TITLE: An interesting integration
QUESTION [8 upvotes]: For any positive integer $n$, let
$$A_n=\idotsint\limits_{\substack{x_1+\cdots+x_n+y_1+\cdots+y_n\leq1\\x_1,\cdots,x_n,y_1,\cdots,y_n\geq0}}\prod_{i,j=1}^n(x_i-y_j)dx_1\cdots dx_ndy_1\cdots dy_n.$$
It is easy to prove that $A_n=0$ when $n$ is odd.
I conjecture that $A_n>0$ when $n$ is even. 
To prove it, I constructed the following real polynomial:
$$\varphi_n(X_1,X_2,\cdots,X_{2n})=\prod_{\substack{1\leq i\leq n\\n+1\leq j\leq 2n}}(X_i-X_j)$$
and let
$$S_{\varphi_n}(X_1,X_2,\cdots,X_{2n})=\sum_{\sigma=[i_1,i_2,\cdots,i_{2n}]\in S_{2n}}f(X_{i_1},X_{i_2},\cdots,X_{i_{2n}}).$$
Here the sum is computed over all permutations $\sigma=[i_1,i_2,\cdots,i_{2n}](\sigma(1)=i_1,\sigma(2)=i_2,\cdots,\sigma(2n)=i_{2n})$ of the set $\{1,2,\cdots,2n\}$ and the set of all such permutations is denoted $S_{2n}.$ 
Then it is easy to see
$$S_{\varphi_n}(a_1,a_2,\cdots,a_{2n})\geq0\ \text{for every}\  (a_1,a_2,\cdots,a_{2n})\in \mathbb{R}^{2n},\text{when}\ n\ \text{is even}$$
$$\Longrightarrow$$
$$A_n=\dfrac{1}{\left(2n\right)!}\idotsint\limits_{\substack{x_1+x_2+\cdots+x_{2n}\leq1\\x_1,x_2,\cdots,x_{2n}\geq0}}S_{\varphi_n}(x_1,x_2,\cdots,x_{2n})dx_1dx_2\cdots dx_{2n}>0,\text{when}\ n\ \text{is even}$$
I believed the former is right, but I can not prove it. So I asked another question here.
Unfortunately, it has been proved to be wrong:
When $n$ is even and $n\geq6$, there exists $(a_1,a_2,\cdots,a_{2n})\in \mathbb{R}^{2n}$ such that $S_{\varphi_n}(a_1,a_2,\cdots,a_{2n})<0$.
Now I have no idea how to prove or disprove my original conjecture:
$$A_n>0\quad\text{when}\ n\ \text{is even}.$$
So I ask for some help.

REPLY [19 votes]: Let's prove a bit more, namely that the integral over $x_i>0, y_j>0, \sum_i x_i+\sum_j y_j=S$ with respect to the $2n-1$-dimensional Lebesgue measure is positive. Note that due to the obvious scaling $x_i,y_j\mapsto tx_i,ty_j$, this integral depends on $S$ in a trivial way (as some pure power function of $S$). Hence, we can just as well introduce any other fast decaying positive weight $w(S)$ and prove that $\idotsint_{(0,+\infty)^{2n}}\prod\dots w(x_1+\dots+x_n+y_1+\dots+\dots y_n)dx_1\dots dx_n dy_1\dots dy_n>0$ instead. Obviously the easiest weight to deal with is just $\exp(-x_1-\dots x_n-y_1-\dots-y_n)$, in which case the integration in $x$ gives $\left(\int_0^\infty P_y(x)e^{-x}\right)^n$ where $P_y(x)=\prod_j(x-y_j)$. For even $n$ this is never negative and certainly not zero for at least some $y$ (say, all $y_j$ equal to each other). Thus the new weighted integral is positive and, thereby, the claim follows.