Timeline for A Cauchy-Schwarz inequality
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 19, 2018 at 18:08 | answer | added | Qiang Zhang | timeline score: 1 | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Nov 27, 2016 at 19:34 | comment | added | Fan Zheng | Sorry but I missed a factor $\bar\kappa-\min\kappa_i$, but fortunately it appears on both sides of the inequality, so it still works. But I guess you may have a different argument in mind because it's really a stretch to call the above argument C-S. | |
| Nov 27, 2016 at 18:16 | comment | added | Lucia | @FanZheng: I see no reason for $K-1\ge n/m$. If half the $\kappa$s are very slightly more than average and half are very slightly less, then $K$ can be pretty close to $1$. | |
| Nov 27, 2016 at 5:14 | comment | added | Fan Zheng | Looks like this is more combinatorial than it appears. Suppose there are m k's larger than average and n k's smaller than average. Then |J|>=m+n and K-1 >= n/m, so the coefficient in the inequality comes out as √(n/m).(m+n). By AM-GM (a special case of C-S!) This is at least √(m/n).2√(mn)=2n, which is what you get from the trivial bound. | |
| S Nov 27, 2016 at 2:13 | history | suggested | Botrance | CC BY-SA 3.0 |
added reference
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| Nov 27, 2016 at 1:46 | comment | added | Lucia | Look at $\widetilde{A} - \kappa A$ and then use Cauchy-Schwarz. | |
| Nov 27, 2016 at 1:26 | review | Suggested edits | |||
| S Nov 27, 2016 at 2:13 | |||||
| Nov 27, 2016 at 1:18 | review | First posts | |||
| Nov 27, 2016 at 1:41 | |||||
| Nov 27, 2016 at 1:17 | history | asked | user101672 | CC BY-SA 3.0 |