TITLE: Slick proof of exponential inequality
QUESTION [7 upvotes]: Today I saw that using taylor series, one can show that $e^x+e^{-x}\leq 2e^{x^2/2}$. Is there a slick proof using some sort of Jensen-type inequality or integral bound?

REPLY [5 votes]: Note that by the Mean Value Theorem, we have
$$
\frac{\tanh(x)}{x}=\mathrm{sech}^2(\xi)\tag{1}
$$
for some $\xi$ between $0$ and $x$. Therefore, for $x\ne0$,
$$
0\lt\frac{\tanh(x)}{x}\lt1\tag{2}
$$
Thus, for $x\ne0$,
$$
\begin{align}
x\frac{\mathrm{d}}{\mathrm{d}x}\left(e^{x-x^2/2}+e^{-x-x^2/2}\right)
&=(x-x^2)e^{x-x^2/2}-(x+x^2)e^{-x-x^2/2}\\
&=e^{-x^2/2}\left(2x\sinh(x)-2x^2\cosh(x)\right)\\
&=2x^2e^{-x^2/2}\cosh(x)\left(\frac{\tanh(x)}{x}-1\right)\\
&\lt0\tag{3}
\end{align}
$$
If $xf'(x)\lt0$ for $x\ne0$, then $f$ has a global maximum at $x=0$. Thus,
$$
e^{x-x^2/2}+e^{-x-x^2/2}\le2\tag{4}
$$
which is the same as what is asked.