Question: 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).

Answer: 
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,](https://www.desmos.com/calculator/shf1vhxyui) 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.