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I've been trying to solve the next problem but I have no idea of how to find the solution:

Find the largest number of positive integers in such a way that any two of them $x$ and $y$ ($x\neq y$) satisfy $$\frac{xy}{100}\leq|x-y|.$$

Thank you for your help!

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  • $\begingroup$ It's unclear what you're asking and you show no work. There are infinitely many $(x,y)$ that satisfy that. For example, let $x\geq 2$ and let $y=1$. $\endgroup$ – user223391 Jul 5 '15 at 6:58
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    $\begingroup$ Great problem, Try using the equivalent inequality $\left| \frac{1}{x} - \frac{1}{y}\right | \ge \frac{1}{100}$. $\endgroup$ – Orest Bucicovschi Jul 5 '15 at 7:00
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    $\begingroup$ @avid19 I just know that if $X$ is a set with the maximum number of positive integers satisfying the conditions given and $x,y\in X$ are distinct and $x<y$ then $\frac{100x}{100-x}\leq y$ and also that there is at most one $x\in X$ such that $x\geq100$ but nothing important $\endgroup$ – user116226 Jul 5 '15 at 7:05
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Let $X=\{x_1,\ldots,x_n\}$ be a set with the largest number of positive integers satisfying the conditions given.

It is clear that $x_k\geq k,$ for each $k\in[[1,n]]$ (where $[[1,n]]:=[1,n]\cap\Bbb Z$), which implies that $$\frac{100k}{100-k}\leq\frac{100x_k}{100-x_k}\leq x_{k+1},$$ for each $k\in[[1,n-1]].$

Note that since the set $[[1,10]]$ fulfills the requirements of the problem, then $n\geq10.$

Now, we have the following:

$(1)$ $$x_{11}\geq\frac{100x_{10}}{100-x_{10}}\geq\frac{100\cdot10}{100-10}=\frac{100}{9}>11\Longrightarrow x_{11}\geq12,$$

$(2)$ $$x_{12}\geq\frac{100x_{11}}{100-x_{11}}\geq\frac{100\cdot12}{100-12}=\frac{150}{11}>13\Longrightarrow x_{12}\geq14;$$ $$\vdots$$ in this way we find that $x_{13}\geq17,$ $x_{14}\geq21,$ $x_{15}\geq27,$ $x_{16}\geq37,$ $x_{17}\geq59$ and $x_{18}\geq144$ (you already proved that at most one $y\in X$ is greater than or equal to $100$) which implies that $10\leq n\leq18.$

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