The derivative of \(\displaystyle{f{{\left({x}\right)}}}={\log{{\left({1}+{x}^{{2}}\right)}}}\) is

\(\displaystyle{f}'{\left({x}\right)}={\frac{{{2}{x}}}{{{1}+{x}^{{2}}}}}\)

and

\(\displaystyle{\left|{\frac{{{2}{x}}}{{{1}+{x}^{{2}}}}}\right|}\leq{1}\)

because

\(\displaystyle{2}{\left|{x}\right|}\leq{1}+{\left|{x}\right|}^{{2}}\)

since

\(\displaystyle{\left({1}−{\left|{x}\right|}\right)}^{{2}}\geq{0}\)

Thus, by the mean value theorem,

\(\displaystyle{\frac{{{\log{{\left({1}+{x}^{{2}}\right)}}}-{\log{{\left({1}+{y}^{{2}}\right)}}}}}{{{x}-{y}}}}={f}'{\left({z}\right)}\)

for \(\displaystyle{z}\) between \(\displaystyle{x}\) and \(\displaystyle{y}\) (assuming \(\displaystyle{x}\ne{y}\) or the inequality is obvious). Then

\(\displaystyle{\left|{\frac{{{\log{{\left({1}+{x}^{{2}}\right)}}}-{\log{{\left({1}+{y}^{{2}}\right)}}}}}{{{x}-{y}}}}\right|}={\left|{f}'{\left({z}\right)}\leq{1}\right|}\)

\(\displaystyle{f}'{\left({x}\right)}={\frac{{{2}{x}}}{{{1}+{x}^{{2}}}}}\)

and

\(\displaystyle{\left|{\frac{{{2}{x}}}{{{1}+{x}^{{2}}}}}\right|}\leq{1}\)

because

\(\displaystyle{2}{\left|{x}\right|}\leq{1}+{\left|{x}\right|}^{{2}}\)

since

\(\displaystyle{\left({1}−{\left|{x}\right|}\right)}^{{2}}\geq{0}\)

Thus, by the mean value theorem,

\(\displaystyle{\frac{{{\log{{\left({1}+{x}^{{2}}\right)}}}-{\log{{\left({1}+{y}^{{2}}\right)}}}}}{{{x}-{y}}}}={f}'{\left({z}\right)}\)

for \(\displaystyle{z}\) between \(\displaystyle{x}\) and \(\displaystyle{y}\) (assuming \(\displaystyle{x}\ne{y}\) or the inequality is obvious). Then

\(\displaystyle{\left|{\frac{{{\log{{\left({1}+{x}^{{2}}\right)}}}-{\log{{\left({1}+{y}^{{2}}\right)}}}}}{{{x}-{y}}}}\right|}={\left|{f}'{\left({z}\right)}\leq{1}\right|}\)