Prove |x|+|y|+|z|+|x+y+z| >= |x+y|+|y+z|+|z+x|

[astaneh]

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).

 

[Steven G]

You need to try a bit harder.

We know that -|x| <= x <= |x|
We also know -|y|<= y <= |y|.

Now what do you get when you add those two inequalities. Show us what you can do from there.

 

[astaneh]

-|x| <= x <= |x| is only true for real numbers for multi-dim vectors or complex numbers cannot compare x to |x|

 

[Steven G]

Why do students post their question in the title and do not include everything in the title!