Do more generalizations of Schur's inequality exist? 
I meet this following problem
  If $$n\ge 3,\sum_{i=1}^{n}\left(\prod_{j\neq i}(a_{i}-a_{j})\right)\ge 0$$
  where $a_{i}$ are real numbers.

when $n=3$, it is Schur's inequality
 so which $n$ such this inequality?
but for  more generalization  form of Schur'.s Inequality exists?

Now I have solve it, this inequality it is only true for $n=3,n=5$ holds.
For $n=3$ then inequality can write as
$$(a_{1}-a_{2})(a_{1}-a_{3})+(a_{2}-a_{1})(a_{2}-a_{3})+(a_{3}-a_{1})(a_{3}-a_{2})=a^2_{1}+a^2_{2}+a^2_{3}-a_{1}a_{2}-a_{1}a_{3}-a_{2}a_{3}\ge 0$$
it is clear true
for $n=5$,WLOG, we assume that
$$a_{1}\ge a_{2}\ge a_{3}\ge a_{4}\ge a_{5}$$
Lemma 1:
$$I=(a_{1}-a_{2})(a_{1}-a_{3})(a_{1}-a_{4})(a_{1}-a_{5})+(a_{2}-a_{1})(a_{2}-a_{3})(a_{2}-a_{4})(a_{2}-a_{5})\ge 0$$
Proof: because
$$I=(a_{1}-a_{2})[(a_{1}-a_{3})(a_{1}-a_{4})(a_{1}-a_{5})-(a_{2}-a_{3})(a_{2}-a_{4})(a_{2}-a_{5})]$$
since
$$a_{1}-a_{3}\ge a_{2}-a_{3}\ge 0$$
$$a_{1}-a_{4}\ge a_{2}-a_{4}\ge 0$$
$$a_{1}-a_{5}\ge a_{2}-a_{5}\ge 0$$
$$a_{1}-a_{2}\ge 0$$
so
$$I\ge 0$$
Use same methods we have
$$(a_{5}-a_{1})(a_{5}-a_{2})(a_{5}-a_{3})(a_{5}-a_{4})+(a_{4}-a_{1})(a_{4}-a_{2})(a_{4}-a_{3})(a_{4}-a_{5})\ge 0$$
and
$$(a_{3}-a_{1})(a_{3}-a_{2})(a_{3}-a_{4})(a_{3}-a_{5})\ge 0$$
so for $n=5$ is hold
for $n=4$, you can take $a_{1}=0,a_{2}=a_{3}=a_{4}=1$
for $n\ge 6$,you can take
$$a_{1}=0,a_{2}=a_{3}=a_{4}=1,a_{5}=a_{6}=\cdots=a_{n}=-1$$
then $$\sum_{i=1}^{n}\left(\prod_{j\neq i}(a_{i}-a_{j})\right)=(-1)^3<0$$
