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Mathematical competitions in Croatia

Number theory Difficulty 4.8 AIME Prove it Croatia

Let $c$ and $d$ be positive divisors of a positive integer $n$ . If $c > d$ , prove that
$c > d + \frac{d^2}{n}.$
(Russia 2011)

Solution

Consider $\frac{n}{d}$ and $\frac{n}{c}$ . These are also positive divisors of $n$ which satisfy $\frac{n}{d} > \frac{n}{c}$ .
We now have
$\frac{n}{d} - \frac{n}{c} \ge 1,$
from which
$c - d \ge \frac{dc}{n} > \frac{d^2}{n}$
follows.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.