$\sum\limits_{i=1}^n \frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod \limits_{j=1}^nx_j}} \ge 1$, for all $x_i>0.$ 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
  $$\sum_{i=1}^n\frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod\limits _{j=1}^nx_j}} \ge
1\,.$$

 A: The inequality $\displaystyle\sum_{i=1}^n\frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod \limits_{j=1}^nx_j}} \ge 1$ is trivial given the claim below.  Of course, the equality occurs if and only if $x_1=x_2=\ldots=x_n$.

Claim: For
   every $i=1,2,\ldots,n$, we have $\displaystyle\frac{x_i}{\sqrt[n]{x_i^n+\left(n^n-1\right)\,\prod\limits_{j=1}^n\,x_j}}
 \geq
 \frac{x_i^{1-\frac{1}{n^n}}}{\sum\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}}$.  The equality holds if and only if
   $x_1=x_2=\ldots=x_n$.

Proof:  The required inequality is equivalent to $$\left(\sum_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^n-x_i^{n\left(1-\frac{1}{n^n}\right)}\geq \left(n^n-1\right)\,x_i^{-\frac{1}{n^{n-1}}}\,\prod_{j=1}^n\,x_j\,.$$
Note that the expansion of $\left(\sum\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^n$ consists of $n^n$ terms of the form $x_{j_1}^{1-\frac{1}{n^n}}x_{j_2}^{1-\frac{1}{n^n}}\cdots x_{j_n}^{1-\frac{1}{n^n}}$, where $j_1,j_2,\ldots,j_n\in\{1,2,\ldots,n\}$.  The product of these terms is equal to $$\left(\prod\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^{n^n}\,.$$  If we take the term $x_i^{n\left(1-\frac{1}{n^n}\right)}$ out of the product, we get the product of $n^n-1$ terms from the expansion of $\left(\sum\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^n-x_i^{n\left(1-\frac{1}{n^n}\right)}$, which is then equal to
$$\frac{\left(\prod\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^{n^n}}{x_i^{n\left(1-\frac{1}{n^n}\right)}}=x_i^{-\frac{1}{n^{n-1}}\left(n^n-1\right)}\,\prod\limits_{j=1}^n\,x_j^{n^n-1}\,.$$
By the AM-GM Inequality,
$$\frac{\left(\sum\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^n-x_i^{n\left(1-\frac{1}{n^n}\right)}}{n^n-1}\geq \left(x_i^{-\frac{1}{n^{n-1}}\left(n^n-1\right)}\,\prod\limits_{j=1}^n\,x_j^{n^n-1}\right)^{\frac{1}{n^n-1}}=x_i^{-\frac{1}{n^{n-1}}}\,\prod\limits_{j=1}^n\,x_j\,,$$
which is what we want.  Hence, the claim is true.  The equality case happens, due to the AM-GM Inequality, if and only if $x_1=x_2=\ldots=x_n$.

How did I get the exponent $1-\frac{1}{n^n}$? 
I assumed it was $k$ at first, and the desired inequality was equivalent to
$$\left(\sum\limits_{j=1}^n\,x_j^{k}\right)^n-x_i^{nk}\geq \left(n^n-1\right)\,x_i^{n(k-1)}\,\prod\limits_{j=1}^n\,x_j\,.$$
Then, the last inequality read
$$\frac{\left(\sum\limits_{j=1}^n\,x_j^{k}\right)^n-x_i^{nk}}{n^n-1}\geq \left(x_i^{-nk}\,\prod\limits_{j=1}^n\,x_j^{n^nk}\right)^{\frac{1}{n^n-1}}\,,$$
the right-hand side of which I wanted to equal $x_i^{n(k-1)}\,\prod\limits_{j=1}^n\,x_j$.  Therefore, $\frac{n^nk}{n^n-1}=1$ and $n(k-1)=-\frac{nk}{n^n-1}$, both of which gave me $k=1-\frac{1}{n^n}$.
A: 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 $\sum\limits_{i=1}^nx_i^{n+1}.$
