For which real numbers $c$ is $\frac{e^x+e^{-x}}{2} \le e^{cx^2}$ for all real numbers $x$?

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}

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 $$ \left\{c:\forall\, x,~\cosh(x)\leq \exp(c x^2)\right\}=\left[\frac{1}{2},+\infty\right). $$