# Mathematica fails with boolean simplification with exponents

I have a truly simple inequality, which I want to prove using Mathematica:

$$a^x \geq 1 ,\quad \quad with \quad 1\leq a \quad and \quad 1\leq x \quad a,x \in R$$

This is obviously true. When I try to solve this in Mathematica using for example:

Assuming[1 <= a && x >= 1, FullSimplify[a^x >= 1]]


the only output I get, is a^x >= 1 and not the desired True. Is Mathematica really not able to solve this simple problem or am I missing something basic here?

Of course, my real inequality is much more complicated, but I broke it down until I identified working with exponents as the problem here.

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Indeed it is strange that FullSimplify[] + Assuming[] fails here; however you might try FullSimplify[Reduce[a^x >= 1 && a >= 1 && x >= 1, {a, x}], a >= 1 && x >= 1] instead. – J. M. May 11 '11 at 12:47
@J. M. Thanks, this works. Could you post this as an answer, so I can accept it? – meep.meep May 11 '11 at 17:59

Note that FullSimplify[] accepts a second argument of constraints/assumptions, so FullSimplify[a^x >= 1, 1 <= a && x >= 1] would be a shorter way to write your snippet. That being said, one way that might work when the first one doesn't would be to use the function Reduce[] in tandem with FullSimplify[]; thus,
FullSimplify[Reduce[a^x >= 1 && a >= 1 && x >= 1, {a, x}], a >= 1 && x >= 1]