Proving that $(1+1/x)\tan^{-1} x > 1$, when $x> 0$ Monotonicity of a function is an  itneresting method to prove inequalities. However, one needs to construct a suitable function $f(x)$ in the given domain. In this, out of several choices one particular form of $f(x)$ may suceed. Here, we prove that  $$(1+1/x)\tan^{-1} x>1, ~\text{when}~ x > 0. \tag{$*$}$$
Out of several options, let us take $$f(x)=\tan^{-1}x -\frac{x}{1+x}.$$
$$\implies f'(x)=\frac{1}{1+x^2}-\frac{1}{(1+x)^2}=\frac{2x}{(1+x^2)(1+x)^2}\ge 0, \forall~ x\ge 0.$$
So, $f(x)$ is an increasing function in $[0,\infty)$, so $f(0)\ge 0$. Hence we get
$$\tan^{-1}x -\frac{x}{1+x}\ge 0 \implies (1+x)\tan^{-1}x\ge x.$$
Finally for $x>0$, we have
$$\implies (1+1/x)\tan^{-1} x >1.$$
It will be interesting to know other options/methods to prove $(*)$.
 A: Dividing both sides by $(1+1/x)$,
$$
\arctan x > \frac x{x+1}
$$
Then we have
$$
\begin{align}
x > \tan \frac x{x+1}
\end{align}
$$
Applying inequalities about $\sin$ and $\cos$ i.e. $\sin \theta < \theta$ and $\cos \theta > 1 - 1/2\theta^2, \forall \theta \in (0,1)$,
$$
\begin{align}
\tan \frac x{x+1} & = \frac{\sin \frac x{x+1}}{\cos \frac x{x+1}} \\
&< \frac{\frac x{x+1}}{1 - \frac12(\frac x{x+1})^2} \\
&< \frac{\frac x{x+1}}{1 - \frac x{x+1}} \\
&= x
\end{align}
$$
A: The inequality is equivalent to
$$
\arctan x > \frac x{x+1}.
$$
Note
$$ \arctan x=\int_0^x \frac1{1+t^2}dt, \ \frac x{x+1}=\int_0^x\frac1{(1+t)^2}dt. $$
Since
$$ {1+t^2}<(1+t)^2 $$
one has
$$ \arctan x=\int_0^x \frac1{1+t^2}dt>\int_0^x\frac1{(1+t)^2}dt=\frac x{x+1}. $$
A: For $x>0$, we have
$$\frac{x}{\sqrt[2]{x^2+1}}<\tan^{-1}(x) <\frac{x}{\sqrt[3]{x^2+1}}$$ which seems to be sufficient.
A: Consider $f(x) = \arctan(x) - \dfrac{x}{x+1}, x \in(0,\infty)$. We have: $f'(x) = \dfrac{1}{x^2+1}-\dfrac{1}{(x+1)^2}>0$. Thus $f$ is strictly increasing for $x > 0$. Hence $f(x) > f(0) = 0 \implies \arctan(x) -\dfrac{x}{x+1} > 0\implies \arctan(x) > \dfrac{x}{x+1}$.
A: In my opinion, using MVT here makes it more rigorous. If you use MVT, you get a constant $0<c<x$ for which $f(x)=f’(c)\cdot x$, but the derivative is always positive as you proved, hence the function will always be strictly greater than 0.
