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?
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?
Fix $\displaystyle 0< \beta < 1$ and consider the function $\displaystyle f(\alpha) = \alpha^\beta + \beta^\alpha = g(\alpha) + h(\alpha)$.Originally Posted by Rebesques
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,
$\displaystyle \alpha^\beta + \beta^\alpha = f(\alpha) \ge f(0)^{1-\alpha}f(1)^\alpha = (1+\beta)^\alpha > 1.$