Inequalities for the Gudermannian function of the type $\operatorname{gd}(x)\operatorname{gd}\left(\frac{1}{x}\right)<\text{upper bound}$, where $x>0$ After I've read the solution of Problem 4327 (see [1]) I wondered if an inequality for a similar upper bound in the RHS is feasible for the known as Gudermannian function 
$$\operatorname{gd}(x)\operatorname{gd}\left(\frac{1}{x}\right)<\text{upper bound}\tag{1}$$
for all $x>0$. Here thus as in the Problem 4327, this RHS is a function of $x$, thus the function $\text{upper bound}=\text{upper bound}(x)$ depends on the real variable $x$.
My proposal is to state an inequality involving the Gudermannian function, a function with a good mathematical content, see for example the Wikipedia encyclopedia Gudermannian function (but feel free if you prefer to explore in your home other special function, I don't know if the case for the Airy function is also feasible or more interesting).

Question. Propose an inequality of the type $(1)$ valid for all positive real numbers $0<x$. If it is difficult to get, but you know how to do the analysis for some interval of positive real numbers, feel free to add your computations. Many thanks.

My belief is thus that isn't feasible to get my inequality $(1)$ using specializations or easy computations from the mentioned problem of Crux Mathematicorum. In particular I've calculated just, using a formula from the encyclopedia Wikipedia (please see also MathWorld, and the cited references in the encyclopedias)
$$\operatorname{gd}(x)\operatorname{gd}\left(\frac{1}{x}\right)=\left(\frac{\pi}{2}-2\arctan(e^{-x})\right)\left(\frac{\pi}{2}-2\arctan(e^{-1/x})\right),$$ and I've not a good/explicit candidate as $\text{upper bound}$.
References:
[1] Problem 4327, Crux Mathematicorum, Vol. 45(3), (March 2019). Canadian Mathematical Society, 2019.
 A: $\newcommand{\gd}{\operatorname{gd}}
\newcommand{\ch}{\operatorname{ch}}
\renewcommand{\th}{\operatorname{th}}$
The answer is 
\begin{equation}
 p(x):=\gd(x)\gd(1/x)\le p(1)=0.74955\dots 
\end{equation}
for all $x>0$. Moreover, $p$ is increasing on $(0,1]$ and decreasing on $[1,\infty)$. 
Since $p(x)=p(1/x)$ for $x>0$, it is enough to show that $p$ is increasing on $(0,1)$. In what follows, assume therefore by default that $x\in(0,1)$. Note that 
\begin{equation}
 \gd(y)=\int_0^y\frac{du}{\ch u}=y\int_0^1\frac{ds}{\ch ys}  
\end{equation}
for $y>0$, where $\ch:=\cosh$. So, 
\begin{align*}
 p'(x)&=\frac1{\ch x}\,\gd(1/x)-\frac1{x^2\ch 1/x}\,\gd(x) \\ 
 &=\frac1{x\ch x}\,\int_0^1\frac{ds}{\ch s/x}
 -\frac1{x\ch 1/x}\,\int_0^1\frac{ds}{\ch xs} \\ 
 &=\frac1{x}\,\int_0^1 ds\,\Big(\frac1{\ch x\ch s/x}-\frac1{\ch(1/x)\ch sx}\Big) \\ 
 &=\frac1{x}\,\int_0^1 \frac{ds}{\ch(1/x)\ch(s/x)}\,\Big(\frac{\ch(1/x)}{\ch x}-\frac{\ch s/x}{\ch sx}\Big)>0, 
\end{align*}
because of 
Lemma For any real $a$ and $b$ such that $a>b>0$, 
\begin{equation}
 r(s):=\frac{\ch as}{\ch bs}
\end{equation}
is increasing in $s>0$. 
So, $p'>0$ and hence $p$ is increasing on $(0,1]$, as claimed. 
Here is the graph $\{(x,p(x))\colon0<x<10\}$: 

It remains to provide 
Proof of the lemma. By rescaling, without loss of generality $b=1$, so that $a>1$. Then
\begin{equation}
 r'(s)\frac{\ch s}{\ch as}=a\th as-\th s>\th as-\th s>0
\end{equation}
for $s>0$, where $\th:=\tanh$. So, $r'(s)>0$, and the lemma follows.  $\Box$
