metric $d(a,b)=2\frac{|a-b|}{a+b},a,b>0$ satisfies triangle inequality

Question

How to show that metric $d(a,b)=2\frac{|a-b|}{a+b},a,b>0$ satisfies triangle inequality? That is, $$d(a,b)+d(b,c)\geq d(a,c)$$ for $a,b,c>0$ My thought: $a$ and $c$ are symmetric, and we can assume $a<c$. Discuss the following three cases respectively:

• $b<a<c$
• $a<b<c$
• $a<c<b$

In each case, we can strip off the absolute symbol and use brute force method to show the inequality holds.

Is there any elegant way to prove this equality?

Answer 1

Let $a\geq b\geq c$.

Hence, $$\frac{a-c}{a+c}-\frac{b-c}{b+c}=\frac{2c(a-b)}{(a+c)(b+c)}\geq0$$ and $$\frac{a-c}{a+c}-\frac{a-b}{a+b}=\frac{2a(b-c)}{(a+b)(a+c)}\geq0.$$ Thus, it's enough to prove that $$\frac{a-b}{a+b}+\frac{b-c}{b+c}\geq\frac{a-c}{a+c}$$ or $$\sum_{cyc}\frac{a-b}{a+b}\geq0$$ or $$\frac{(a-b)(a-c)(b-c)}{\prod\limits_{cyc}(a+b)}\geq0,$$ which is obvious.