# How to prove $|e^{x_1} -e^{x_2}| \leq \frac{1}{2} |x_1 -x_2|(e^{x_1} +e^{x_2})$?

How to prove $|e^{x_1} -e^{x_2}| \leq \frac{1}{2} |x_1 -x_2|(e^{x_1} +e^{x_2})$? Should we use the convexity of exponent function and its derivative?

By the mean value theorem,$$\left|\frac{e^{x_1}-e^{x_2}}{x_1-x_2}\right|=e^x,$$for some $x$ between $x_1$ and $x_2$. Since $\exp$ is convex,$$e^x\leqslant\frac{e^{x_1}+e^{x_2}}2.$$
Of course, I am assuming that $x_1\neq x_2$. If they're equal, the statement is trivial.
• @max_zorn $\left|\frac{e^{x_1}-e^{x_2}}{x_1-x_2}\right|\leqslant\frac{e^{x_1}+e^{x_2}}2\iff|e^{x_1}-e^{x_2}|\leqslant\frac12|x_1-x_2|(e^{x_1}+e^{x_2})$. – José Carlos Santos Jan 5 '18 at 7:18