Proving that $\frac{e^x + e^{-x}}2 \le e^{x^2/2}$ 
Prove the following inequality:
  $$\dfrac{e^x + e^{-x}}2 \le e^{\frac{x^2}{2}}$$

This should be solved using Taylor series.
I tried expanding the left to the 5th degree and the right site to the 3rd degree, but it didn't help.
Any tips?

Obscene overkill: since
$$ \cosh(z)=\prod_{n\geq 0}\left(1+\frac{4z^2}{(2n+1)^2\pi^2}\right) $$
we have
$$ \log\cosh(z)\leq \sum_{n\geq 0}\frac{4z^2}{(2n+1)^2\pi^2} =\frac{z^2}{2}$$
and the claim follows by exponentiating both sides. In other terms, it is sufficient to integrate both sides of $\tanh(z)\leq z$ over $[0,x\in\mathbb{R}^+]$.
