<s>[INST] 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.
 [/INST] We can write the term of interest as
$$\left|\log(1+x^2)=\log(1+y^2)\right|=\left|\int_y^x \frac{2t}{1+t^2}\,dt\right|$$
Now, the integrand $f(t)=\frac{2t}{1+t^2}$ attains its maximum value of $1$ when $t=1$.  Therefore, the inequality follows immediately from the mean value theorem.  
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