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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.

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Check out…, 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
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 $c=1$.

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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

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