Prove that for any real numbers $a,b$ we have $\lvert \arctan a−\arctan b\rvert\leq \lvert a−b\rvert$. Prove that for any real numbers  $a,b$ we have  $\lvert \arctan a−\arctan b\rvert\leq \lvert a−b\rvert$. This should have an application of the mean value theorem.

Let $f(x)=\arctan x$, then by mean value theorem there exist $c\in (a,b)$ such that
$$|f(a)-f(b)|=f^{\prime}(c) |a-b|$$
$$f^{\prime}(x)=\frac{1}{1+x^2}\leq1$$
So $f^{\prime}(c)\leq1$ and $|\arctan a-\arctan b|\leq |a-b|$.
