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.