I have to show $ T : X \to X$ given by $x \mapsto x/2 + 1/x$ is a contraction map, where $X = \{x \in R : x \ge 1 \}$ and find the smallest contraction constant.
I have worked out that $|T(x) - T(y)| \le |1/2 - 1/(xy)| | x - y|.$ I'm stuck with how to go about finding the smallest constant $\alpha \in (0, 1)$ such that $|T(x) - T(y)| \le \alpha | x - y|$ holds for every $x, y$ in $X$. Please help.
Think about the mean value theorem. For any $x <y$, $T(x) = T(y) + T'(c)(x-y)$ for some $c$ between $x$ and $y$. Then $T'(c) = \frac{1}{2} - \frac{1}{c^2}$. What's the largest $T'(c)$ could possibly be?