Prove the inequality $\frac{e^x+e^{-x}}{2} \leq e^{x^2/2}$ for all real numbers $x$. How do I prove what's written in the title? I was able to get an incomplete proof for the case $x>2$. Here's my try: Use $e^x = \sum_{j=1}^{\infty} \frac{x^j}{j!}$.  Now we can see that if $x$ is a real number, then: $$e^{x^2/2}-e^x/2-e^{-x}/2=\sum_{j=1}^\infty \frac{\frac{x^{2j}}{2^{j-1}}-x^j-(-x)^j}{2j!}$$ and we need to show this is positive. Now, if $j$ is odd, then the numerator is just $x^{2j}/2^{j-1}$ which is always greater than zero. But if $j$ is even, multiply the numerator by $2^{j-1}$, and check if the result is positive. The result is $x^{2j} -2^{j} \cdot x^j=x^{j}(x^j-2^j)$ So if $x$ is greater then $2$, we get a positive result, but otherwise, we don't. So how do I  continue for other values of $x$?

Let $f(x)=\frac{x^2}{2}-\ln\frac{e^x+e^{-x}}{2}.$
Since $f$ is an even function, it's enough to prove that $f(x)\geq0$ for $x\geq0.$
We see that $$f'(x)=x-\frac{e^x-e^{-x}}{e^x+e^{-x}}$$ and
$$f''(x)=1-\frac{(e^x+e^{-x})^2-(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=\frac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2}\geq0.$$
Id est, $$f'(x)\geq f'(0)=0,$$
$$f(x)\geq f(0)=0$$ and we are done!
