TITLE: Inequality in a bounded real sequence
QUESTION [21 upvotes]: Prove or disprove that for any bounded real sequence $\{x_n\}_{n\in\mathbb{N}}$ there exist two distinct natural numbers $u,v$ such that:
$$|x_u-x_v|\cdot|u-v|\leq 1.$$

REPLY [14 votes]: Here is a counter-example using Hurwitz's theorem
Let $\phi = \frac{1+\sqrt 5}2$ be the golden ration.
Since the inequality $|\phi - p/q| < 1/3q^2$ has finitely many solution for integers $p,q$,
let $N$ be an integer greater than all the $|p|$ occurring in those solutions.
Let $x_n = 3n\phi \pmod {3N}$
Suppose you find distinct integers $u$ and $v$ such that $|x_u - x_v||u-v| < 1$.
Then there are integers $p_u$ and $p_v$ such that $x_u = 3u\phi - 3Np_u$ and $x_v = 3v\phi - 3Np_v$. So $|3(u-v)\phi-3N(p_u-p_v)||u-v| < 1$, and $|\phi - N(p_u-p_v)/(u-v)| < 1/3(u-v)^2$. Hence $(N(p_u-p_v), u-v)$ is one of the finitely many solutions to the inequality above, hence $N > |N(p_u-p_v)|$. This implies $p_u = p_v$, and we get $|\phi| < 1/3(u-v)^2 \le 1/3 < \phi$, which is a contradiction.