Let be $a$ be positive real number. Prove that there are no real numbers $b$ and $c$, with $b < c$, so that for any distinct numbers $x, y \in (b, c)$ we have $|\frac{x+y} {x-y}| \leq a$.
Problem
Source: Romania 3rd JBMO TST 2023 P4
Tags: algebra, absolute value
20.05.2023 23:41
This is not suited at all for Romanian JBMO TST P4 because it is an analysis problem and it is trivial. We have to show that for every constant $a$ and open interval $I$ there exists $x,y\in I$ such that $\left|\frac{x+y}{x-y}\right|>a$. For this we can just take any $y\in I$ and since $\lim_{x\to y} \left|\frac{x+y}{x-y}\right|=\infty$ we can find $x\in I$ close enough to $y$ to satisfy the above inequality.
14.01.2025 14:21
Can anyone translate the official solutions and post it here? I think it is too easy or the wording is not correct…
14.01.2025 19:09
The wording is correct - the problem is indeed easy. Below is a translation of the official solution by ChatGpt, checked by me.
14.01.2025 19:15
Literally just take x to be extremely close to y.