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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?

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You have to first apply triangle in equality for $d_1$ and $d_2$ and then use Minkowski's inequality.

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  • $\begingroup$ Sir, thanks for the answer. ( Yes I forget to write after applying triangle inequality for $d_1$, $d_2$) $\endgroup$ – Akash Patalwanshi Jul 18 '19 at 5:29
  • $\begingroup$ Sir, I am still facing the problem :-( $\endgroup$ – Akash Patalwanshi Jul 18 '19 at 6:36
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    $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Jul 18 '19 at 6:41

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