TITLE: How do I prove that $\exp(\frac{h}{1+h})\leq 1+h$?
QUESTION [7 upvotes]: 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.

REPLY [8 votes]: 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 $h > -1$,
$$
1+h \geq \exp\left(\frac{h}{1+h}\right) \>.
$$