An inequality for two independent identically distributed random vectors in a normed space 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. 
 A: 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.
A: fedja: This is a very nice and elegant answer, thank you! (Have you seen/used such a norm ever before?) Before accepting your answer, I'd like to wait a bit, so that other participants feel more encouraged to present other answers. 
In fact, let me give here a modification of your answer. I noticed that for $N=3$ your norm of $v=(v_1,\dots,v_N)$ is $\sum_i|v_i-\frac12\,\sum_1^N v_j|$. So, I thought such a modified norm will work as well, for all $N\ge3$, perhaps with some other factor in place of $\frac12$. This does not seem to work, though. [Addendum: Of course, since the conjectured inequality in my question holds when $B$ is one-dimensional, it was silly for me to think even for a moment that such an $\ell_1$ norm as $\|v\|:=\sum_i|v_i-\frac12\,\sum_1^N v_j|$ could possibly provide a counterexample.]
However, if I am not mistaken, if $\|v\|:=\max_i|v_i-\frac14\,\sum_1^N v_j|$ for $v=(v_1,\dots,v_N)$, then 
$\|e_1\|=3/4$, $\|e_1+e_2\|=1/2$, and $\|e_1-e_2\|=1$, so that your main inequality $(*)$ holds for all $N\ge5$. 
Only the dimensions $N=3,4$ now remain to be considered. 
A: I'm not in my best shape at the moment, so, please, check thoroughly what is written below.
The answer is "No".
The distribution does not matter much, but the norm does. So, we want to take $N\ge 3$ standard basis vectors $e_n$ and see if there is a chance to create a permutation invariant norm in $\mathbb R^N$ such that 
$$(*)\qquad N(N-1)\|e_1+e_2\|+2N\|e_1\|< N(N-1)\|e_1-e_2\|.$$
Now, just define the norm of $v$ as the infimum of $\sum_{i< j}|a_{i,j}|$ with  $v=\sum_{i< j}a_{ij}(e_i+e_j)$. Then, obviously, $\|e_1\|\le 3/2$ ($2e_1=(e_1+e_2)+(e_1+e_3)-(e_2+e_3)$), $\|e_1+e_2\|\le 1$. However, if
$$
e_1-e_2=\sum_{i< j}a_{i,j}(e_i+e_j)\,,
$$
then
$2=\sum_{k>2}(a_{1k}-a_{2k})\le\sum_{i<j}|a_{ij}|$, so $\|e_1-e_2\|\ge 2$ and if $N$ is large, the inequality holds.
