Timeline for Existence of solution to these inequalities
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| S Jan 31, 2017 at 19:28 | history | bounty ended | Abhishek Parab | ||
| S Jan 31, 2017 at 19:28 | history | notice removed | Abhishek Parab | ||
| Jan 31, 2017 at 19:28 | vote | accept | Abhishek Parab | ||
| Jan 30, 2017 at 17:11 | history | edited | Abhishek Parab | CC BY-SA 3.0 |
Added a different, corrected question.
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| Jan 29, 2017 at 11:16 | comment | added | Pietro Majer | I put it as an answer, to explain myself more clearly. | |
| Jan 29, 2017 at 11:13 | answer | added | Pietro Majer | timeline score: 8 | |
| Jan 29, 2017 at 3:07 | comment | added | Abhishek Parab | The LHS varies according to the permutation $\tau$ so it's not clear why the LHS mean should overpower the RHS. As I said in the remark, the $b_i$'s cannot be independent of $\tau$. | |
| Jan 29, 2017 at 1:07 | comment | added | Pietro Majer | Why you can't take $b_i:=n-i$? The mean of the 2i numbers on the LHS of eq1 is larger than the mean of all n numbers for i in $\Delta$... what am I missing? | |
| S Jan 29, 2017 at 0:09 | history | bounty started | Abhishek Parab | ||
| S Jan 29, 2017 at 0:09 | history | notice added | Abhishek Parab | Draw attention | |
| Jan 26, 2017 at 22:03 | comment | added | LSpice | Of course this is possible when $\tau$ is the long element, since then $\Delta(\tau)$ is empty. | |
| Jan 26, 2017 at 15:44 | history | edited | Abhishek Parab | CC BY-SA 3.0 |
deleted 3 characters in body
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| Jan 25, 2017 at 22:34 | history | asked | Abhishek Parab | CC BY-SA 3.0 |