How to prove $|\arctan(2b)-\arctan(2a)| \le 2|b-a|$ for any real numbers $a, b$ I need to prove the following inequality:
$$|\arctan(2b)-\arctan(2a)| \le 2|b-a|$$
If $a=b$, this is obvious. So,  without loss of generality, I assumed $a < b$.
I defined $f(x) =\arctan(2x)$ and used the mean value theorem in $[a,b]$ and some $a < c < b$:
$$\arctan(2b)-\arctan(2a)|=|f'(c)(b-a)|=\bigg|\frac{2}{1+4c^2}\cdot(b-a)\bigg|=\frac{1}{1+4c^2}\cdot2|b-a|$$
Now, I need to show somehow that $\frac{1}{1+4c^2}\le1$, but I don't know how.
In addition, if there is another way to solve this, I will be happy to know.

$$1+4c^2\geq 1\implies \frac{1}{1+4c^2}\leq 1.$$
