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.

  • $\begingroup$ How does the inequality follow? $\endgroup$ – max_zorn Jan 5 '18 at 1:37
  • 1
    $\begingroup$ @max_zorn $\left|\frac{e^{x_1}-e^{x_2}}{x_1-x_2}\right|\leqslant\frac{e^{x_1}+e^{x_2}}2\iff|e^{x_1}-e^{x_2}|\leqslant\frac12|x_1-x_2|(e^{x_1}+e^{x_2})$. $\endgroup$ – José Carlos Santos Jan 5 '18 at 7:18
  • $\begingroup$ Yes, thanks for the detail! $\endgroup$ – max_zorn Jan 5 '18 at 7:34

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