Prove inequality so I can use Banach fixed point thrm

Question

I have the function $\varphi(x)=\sqrt{2-\sqrt{4-x^2}}$. I want to prove that $d(\varphi(x),\varphi(y)\leq k\cdot d(x,y)$ for some $k,\ \forall x,y$. I thoight it would be simple, just by saying that
$\left|\sqrt{2-\sqrt{4-x^2}}-\sqrt{2-\sqrt{4-y^2}}\right|\leq k\cdot |x-y|$ and for x>y I started to compare only $\sqrt{2-\sqrt{4-x^2}}-\sqrt{2-\sqrt{4-y^2}}=x-y$. But I tried everything, rationalizing, multiplying by the compose, squaring, and I don't come any further. I would be really glad if someone could tell me if there is a trick to it or something.
This is as far as I go...
$\frac{\sqrt{4-y^2}-\sqrt{4-x^2}}{\sqrt{2-\sqrt{4-x^2}}+\sqrt{2-\sqrt{4-y^2}}}=x-y $
$\frac{x^2-y^2}{(\sqrt{2-\sqrt{4-x^2}}+\sqrt{2-\sqrt{4-y^2}})(\sqrt{4-y^2}+\sqrt{4-x^2})}=x-y \Leftrightarrow \frac{x+1}{(\sqrt{2-\sqrt{4-x^2}}+\sqrt{2-\sqrt{4-y^2}})(\sqrt{4-y^2}+\sqrt{4-x^2})}=1 $

Answer 1

[edited]

The domain of $\varphi$ is $[-2,2]$ and its only fixed point is $z=0$. If I understood, your question concerns the convergence of $x_{n+1} = \varphi(x_n), \quad x_0 \in [-2,2]$, to the fixed point $z=0$.

You can start by noting that $\varphi([-2,2]) = [0,\sqrt{2}]$ and so, in terms of convergence, we can suppose right form the start that this is all about the interval $I=[0, \sqrt{2}]$.

The sequence $(x_n)$ is bounded (all terms belong to $I$) and decreasing (assuming, without loss of generality that $x_0 \in I$) and so it is convergent. There is no need for the fixed point theorem.

if you really want to use the fixed point theorem, you can still use this idea that the original domain can be restricted and proof the boundness of $\varphi'$ in $I$. This way you can conclude that $\varphi$ is Lipschitz on $I$, with $k = \frac{1}{2 \sqrt{2-\sqrt{2}}}<1$.