$\large f(x)=\tan^{-1}(\frac{1}{x^2}) -\ln(x^2+1)$
$\large if 1\leq x < y$ $\large \text { and } 1-\frac{2x}{1+x^2}\geq 0$
Demonstrate that
$ \large |f(x)-f(y)|\leq 2|x-y|$
I managed to get a quadratic équation in term of $c$ through the expression $f'(c)=\frac{f(y)-f(1)}{y-1}$. then I determined $f'(x)$ and made the substitution with c then compared both $f'(c)$ But I did not manage to get further.
I don't think the result is true, since $f'(1) \approx -2.74$ which is greater than $2$ in absolute value. So for instance, if we take $x=1$ and $y=1.1$, we have \begin{align*} f(x)&=0.298\\ f(y)&=0.033\\ |f(x)-f(y)|&=0.265\\ |x-y|&= 0.1 \end{align*} so $|f(x)-f(y)|>2|x-y|$.