Show that $|T(x) - T(y)| \lt |x-y|$ when $x \neq y$ but the mapping has no fixed points.

Question

Consider $X = \{ x \in \mathbb{R} \mid 1 \le x \lt \infty \}$, taken with the usual metric of the real line, and $T \colon X \to X$ defined by $x \mapsto x +x^{-1}$. Show that $|T(x) - T(y)| \lt |x-y|$ when $x \neq y$ but the mapping has no fixed points.

Any help with this question would be greatly appreciated as I really am having trouble just figuring out where to start. Thank you.

Answer 1

Hint:

Let $f(x) = x + 1/x$, caculate $|f(x) - f(y)|$;

If $f(x) = x $ for some $x$, then what happnes to equation?