TITLE: $\sum\limits_{i=1}^n \frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod \limits_{j=1}^nx_j}} \ge 1$, for all $x_i>0.$
QUESTION [11 upvotes]: Can you prove the following new inequality? I found it experimentally.

Prove that, for all $x_1,x_2,\ldots,x_n>0$, it holds that
  $$\sum_{i=1}^n\frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod\limits _{j=1}^nx_j}} \ge
1\,.$$

REPLY [2 votes]: By Holder $$\left(\sum_{i=1}^n\frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod\limits_{j=1}^nx_j}}\right)^n\sum_{i=1}^nx_i\left(x_i^n+(n^n-1)\prod\limits_{j=1}^nx_j\right)\geq\left(\sum_{i=1}^nx_i\right)^{n+1}$$ and it's enough to prove that:
$$\left(\sum_{i=1}^nx_i\right)^{n+1}\geq\sum_{i=1}^nx_i^{n+1}+(n^n-1)\prod\limits_{j=1}^nx_j\sum_{i=1}^nx_i,$$ which is true by Muirhead because the term $\prod\limits_{j=1}^nx_j\sum\limits_{i=1}^nx_i$ is the smallest after deleting of $\sum\limits_{i=1}^nx_i^{n+1}.$