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Dec 3, 2019 at 13:47 answer added Doris timeline score: 1
Aug 16, 2018 at 11:29 comment added Intelligenti pauca I found a proof, different from that given by Alex Ravsky, in this paper by Dietrich Morgenstern (2000), where the inequality is said to be a conjecture by Walter Deuber (1998).
Aug 1, 2017 at 6:34 history edited Alex Ravsky
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Aug 1, 2017 at 6:29 answer added Alex Ravsky timeline score: 8
Nov 14, 2016 at 21:07 comment added Alex Ravsky Computer simulation suggests that the inequality holds for each $n\ge 1$.
Nov 5, 2015 at 21:16 vote accept mathlove
Sep 19, 2015 at 21:34 history edited mathlove
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S Aug 19, 2015 at 11:58 history bounty ended CommunityBot
S Aug 19, 2015 at 11:58 history notice removed CommunityBot
Aug 12, 2015 at 0:24 history tweeted twitter.com/#!/StackMath/status/631259899410968576
S Aug 11, 2015 at 10:47 history bounty started mathlove
S Aug 11, 2015 at 10:47 history notice added mathlove Draw attention
Aug 6, 2015 at 10:33 answer added Intelligenti pauca timeline score: 1
Aug 5, 2015 at 19:18 history edited mathlove CC BY-SA 3.0
added that I used the triangular inequality for n=2 case
Aug 5, 2015 at 16:43 history edited mathlove
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Aug 5, 2015 at 16:25 comment added mathlove @TravisJ: Thank you for your comment. All I can obtain from $(1)$ is $(2)$. (I added the details.)
Aug 5, 2015 at 16:23 history edited mathlove CC BY-SA 3.0
added how I obtained (2)
Aug 5, 2015 at 14:00 comment added TravisJ The triangle inequality cannot work (at least not naively) since if you have $n$ red and $n$ blue points you have $2\binom{n}{2}=n^2-n$ monochromatic distances and you have $n^2$ bi-colored distances. In order to naively apply the triangle inequality you'd need $2(n^2-n)\approx 2n^2$ bi-colored distances.
Aug 5, 2015 at 13:35 history edited mathlove CC BY-SA 3.0
added some explanations
Aug 5, 2015 at 13:05 history asked mathlove CC BY-SA 3.0