According to [this comment](https://mathoverflow.net/questions/444490/ask-for-a-proof-of-an-inequality-involving-the-bernoulli-numbers#comment1147661_444490) and [this comment](https://mathoverflow.net/questions/444490/ask-for-a-proof-of-an-inequality-involving-the-bernoulli-numbers#comment1147671_444490), a positive answer to [this recent question](https://mathoverflow.net/q/444490/36721) (about Bernoulli numbers) would be sufficient to prove the following:



> 
> $r:=f/g$ is increasing on $(0,\pi/2)$ from $5/6$ to $1$, where
> \begin{equation}
>  f(x):=\ln\frac{\tan x}x,\quad g(x):=\ln\frac{\tan x-x}{x^3/3}. 
> \end{equation}
> 
> 
> 


However, the mentioned sufficient condition seems much, much harder to prove than the actually desired highlighted claim above. In the answer below, the highlighted claim will be proved, using a so-called l'Hospital-type rule for monotonicity.



On the monotonicity of the ratio of two logarithmic expressions
The values $5/6$ and $1$ for $r(0+):=\lim\_{x\downarrow0}r(x)$ and $r(\frac\pi2-):=\lim\_{x\uparrow\pi/2}r(x)$ are easy to get by the l'Hospital rule for limits.


It remains to show that $r=f/g$ is increasing on $(0,\pi/2)$.


Note that $f(0+):=\lim\_{x\downarrow0}f(x)=0$ and $g(0+)=0$. Next, for real $t>0$ we have $0<\arctan t<t$ and
\begin{equation}
G\_1(t):=g'(\arctan t)\frac{(t-\arctan t) \arctan t}{3 + t^2}
=\arctan t-\frac{3t}{3+t^2}. 
\end{equation}
In what follows, by default, $t\in(0,\infty)$.


Next, $G\_1(0)=0$ and $G'\_1(t)=4 t^4/((1 + t^2) (3 + t^2)^2)>0$. So, $G\_1>0$ (on $(0,\infty)$) and hence $g'>0$ (on $(0,\pi/2)$). So, by the l'Hospital-type rule for monotonicity given by [Proposition 4.1](https://www.emis.de/journals/JIPAM/images/157_05_JIPAM/157_05.pdf), it is enough to show that the "derivative ratio"
\begin{equation}
 r\_1(t):=\frac{f'(\arctan t)}{g'(\arctan t)}=\frac{f\_1(t)}{g\_1(t)}
\end{equation}
is increasing (in $t\in(0,\infty)$), where
\begin{equation}
 f\_1(t):=\frac{(t - \arctan t) (-t + (1 + t^2) \arctan t)}{1 + t^2}, 
\end{equation}
\begin{equation}
 g\_1(t):=\frac{t (-3 t + (3 + t^2) \arctan t)}{1 + t^2}. 
\end{equation}


Note that
\begin{equation}
G\_2(t):=g\_1'(t)\frac{(1 + t^2)^2}{3 + t^4}=\frac{t (-3 + t^2)}{3 + t^4} +\arctan t, 
\end{equation}
so that $G\_2(0)=0$ and $G'\_2(t)=(8 t^4 (3 + t^2))/((1 + t^2) (3 + t^4)^2)>0$. So, $G\_2>0$ and hence $g\_1'>0$.
Also, $f\_1(0+)=0$ and $g\_1(0+)=0$.
So, by the same l'Hospital-type rule for monotonicity, it is enough to show that the second "derivative ratio"
\begin{equation}
 r\_2(t):=\frac{f\_1'(t)}{g\_1'(t)}=\frac{f\_2(t)}{g\_2(t)}
\end{equation}
is increasing, where
\begin{equation}
 f\_2(t):=\frac{t^2 (t + (-1 + t^2)\arctan t)}{3 + t^4}, 
\end{equation}
\begin{equation}
 g\_2(t):=\frac{t (-3 + t^2) + (3 + t^4) \arctan t}{3 + t^4}. 
\end{equation}
Next, $g'\_2(t)=(8 t^4 (3 + t^2))/((1 + t^2) (3 + t^4)^2)>0$ and $f\_2(0+)=0=g\_2(0+)$.


So, using once again the same l'Hospital-type rule for monotonicity, we see that it is enough to show that the third "derivative ratio"
\begin{equation}
 r\_3(t):=\frac{f\_2'(t)}{g\_2'(t)}
 =\frac{3 t + 6 t^3 - t^5 + (-3 + 3 t^2 + 7 t^4 + t^6) \arctan t}{4 t^3 (3 + t^2)}
\end{equation}
is increasing. Let
\begin{equation}
 r\_{31}(t):=r'\_3(t)\frac{4 t^4 (3 + t^2)^2}{(3 + t^4) (9 + 2 t^2 + t^4)}
 =\frac{t (-9 + t^2)}{9 + 2 t^2 + t^4} + \arctan t. 
\end{equation}
Then $r\_{31}(0)=0$ and $r'\_{31}(t)=(32 t^4 (3 + t^2))/((1 + t^2) (9 + 2 t^2 + t^4)^2)>0$, so that $r\_{31}>0$ and hence $r\_3$ is indeed increasing. $\quad\Box$


