Lipschitz constant. How to show this inequality?

Question

Show that $f_{n}:[0,1]\rightarrow \mathbb{R}$, $f_{n}=\sqrt{x+1/n}$, has $\frac{\sqrt{n}}{2}$ as lipschitz constant.

That is, show that $\left|\sqrt{x+1/n}-\sqrt{y+1/n}\right|\leq \frac{\sqrt{n}}{2}|x-y| ,\forall x.y\in[0,1]$ and $n\in \mathbb{N}$.

I've tried it in various ways, using the norm of the difference of $| x-y |\leq 1$, triangular inequality, other enhancements but I got nowhere.

Answer 1

it is $$\sqrt{x+1/n}-\sqrt{y+1/n}=\frac {x+1/n-y-1/n}{\sqrt{x+1/n}+\sqrt{y+1/n}}$$ and therefore $$\frac{|x-y|}{\sqrt{x+1/n}+\sqrt{y+1/n}}\le \frac{|x-y|}{2\sqrt{\frac{1}{n}}}$$

Answer 2

Let $f(x) = \sqrt{x+1/n}$.

$$f'(x)= \frac{1}{2\sqrt{x+1/n}}=\frac{\sqrt{n}}{2\sqrt{xn+1}} \leq \frac{\sqrt{n}}{2}$$

Answer 3

If $x,y\in [0,1]$, then $\sqrt{x + \frac{1}{n}} + \sqrt{y + \frac{1}{n}} \ge \sqrt{\frac1{n}} + \sqrt{\frac1{n}} = \frac{2}{\sqrt{n}}$. Using the identity

$$\lvert \sqrt{x + \frac{1}{n}} - \sqrt{y + \frac{1}{n}}\rvert = \frac{\lvert x - y\rvert}{\sqrt{x + \frac{1}{n}} + \sqrt{y + \frac{1}{n}}}$$ obtained by rationalizing the numerator, you can now prove the result.

Answer 4

Hint: Use the mean value theorem.