I am having a hard time proving this proposition:
Show that for $x_i, n \in \mathbb{N}$, where $x_i\geq 2$ and for any real number $r>1$,
if $\prod\limits_{i=1}^{n}\left(1-\frac{1}{x_i}\right) \leq \left(1-\frac{1}{nr}\right)^{n}$, then $\exists i$ such that $x_i\leq nr$.
How do I even begin my attack on this problem? Thanks in advance!
Disregard the first paragraph, it was written before your edit.
Does $nr$ have to be an integer? Because if not we can say $x_1 = x_2 = 1, n=2, nr=\frac{1}{3}$, which would make the inequality true but there wouldn't be an $x_i \le nr$. Hagen's claim for negative $r$ above also stands.
Assume $x$, $n$, and $r$ are positive integers. Then $1-\frac{1}{x_i} \ge 0$ and $1-\frac{1}{nr} \ge 0$. If all $x_i \gt nr$, then every $1-\frac{1}{x_i}$ would be greater than $(1 - \frac{1}{nr})$. If you have a number of nonnegative multiplicands that are all greater than an equal number of other multiplicands, then the product of the formers will be greater than that of the latters. This fails the condition that $\prod_{i=1}^{n}(1-\frac{1}{x_i}) \le (1-\frac{1}{rn})^n$, therefore the assumption that all $x_i \gt nr$ is false, proving what you need.