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?
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]
Assuming[a > 0 && x > 1,
Reduce[x^a >= 1 && a > 0 && x > 1] // Simplify]
(* True *)
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