Show that $\sqrt{x+2}$ is a contraction Let $f:X \rightarrow X$ where $X=[0,\infty)$ be defined as $f(x)=\sqrt{x+2}$. I have to show that this mapping is a contraction and find its unique fixed point. The second part is easy: by the CMT, it has a unique fixed point in $X$ and it is $x^\ast = 2$. 
For $f$ being a contraction I wts the following: $ \exists \beta \in [0,1)$ such that
$\mid\sqrt{x+2}-\sqrt{y+2}\mid \leq \beta \mid x-y \mid, \ (\forall x,y\geq0)$ 
Since 
$\mid\sqrt{x+2}-\sqrt{y+2}\mid = \dfrac{\mid x-y \mid}{\sqrt{x+2}+\sqrt{y+2}} $
I'm tempted to set $ \beta = \dfrac{1}{\sqrt{x+2}+\sqrt{y+2}}$ but $\beta$ cannot depend on $x$ or $y$.... Any ideas about how to proceed? Thanks!
 A: Note that for all $x,y\geqslant 0$
$$\frac{1}{\sqrt{x+2}+\sqrt{y+2}}\leqslant \frac{1}{\sqrt{2}+\sqrt{2}}=\frac{\sqrt{2}}{4}$$
A: hint
Let $x,y\in [0,+\infty)$.
By MVT
$$f(x)-f(y)=(x-y)f'(c)$$
where $$0\le x<c<y.$$
$$f'(c)=\frac{1}{2\sqrt{c+2}}\le \frac{1}{2\sqrt{2}}$$
A: As per Salahamam's solution, to show a function is a contraction, it is sufficient to show that its derivative has $|f'(x)|<1$.
Prop. If $f(x)$ is a differentiable function function with $|f'(x)|<1$ for all $x$, then $f(x)$ is a contraction. 
Proof. Let $x,y \in \mathbb{R}$. By the Mean Value Theorem, we have $|f(x)-f(y)|= |f'(c)(x-y)|= |f'(c)||x-y|$ for some $c$ between $x$ and $y$. But as $|f'(c)|<1$ by assumption, we must have 
$$|f(x)-f(y)|=|f'(c)||x-y| < 1 \cdot |x-y|=|x-y|.$$
 Therefore, $f$ is a contraction. 
Note that the converse is false as contractions need not be differentiable.
So in your case, you only need show that $\sqrt{x+2}$ has bounded derivative. But $\dfrac{d}{dx} \; \sqrt{x+1}= \dfrac{1}{2\sqrt{x+2}}$ which is at most $\frac{1}{2\sqrt{2}}$ on the interval $[0,\infty)$.
