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I am having difficulty proving the following inequality: $$ \sqrt[n]{n} < 1 + \sqrt{\frac{2}{n}} \quad \text{for all positive integers}\,\,\, n. $$ I am trying to use mathematical induction but I am having trouble going from the left side to the right side (in the induction step) and vice versa.

I also tried to take the power of $n$ on both sides and somehow use the binomial expansion but I am having difficulty with that. Thank you!

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  • $\begingroup$ What is the domain of $n$? $\endgroup$ – Yiyuan Lee Mar 13 '14 at 15:09
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    $\begingroup$ Induction would be difficult at best. Binomial expansion should yield the result quickly, what problems had you with that? $\endgroup$ – Daniel Fischer Mar 13 '14 at 15:09
  • $\begingroup$ $n \in N$, it should be $n \geq 1$. I think it was more of the algebra and showing that it must be the case that the sum was greater than n $\endgroup$ – InsigMath Mar 13 '14 at 15:11
  • $\begingroup$ See also: math.stackexchange.com/q/1436663 $\endgroup$ – Martin Sleziak Sep 16 '15 at 14:40
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We shall show that $\,\Big(1+\sqrt{\frac{2}{n}}\,\Big)^n>n$.

For $n=1$ it is obvious.

Assume that $n\ge 2$, then according to the Binomial Formula: \begin{align} \left(1+\sqrt{\frac{2}{n}}\right)^n&=\binom{n}{0}+\binom{n}{1}\sqrt{\frac{2}{n}}+\binom{n}{2}\left(\sqrt{\frac{2}{n}}\right)^{\!2}+\cdots+\binom{n}{n}\left(\sqrt{\frac{2}{n}}\right)^{\!n} \\ &\ge 1+n\sqrt{\frac{2}{n}}+\frac{n(n-1)}{2}\cdot\frac{2}{n}=n+\sqrt{2n}>n. \end{align}

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  • $\begingroup$ How did you get any of that if you don't mind me asking (I mean the stuff right of the inequality) $\endgroup$ – InsigMath Mar 13 '14 at 15:12
  • $\begingroup$ @InsigMath ''according to the binomial theorem'' $\endgroup$ – OBDA Mar 13 '14 at 15:12
  • $\begingroup$ @InsigMath: The first thing that came to my mind was to show that $$\left(1+\sqrt{\frac{2}{n}}\right)^n>n.$$ And after that the only available tool is the binomial expansion. $\endgroup$ – Yiorgos S. Smyrlis Mar 13 '14 at 15:15
  • $\begingroup$ Yeah but the binomial expansion is a very large expression, you clearly did a lot of bounding at some point. $\endgroup$ – InsigMath Mar 13 '14 at 15:19
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    $\begingroup$ Sorry I understand now! Thank you! $\endgroup$ – InsigMath Mar 13 '14 at 15:22

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