Given seven real numbers. Show that we can always choose two of them, say $a, b$ such that $0 < \dfrac{a-b} {ab+1}< \sqrt{3}$
I think this is one of pigeon hole principle problem. I have tried to think further about the cases for some days, but didn't found any logical sense answer. Instead of plugging random real numbers, do you have any argument to prove it?? Please help
Let the numbers be $\{x_i\}_{i=1}^7$ Then there exists seven angles $\{\theta_i\}_{i=1}^7$ such that
$x_i = \tan \theta_i$ and $-\dfrac 12\pi < \theta_i < \dfrac 12 \pi$.
Break the interval, $(--\dfrac 12\pi, \dfrac 12\pi)$ into six, $30^\circ$ - wide, intervals. By the PHP one of those intervals must contain two of the seven $\theta_i s$...