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?

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    $\begingroup$ Case (1) is reminiscent of en.wikipedia.org/wiki/Amitsur%E2%80%93Levitzki_theorem $\endgroup$ – Suvrit Dec 20 '15 at 15:56
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    $\begingroup$ @Suvrit Case (1) is completely obvious: if we interchange sets $\{x_1,\dots,x_n\},\{x_{n+1},\dots,x_{2n}\}$, the product changes its sign. $\endgroup$ – Fedor Petrov Dec 20 '15 at 16:34
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    $\begingroup$ @FedorPetrov thanks, it is indeed obvious. I had just remarked about that to make a connection with a well-known symmetric polynomial identity :-) $\endgroup$ – Suvrit Dec 20 '15 at 17:28
  • $\begingroup$ related question mathoverflow.net/questions/107526/… $\endgroup$ – Abdelmalek Abdesselam Mar 26 '16 at 16:42

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 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.

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  • $\begingroup$ Fedor Petrov: Thank you very much for your help! $\endgroup$ – user173856 Dec 22 '15 at 1:53

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