Mathematica fails with boolean simplification with exponents

Question

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.

Answer 1

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]

is one possible solution.