TITLE: cosh x inequality
QUESTION [42 upvotes]: While reading an article on Hoeffding's Inequality, I came across a curious inequality. Namely
$$\cosh x \leq e^{x^2/2} \quad \forall x \in \mathbb{R}$$
I tried many ways to prove it and finally, the Taylor series approach worked:
$$e^x = 1 + x + \frac{x^2}{2!} + \cdots$$
$$e^{-x} = 1 - x + \frac{x^2}{2!} - \cdots$$
 Adding the two and dividing by 2 (This operation being justified as both series converge),
we get
$$\cosh x  = 1 + \frac{x^2}{2!}+ \frac{x^4}{4!} + \cdots$$ 
Expanding $e^{x^2/2}$ yields
$$e^{x^2/2}  = 1 + \frac{x^2}{2!}+ \frac{x^4}{4\times 2!} + \cdots$$ 
If you do a term by term comparison, you get the desired result. 
My question is: Is there another more "Cute"/elegant way to get this result? If so, what is it? I tried using Jensen's Inequality but that didn't help. Also I searched for this inequality using the keywords "cosh x" and "inequality", but didn't get it. 
Any ideas will be appreciated.

REPLY [8 votes]: Noticing that $\frac{d}{dx}\tanh x = 1 -\tanh^2x$, note that:

$1-\tanh^2 u\le1\implies\int_0^t(1-\tanh^2 u)du\le\int_0^tdu\implies\tanh t \le t$
$\int_0^x \tanh t dt\le\int_0^xtdt\implies\log\cosh x \le \frac{1}{2}x^2\implies\cosh x\le e^{\frac{1}{2}x^2}$