Hyperdiabolical

If $\cosh(x) \le e^{ax^{2}}$ , for all $x$ , then the minimum of $a$ would be when they intersect once.

So, $f(x)=\cosh(x) - e^{ax^{2}}$ has to have only $1$ real root. Since both equations, $\cosh(x)$ and $e^{ax^{2}}$ are symmetrical to the line $x=0$ , and they have a shape similar to parabolas, the $1$ root is likely to be when $x=0$ .

Well, substituting $x=0$ in the equation $\cosh(x) - e^{ax^{2}}=0$ to solve for $a$ isn't going to do much help, so why not differentiate the equation and substitute $x=0$

$f'(x)=\frac { d }{ dx } \cosh (x)-e^{ ax^{ 2 } }=\sinh (x)-2axe^{ ax^{ 2 } }$

As you can see, substituting $x=0$ in this equation isn't much help either, so let's differentiate one more time:

$f''(x)=\cosh (x)-2ae^{ax^2}(2ax^2+1)$

Now, substituting $x=0$ and evaluating $f''(x)=0$

$\cosh (0)-2ae^{ 0 }(0+1)=0$ $a=\boxed{0.5}$

We have $e^{ax^{2}} \geq \cosh(x)$ Since natural logarithm is an increasing function and both $e^{ax^{2}}$ and $\cosh[x]$ have positive value, then we have $a \geq \frac{1}{x^{2}}\log(\cosh(x))$

Since the inequality is true for all $x \in R$ , then the minimum value of $a$ is $a = \max\left(\frac{1}{x^{2}}\log(\cosh(x))\right) = \frac{1}{2}$ .

If the inequality holds true for all real $x$ , then

$0 \le \lim_{x \rightarrow 0} \dfrac{e^{ax^{2}} - \frac{1}{2}(e^{x} + e^{-x})}{x^{2}} =$

$\lim_{x \rightarrow 0} \dfrac{(1 + ax^{2} + \cdots) - (1 + \frac{1}{2}x^{2} + \cdots)}{x^{2}} = a - \dfrac{1}{2}.$

We would thus require that $a \ge \dfrac{1}{2}$ for the inequality to hold true for all $x$ .

Now for $a \ge \dfrac{1}{2}$ we have that

$\cosh(x) = \dfrac{e^{x} + e^{-x}}{2} = \displaystyle\sum_{n=0}^{\infty} \dfrac{x^{2n}}{(2n)!} \le \sum_{n=0}^{\infty} \dfrac{x^{2n}}{2^{n}n!} = e^{\frac{1}{2}x^{2}} \le e^{ax^{2}}.$

Thus the inequality holds true if and only if $a \ge \dfrac{1}{2} = \boxed{0.5}.$