How would I prove (or disprove) $f(x)=\frac{1}{x}$ is not a contraction on $([1,\infty),|.|)$?
So I computed the derivative of $f(x)$ and realised that the modulus of the derivative is $1$ at $x=1$ so I deduced that this is not a contraction but I wanted to prove it in a more concrete way. So I chose to prove this by contradiction and tried to focus on the neighbourhood of $1$.
My attempt:
Assume for a contradiction that this is a contraction map then there exists $K\in(0,1)$ such that $|\frac{1}{x}-\frac{1}{y}|\leq K|x-y|$. Let $x=1$ so the inequality reduces to $1-\frac{1}{y}\leq K(y-1)$. So the problem boils down to finding a $y\in [1, \infty)$ such that the aforementioned inequality fails to hold. Moreover I found two intersections of $(1-\frac{1}{y})$ and $K(y-1)$ are $y=1$ and $y=\frac{K}{2}$ but I am not sure how to find such $y$ efficiently, or perhaps I am proving something that is incorrect?