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?

$$e^{|x|}-1\leq |x|e^{|x|}\iff 1-e^{-|x|}\le|x|\iff\int_0^{|x|}e^{-t}\,dt\le\int_0^{|x|}1\,dt$$
which is clearly true because $e^{-t}\le1$ for $t\ge0$
