# Is $\arctan(f(x))\leq f(x)$ whenever $|f(x)|<1\ \ \forall x\in\mathbb{R}$?

Is the following always true? $$\arctan(f(x))\leq f(x),\ \ \text{where}\ \ |f(x)|<1\ \ \forall x\in\mathbb{R}$$

For example, is it correct to say that $$\arctan\left(\frac{\sin x}{\sqrt{x+1}}\right)\le\frac{\sin x}{\sqrt{x+1}}\ \ \forall x\in\mathbb{R},$$ since $$\arctan t=t-\frac{t^3}{3}+\frac{t^5}{5}-\dots$$, where each term is not greater than the one before?

I'm confused because in the example above $$\arctan$$'s argument can be negative (i.e., $$\sin x$$ can be negative).

If $$0\leq f(x) <1$$, yes, of course. Since $$\frac{d}{dy}\arctan y=\frac{1}{1+y^2}\leq1$$ (and $$\arctan(0)=0$$), you'll have $$\arctan y \leq y, \quad\text{for }\ y\geq0.$$ Clearly, the situation is reversed if $$y<0$$, because both functions are negative. You can say that $$|\arctan f(x)|\leq|f(x)|$$ with no restrictions on $$f(x)$$, though.
The claim is false, since whenever $$\xi<0$$, $$\arctan(\xi)>\xi$$. On the other hand, if we include absolute values, the inequality $$(\forall \xi\in[-1,1])\quad|\arctan(\xi)|\leq|\xi|$$ is true.
If you derive the function $$x\mapsto \text{Arctan} (x) -x$$, you get that:
• if $$x>0$$ then $$\text{Arctan} (x) < x$$
• but if $$x<0$$ then $$\text{Arctan} (x) > x$$.