Assuming x,y,z are complex numbers, or vectors, prove
|x|+|y|+|z|+|x+y+z| >= |x+y|+|y+z|+|z+x|
I've tried replacing x+y, y+z, z+x with a, b, c respectively to see if it is any easier to prove it and this is the new form:
2(|a|+|b|+|c|) <=|a+b+c|+|a+b-c|+|b+c-a|+|c+a-b|
My hunch is that showing that $d(r,s)=|r|+|s|-|r+s| is a metric space might help using d(r,s) <= d(r,t)+d(t,s).
|x|+|y|+|z|+|x+y+z| >= |x+y|+|y+z|+|z+x|
I've tried replacing x+y, y+z, z+x with a, b, c respectively to see if it is any easier to prove it and this is the new form:
2(|a|+|b|+|c|) <=|a+b+c|+|a+b-c|+|b+c-a|+|c+a-b|
My hunch is that showing that $d(r,s)=|r|+|s|-|r+s| is a metric space might help using d(r,s) <= d(r,t)+d(t,s).