Slick proof of exponential inequality 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?
 A: First, since both sides are even functions, it is sufficient to prove that $e^x+e^{-x} \leq 2e^{x^2/2}$ for all $x \geq 0$. Then we can change the problem to the equivalent problem of proving that for $x \geq 0$
$$
u(x) \equiv \frac{1}{e^x+e^{-x}} \geq \frac{1}{2}e^{-x^2/2} \equiv v(x)$$
We start by observing that $u(0) = v(0).$
Now in the immortal words of William Feller, let's pass to the logarithm:  We will aim at showing that for $\forall x > 0$, 
$$\ln u(x) > \ln v(x)$$.
To do this, we will start with $\ln u(0) > \ln v(0)$ and note that
$$
\frac{d}{dx} \ln u(x) = \frac{1}{u} d\frac{du(x)}{dx} = -\frac{e^x-e^{-x}}{e^x+e^{-x}} = -\frac{e^x+e^{-x}}{e^x+e^{-x}} +\frac{2e^{-x}}{e^x+e^{-x}} 
$$
$$
\left. \frac{d}{dx} \ln u(x) \right|_{x=0} =0 
$$
$$
\frac{d}{dx} \ln v(x)=\frac{1}{v} d\frac{dv(x)}{dx} = -x$$
$$
\left. \frac{d}{dx} \ln v(x) \right|_{x=0} =0 $$
$$
\frac{d^2(\ln u(x))}{dx^2} = -\, \frac{4}{(e^x+e^{-x})^2}> -1
$$
$$
\frac{d^2(\ln v(x))}{dx^2} = -1$$
So the functions $\ln u(x)$ and $\ln v(x)$, along with their derivatives, match at $x=0$, and for all $x>0$ the second derivative of $\ln u(x)$ is greater than the second derivative of $\ln v(x)$. Thus 
$$\ln u(x) \geq \ln v(x) \Rightarrow u(x) \geq v(x) \Rightarrow \frac{1}u(x) \leq \frac{1}{v(x)} \Rightarrow e^x + e^{-x} \leq e^{-x^2/2}
$$
which is what we set out to prove. By using the second derivative we effectively used the concavity of the function $\ln u - \ln v$.
A: 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.
A: $y = 2 e^{x^2/2}$ satisfies the d.e. $y' = x y$ with $y(0) = 2$, while $z = e^{x} + e^{-x} = 2 \cosh(x)$ satisfies $z' = \tanh(x) z$ with $z(0)=2$.  Since $x \ge \tanh(x) $ for $x \ge 0$,  Gronwall's inequality does the rest.
A: As @flawr points out, the L.H.S. is $\displaystyle \cosh x = \frac{e^x+e^{-x}}{2}$ has an infinite product representation:
$$\cosh x = \prod\limits_{n=1}^{\infty} \left(1+\frac{4x^2}{\pi^2(2n-1)^2}\right) \le \exp \sum\limits_{n=1}^{\infty}\frac{4x^2}{\pi^2(2n-1)^2} = e^{x^2/2}$$
