How to prove $ \frac{e^{x}+e^{-x}}{2} \le e^{\frac{x^2}{2}} $? Let $x\in \mathbb{R}$, how to prove
$$
\frac{e^{\large x}+e^{\large-x}}{2} \le e^{\large\frac{x^2}{2}}
$$
 A: Since $(2n)!\geq 2^n n!$ for $n\geq0$ we have
$$
\frac{e^x+e^{-x}}{2}=\sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}\leq 
\sum_{n=0}^\infty \frac{x^{2n}}{2^n n!}=e^{x^2/2}.
$$
A: Your inequality is equivalent to $$\log(\cosh(x))\leq \frac{x^2}{2},$$
where $\cosh(x)=\frac{e^x+e^{-x}}{2}$. To prove it for $x\geq 0$ (which is obviously sufficient) use the following: if $f(0)=g(0)$, $f'(0)=g'(0)$ and $f''(x)\leq g''(x)$ for $x\geq 0$, then $f(x)\leq g(x)$ for $x\geq 0$. I hope you can take it from here.
A: Consider the expression $$A=e^z-\frac{e^x+e^y}{2}$$ Now, use the Taylor series for $e^t$ and apply it to each of the terms; replace successively $t$ by $x$, $-x$ and $\frac{x^2}{2}$. You will notice that only even powers are left and that all coefficients are positive. As a result, the beginning will be $$A=\frac{x^4}{12}+\frac{7 x^6}{360}+\frac{13 x^8}{5040}+\frac{59
   x^{10}}{226800}+\frac{5197 x^{12}}{239500800}+O\left(x^{13}\right)$$
Added later
Following the good comment from Did, you must notice that, for any power of $x$, the corresponding coefficient is positive (as shown by Omran Kouba in his answer).
