Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 7 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 $0<f(y)-f(x)=\displaystyle(y-x)(1-\frac1{xy})<y-x$.

share|cite|improve this answer

Use the mean value theorem. See Example 3.4 in

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.