Answer
$e^x\geq 1+x$ for all x
Work Step by Step
The Taylor series for $e^{x}$ with $r$ in $(-\infty,\infty)$ is
$e^x=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+...+\frac{1}{n!}x^n+...$
$e^x=(1+x)+(\frac{x^2}{2!}+\frac{x^3}{3!})+...+\frac{1}{n!}x^n+...$
$e^x-(1+x)\geq0$
(0 represents $(\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{1}{n!}x^n+...)$
$e^x\geq 1+x$
Hence, we can suggest that $e^x\geq 1+x$ for all x
