Prove that $\sum_{i=1}^n\frac{x_i}{\sqrt[nr]{x_i^{nr}+(n^{nr}-1)\prod_{j=1}^nx^r_j}} \ge 1$ for all $x_i>0$ and $r \geq \frac{1}{n}$. 
Prove that, for all $x_1,x_2,\ldots,x_n>0$ and $r \geq \frac{1}{n}$, it holds that
$$\sum_{i=1}^n\frac{x_i}{\sqrt[nr]{x_i^{nr}+(n^{nr}-1)\prod \limits_{j=1}^nx^r_j}} \ge 1.$$

This is a slightly modified version of my earlier question here (which corresponds to the case $r=1$).
The case $n=3$ and $r=\frac{2}{3}$ can be reduced to the inequality problem of IMO 2001 (Problem 2).
The case $r=\frac{1}{n}$ can be reduced to
$$\sum_{i=1}^n\frac{x_i^{n-1}}{x_i^{n-1}+(n-1) \prod_{j\neq i}x_j}\geq \sum_{i=1}^n\,\frac{x_i^{n-1}}{x_i^{n-1}+\sum_{j\neq i}\,x_j^{n-1}}=1\,.$$
(Note that we have an equality case when $n=2$ and $r=\frac1n$ here, regardless of the values of the variables $x_i$.)

The following reasoning can give a solution.
Let $nr=\alpha$.
Thus, $\alpha\geq1$ and we need to prove that:
$$\sum_{i=1}^n\frac{x_i}{\left(x_i^{\alpha}+(n^{\alpha}-1)\prod\limits_{j=1}^nx_j^{\frac{\alpha}{n}}\right)^{\frac{1}{\alpha}}}\geq1.$$
Now, by Holder $$\left(\sum_{i=1}^n\frac{x_i}{\left(x_i^{\alpha}+(n^{\alpha}-1)\prod\limits_{j=1}^nx_j^{\frac{\alpha}{n}}\right)^{\frac{1}{\alpha}}}\right)^{\alpha}\sum_{i=1}^nx_i\left(x_i^{\alpha}+(n^{\alpha}-1)\prod\limits_{j=1}^nx_j^{\frac{\alpha}{n}}\right)\geq\left(\sum_{i=1}^nx_i\right)^{\alpha+1}.$$
Thus, it's enough to prove that:
$$\left(\sum_{i=1}^nx_i\right)^{\alpha+1}\geq\sum_{i=1}^nx_i^{\alpha+1}+(n^{\alpha}-1)\prod\limits_{j=1}^nx_j^{\frac{\alpha}{n}}\sum_{i=1}^nx_i,$$
which is true by Muirhead for any integer $\alpha\geq n$.
Now, let $\sum\limits_{i=1}^nx_i$ be a constant and $\sum\limits_{i=1}^nx_i^{\alpha+1}$ be a constant.
Thus, by the Vasc's EV Method (https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf Corollary 1.8 b )
$\prod\limits_{j=1}^nx_j$ gets a maximal value for equality case of $n-1$ variables.
Since the last inequality is homogeneous, it's enough to assume $x_2=...=x_n=1$ and $x_1=x$,
which gives something easier.
