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I tried to solve the inequality

$$n\left(\frac{n}{n+1}\right)^n>1,\quad n\in\Bbb N$$

with this code

Solve[x*(x/(1 + x))^x > 1 && x > 0, x, Integers]

but the program said

This system cannot be solved with the methods available to Solve.

My attempt using 'NSolve' failed too. My question: there is a way to solve the inequality?

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3 Answers 3

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Reduce works like a charm:

Reduce[x*(x/(1 + x))^x > 1 && x > 0, x, Integers]

x ∈ Integers && x >= 3

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    $\begingroup$ Yes, this is exactly what I want, thank you. $\endgroup$
    – Masacroso
    Jan 25, 2017 at 20:42
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    $\begingroup$ When I tried this myself I forgot the $x > 0$ constraint, and so MM returned an error. Now I'm kicking myself. Nicely done. $\endgroup$
    – Michael Seifert
    Jan 25, 2017 at 20:48
  • $\begingroup$ Actually I was a little bit dissappointed that i had to add the constraint. I'd thought that mathematica itself would plug me the solution out with the additional constraint since the function is anyway complex between [-1,0] and everything below -1 is negative. But thanks. $\endgroup$
    – Julien Kluge
    Jan 25, 2017 at 20:59
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One approach is to use FindRoot for the equality:

FindRoot[n (n/(n + 1))^n == 1, {n, 2.5}]

{n -> 2.29317}

So any n greater than 2.29 will have the correct direction in the inequality, i.e., n=3 would be the smallest integer.

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If you only need to know a limited number of examples that solve the inequality, then FindInstance will work:

FindInstance[x (x/(1 + x))^x > 1, x, Integers, 2]
(* {{x -> 278}, {x -> 358}} *)

The 2 in the above code tells Mathematica to find two instances. The instances that Mathematica finds are pretty unpredictable (if you only ask it for one instance, it returns x -> 16), and it takes longer than one might want to run; but depending on the precise circumstances, it could be an option.

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