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 $x$?
 A: Your computation is flawed. The expansion of $\cosh{x}$ is $\sum_{j \geq 0}{\frac{x^{2j}}{(2j)!}}$, but the expansion of $e^{x^2/2}$ is $\sum_{j \geq 0}{\frac{x^{2j}}{2^j \cdot j!}}$. 
So you just need to prove that $j!2^j < (2j)!$ for each $j$. 
A: 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!
A: Since both functions are even, we can focus on where $x$ is nonnegative. The inequality $e^x\geq 1+x$ is a little weak here. However the inequality
$$e^x\geq 1+x+\frac{x^2}{2}$$
is of help here.
It suffices to show
$$\cosh(x)\leq 1+\frac{x^2}{2}+\frac{x^4}{8}\leq e^{x^2/2}$$
which is much more approachable with derivatives. Of course, this doesn't hold for all $x$ due to exponential growth. However, it's fairly easy to show that it holds on $[0,1.5]$.
The inequality
$$\cosh(x)\leq 1+\frac{x^2}{2}+\frac{x^4}{8}+\frac{x^6}{96}$$
holds on a much larger set--- a little more than $[0,10]$. This is just as simple as the preceding approach.
A: Notice:
$$\frac12(e^x+e^{-x})<\frac12(e^x+1)$$
and for $x>2.11$ that $e^\frac{x^2}{2}>e^x+1$
See if you can show it holds for $0<x<2.11$ (Note, negativity is irrelevant as both sides are even functions)
A: Noting that $\cosh x = \frac{e^x + e^{-x}}{2}$, you may proceed as follows:


*

*Note that $\cosh (-x) = \cosh x$ and $e^{\frac{(-x)^2}{2}} = e^{\frac{x^2}{2}}$

*$ \Rightarrow$ It is enough to show the inequality for $x \geq 0$.

*For $x= 0$ you have equality.

*$ \Rightarrow$ It is enough to show that $\left(e^{\frac{x^2}{2}} - \cosh x \right)' \geq 0$ for $x>0$.

*$\left(e^{\frac{x^2}{2}}\right)' = xe^{\frac{x^2}{2}} = \sum_{n=0}^{\infty}\frac{x^{2n+1}}{2^n\cdot n!}$

*$\left(\cosh x \right)' = \sinh x = \sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}$
$$\frac{1}{2^n\cdot n!} \geq \frac{1}{(2n+1)!}\Leftrightarrow \frac{(n+1)\cdots (2n+1)}{2^n} \geq 1 \mbox{ True!}$$
From this the inequality follows.
