Is sum of two metrics a metric?

Question

The production of two metrics is a metric also. It's googled easy. But what's about a sum? As I can see sum is metric, as the triangle inequality of metric sum is the consequence of the inequality feature and two other axioms looks obvious. Is it correct?

Answer 1

In short, yes. Your sketched argument is correct for the triangle inequality (we're just adding two valid inequalities to get a third). One does also need to remark that the sum of two numbers $\ge 0$ can only be $0$ if both summands are $0$ (and then we apply the axiom for the two constituent metrics after that).

Answer 2

For sum of metrics it's not hard exercise I think:

1. Let $\rho(x,y) = \rho_1(x,y) + \rho_2(x,y)$ and $\rho_1, \rho_2$ are metrics.

• x=y: $\rho(x,y)=\rho_1(x,y)+ \rho_2(x,y)=0+0=0$
• $\rho(x,y)=\rho_1(x,y)+ \rho_2(x,y)=\rho_1(y,x)+\rho_2(y,x)=\rho(y,x)$
• We know that $\rho_1(x,y)\leq\rho_1(x,z)+ \rho_1(z,y)$ and $\rho_2(x,y)\leq\rho_2(x,z)+ \rho_2(z,y)$ It means that $\rho_1(x,y)+ \rho_2(x,y) \leq\rho_1(x,z)+\rho_1(z,y)+\rho_2(x,z)+\rho_2(z,y)= \rho_1(x,z)+\rho_2(x,z)+\rho_1(z,y)+\rho_2(z,y) = (\rho_1(x,z)+\rho_2(x,z))+(\rho_1(z,y)+\rho_2(z,y))=\rho(x,z)+\rho(z,y)$