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.

Consider $f(x) = x+\mathrm{e}^{x^2}-\mathrm{e}^x$. Then $f'(x) = 1+ 2x\mathrm{e}^{x^2}-\mathrm{e}^x$ and 
$$f''(x) = 2(1+2x^2)\mathrm{e}^{x^2}-\mathrm{e}^x\ge \mathrm{e}^xg(x),
$$ 
with $g(x)=2\mathrm{e}^{x^2-x}-1$. Since $x^2-x\ge-\frac14$, $g(x)\ge2\mathrm{e}^{-1/4}-1>0$ for every $x$, hence $f''$ is positive everywhere and $f'$ is increasing. Since $f'(0)=0$, $f'(x)<0$ if $x<0$ and $f'(x)>0$ if $x>0$. Thus, the function $x\mapsto f(x)$ is decreasing on $x\le0$ and increasing on $x\ge0$. Since $f(0)=0$, we are done.
