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

Let us evaluate Taylor series of $\displaystyle e^{\frac{x^2}{2}}-\frac{e^x+e^{-x}}{2}$ near $x=0$. As we know $\displaystyle e^x=\sum_{k=0}^{\infty}\frac{1}{k!}x^k$. The radius of convergence is $\infty$, thus we can write $\displaystyle e^{\frac{x^2}{2}}=\sum_{k=0}^{\infty}\frac{1}{k!}\left(\frac{x^2}{2}\right)^k=\frac{1}{2^k k!}x^{2k}$ and $\displaystyle e^{-x}=\sum_{k=0}^{\infty}\frac{1}{k!}(-x)^k=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}x^k$. Furthermore, $$\cosh(x)=\frac{e^x+e^{-x}}{2}=\sum_{k=0}^{\infty}\frac{1+(-1)^k}{k!}x^k=\sum_{k=0}^{\infty}\frac{1}{(2k)!}x^{2k}$$
Finally $$e^{\frac{x^2}{2}}-\frac{e^x+e^{-x}}{2}=\sum_{k=0}^{\infty}\left(\frac{1}{2^k k!}-\frac{1}{(2k)!}\right)x^{2k}$$
It is easy to show by induction that $\forall \ n\in\mathbb{N}:\ 2^n\cdot{n!}<(2n)!$.
