Timeline for An irresistible inequality
Current License: CC BY-SA 3.0
11 events
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Jul 10, 2016 at 14:55 | comment | added | Fedor Petrov | If you replace $e$ to $t=e^a$, then multiply by $a^2$ and denote $ax+a=y$, you get $g(y-a)+y^2/(e^y+1)^2$. | |
Jul 10, 2016 at 13:59 | comment | added | T. Amdeberhan | @FedorPetrov: very nice argument, thanks. Will see how to adopt this to the case (2), e replaced by t, or someone else does. I'm also curious if there is any other approach. | |
Jul 8, 2016 at 13:26 | comment | added | Francesco Polizzi | @FedorPetrov: well, the way in which the question is asked (for instance, the imperative tone and the last remark) gives the impression that it is a challenge for the MO users, and this kind of questions are surely non-acceptable here. Rewording the question in the sense you suggest might be a possibility. | |
Jul 8, 2016 at 13:20 | comment | added | Fedor Petrov | @FrancescoPolizzi sorry, this text was written before the question became officially considered an off-topic, or simultaneously. If you find it reasonable, I may remove it and send to OP elsewhere (I know him personally). I think, if the question is supposed to mean something like 'please explain this fact arising in my research with a short and elegant argument', it is suitable on MO, there are many such questions. | |
Jul 8, 2016 at 13:09 | comment | added | Francesco Polizzi | As a rule of thumb, one should not answer off-topic questions (for instance, in order not to encourage them). | |
Jul 8, 2016 at 12:54 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
added 1437 characters in body
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Jul 8, 2016 at 12:51 | history | undeleted | Fedor Petrov | ||
Jul 8, 2016 at 9:07 | history | deleted | Fedor Petrov | via Vote | |
Jul 8, 2016 at 9:06 | comment | added | Fedor Petrov | Oh, it is plus in denominator, not minus. Sorry. | |
Jul 8, 2016 at 8:39 | comment | added | T. Amdeberhan | The second summand is not decreasing. | |
Jul 8, 2016 at 8:37 | history | answered | Fedor Petrov | CC BY-SA 3.0 |