or Euclidean.

REPLY [6 votes]: To get dimension 4, I think the norm 
\begin{equation} \|a\| =  \max_{\{i,j\}}  |a_i-a_j| \vee \|a\|_\infty,
\end{equation}
where $a=(a_1,a_2,a_3,a_4)$, 
works. Again $X$ samples the unit vector basis uniformly.
EDIT: It looks like a variation takes care of dimension 3.  Use again
\begin{equation} \|a\| =  \max_{\{i,j\}}  |a_i-a_j| \vee \|a\|_\infty,
\end{equation}
where $a=(a_1,a_2,a_3)$. This time, let $X$ sample $e_1, e_2, e_3 $ and $-(e_1+e_2+e_3)/2$ uniformly. The first three vectors have norm 1 and the last one norm $1/2$.  In the (unlikely) event that my arithmetic is correct, the expectation of $\|X+Y\|$ is $19/16$ and the the expectation of $\|X-Y\|$ is $21/16$.
If this is correct, the only remaining thing is whether the inequality is true for  random variables that take on three non zero values.