Why is $\frac{(e^x+e^{-x})}{2}$ less than $e^\frac{x^2}{2}$? I have read somewhere that this equality holds for all $x \in \mathbb {R}$. Is it true, and if so, why is that?
 $$\frac{(e^x+e^{-x})}{2} \leq e^\frac{x^2}{2}$$

We may prove that:
$$ \log\cosh(x) \leq \frac{x^2}{2} \tag{1}$$
by integrating the inequality:
$$ \forall u\geq 0,\qquad \tanh(u)\leq u\tag{2} $$
over the interval $(0,|x|)$. $(2)$ is a simple convexity inequality.
