Timeline for Is $\iiint_{[0, 1]^3} \lvert f(x)+f(y)+f(z)\rvert\, dx\, dy\, dz \ge \int_0^1 \lvert f(x)\rvert\, dx$?
Current License: CC BY-SA 4.0
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| Feb 7, 2021 at 0:20 | comment | added | Iosif Pinelis | @fedja : That is another great answer of yours! I think your inequality for $X$ and $Y$ needs to be widely known -- so, perhaps you can consider publishing it. | |
| Feb 6, 2021 at 13:26 | comment | added | fedja | @IosifPinelis I added to my answer the endgame for your approach (and an additional question). Enjoy! | |
| Feb 6, 2021 at 2:14 | comment | added | fedja | @inequality It surely does. Moreover, we can get a better constant factor than $1$ on the RHS though to find the best one may be quite a headache. | |
| Feb 6, 2021 at 0:46 | comment | added | math110 | Thanks,so for my question when $n=3$ not hold for any $f?$ | |
| Feb 5, 2021 at 17:20 | comment | added | Iosif Pinelis | @fedja : Your inequality for $X$ and $Y$ is very nice! | |
| Feb 5, 2021 at 16:22 | comment | added | fedja | You are overthinking it a bit. :-) Indeed, you can immediately reduce the problem to a 2-value case as you did. But you don't need to square anything to check the resulting inequality. An alternative (if you do not want to bother with binomial coefficients) is to generalize the statement to $E|X+Y|\ge \min(E|X|,E|Y|)$ when $EX$ and $EY$ have the same sign, prove that, and induct. I have to teach now. I'll post later unless somebody beats me to it. | |
| Feb 5, 2021 at 15:30 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 23 characters in body
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| Feb 5, 2021 at 15:16 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |