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

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

Mathematics
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why 9 of them? does it fail with 8 of them?
pigeon-hole?

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sqrt(2) - 1 is less then 1/2
good luck
\[\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)
@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!
That idea suddenly came to my mind with \(\Large\sf{\tan{\frac{\pi}{8}}=\sqrt{2}-1}\)

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