Prove the inequalities $|e^{x}-1|\leq e^{|x|}-1\leq |x|e^{|x|}$ Prove that $|e^{x}-1|\leq e^{|x|}-1\leq |x|e^{|x|}$ for all $x\in \mathbb{C}$. I did this by Maclaurin series of $e^x$,
$$|e^{x}-1|\leq |x|+\frac{|x|^2}{2!}+\frac{|x|^3}{3!}+\mathcal{O}(|x|^{4})=e^{|x|}-1 \\
\leq |x|^2+\frac{|x|^3}{2!}+\frac{|x|^4}{3!}+\mathcal{O}(|x|^{5})=|x|e^{|x|}.$$
My teacher disliked the last inequality, and I don't know why. Is there a way to fix this problem?
 A: $$e^{|x|}-1\leq |x|e^{|x|}\iff 1-e^{-|x|}\le|x|\iff\int_0^{|x|}e^{-t}\,dt\le\int_0^{|x|}1\,dt$$
which is clearly true because $e^{-t}\le1$ for $t\ge0$
A: Reason why one's argument is invalid:
The last inequality shows something untrue:
$$\frac{|x|^3}{2!}+\frac{|x|^4}{3!}+\mathcal{O}(|x|^{5})=|x|e^{|x|}-|x|$$
If one try to use $$e^{|x|}-1\leq e^{|x|} \leq |x|e^{|x|}$$
It is invalid because the last inequality when $x = 0.5$ does not hold.
Here is my solution:
Let u = |x| ($u\geq 0$), then the inequality
$$e^{|x|}-1\leq |x|e^{|x|}\\ \iff e^u-1 \leq ue^u$$ 
One's goal is to show $e^u-1 \leq ue^u$ holds for all $u\geq0 $
Designate function $f(x) := e^x-xe^x-1$ $ x \in [0,+\infty]$
$f'(x) = e^x-(e^x+xe^x) = -xe^x$ 
Which means $f'(x)\leq0$ for all  $ x \in [0,+\infty]$
Therefore f(x) is decreasing on $ [0,+\infty] \iff$ $f(0)$ is maximum over interval$ [0,+\infty] $
Since $f(0) = 0$, $f(x) = e^x-xe^x-1 \leq 0 $ for all $x\in[0,+\infty]$
=> $e^x-1 \leq xe^x$ which is one's goal.
Qed.
A: Well, you could use some steps in between:
\begin{align*}
e^{| x| } - 1 = \sum_{n=1}^\infty \frac{|x|^n}{n!} = | x | \sum_{n=1}^\infty \frac{|x|^{n-1}}{(n-1)!} \frac{1}{n!} \le |x| \sum_{n=1}^\infty \frac{|x|^{n-1}}{(n-1)!} = |x| e^{|x|}
\end{align*}
as $\frac{1}{n!} \le 1$ for all $n \ge 1$.
Note that after the inequality we have start with a  factor $|x|$ instead of $|x|^2$ as the Maclaurin series of $\exp(|x|)$ start with a $1$ and not with $|x|$.
