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I need some help to prove this inequality:

$\ (1 + \frac{1}{a_n})^{a_n} \le (1 + \frac{1}{\lfloor{a_n}\rfloor})^{\lfloor{a_n}\rfloor + 1} $

where $\ n \in \mathbb{N} $ and$\ a_n \in \mathbb{R}$ with$\ a_n > 1 $ is any element of a sequence.

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closed as off-topic by Henrik, kingW3, John B, zhoraster, Watson Dec 27 '16 at 21:01

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    $\begingroup$ I don't get it. There is no further conditions on $a_n$? I mean, e.g. with $a_3=8$ it doesn't work. $\endgroup$ – md5 Dec 27 '16 at 16:29
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    $\begingroup$ I don't get why you need the floor function if $n$ is a natural number. $\endgroup$ – scott Dec 27 '16 at 16:29
  • $\begingroup$ If $\lfloor n\rfloor =n$, why even include it in the inequality? $\endgroup$ – scott Dec 27 '16 at 16:31
  • $\begingroup$ sorry, I miss a_, in the second part of inequality $\endgroup$ – Stefano Balzarotti Dec 27 '16 at 16:32
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    $\begingroup$ I believe you'll also need $a_n > 1$ for this to hold. $\endgroup$ – TastyRomeo Dec 27 '16 at 16:34
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An Elementary Answer

Since $\lfloor x\rfloor+1\ge x$, we can use Bernoulli's Inequality: $$ \begin{align} \left(1+\frac1{\lfloor x\rfloor}\right)^{\lfloor x\rfloor+1} &=\left[\left(1+\frac1{\lfloor x\rfloor}\right)^{\large\frac{\lfloor x\rfloor+1}x}\right]^{\normalsize\,x}\\ &\ge\left(1+\frac{\lfloor x\rfloor+1}{\lfloor x\rfloor x}\right)^{\normalsize x}\\[9pt] &\ge\left(1+\frac1x\right)^x \end{align} $$ for $x\ge1$, so that $\frac1{\lfloor x\rfloor}$ exists.


Notes on Bernoulli's Inequality

At the end of this answer, Bernoulli's Inequality is proven for integer exponents using only induction.

At the end of this answer, the integer version is extended to the rational version using only induction.

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  • $\begingroup$ Thanks, for sure this is a good solution, but the exercise ask to use the inequality with floor function. Furthermore the exercise in my book is before the chapter of derivative, so I think this is not the expected solution. So I give you an upvote, but I can't accept as an answer. $\endgroup$ – Stefano Balzarotti Dec 27 '16 at 18:26
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    $\begingroup$ For $a_n\ge1$, set $x=a_n$ and $y=\lfloor a_n\rfloor$, and you get the inequality you ask about $\endgroup$ – robjohn Dec 27 '16 at 18:30
  • $\begingroup$ Thanks for explanation, but I prefer to wait some time to see if someone can provide a solution without the use of derivatives. Otherwise I'll accept your answer. $\endgroup$ – Stefano Balzarotti Dec 27 '16 at 18:38
  • $\begingroup$ Thanks, I suppose the exercise requires the use of Bernoulli's Inequality. $\endgroup$ – Stefano Balzarotti Dec 27 '16 at 18:46
  • $\begingroup$ @StefanoBalzarotti: If you want an answer not using derivatives, it would be nice to specify that in the question. I have added an answer using only Bernoulli's Inequality. $\endgroup$ – robjohn Dec 27 '16 at 18:46

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