Showing that $d(x,y)=|(1/x)-(1/y)|$ is a metric

Question

I am tasked with showing that $d(x,y)=|(1/x)-(1/y)|$ is a metric for all $x,y$ in $(0,\infty)$.

It is obvious that $d(x,y)=0$ iff $x=y$, and also that $d(x,y)=d(y,x)$. However, I am totally stumped on proving the triangular inequality. Any hints?

Answer 1

More generally, if $X$ is a set, and $(Y,\delta)$ a metric space and $f\colon X\to Y$ an injective function, then $d(x,y):=\delta(f(x),f(y))$ is a metric on $X$.

This be be easier to show than your the special case.

Answer 2

In general way, $d(x,y)=|f(x)-f(y)|$, for any injective "real" map $f$, is a metric.

For the triangular inequality, it follows from the triangular inequality for real numbers. In fact, we have for all $x,y,z \in \mathbb{R}$ we have $$d(x,z)=|f(x)-f(z)|=|(f(x)-f(y))+ (f(y)-f(z))|\leq |f(x)-f(y)|+|f(y)-f(z)|=d(x,y)+d(y,z).$$ In your case $f(x)=1/x$. More general case is discussed in the above answer.