1. ## an inequality

Here's a little troublesome inequality: Prove that

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

I would appreciate a shorter solution, but with less theory involved... Asking 2 much?

2. ## :(

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?

3. 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.

4. Originally Posted by Rebesques
Prove that
$\displaystyle \alpha^\beta+\beta^\alpha> 1, \forall \ 0<\alpha,\beta<1.$
Fix $\displaystyle 0< \beta < 1$ and consider the function $\displaystyle f(\alpha) = \alpha^\beta + \beta^\alpha = g(\alpha) + h(\alpha)$.

Then f(0) = 1, $\displaystyle f(1) = 1+\beta$, 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,
$\displaystyle \alpha^\beta + \beta^\alpha = f(\alpha) \ge f(0)^{1-\alpha}f(1)^\alpha = (1+\beta)^\alpha > 1.$

5. ## ?

Ellegant, short, not heavily theoretic; Excellent

Thanks!