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Here is a simple example of an inequality which I cannot get Mathematica to state as True:

FullSimplify[x^a >= 1, Assumptions -> {a > 0, x > 1}]

Is there a case for which the inequality is false that I haven't ruled out by the assumptions?

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I think using Reduce would be a better approach, although it's not completely straightforward:

Reduce[x^a < 1 && a > 0 && x > 1, Reals]

False

(updated with another approach using Resolve)

Another possibility is to note that your question is basically the same as proving that ForAll[{x, y}, x>1 && a>0, x^a>1] is True. Equivalently, we can attempt to prove that the negation, !ForAll[{x, y}, x>1 && a>0, x^a>1], is False:

Resolve[
    !ForAll[{x, y}, x>1 && a>0, x^a>1],
    Reals
]

False

Unfortunately, a direct Resolve doesn't work:

Resolve[
    ForAll[{x, y}, x>1 && a>0, x^a>1],
    Reals
]

ForAll[{x}, x > 1 && a > 0, x^a > 1]

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  • $\begingroup$ Why won't Reduce[x^a > 1 && a > 0 && x > 1, Reals] return TRUE? $\endgroup$ – Angus Were Jan 25 '18 at 22:08
  • $\begingroup$ Nevermind about my follow-on comment. I understand now why Reduce has to be stated as a False statement. Thanks for your help. $\endgroup$ – Angus Were Jan 25 '18 at 22:17
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Assuming[a > 0 && x > 1, 
 Reduce[x^a >= 1 && a > 0 && x > 1] // Simplify]

(* True *)
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You can apply Log to both sides of the equation. This changes x^a>=1 into the equivalent inequality Log[x^a] >= Log[1]=0. FullSimplify works fine on the Log'ed version:

FullSimplify[Log[x^a] >= 0, {a > 0, x > 1}]
True
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