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I have been really struggling with this problem ... please help! Let a,b be real numbers. If $0<a<1, 0<b<1, a+b=1$, then prove that $a^{2b} + b^{2a} \le 1$

What I have thought so far: without loss of generality we can assume that $a \le b$, since $a^{2b} + b^{2a}$ is symmetric in $a$ and $b$. This gives us $0<a \le 1/2, 1/2 \le b<1$. But then I am stuck. I also thought of solving by Lagrange's multiplier method, but it produces huge calculations.

Any help is welcome :)

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This very question cropped up on mathoverflow a few weeks ago. – Olivier Bégassat May 18 '13 at 20:53
@Olivier: Can you give the link to the mathoverflow question page? – mathmansujo May 18 '13 at 20:57
1… from 2010. – Will Jagy May 18 '13 at 21:01
Looking at the MO thread...what kind of professor left such problems as homework? Maybe it is because you arrived at the classroom late like George Dantzig did ( and your professor happened to have the habit of leaving hard problems on the blackboard like Prof Neyman? – Shuhao Cao May 18 '13 at 22:02
up vote 3 down vote accepted

Not having a good day with websites. I have downloaded what seems to be the source of the question, a 2009 paper by Vasile Cirtoaje which is about 14 pages. Then a short answer, in a four page document by Yin Li, probably from the same time or not much later. The question was posted on MO by a selfish guy who knew the status of the problem but was hoping for a better answer, a complete answer was also given there in 2010 by fedja,

I have both pdfs by Cirtoaje and Li, email me if you cannot find them yourself.

This is not a reasonable homework question, so I would like to know more of the story, what course for example.


Yin Li of Binzhou University, Shandong, 2009 or 2010, excerpts I believe he just spelled Jensen's incorrectly, see JENSEN

enter image description here


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This is not actually a homework problem, my teacher found this problem from some other student and tried for himself but could not solve it, so he gave it to me to have a look. Neither he, nor I had any idea about its level, since it is so simple looking. Now I have clear idea about its hardness, though :). – mathmansujo May 19 '13 at 4:31

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