proof of inequality $e^x\le x+e^{x^2}$ Does anybody have a simple proof this inequality 
$$e^x\le x+e^{x^2}.$$
Thanks.

For $x \geq 0$,
$$
e^{x^2 - x} + x e^{-x} \geq 1 + (x^2 - x) + x (1-x) = 1 \> .
$$
For $x < 0$, let $y = - x > 0$, whence,
$$
e^{-x^2}(e^x - x) = e^{-y^2} (e^{-y} + y) \leq e^{-y^2} (1 - y + y^2/2 + y) \leq e^{-y^2}(1+y^2) \leq e^{-y^2} e^{y^2} = 1 \> .
$$

Notice that we've only used the basic facts that $e^x \geq 1 + x$ for all $x \in \mathbb{R}$ and that $e^{-x} \leq 1 - x + x^2/2$ for $x \geq 0$, both of which are trivial to derive by simple differentiation, similar to Didier's approach.
