Frank Guyton

2022-01-03

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

Annie Gonzalez

The derivative of $f\left(x\right)=\mathrm{log}\left(1+{x}^{2}\right)$ is
${f}^{\prime }\left(x\right)=\frac{2x}{1+{x}^{2}}$
and
$|\frac{2x}{1+{x}^{2}}|\le 1$
because
$2|x|\le 1+{|x|}^{2}$
since
${\left(1-|x|\right)}^{2}\ge 0$
Thus, by the mean value theorem,
$\frac{\mathrm{log}\left(1+{x}^{2}\right)-\mathrm{log}\left(1+{y}^{2}\right)}{x-y}={f}^{\prime }\left(z\right)$
for $z$ between $x$ and $y$ (assuming $x\ne y$ or the inequality is obvious). Then
$|\frac{\mathrm{log}\left(1+{x}^{2}\right)-\mathrm{log}\left(1+{y}^{2}\right)}{x-y}|=|{f}^{\prime }\left(z\right)\le 1|$

Travis Hicks

HINT: using the mean value theorem for the function
$f\left(t\right)=\mathrm{ln}\left(1+{t}^{2}\right)$
we get
$f\left(x\right)-f\left(y\right)=\frac{2t}{1+{t}^{2}}\left(x-y\right)$
and we have
$|\frac{2t}{1+{t}^{2}}|\le 1$

Vasquez

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

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