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I have come across the inequality $$\exp\left(\frac{h}{1+h}\right)\leq 1+h,\quad\forall h>-1,$$ on http://functions.wolfram.com/ElementaryFunctions/Exp/29/.

I would like some help proving this. A straightforward expansion of the exponential doesn't seem to yield anything.

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How similar it is to Bernoulli's inequality! –  Babak S. Jan 21 '13 at 13:51
    
Hint: $\exp(-y) \geq 1-y$. –  cardinal Jan 21 '13 at 14:08

2 Answers 2

up vote 7 down vote accepted

For all $y \in \mathbb R$, $e^{-y} \geq 1 - y$. So, taking $y = h / (1+h$), we have $$ \exp\left(-\frac{h}{1+h}\right) \geq 1 - \frac{h}{1+h} = \frac{1}{1+h} \>, $$ and so, for $h > -1$, $$ 1+h \geq \exp\left(\frac{h}{1+h}\right) \>. $$

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Thank you for your help! –  JW1986 Jan 22 '13 at 9:20

Write $(1+h)\ln (1+h)$ as Maclaurin series $$(1+h)\ln (1+h)=h+\dfrac{h^2}{2(1+\xi)}, \text{$\xi$ between 0 and $h$}$$ so, $(1+h)\ln (1+h)\geq h\Rightarrow \ln (1+h)\geq \dfrac{h}{1+h}\Rightarrow 1+h\geq \exp(\dfrac{h}{1+h})$

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