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Jan 3, 2013 at 13:32 comment added Shahrooz Dear Fedja, your solution is nice. Before, I saw the below inequality that is very similar to this one: $$x,y>0, x+y=1 \arrow x^y+y^x\leq sqrt(2)$$. Now, I can see the solution of this inequality.
Nov 27, 2012 at 19:09 comment added jbc Since I can't edit the above comment, I will have to add a missing bracket in an additional comment. The above estimate holds outside the crucial interval $[\frac 12,1]$, otherwise the bound is $1$. (There are also $2$ end brackets missing).
Nov 27, 2012 at 17:41 comment added jbc With respect, I would suggest that the following form of the inequality is more interesting for several reasons: $f(a,b)\leq \frac{k^k}{2^{k-1}}$ if $a+b=k$ ($f$ denotes, of course, the expression to be maximised It includes the cases $a+b=\frac 12$ and $a+b=1$, is more precise and displays more interesting behaviour (the dependence of the maximum on $k$ and the nature of the point where it is assumed.
Apr 13, 2010 at 15:22 comment added Sunni @fedja: That is good. For me, I learned a good proof, that is all I want for posting this problem.
Apr 13, 2010 at 14:26 comment added fedja Thanks. Well, anybody who is interested can find it here, so I do not think it makes much sense to submit it anywhere else. I'll, probably, just send a PM to Vasile.
Apr 13, 2010 at 14:22 history edited fedja CC BY-SA 2.5
added the bottom line proof
Apr 13, 2010 at 13:26 comment added Sunni @fedja: I think your proof meets the criteria. If you are interested, you may submit your proof featuring solving a conjecture in <On Some Inequalities With Power-Exponential Functions> by Vasile Cirtoaje.
Apr 13, 2010 at 13:21 vote accept Sunni
Apr 13, 2010 at 4:26 comment added Tom Leinster Bravo !
Apr 13, 2010 at 3:50 history edited fedja CC BY-SA 2.5
fixed the proof
Apr 12, 2010 at 17:28 history edited fedja CC BY-SA 2.5
deleted 107 characters in body
Apr 12, 2010 at 17:17 history answered fedja CC BY-SA 2.5