Timeline for Is there a good reason why $a^{2b} + b^{2a} \le 1$ when $a+b=1$?
Current License: CC BY-SA 2.5
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| Apr 13, 2010 at 4:35 | comment | added | fedja | Since $e^{at}t^t$ is convex for all $a\in\mathbb R$ (a boring but straightforvard computation shows that the second derivative is $[(a+\log t+1)^2+\frac 1t]e^{at}t^t$), including the point $(x,y)$ into the family $(xt,yt)$, we see that it is enough to check the boundary curves. I've done the upper one already. Anybody wants to try the lower one? | |
| Apr 13, 2010 at 2:40 | history | edited | Will Jagy | CC BY-SA 2.5 |
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| Apr 12, 2010 at 22:42 | history | edited | Will Jagy | CC BY-SA 2.5 |
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| Apr 12, 2010 at 22:28 | history | answered | Will Jagy | CC BY-SA 2.5 |