# Prove that |\log(1+x^2)-\log(1+y^2)|\leq|x-y|

Prove that $$\displaystyle{\left|{\log{{\left({1}+{x}^{{2}}\right)}}}-{\log{{\left({1}+{y}^{{2}}\right)}}}\right|}\leq{\left|{x}-{y}\right|}$$

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Annie Gonzalez
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
$$\displaystyle{\left|{\frac{{{\log{{\left({1}+{x}^{{2}}\right)}}}-{\log{{\left({1}+{y}^{{2}}\right)}}}}}{{{x}-{y}}}}\right|}={\left|{f}'{\left({z}\right)}\leq{1}\right|}$$
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Travis Hicks
HINT: using the mean value theorem for the function
$$\displaystyle{f{{\left({t}\right)}}}={\ln{{\left({1}+{t}^{{2}}\right)}}}$$
we get
$$\displaystyle{f{{\left({x}\right)}}}-{f{{\left({y}\right)}}}={\frac{{{2}{t}}}{{{1}+{t}^{{2}}}}}{\left({x}-{y}\right)}$$
and we have
$$\displaystyle{\left|{\frac{{{2}{t}}}{{{1}+{t}^{{2}}}}}\right|}\leq{1}$$
Vasquez

We can write the term of interest as
$$|\log(1+x^2)=\log(1+y^2)|=|\int^x_y\frac{2t}{1+t^2}dt|$$
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.