TITLE: For which real numbers $c$ is $\frac{e^x+e^{-x}}{2} \le e^{cx^2}$ for all real numbers $x$?
QUESTION [11 upvotes]: 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?

REPLY [2 votes]: In fact we have the following:

The method presented in the proposal of the problem proves that any $c\geq \frac{1}{2}$ works.
The solution of Leucippus proves that no $c< \frac{1}{2}$ works. So, these are not alternative solutions, but complementary ones. and the conclusion is that
$$
\left\{c:\forall\, x,~\cosh(x)\leq \exp(c x^2)\right\}=\left[\frac{1}{2},+\infty\right).
$$