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I'm trying to prove this inequality. Let n be a positive integer. Prove that: $$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{2n-1}\ge n\sqrt[n]{2}-n$$ I've tried doing it with algebraic, geometric, and harmonic mean, the integral bound (and realised both sides are decreasing and going towards $\log 2$), but none have worked. I'd be grateful if anyone could show me the trick needed.

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  • $\begingroup$ Have you tried using induction? $\endgroup$ – Hawk Jan 24 '14 at 10:41
  • $\begingroup$ @Hawk: have you tried using induction for this problem? Unless I am missing something obvious, this is a red herring. $\endgroup$ – robjohn Jan 24 '14 at 15:27
  • $\begingroup$ @robjohn Obviously you are not missing anything...and I did not mean to use induction directly to the problem but after some manipulations because if we use induction here, then we would have to prove yet another inequality. $\endgroup$ – Hawk Jan 24 '14 at 18:14
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Write this as $$2^{1/n} \le \frac{\left(1 + \frac{1}{n}\right) + \dots + \left(1 + \frac{1}{2n-1}\right)}{n}.$$ The result follows from AM-GM once you show that (for example by induction) $$\left(1 + \frac{1}{n}\right) \left(1 + \frac{1}{n+1}\right) \dots \left(1 + \frac{1}{2n-1} \right) = 2.$$

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  • $\begingroup$ How did you ever come up with doing AM-GM on the numbers $1+\frac1k$? I know, once you say their product is $2$ it becomes obvious, but you should already know this trick to be able to invent this, no? $\endgroup$ – punctured dusk Jan 24 '14 at 10:50
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    $\begingroup$ @barto: Well, the power $1/n$ hints at AM-GM. To use AM-GM you want some approximately same size numbers. From this point of view distributing the $n$ evenly among the $n$ terms makes sense. $\endgroup$ – J. J. Jan 24 '14 at 10:52
  • $\begingroup$ Hmm, yes. Thanks for explaining your reasoning behind it. $\endgroup$ – punctured dusk Jan 24 '14 at 10:54
  • $\begingroup$ Wow, thanks. I would never have guessed that you need to split the n into n 1s. Great idea! $\endgroup$ – user123506 Jan 24 '14 at 11:13
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Since $\log(1+x)$ is concave, Jensen's inequality says that $$ \begin{align} \frac1n\log(2) &=\frac1n\sum_{k=n}^{2n-1}\log\left(1+\frac1k\right)\\ &\le\log\left(1+\frac1n\sum_{k=n}^{2n-1}\frac1k\right)\\ \end{align} $$ which is equivalent to $$ 2^{1/n}\le1+\frac1n\sum_{k=n}^{2n-1}\frac1k $$ and therefore, $$ n2^{1/n}-n\le\sum_{k=n}^{2n-1}\frac1k $$

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