Maximum of derivative of a smooth cutoff function

Question

Let $a,b\in\mathbb{R}$ with $a<b$, and let $h:\mathbb{R}\to\mathbb{R}$ be the smooth cutoff function defined by \begin{align} h(x):=\frac{f(b-x)}{f(b-x)+f(x-a)} \end{align} where $f:\mathbb{R}\to\mathbb{R}$ is the smooth function \begin{align} f(x)=\left\{ \begin{array}{ccc} e^{-1/x} & \text{if} & x>0 \\ 0 & \text{if} & x\leq 0 \end{array} \right. \end{align} Thus $h\equiv 1$ in $(-\infty,a]$ and $h\equiv 0$ in $[b,\infty)$.

In this question I am concerned about the maximum of $|h'|$, where $h'$ is the derivative of $h$. Explicitly:

My question

How to show that $|h'|$ attains the maximum at the point $x_c=(a+b)/2$, the center of $(a,b)$, and \begin{align} |h'(x_c)|=\frac{C}{(b-a)^2} \end{align} for some constant $0<C<\infty$?

---

My attempt

1. Clearly $h'\equiv 0$ outside the interval $(a,b)$, so in the following I will focus only on the interval $(a,b)$. In this interval we have \begin{align} h(x)=\frac{\exp\left(-\frac{1}{b-x}\right)}{\exp\left(-\frac{1}{b-x}\right) +\exp\left(-\frac{1}{x-a}\right)} \end{align} It can be computed that $\forall x\in(a,b)$, \begin{align} h'(x)=-h(x)(1-h(x))\left[\frac{1}{(b-x)^2}+\frac{1}{(x-a)^2}\right]\qquad\text{and}\qquad |h'(x)|=-h'(x) \end{align} I then proceed in the standard calculus way, start by finding the critical point of $-h'$. Denote the term inside the square bracket by $f(x)$ (i.e. $h'=-h(1-h)f$). It can also be computed that \begin{align} -h''=2h(1-h)\left[\frac{f'}{2}+\left(h-\frac{1}{2}\right)f^2\right] \end{align} Since $h(1-h)>0$ in $(a,b)$, thus $-h''(x)=0$ iff \begin{align} \frac{f'(x)}{2}+\left(h(x)-\frac{1}{2}\right)f(x)^2=0 & & (*) \end{align} However, I am unable to solve this equation.

2. Plotting the graph of $-h'$ we can observe that it looks like a bell shaped centered at $x_c$. In particular, it suggests that $-h'$ is increasing in $(a,x_c)$ and decreasing in $(x_c,b)$. So I next try to show this instead, by showing \begin{align} -h''>0\text{ in }(a,x_c)\qquad\text{while}\qquad-h''<0\text{ in }(x_c,b) & & (**) \end{align} This looks slightly more doable, as now we don't need to solve equation but only need to do estimate. But it turns out to be equally difficult to estimate the LHS of $(*)$. For example, in $(a,x_c)$, while $h-1/2>0$, one has on the other hand $f'<0$, and so to establish that L.H.S. of (*) $>0$, one has to show that $|(h-1/2)f^2|>|f'/2|$, but this seems to also not easy as well.

3. The claim $(**)$ is also equivalent to saying that $h$ is concave down in $(a,x_c)$ and is concave up in $(x_c,b)$. This is also suggested by the graph of $h$. For example, plotting the part of it over $(a,x_c)$, it does look like lying above the straight line segment connecting $(a,h(a))$ and $(x_c,h(x_c))$, which is the graph of some linear function, says $\ell_1$. So I also thought of trying to tackle the problem in terms of convexity. Note that $h(x_c)=1/2$, so $\ell_1$ is given by \begin{align} \frac{\ell_1(x)-1}{x-a}=\frac{\frac{1}{2}-1}{x_c-a}\qquad\Longleftrightarrow\qquad \ell_1(x)=1-\frac{x-a}{b-a}=\frac{b-x}{b-a} \end{align} and so by definition of concave-down function, we want to show that $\forall x\in(a,x_c)$, \begin{align} h(x)\geq\ell_1(x)\qquad\Longleftrightarrow\qquad \frac{\exp\left(-\frac{1}{b-x}\right)}{\exp\left(-\frac{1}{b-x}\right) +\exp\left(-\frac{1}{x-a}\right)}\geq\frac{b-x}{(b-x)+(x-a)} \end{align} This looks like something that can be deduced directly from a more general result, but I have no idea what will it be. I'm stuck again.