Take the 2-minute tour ×
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.

I need to prove that $|x^{1/n}-y^{1/n}|\leq c|x-y|^{1/n}$, where $x,y\in [0,+\infty)$, $n\in\mathbb{N}-\{0,1\}$, and $c=2^{(n-1)/n}$.

I tried a lot ways for proving cases $n=2$ and $n=3$, but are quite different, I don't know how to prove the general case.

I've done case $n=2$ in the following way. I was prove that $|\sqrt{x}-\sqrt{y}|\leq \sqrt{|x-y|}$ squaring both sides when $x\geq y$.

The case $n=3$ I've done in a similar way. I was prove that $|\sqrt[3]{x}-\sqrt[3]{y}|\leq \sqrt[3]{|x-y|}$ cubing both sides, but is quite different for the case $n=2$. I don't know if it is possible prove using induction with a binomial expansion. I only prove both cases $n=1$, $n=2$ using $c=1\leq 2^{(n-1)/n}$, therefore I proved a better inequality for the first two cases.

share|improve this question
1  
Check out math.stackexchange.com/questions/143173/…, then try some change of variable to the inequality in above thread to get the inequality you want. –  Shuhao Cao May 13 '12 at 1:57
    
Is so similar inequalitty. Thanks I'll try it by this method. –  Gastón Burrull May 13 '12 at 2:18

1 Answer 1

up vote 1 down vote accepted

Suppose $a,b\geq 0$. Then expanding $(a+b)^n$ with the binomial formula gives $$a^n+b^n\leq (a+b)^n.\tag1$$

Without loss of generality, assume that $x\geq y$. Substitute $a=x^{1/n}-y^{1/n}$ and $b=y^{1/n}$ into (1) and take the $1/n$ th power to get your inequality with $c=1$.

share|improve this answer
    
Really easy your answer. Thanks a lot!!! –  Gastón Burrull May 13 '12 at 2:03
    
Glad to be of help. –  Byron Schmuland May 13 '12 at 2:07

Your Answer

 
discard

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.