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](https://mathoverflow.net/questions/226568/an-interesting-inequality).
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.



An interesting integration
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. 


