Prove that $\sqrt x$ is Lipschitz on $[1, \infty)$ Prove that $\sqrt x$ is Lipschitz on $[1, \infty)$
I want to show that $|f(x) - f(y)| \leq L |x - y|$
So $|\sqrt x  - \sqrt y| = \frac{|x - y|}{\sqrt x + \sqrt y} \leq \frac{1}{2}|x - y|$.
I can see that to get $|\sqrt x  - \sqrt y| = \frac{|x - y|}{\sqrt x + \sqrt y}$ I just multiply the first expression by its conjugate. But I don't really understand how $\frac{|x - y|}{\sqrt x + \sqrt y} \leq \frac{1}{2}|x - y|$ How does $\frac{1}{\sqrt x + \sqrt y} \leq \frac{1}{2}$?

We know $x,y \in [1, \infty)$ thus both $x,y \geq 1$.
Now we can see $$\sqrt{x} + \sqrt{y} \geq 1 + 1 =2$$
and so we know $$\frac{1}{\sqrt{x} + \sqrt{y}} \leq \frac{1}{2}$$
because both values are positive.
Hence all that remains for us is to multiply through by $|x-y|$ and we get $$\frac{|x-y|}{\sqrt{x} + \sqrt{y}} \leq \frac{|x-y|}{2}$$
