# Is triangle inequality for product space can follow from Minkowski inequality?

Below question is asked so many times, just like here but I need to know that can we prove triangle inequality by using Minkowski inequality for sum.

Question: let $$X=X_1\times X_2$$ be Cartesian product of two metric spaces $$(X_1,d_1)$$ and $$(X_2,d_2)$$. Is the function $$d$$ defined by

$$d(x,y)=\sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}$$ is metric on $$X$$?

My attempt: we need to prove,

$$d(x,y)≤d(x,z)+d(z,y)$$

i.e. $$\sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}≤ \sqrt{d_1(x_1,z_1)^2+d_2(x_2,z_2)^2} +\sqrt{d_1(z_1,y_1)^2+d_2(z_2,y_2)^2}$$

Which follows directly from Minkowski's inequality for sum. Because,

$$d_1(x_1,y_1),d_1(x_2,y_2),d_1(x_1,z_1)d_1(x_2,y_2),d_1(z_1,y_1),d_2(z_2,y_2)$$ are positive real numbers.

is am i correct?

You have to first apply triangle in equality for $$d_1$$ and $$d_2$$ and then use Minkowski's inequality.
• Sir, thanks for the answer. ( Yes I forget to write after applying triangle inequality for $d_1$, $d_2$) – Akash Patalwanshi Jul 18 '19 at 5:29
• Consider the vectors $(d_1(x_1,z_1),d_2(x_2,z_2)$ and $(d_1(z_1,y_1),d_2(z_2,y_2)$. If these are $(a,b)$ and $(c,d)$ then $\|(a,b)+(c,d)\| \leq \|(a,b)\|+\|(c,d\|$ where $\|.\|$ denotes the Euclidean norm on $\mathbb R^{2}$. This gives the answer immediately. – Kavi Rama Murthy Jul 18 '19 at 6:41