International Mathematical Arhimede Contest

Denote by
$$f_n(x_1, x_2, \dots, x_n) = \sum_{i=1}^{n} \prod_{j \neq i} (x_i - x_j)^p = (x_1 - x_2)^p (x_1 - x_3)^p \dots (x_1 - x_n)^p$$
$$+(x_2-x_1)^p(x_2-x_3)^p\dots(x_2-x_n)^p+\dots+(x_n-x_1)^p(x_n-x_2)^p\dots(x_n-x_{n-1})^p$$
Since $f_2(x_1, x_2) = (x_1 - x_2)^p + (x_2 - x_1)^p = 0$ for all $x_1, x_2 \in \mathbb{R}$, then for $n = 2$ the statement holds and equality occurs. Let $n \ge 3$. Then
$$\begin{aligned} f_n(x_1, x_2, a, a, \dots, a) &= (x_1 - x_2)^p (x_1 - a)^{p(n-2)} + (x_2 - x_1)^p (x_2 - a)^{p(n-2)} \\ &= (x_1 - x_2)^p \left[ (x_1 - a)^{p(n-2)} - (x_2 - a)^{p(n-2)} \right] \end{aligned}$$
for all $x_1, x_2 \in \mathbb{R}$. Observe that $f_n(x_1, x_2, a, a, \dots, a) \ge 0$ when the function $u_a(x) = (x - a)^{p(n-2)}$ is an increasing function. It occurs when $p(n-2)$ is an odd number from which follows that $n$ must be an odd number too.
Now we consider the case when $n \ge 7$ is an odd number. Then, should be
$$f_n(x_1, a, a, a, b, \dots, b) = (x_1 - a)^{3p} (x_1 - b)^{p(n-4)} \ge 0,$$
for all $x_1, a, b \in \mathbb{R}$. But, the preceding inequality does not hold when $x_1 = \frac{a+b}{2}$ and $a \ne b$. So, we have to analyze the cases $n = 3$ and $n = 5$.
(1) For $n = 3$ we may suppose WLOG that $x_1 \ge x_2 \ge x_3$. Let $g(x_1, x_2, x_3) = (x_3 - x_1)^p (x_3 - x_2)^p$ and $u(x) = (x - x_3)^p$, $x \ge x_3$. Then, we have
(i) $u$ is increasing.
(ii) $g(x_1, x_2, x_3) \ge 0$.
On account of (i) and (ii), we have that
$$\begin{aligned} f(x_1, x_2, x_3) &= (x_1 - x_2)^p \left[ (x_1 - x_3)^p - (x_2 - x_3)^p \right] + (x_3 - x_1)^p (x_3 - x_2)^p \\ &= (x_1 - x_2)^p \left[ u(x_1) - u(x_2) \right] + g(x_1, x_2, x_3) \ge 0 \end{aligned}$$
and the statement holds for $n = 3$.
(2) For $n = 5$ we also suppose that $x_1 \ge x_2 \ge x_3 \ge x_4 \ge x_5$. Let
$$h(x_1, x_2, x_3, x_4, x_5) = (x_3 - x_1)^p (x_3 - x_2)^p (x_3 - x_4)^p (x_3 - x_5)^p$$
and let $v(x) = (x - x_3)^p (x - x_4)^p (x - x_5)^p$, $x \ge x_3$ and $w(x) = (x - x_1)^p (x - x_2)^p (x - x_3)^p$, $x \le x_3$. Observe that $h(x_1, x_2, x_3, x_4, x_5) \ge 0$ and $v$ and $w$ are increasing. Since
$$\begin{aligned} f(x_1, x_2, x_3, x_4, x_5) &= (x_1 - x_2)^p [v(x_1) - v(x_2)] \\ &\quad + (x_4 - x_5)^p [w(x_4) - w(x_5)] + h(x_1, x_2, x_3, x_4, x_5), \end{aligned}$$
then $f(x_1, x_2, x_3, x_4, x_5) \ge 0$. We conclude that the natural numbers for which the statement follows are $n = 2, n = 3$ and $n = 5$, respectively. $\square$