Im trying to check if $d_1(x,y)=2|x-y|$ and $d_2(x,y)=|x-y|^2$ are metric spaces.
Im just not sure how to proceed with checking the triangle inequality property $d(x,y)\le d(x,z)+d(z,y)$. Is what I did below sufficient?
$d(x,y) = 2|x-y| =2|x-z+z-y| \le 2|x-z|+2|z-y| = d(x,z)+d(z,y)\\ \text{and} \\ d(x,y)=|x-y|^2=|x-z+z-y|^2\le |x-z|^2+|z-y|^2=d(x,z)+d(z,y)?$
Also is $d(x,y)=|\arctan(x)-\arctan(y)|$a metric space? Can I write
$d(x,y)=|\arctan(x)-\arctan(y)| = |\arctan(x)-\arctan(z)+\arctan(z)-\arctan(y)| \le |\arctan(x)-\arctan(z)|+|\arctan(z)-\arctan(y)|$