Timeline for Is $k(|a_1|+|a_2|+...+|a_n|) \le |b_1|+|b_2|+...+|b_n|+k|S|$ right?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 24, 2019 at 1:01 | vote | accept | Đào Thanh Oai | ||
| Jun 23, 2019 at 16:06 | answer | added | Fedor Petrov | timeline score: 2 | |
| Jun 23, 2019 at 14:39 | comment | added | Đào Thanh Oai | @GerryMyerson After I corrected, Maybe is not trivially true. | |
| Jun 23, 2019 at 13:47 | comment | added | Đào Thanh Oai | My computer is breakdown. I wrote this question by my mobilephone. I have corrected. Thanks You all. | |
| Jun 23, 2019 at 13:43 | history | edited | Đào Thanh Oai | CC BY-SA 4.0 |
added 20 characters in body
|
| Jun 23, 2019 at 13:05 | review | Close votes | |||
| Jun 26, 2019 at 21:48 | |||||
| Jun 23, 2019 at 12:42 | comment | added | Gerry Myerson | You have specified $a_i>0$. So $|a_1|+\cdots+|a_n|=a_1+\cdots+a_n=S=|S|$, and your inequality is trivially true. | |
| Jun 23, 2019 at 12:15 | comment | added | Fedor Petrov | Hm, $a_i$ are positive? | |
| Jun 23, 2019 at 12:03 | history | asked | Đào Thanh Oai | CC BY-SA 4.0 |