The real function $$f(x) = \ln\cosh x$$ loosely resembles the shape of a hyperbola
$$h_c(x) = \sqrt{c^2+x^2} - c$$
and my question is what's the best $c$ in terms of the maximum norm, i.e. $$ \text{Let } c \text{ be such that } \|f-h_c\|_{\max} \stackrel!=\min \tag 1 $$ For large $|x|$ we have that $f(x)\approx|x|-\ln2$ so that $c=\ln 2$ is a choice with $$ \lim_{x\to\pm\infty}(f(x) - h_{\ln2}(x)) = 0 $$ However it's $h_{\ln2}(x)\geqslant f(x)$ (with equality for $x=0$) so that $c=\ln2$ is not optimal with respect to $(1)$: A better hyperbola would trade positive error against negative error equally so that
$$\sup (f-h_c) = -\inf(f-h_c)$$
presumably with $c\in[\ln 2, 1]$ somewhere $c\approx 0.8$. So what's the best $c$? One idea it to use $c>\ln2$ so that
$$ \lim_{x\to\pm\infty}(f(x) - h_c(x)) = c-\ln2=\delta > 0 $$ and find $x_0$ such that
$$-\delta = \inf (f-h_c) = f(x_0)-h_c(x_0)$$ so that $x_0$ is a local minimum of $f-h_c$. For the local minimum, taking derivative and determine $x_0$ and $c$ such that:
$$f'(x_0)-h'_c(x_0) = \tanh x_0 - \frac{x_0}{\sqrt{c^2+x_0^2}} \stackrel!= 0 \tag 2$$ and simultaneously $$ f(x_0)-h_c(x_0) = \ln\cosh x -\sqrt{c^2+x_0^2}+c\stackrel!= \ln2 -c \tag 3 $$ so it's a system of two equations for two variables $c$ and $x_0\neq0$ where $x_0$ is only determined up to sign. But I don't know how to preceed from here.
Addendum: The current choice of $h_c$ is such that $h_c(0) = 0$. As an extension of the problem, this requirement could be dropped so that the available space becomes a 2-parameter family of functions $h_{c,d}(x)=\sqrt{c^2+x^2}-d$.
Addendum 2: To solve the system (2)+(3): Using (2), express $c$ in terms of $x$:
$$\sqrt{c^2+x^2}=\frac x{\tanh x}\quad\text{ and }\quad c = |x|\sqrt{\frac{1}{\tanh^2 x}-1} \tag 4$$ Drop these into (3) to eliminate $c$ and to determine $x$:
$$\ln\cosh x-\frac{x}{\tanh x}+2|x|\sqrt{\frac{1}{\tanh^2 x}-1}-\ln 2 \stackrel!= 0 \tag 5$$
Numerical (Newton-Raphson) approximation of (5) yields $|x|\approx1.317$, which can be plugged back into (4) to get $c$:
$$c\approx0.7603459963 \quad\text{ and }\quad \|f-h_c\|_\max\approx 0.0671$$ where the original distance was $\|f-h_{\ln 2}\|_\max\approx 0.10387$, so that there is an improvement.
Addendum 3: Finding a more general hyperbola $h(x)=\sqrt{c^2+x^2}-d$ as of Addendum 1 is no more complicated than the computation above, so here it is: In order to establish an equi-oscillation property, make the ansatz
$$\begin{align} f(0) - h(0) &= d-c \stackrel!= \delta \tag{6.1}\\ f(\infty) - h(\infty) &= d-\ln2 \stackrel!= \delta \tag{6.2}\\ f(x_0) - h(x_0) &\stackrel!= -\delta \tag{6.3}\\ \end{align}$$ From (6.1) and (6.2) it follows that $c=\ln 2\approx0.69315$. Then differentiate (6.3) to find $x_0$ as a local extremum. The derivative of (6.3) is (2), but this time for an already known $c=\ln 2$. Solving it yields $x\approx1.5366$ and then combine (6.3) with (6.1) to find $d$ as
$$d = \frac12\big(c-f(x)+\sqrt{c^2+x^2}\big)$$ Plugging in the values yields
$$d\approx 0.74508129\quad\text{ and }\quad \|f-h_{c,d}\|_\max \approx 0.052$$ so this proximal hyperbola with its distance of 0.052 from $f$ is again closer than the 0.067 from Addendum 2.
This is not an answer.
Numerical results
Consider the norm $$\Phi(c,k)=\int_0^{10^k} \Big[\sqrt{c^2 +x^2}-c -\log (\cosh (x))\Big]^2\,dx$$ and minimize it with respect to $c$ for a given $k$
$$\left( \begin{array}{cc} k & \Phi_{\text{min}} & c_{\text{opt}} \\ 1 & 0.01092 & 0.753021 \\ 2 & 0.02975 & 0.703006 \\ 3 & 0.03715 & 0.694634 \\ 4 & 0.03898 & 0.693350 \\ 5 & 0.03933 & 0.693173 \end{array} \right)$$
Trying to solve $(2)$
$$\tanh (x)-\frac{x}{\sqrt{c^2+x^2}}=0$$ For large $x$, we have $$\tanh(x) = 1 - 2 e^{-2x} + o(e^{-2x})$$ So, we have $$e^{-2x} \sim \frac{\sqrt{c^2+x^2}-x}{2 \sqrt{c^2+x^2}}=\frac{c^2}{4 x^2}+O\left(\frac{1}{x^4}\right)\implies x \sim -W\left(-\frac{c}{2}\right)$$ which would require $c \leq \frac 2e$ but this does not allow to find a solution for $(3)$.
So, we would need to use $$\tanh(x) = 1 - 2 e^{-2x}+ 2 e^{-4x}+ o(e^{-6x})$$ and solve for $X=e^{-2x}$ the equation $$1-2X+2X^2 =\frac{c^2}{4 x^2}$$