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What is an upper-bound on $x$, given that $x < k \ln(1+x)$? I believe that the solution is something of the form $\mathcal{O}(k \ln k)$ but I am unable to prove this.

This is my first encounter with such an inequality and any pointers to relevant techniques will be helpful. Thanks.

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  • $\begingroup$ Are you interested in the case when $k \to \infty$? $\endgroup$ – Antonio Vargas Feb 15 '15 at 16:58
  • $\begingroup$ I am not certain about it. Since in the cases I saw, $k$ is of the form $\frac{1}{\epsilon}$ (where $\epsilon$ is a sufficiently small error quantity) I believe what you say might be true. Can you provide a solution for that case? $\endgroup$ – Enigman Feb 15 '15 at 17:13
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For $k>1$ let $X(k)$ be the unique positive solution of $x=l\log(1+x)$. Then $$ k\,(\log k+\log\log k)<X(k)<k\,\Bigl(\log k+\log\log k+\frac{1}{\sqrt{\log k}}\Bigr). $$ I will prove the first inequality. We want to prove that $$ k\,(\log k+\log\log k)<k\log(1+k\,(\log k+\log\log k)). $$ This is equivalent to $$ \log k+\log\log k<\log(1+k\,(\log k+\log\log k)), $$ which in turn is equivalent to the trivial inequality $$ k\log k<1+k\,(\log k+\log\log k). $$ The other inequality can be proved in a similar way, with a little more work.

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