Show for every $x_{1}<x_{2}$ that $\arctan x_{2}-\arctan x_{1}<x_{2}-x_{1}$.

Show $\arctan x_{2}-\arctan x_{1}<x_{2}-x_{1}$ when $x_1<x_2$

I think that the mean value theorem can help us to solve this inequality. But I still don't know how to continue the example.

Since $x_1<x_2$, using the intermediate mean value theorem on the interval $[x_1,x_2]$, there exists some $c\in (x_1,x_2)$ such that $\frac{\tan^{-1}x_2-\tan^{-1}x_1}{x_2-x_1}=d/dx( \tan^{-1} c)=\frac{1}{1+c^2}<1$. This proves the assertion.

Sketch: Apply mean value theorem to $\arctan x$ in $[x_1,x_2]$, and observe that the derivative is bounded by $1$.