How to prove $|\arctan(2b)-\arctan(2a)| \le 2|b-a|$ for any real numbers $a, b$

Question

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.

Answer 1

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

Answer 2

You made the most difficult step! Just $1 + 4 c^2 \ge 1$, so $\frac{1}{1 + 4 c^2} \le 1$.

Answer 3

If you want to avoid the mean value theorem, we have that

$$|\arctan(2b) - \arctan(2a)| = \left|\arctan\left(\frac{2b-2a}{1+4ab}\right) \right| \leq \left| \frac{2b-2a}{1+4ab}\right| \leq 2|b-a|$$

using the tangent angle subtraction formula and $\arctan x \leq x$ for $x \geq 0$