TITLE: Prove $|a+b|+|a-b| \geq |a|+|b|$
QUESTION [11 upvotes]: I am fighting with this proof-writing problem for a while. The statement says $$|a+b|+|a-b| \geq |a|+|b|.$$
I know the triangle inequality which says$$|a+b| \leq |a|+|b|.$$ 
How can I use this inequality to prove the statement above? I am adding the distance between $a$ and $b$ to $|a+b|$ which then turns the triangle inequality to another direction. How can I prove this in a mathematical way?

REPLY [4 votes]: I'm following up on what said potato about complex numbers:
Let $a$ and $b$ be two complex numbers. In the complex plane the points with affixes $0$, $a$, $b$ and $a+b$ form a parallelogram. Two diagonals of a parallelogram have a total length higher than the sum of the lentgh of two adjacent sides (use triangular inequality in two adjacent triangles). This means that: $|a-b| + |a+b| \geq |a| + |b|$.
In particular this holds if $a$ and $b$ are real (the parallelogramm is flat then).
This could be generalized to vectorial normed spaces. Hope this helps.