**Question:** let $x\_{i}>0$ $(i=1,2,\cdots,n)$, such that $x\_{i}\neq x\_{j},\forall i\neq j$, find all real numbers $p$ that satisfy the 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 solved it as $p\le 0$ 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



How to show this symmetric function inequality
The conjecture for $n=4$ is correct. For general $n$ and any $x\_1,\ldots,x\_n$,
$$
f(p) := \sum\_{i=1}^{n}\dfrac{x^p\_i}{\prod\_{j\neq i}(x\_i-x\_j)}
$$
is positive for $p > n-2$ and switches sign at $p=0,1,2,\ldots,n-2$
and nowhere else. For example:  

If $n=5$ then $f(p) \geq 0$ iff
$p \geq 3$ or $1 \leq p \leq 2$ or $p \leq 0$;  

if $n=6$ then $f(p) \geq 0$ for $p \in [0,1]$, $[2,3]$, or $[4,\infty)$;  

if $n=7$ then $f(p) \geq 0$ for
$p \in (-\infty,0]$, $[1,2]$, $[3,4]$, or $[5,\infty)$;  

etc.


The key fact is the following exponential analogue of "Descartes' law of sines":


**Proposition.** *Let $x\_1,\ldots,x\_n$ be pairwise distinct
positive real numbers, and $c\_i$ any real numbers not all zero.
Then the function $f(p) := \sum\_{i=1}^n c\_i^{\phantom.} x\_i^p$
has at most $n-1$ real zeros, counted with multiplicity.*


*Proof*: Induction on $n$. For $n=1$ the function $f$ is a nonzero multiple of
$x\_1^p$, which is never zero; this establishes the base case.
Now let $n>1$, and assume we have proved the case $n-1$ of the Proposition.
Let $f\_1(p) = x\_n^{-p} f(p) = \sum\_{i=1}^n c\_i (x\_i/x\_n)^p$, which is
of the same form as $f$ with $x\_n=1$ and has the same zeros and multiplicities.
Then the inductive hypothesis applies to the derivative
$$
f'\_1(p) = \sum\_{i=1}^{n-1} (c\_i \log (x\_i/x\_n)) \, (x\_i/x\_n)^p;
$$
so $f'\_1$ has at most $n-2$ zeros, counted with multiplicity.
By Rolle's Theorem it follows that $f\_1$ has at most $n-1$ zeros,
counted with multiplicity. This completes the induction step and the proof.
**QED**


It remains to check that our $f$ vanishes at $p=0,1,\ldots,n-2$;
then we can use our Proposition to deduce that each of these $n-1$ zeros
is simple and there are no others. This can be done by recognizing
$f(p)$ as the sum of the residues of the differential
$x^p \, dx \left/ \prod\_{j=1}^n (x-x\_j) \right.$
(which is regular at $x=\infty$ for integers $p \leq n-2$).
Alternatively we can relate $f(p)$ to
a Schur-Vandermonde determinant with a repeated row.


**EDIT** here's a more self-contained argument that
$f(0)=f(1)=\cdots=f(n-2)=0$. Let $P(x)$ be any polynomial with $\deg P < n$.
Then $P(x) \left/ \prod\_{j=1}^n (x-x\_j) \right.$ has a partial-fraction
decomposition $\sum\_{i=1}^n c\_i / (x-x\_i)$. Taking $x \to x\_i$
(or multiplying through by $\prod\_{i=1}^n (x-x\_i)$ and setting $x = x\_i$),
we see that $c\_i = P(x\_i) \left/ \prod\_{j\neq i}(x\_i - x\_j) \right.$.
On the other hand, taking $x \to \infty$
(or multiplying through by $\prod\_{i=1}^n (x-x\_i)$ and comparing $x^{n-1}$ coefficients),
we see that the $x^{n-1}$ coefficient of $P$ is $\sum\_{i=1}^n c\_i$.
In particular if $P(x) = x^p$ for some nonnegative integer $p \leq n-2$ then
$f(p) = \sum\_{i=1}^n c\_i = 0$. $\Box$


