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

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amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1@rational @amistre64 @AngusV

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0why 9 of them? does it fail with 8 of them?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0sqrt(2)  1 is less then 1/2

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

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0@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!

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1That idea suddenly came to my mind with \(\Large\sf{\tan{\frac{\pi}{8}}=\sqrt{2}1}\)
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