Question: let $x_{i}>0(i=1,2,\cdots,n)$,such $x_{i}\neq x_{j},\forall i\neq j$, find the all real number $p$.we have this following inequality $$\sum_{i=1}^{n}\dfrac{x^p_{i}}{\prod_{j\neq i}(x_{i}-x_{j})}\ge 0$$

when $n=3$, I have solve it.the $p\le 0$ or $p\ge 1$,because we only use this $$x^p\ge y^p+p(x-y)y^{p-1},x>y>0$$so WLOG $x>y>z$ we have $$\sum_{cyc}\dfrac{x^p}{(x-y)(x-z)}\ge\dfrac{y^p+p(x-y)y^{p-1}}{(x-y)(y-z)}+\dfrac{y^p}{(y-z)(y-x)}+\dfrac{y^p+p(z-y)z^{p-1}}{(z-x)(z-y)}=0$$

For $n=4$ I tried some values, Now I conjecture $p\ge 2$ or $0\le p\le 1$,and I can't prove it

In general,Now I can't find the $p$.Thanks