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I am trying to find the solution to the inequality $t^{-1/n}<1$ when $n$ is an integer and $t$ is positive, however when using

Reduce[t^(-1/n) >= 1, t]

it returned an error message stating that reduce was not able to solve this inequality (Mathematica 12.0). I also tried

Assuming[Element[n, Integers] && n > 0, Reduce[t^(-n^(-1)) <= 1, t]]

If I use

f[n_] := Reduce[t^(-1/n) <= 1, t]; 
f /@ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

I get the correct intervals of $t\geq1$. How can I show this for every $t>0$ and $n\geq1$ using Mathematica.

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  • $\begingroup$ wolfram alpha can solve this. $\endgroup$ – Xminer Nov 19 '19 at 2:47
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    $\begingroup$ Mathematica is not yet capable of reading minds, so you need to explicitly tell it your assumptions: FullSimplify[Reduce[t^(-1/n) < 1 && t > 1 && n ∈ Integers && n > 0, t], t > 1 && n ∈ Integers && n > 0] $\endgroup$ – J. M. will be back soon Nov 19 '19 at 2:52
  • $\begingroup$ But you are basically implementing the answer and checking whether it is true. What if the answer was not available or not immediate? That is why I posed the question. $\endgroup$ – DMH16 Nov 19 '19 at 2:58
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You state that n is an integer and that t is positive. Include both of these constraints in Reduce

Reduce[t^(-1/n) <= 1 && Element[n, Integers] && t > 0, t] // Simplify

(* t^(-1/n) ∈ Reals && 
 n ∈ Integers && ((n <= -1 && 0 < t <= 1) || (n >= 1 && t >= 1)) *)
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