How to show $1 - x\le e^{-x} \le \frac{1}{1 + x}$ for $|x| \le 1$

Question

I want prove that for $|x|<1$ we have:

$$1 - x\le e^{-x} \le \frac{1}{1 + x}.$$

Any help will be appreciated!

Answer 1

$e^x\geq x+1$ for any $x\in\mathbb{R}$ just follows from the convexity of the exponential function ($y=x+1$ is the equation of the tangent line at $x=0$), hence by renaming $x$ as $-z$ we get $e^{-z}\geq 1-z$ and by reciprocating both sides of the previous inequality we get (for any $|x|<1$) $e^{-x}\leq\frac{1}{x+1}$ as wanted.

Answer 2

It's just $e^x\geq1+x$ only, which is obvious even for all real value of $x$.

For example, let $f(x)=e^x-x-1$.

Thus, $f'(x)=e^x-1$, which says $x_{min}=0$ and we are done!