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I have attempted as follows: $|f(x) - f(y)| = |x + 1/x - y - 1/y|$

$\leq |x - y| + |1/x - 1/y|$

Struck here. Any help.

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2 Answers 2

up vote 6 down vote accepted

Suppose $x,y>1$. If $x=y$ then both sides are 0 in your inequality, so I assume $x\ne y$.

Say that $x>y$. Then $f(x)-f(y)=(x-y)+(1/x-1/y)=\displaystyle (x-y)(1-\frac1{xy})$. Now, the fraction $1/(xy)$ is positive, but strictly less than 1, as $1<x,y$. Then the product is clearly positive but strictly less than $x-y$.

Similarly, if $y>x$, we get $0<f(y)-f(x)=\displaystyle(y-x)(1-\frac1{xy})<y-x$.

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Use the mean value theorem. See Example 3.4 in http://www.math.uconn.edu/~kconrad/blurbs/analysis/contractionshort.pdf.

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