Let $\mathbb{R}$ be the real field. For any homogeneous polynomial $f(X\_1,\cdots,X\_n)$ in $\mathbb{R}[X\_1,\cdots,X\_n]$, we use $S\_f(X\_1,\cdots,X\_n)$ to denote the following homogeneous symmetric polynomial:
$$S\_f(X\_1,\cdots,X\_n)=\sum\_{\sigma=[i\_1,\cdots,i\_n]\in S\_n}f(X\_{i\_1},\cdots,X\_{i\_n}).$$
Here the sum is computed over all permutations $\sigma=[i\_1,\cdots,i\_n](\sigma(1)=i\_1,\cdots,\sigma(n)=i\_n)$ of the set $\{1,\cdots,n\}$ and the set of all such permutations is denoted $S\_n.$ We say $f(X\_1,\cdots,X\_n)$ is good if 
$$S\_f(a\_1,\cdots,a\_n)\geq 0$$
for every $(a\_1,\cdots,a\_n)\in \mathbb{R}^n.$
For example, when $n=3$, we have $$f(X\_1,X\_2,X\_3)=X\_3^2-X\_1X\_2$$ is good because$$S\_f(X\_1,X\_2,X\_3)=(X\_1-X\_2)^2+(X\_2-X\_3)^2+(X\_3-X\_1)^2.$$
For any $n\geq 1$, define the homogeneous polynomial of degree $n^2$ in $\mathbb{R}[X\_1,X\_2,\cdots,X\_{2n}]$ as follow:$$\varphi\_n(X\_1,X\_2,\cdots,X\_{2n})=\prod\_{\substack{1\leq i\leq n\\n+1\leq j\leq 2n}}(X\_i-X\_j).$$
I conjecture that $\varphi\_n$ is good for any $n\geq 1$, and for this conjecture I have got the following simple results:


$(1)$It is easy to proof that $S\_{\varphi\_n}(X\_1,X\_2,\cdots,X\_{2n})=0$ when $n$ is odd;


$(2)$When $n$ is even,$$S\_{\varphi\_2}(X\_1,X\_2,X\_3,X\_4)=4[((X\_1-X\_2)(X\_3-X\_4))^2+((X\_1-X\_3)(X\_2-X\_4))^2+((X\_1-X\_4)(X\_2-X\_3))^2].$$
So the conjecture is right for $n=2$.


I can not proof it any more for $n\geq 4$, but I believe that the conjecture is right. Would you please give me some help?



An interesting inequality
Alas, this is false at least for even $n\geqslant 6$, I do not know about $n=4$.


A similar question about symmetrization of $$f\_{a,b}:=\prod\_{1\leq i\leq a,a+1\leq j \leq a+b} (X\_i-X\_j)$$
may be asked (of course, it equals to 0 if $ab$ is odd). I asked this for $a=1$, $b=n-1$ [here](https://mathoverflow.net/questions/41666/sum-of-derivatives-in-roots-of-a-polynomial-of-odd-degree) and it turned out to fail for odd $n\geq 7$. 


Now for your question. Take even $n\geq 6$ and $n-1$ variables equal taking very-very large value $M$, and some $n+1$ variables $X\_1,\dots,X\_{n+1}$. Then if we divide the value $$f\_{n,n}(M,M,\dots,M,X\_{1},X\_{2},\dots,X\_{n+1})$$ by $M^{n(n-1)}$, it tends to 
$$f\_{1,n}(X\_1,\dots,X\_{n+1}),$$
therefore if symmetrization of the latter is negative, symmetrization of the former is negative for large enough $M$.


It would be interesting to describe all pairs $(a,b)$ for which inequality holds.


