Find the smallest real $c$ such that $(e^x + e^{-x})/(2e^{cx^2}) \leq 1$ for all $x \in \Bbb{R}$

Question

We need to find the smallest real $c$ such that $\dfrac{e^x + e^{-x}}{2e^{cx^2}} \leq1$ $\forall x$.

My initial approach was to consider the function $g(x)=e^x + e^{-x} -2e^{cx^2}$. Since $g(0)=0$, If we find $c$ such that $g'(x)\leq0$ when $x>0$ and $g'(x)\geq0$, when $x<0$, Then we'll have $g(x)\leq g(0)\ \forall x$. We can write this as: $\dfrac{g'(x)}{x}\leq0$. I tried simplifying this using the Maclaurin expansion of $e^a$, but couldn't find a compact method to find a bound for $c$.

I also couldn't understand the supposed "solution" to this problem: let $h(x)=e^{cx^2}-\dfrac{e^x + e^{-x}}{2}$. Since $h(x)\geq0, \dfrac{h(x)}{x^2}\geq0$. Taking limits: $\lim _{x\to 0} \dfrac{h(x)}{x^2} \geq0$ , so we get $(2c-1) \geq 0$ and thus minimum value of $c$ is $0.5$.

I don't have any idea why this solution makes sense: what was the need to divide by $x^2$? And how does the behaviour of our new function $\dfrac{h(x)}{x^2}$ near $x=0$ have any relationship with the lower bound of $c$?

This is from a high-school exam, so this problem can be solved by using elementary methods only.

Answer 1

The use of either $h(x)$ or your $g(x)$ are similar in the fact that they rewrite the original expression into a simpler function. I will work with your expression, but the $h(x)$ is the same.

You have $g(x)=e^x + e^{-x} -2e^{cx^2}$. You can immediately see that $g(-x)=g(x)$. You noticed that $g(0)=0$. And you want $g(x)\le 0$ for any $x$. If $c\le0$ this is not possible (look at the behavior at $\pm\infty$). So we know that $c>0$. With that, $g(\pm\infty)=-\infty$. You took the first derivative of $g(x)$ and you get $$g'(x)=e^x-e^{-x}-4cxe^{cx^2}$$ Notice that $g'(0)=0$. It means that $x=0$ is either a minimum, a maximum,or an inflection point. You want it to be a maximum. Why? If it's not, at least on one side you have $g(x)>0$. So what's the condition for the maximum? $g''(0)\lt0$. If you plug in, you will get $c=0.5$. Similarly, the equivalent of this is to say that $g(x)\approx Ax^2$ in the limit of small $x$. That's why you get $$A=\lim_{x\to 0}\frac{g(x)}{x^2}$$ If $A<0$ you have a maximum

Answer 2

For any real $c$, the map $f_c(x)= \dfrac{e^x + e^{-x}}{2e^{cx^2}}$ is even. So it is sufficient to study the inequality $f_c(x) \le 1$ on $[0,\infty)$.

Around $0$, you have

$$f_c(x) = \frac{1+x^2/2+o(x^2)}{1+cx^2+o(x^2)}$$

hence the desired inequality will only stand for all $x$ if $c \ge 1/2$.

Conversely for $c \ge 1/2$ and $x \ge 0$ we have

$$\begin{aligned} h_c(x)&=e^{cx^2} - \cosh x = \sum_{k=0}^\infty\frac{c^k x^{2k}}{k!} - \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\\ &\ge \sum_{k=0}^\infty \left( \frac{1}{2^k k!} - \frac{1}{(2k)!}\right)x^{2k} \ge 0 \end{aligned}$$

as $(2k)! \ge 2^k k!$ for $ k$ integer.

So we’re done. The answer is $c \ge 1/2$.