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?


3 Answers 3


Reduce works like a charm:

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

x ∈ Integers && x >= 3

  • 1
    $\begingroup$ Yes, this is exactly what I want, thank you. $\endgroup$
    – Masacroso
    Commented Jan 25, 2017 at 20:42
  • 1
    $\begingroup$ When I tried this myself I forgot the $x > 0$ constraint, and so MM returned an error. Now I'm kicking myself. Nicely done. $\endgroup$ Commented 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. $\endgroup$ Commented Jan 25, 2017 at 20:59

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.