how do we get this inequality: $\frac{|x-y|}{\sqrt{x}+\sqrt{y}} \leq \frac{1}{2}|x-y|$?

Asked May 28 '14 at 13:05

Active May 28 '14 at 13:05

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To show that $f(x)=\sqrt{x}$ is Lipshitz continuous,we do it like that:

$$|f(x)-f(y)|=|\sqrt{x}-\sqrt{y}|=\frac{|x-y|}{\sqrt{x}+\sqrt{y}} \leq \frac{1}{2}|x-y|$$

But...how do we get this inequality: $\frac{|x-y|}{\sqrt{x}+\sqrt{y}} \leq \frac{1}{2}|x-y|$ ?Do we knoe that $\sqrt{x}+\sqrt{y} \geq 2$ ?

asked May 28 '14 at 13:05

evinda

4

$f$ is not Lipschitz continuous on $[0, 1]$. I believe you're taking $x, y \in [1, \infty)$. – Michael Albanese May 28 '14 at 13:08
A ok...!!In ny notes,there is no interval,so I thought that it would be $[0,+\infty)$,but maybe it is meant $x \geq 1$.. – evinda May 28 '14 at 13:11

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