Prove that $|\log(1 + x^2) - \log(1 + y^2)| \le |x-y|$ I need to show that $ \forall x,y \in \mathbb R, |\log(1 + x^2) - \log(1 + y^2)| \le |x-y|$
I tried using concavity of log function: $\log(1 + x^2) - \log(1 + y^2)=\log(\frac{1 + x^2}{1 + y^2})=\log(\frac{x^2y^2}{(1 + y^2)y^2}+\frac{1}{1 + y^2}) \ge \frac{2(\log(x)-\log(y))}{1+y^2}$
Also the middle value theorem: $0<\frac{\log(1 + x^2) - \log(1 + y^2)}{x^2-y^2} <1$
But both attempts has not led too far.

The derivative of $f(x)=\log(1+x^2)$ is
$$
f'(x)=\frac{2x}{1+x^2}
$$
and
$$
\left|\frac{2x}{1+x^2}\right|\le1
$$
because
$$
2|x|\le 1+|x|^2
$$
since
$$
(1-|x|)^2\ge0
$$
Thus, by the mean value theorem,
$$
\frac{\log(1+x^2)-\log(1+y^2)}{x-y}=f'(z)
$$
for $z$ between $x$ and $y$ (assuming $x\ne y$ or the inequality is obvious). Then
$$
\left|\frac{\log(1+x^2)-\log(1+y^2)}{x-y}\right|=|f'(z)|\le 1
$$
