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

Frank Guyton 2022-01-03 Answered
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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Expert Answer

Annie Gonzalez
Answered 2022-01-04 Author has 2153 answers
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
Answered 2022-01-05 Author has 1627 answers
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
\(\displaystyle{\left|{\frac{{{2}{t}}}{{{1}+{t}^{{2}}}}}\right|}\leq{1}\)
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Vasquez
Answered 2022-01-11 Author has 8850 answers

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.

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