Question

An advanced challenge for math enthusiasts:Given any 9 reals show that there will be at least 2 distinct reals a, b such that \(\left|\frac{a-b}{1+ab}\right|\leq\sqrt{2}-1\)

Answer 1

@rational @amistre64 @AngusV

Answer 2

why 9 of them? does it fail with 8 of them?

Answer 3

pigeon-hole?

Answer 4

sqrt(2) - 1 is less then 1/2

Answer 5

@mathmate si

Answer 6

good luck

Answer 7

@amistre64 yes

Answer 8

\[\Large\sf{\text{Here is the answer:}\\
\forall x \in \mathbb{R}\ \exists\ \theta\in(-\frac{\pi}{2},\frac{\pi}{2})\ such\ that\\
x=tan\theta\\
\text{For 9 reals there'll be 9 corresponding }\theta s\\
\text{Let's divide the interval }(-\frac{\pi}{2},\frac{\pi}{2})\\
\text{8 equal parts. So out of 9 }\theta s\\
\text{there'll be at least 2 such that}\\
\theta_1-\theta_2<\frac{\pi}{8}\\
\Rightarrow tan(\theta_1-\theta_2)<tan\frac{\pi}{8}\\
\Rightarrow}\left|\frac{a-b}{1+ab}\right|<\sqrt{2}-1\]

Answer 9

@amilapsn
Brilliant. I went as far as splitting the circle in eight parts, but went nowhere. Didn't think of mapping to tan theta.
Thank you!

Answer 10

That idea suddenly came to my mind with \(\Large\sf{\tan{\frac{\pi}{8}}=\sqrt{2}-1}\)