# Inequality with absolute value.

Show that $\forall a,b\in \mathbb{R}$: $$\left| \frac{a}{1+a^2} - \frac{b}{1+b^2} \right| \leq |a-b|$$ Being honest, I do not know where to start (apart from common denominator form) and would appreciate any help. Thanks in advance!

Note that $f(x)=\dfrac{x}{1+x^2}$ has derivative $f'(x)=\dfrac{1-x^2}{(1+x^2)^2}$, and one can show that $|f'(x)|\le1$.
$a^2(1-\dfrac{1}{(a^2+1)^2})+b^2(1-\dfrac{1}{(b^2+1)^2}) \ge 2ab(1-\dfrac{1}{(1+a^2)(1+b^2)}) \\ \iff \dfrac{a^4(a^2+2)}{(1+a^2)^2}+\dfrac{b^4(b^2+2)}{(1+b^2)^2} \ge 2ab\dfrac{a^2b^2+a^2+b^2}{(1+a^2)(1+b^2)} \\ \iff \dfrac{a^4(a^2+2)}{(1+a^2)^2}+\dfrac{b^4(b^2+2)}{(1+b^2)^2} \ge(a^2+b^2)\dfrac{a^2b^2+a^2+b^2}{(1+a^2)(1+b^2)} \\ \iff \dfrac{a^2(a^2-b^2)}{(1+a^2)^2(1+b^2)}- \dfrac{b^2(a^2-b^2)}{(1+b^2)^2(1+a^2)} \ge 0 \\ \iff (a^2-b^2)^2 \ge 0$