How do I prove that $\exp(\frac{h}{1+h})\leq 1+h$?

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.

-
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

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 $h > -1$, $$1+h \geq \exp\left(\frac{h}{1+h}\right) \>.$$
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, $(1+h)\ln (1+h)\geq h\Rightarrow \ln (1+h)\geq \dfrac{h}{1+h}\Rightarrow 1+h\geq \exp(\dfrac{h}{1+h})$