Assuming $x,y,z$ are complex numbers, or vectors, prove $$\|x\|+\|y\|+\|z\|+\|x+y+z\| \ge \|x+y\|+\|y+z\|+\|z+x\|$$
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: $$2(\|a\|+\|b\|+\|c\|)\le\|a+b+c\|+\|a+b-c\|+\|b-a+c\|+\|b-a-c\|$$
My hunch is that showing that $d(r,s)=\|r\|+\|s\|-\|r+s\|$ is a distance might help using $d(r,s)\le d(r,t)+d(t,s)$.
I found these links that seem to have the solution: https://mathoverflow.net/questions/167685/absolute-value-inequality-for-complex-numbers
After squaring of the both sides we need to probe that: $$\sum_{cyc}\left(\|x\|\|x+y+z\|+\|y\|\|z\|\right)\geq\sum_{cyc}\|x+y\|\|x+z\|,$$ which is true by the triangle inequality for complex numbers: $$\|x\|\|x+y+z\|+\|y\|\|z\|=\|x^2+xy+xz\|+\|yz\|\geq$$ $$\geq\|x^2+xy+xz+yz\|=\|x+y\|\|x+z\|.$$ It seems that for vectors it's not true.