How to show $L^a-l^a \ge (L-l)^a$?

Question

Is it possible to show that $L^a-l^a \ge (L-l)^a$ (or the opposite), where $l \in [0,L]$ and $0<a<1$?

Thanks a lot!

This question seems to be related: (And maybe also other posts linked there.) — Martin Sleziak, Apr 01 '16 at 11:10
Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. — Martin Sleziak, Apr 01 '16 at 11:13

score 3 · Accepted Answer · answered Apr 01 '16 at 11:48

Hint:

With $\lambda = l/L \in [0, 1], \qquad (1-\lambda)^a \geqslant 1-\lambda^a $ is true, as $f(\lambda) = (1-\lambda)^a + \lambda^a$ is concave for $a \in (0, 1)$ and hence its minima have to be when $\lambda$ takes extreme values in the interval.

How to show $L^a-l^a \ge (L-l)^a$?

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