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?

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.
