I want to evaluate an inequality of the form $x^a > x$ under the assumptions $x > 1$ and $a > 1$. I expect this inequality to be true, but

Simplify[x^a > x, x > 1 && a > 1 && x \[Element] Reals && a \[Element] Reals]

returns the inequality as is, and

Reduce[{x^a - x >= 0, x > 1 && a > 1}]

gives an error ("This system cannot be solved with the methods available to Reduce").

I am not sure what I am missing here and would appreciate any suggestions on how I could evaluate such inequality to be true.

  • $\begingroup$ Since everything is positive (hence real) you can use Log on each side of the inequality: Simplify[Log /@ (x^a > x), x > 1 && a > 1] which evaluates to True $\endgroup$ – Bob Hanlon Apr 3 '17 at 23:07
  • $\begingroup$ @BobHanlon I am sure the OP would have no difficulty proving the result, his question is why Reduce[] has issues. $\endgroup$ – Igor Rivin Apr 4 '17 at 0:11
  • $\begingroup$ @IgorRivin - I believe that the question was literally "any suggestions on how I could evaluate such inequality to be true" which I interpret as looking for a workaround. $\endgroup$ – Bob Hanlon Apr 4 '17 at 3:51
  • 1
    $\begingroup$ Thank you for the comments. Bob's suggestion is indeed useful for me, so I appreciate it greatly! As Igor points out, knowing the underlying cause of this behavior would be also helpful to deal with similar issues if they arise, so I would welcome any further insights. $\endgroup$ – user36357 Apr 4 '17 at 7:34

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