# an inequality

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• Jul 25th 2005, 08:21 AM
Rebesques
an inequality
Here's a little troublesome inequality: Prove that

$\alpha^\beta+\beta^\alpha> 1,
\forall \ 0<\alpha,\beta<1.$

I would appreciate a shorter solution, but with less theory involved... Asking 2 much? :(
• Jul 25th 2005, 09:40 AM
Rebesques
:(
Just noticed, this is the wrong place for the topic to be... Could one of the administrators please move this thread to the "inequalities" section? :(
• Jul 25th 2005, 10:23 AM
MathGuru
The inequalities forum is for elementary school math (for younger students). If you don't want it in the university level algebra forum, let me know which section suits it the best.
• Jul 30th 2005, 07:34 PM
hpe
Quote:

Originally Posted by Rebesques
Prove that
$\alpha^\beta+\beta^\alpha> 1,
\forall \ 0<\alpha,\beta<1.$

Fix $0< \beta < 1$ and consider the function $f(\alpha) = \alpha^\beta + \beta^\alpha = g(\alpha) + h(\alpha)$.

Then f(0) = 1, $f(1) = 1+\beta$, and both g(x) and h(x) are log-concave (that is, their logarithms are concave down). It is known that the sum of log-concave functions is also log-concave, so f is log-concave. As a consequence,
$\alpha^\beta + \beta^\alpha = f(\alpha) \ge f(0)^{1-\alpha}f(1)^\alpha = (1+\beta)^\alpha > 1.$
• Jul 31st 2005, 12:57 AM
Rebesques
?
Ellegant, short, not heavily theoretic; Excellent :)

Thanks!