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?
 $$\frac{(e^x+e^{-x})}{2} \leq e^\frac{x^2}{2}$$

The taylor series on the LHS is $$\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$$
The taylor series on the RHS is $$\sum_{n=0}^{\infty} \frac{(x^2/2)^n}{n!} = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!!}$$
Where $(2n)!!$ is the double factorial $(2 \times 4 \times \ldots \times 2n)$. It is easy to then see that $(2n)!! \leq (2n)!$, namely since the latter has the extra factor of $(2n-1)!! = (1 \times 3 \times \ldots \times (2n-1))$ multiplying it.
