any solution?

Induction does the trick. Let $f(n,k)$ denote the proposition to be proven. Note that $f(n,1)$ is clearly true for any $n\geqslant 3$. Note further that $f(n,n)$ is true because $$ \sum_{i=1}^{n}\left(x_i^{n-2}\prod_{\substack{j\neq i}}\frac{1}{x_i-x_j}\right) = 0. $$ To finish the base case $n=3$, note that $f(3,2)$ is true, as it reduces to $x_2x_3 \geqslant x_1x_3$, which is true since $x_1>x_2>0\geqslant x_3$. All that is left is to realise that for $2\leqslant k \leqslant n-1$, $f(n,k)$ reduces to $f(n-1,k) \land f(n-1,k-1)$ as follows: \begin{align*} \sum_{i=1}^{k}\left(x_i^{n-2}\prod_{\substack{1\leqslant j \leqslant n\\ j\neq i}}\frac{1}{x_i-x_j}\right) &\geqslant \sum_{i=1}^{k}\left(x_i^{n-2}\prod_{\substack{1\leqslant j \leqslant n,\\ j\neq i}}\frac{1}{x_i-x_j}\right) - \sum_{i=1}^{k}\left(x_i^{n-3}\prod_{\substack{1\leqslant j \leqslant n-1,\\ j\neq i}}\frac{1}{x_i-x_j}\right)\cdot \frac{x_k}{x_k-x_n} \\ &= \sum_{i=1}^{k}\left(x_i^{n-3}\prod_{\substack{1\leqslant j \leqslant n-1,\\ j\neq i}}\frac{1}{x_i-x_j}\cdot \left(\frac{x_i}{x_i-x_n}-\frac{x_k}{x_k-x_n}\right)\right) \\ &= \frac{-x_n}{x_k-x_n}\cdot\sum_{i=1}^{k-1}\left(x_i^{n-3}\prod_{\substack{1\leqslant j \leqslant n,\\ j\neq i, j\neq k}}\frac{1}{x_i-x_j}\right)\geqslant 0. \end{align*}