I have tried out induction, there are some Formulars of phi that I have understood. For example $\varphi=1+\varphi^{-1}$ and $\varphi^2-\varphi=1$.
This is my attempt so far. I want to Show the inequality because then $x_n\rightarrow\varphi$ directly follows:
Inductionbase:
$\varphi-1=\varphi^{-1}$
The inductionbase is proved. Now the inductionstep:
$\varphi^{-(n+1)}=\varphi^{-1}\varphi^{-n}\overset{\text{IH}}{\geq}\varphi^{-1}|\varphi-x_{n-1}|$
What can I do now? I am stuck
For $n=0$, the inequality is obviously true. To make the induction work, suppose that $$ |\varphi-x_k|\leq \varphi^{-(k+1)}\Leftrightarrow \varphi-\varphi^{-(k+1)} \leq x_k \leq \varphi+\varphi^{-(k+1)}. $$ We then have $$ |\varphi-x_{k+1}|=|\varphi-1-x_{k}^{-1}|=|\varphi^{-1}-x_{k}^{-1}|. $$ Here things starts to get tricky because the thing inside abosolute value can be either positive or negative. When it is positive, $$ |\varphi^{-1}-x_{k}^{-1}|=\varphi^{-1}-x_{k}^{-1}\leq \varphi^{-1}-\frac{1}{\varphi+\varphi^{-(k+1)}}=\varphi^{-1}-\varphi^{-1}\frac{1}{1+\varphi^{-(k+2)}}\\ =\varphi^{-1}(1-(1+\varphi^{-(k+2)})^{-1})=\varphi^{-1}(1-1+\varphi^{-(k+2)}-\varphi^{-2(k+2)}+\ldots)\leq \varphi^{-1}\varphi^{-(k+2)} \leq\varphi^{-(k+2)}. $$
Therefore the inequality is also true for $n=k+1$.