Among any $9$ real numbers, two satisfy $0 < \frac{x-y}{1+xy} < \sqrt{2} - 1$

Question

Prove that among any nine given real numbers there are two numbers $a, b$ with the property: $$0 < \dfrac{x-y}{1+xy} < \sqrt{2} - 1$$

The fraction looks like $\tan(a-b)$, but I don't see if it can help. I don't have any other idea, I'm stuck. I also tried by reducing to absurd, but I didn't reach far. Could someone give me a clue?

Answer 1

Outline: let the nine real numbers be $r_1,\dots,r_9$. Define $a_1=\arctan r_1, \dots, a_9=\arctan r_9$. Prove that there exist $1\le i<j\le 9$ satisfying $|a_i-a_j| < \frac\pi8$ (hint: pigeonhole principle). Then consider $\tan(a_i-a_j)$.

Answer 2

Hint :

Divide the half-plane formed by first and fourth quadrants into eight regions by drawing lines emanating from origin at equal angles.

Now argue that some two of given nine numbers must lie in same region. From this you can conclude the given inequality.