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](http://arxiv.org/abs/1506.00537). In particular, the inequality in question holds if $B$ is two-dimensional or Euclidean. 



An inequality for two independent identically distributed random vectors in a normed space
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.


