How can I prove that this function: $ f: X \rightarrow X \text{ with } X = [1, \infty)$ (The metric is the absolute value)
$f(x) = x+\frac{1}{x}$ is not a contraction mapping? I think it isn't because I'd have to find an $L<1$ such that $d(f(x), f(y)) \leq L \cdot d(x, y)$ for all $x,y \in X.$ I don't think that's possible because for every L that I can find, there are always still $x,y \in X$ for which the inequality won't hold.
However, I was also not able to pick $x,y$ to prove the negation.
I have come so far: $$d(f(x), f(y)) = |x-y + \frac{1}{x}-\frac{1}{y}| = |(x-y)(1-\frac{1}{xy})| = (1-\frac{1}{xy})\cdot d(x,y)$$
Thank you very much!