For which real numbers $c$ is $\frac{e^x+e^{-x}}{2} \le e^{cx^2}$ for all real numbers $x$? This question comes from the 1980 Putnam exam. My work is shown below.

For all integers $n \ge 1$, \begin{align}
(2n)!&=n!\cdot\underbrace{(n+1)(n+2)(n+3)\cdots(2n-2)(2n-1)(2n)}_{n \text{ terms}} \\ 
&\ge n! \cdot \underbrace{2 \cdot 2 \cdot 2\cdots 2\cdot2 \cdot 2}_{n \text{ terms}} \\ &=2^nn! \\ \implies \frac{1}{(2n)!} &\le \frac{1}{2^nn!}.\end{align} So we have \begin{align}
\frac{e^x+e^{-x}}{2} = \cosh(x) &= \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \\ &\le \sum_{n=0}^{\infty}  \frac{x^{2n}}{2^nn!} \\ &= \sum_{n=0}^{\infty}  \frac{\left(\frac{x^2}{2} \right)^n}{n!} \\ &= e^{\frac{1}{2}x^2}
\end{align}
Therefore, $\boxed{c \ge \frac{1}{2}}$.
How do I know that it is $\ge$, not $\le$? Other than that, how did I do on my work?

The method presented in the proposal of the problem is correct. 
An alternate method is:
\begin{align}
\cosh(x) \approx 1 + \frac{x^{2}}{2} + \mathcal{O}(x^{4})
\end{align}
and 
\begin{align}
e^{c x^{2}} \approx 1 + c x^{2} + \mathcal{O}(x^{4})
\end{align}
for which the inequality 
\begin{align}
e^{c x^{2}} \geq \cosh(x)
\end{align}
leads to 
\begin{align}
c \geq \frac{1}{2}.
\end{align} 
