Inequality with floor function

Question

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.

Answer 1

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.

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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.