Prove that (triangle inequality)

Let a, b, c be real numbers, prove that:
|a| + |b| + |c| − |a + b| − |b + c| − |c + a| + |a + b + c| ≥ 0

I believe you have to use the triangle inequality (|a + b| ≤ |a| + |b|), yet I have no idea on how to go around it. Here's my "progress":
|a| + |b| + |c| − |a + b| − |b + c| − |c + a| + |a + b + c| ≥ 0
|a| + |b| + |c| + |a + b + c| − (|a + b| + |b + c| + |c + a|) ≥ 0
|a| + |b| + |c| ≥ |a + b| + |b + c| + |c + a| − |a + b + c|

LHS is always positive, and − |a + b + c| is always negative, so it is sufficient to prove that:
|a| + |b| + |c| ≥ |a + b| + |b + c| + |c + a| ?

I'm stuck, my textbook is by far too fast-paced. This is my first exposure to rigorous proofs, etc. Can anyone help me?

The Triangle Inequality says that:
|x + y| ≤ |x| + |y| <--[Ineq. 1]

So if we multiply each term by -1 and flip the inequality sign, we obtain:
−|x + y| ≥ −|x| − |y| <--[Ineq. 2]

Thus:
|a| + |b| + |c| − |a + b| − |b + c| − |c + a| + |a + b + c|
≥ |a| + |b| + |c| − |a| − |b| − |b| − |c| − |c| − |a| + |a + b + c| <--[By Ineq. 2]
= −|a| − |b| − |c| + |a + b + c| <--[Cancel out like terms.]
= −(|a| + |b|) − |c| + |a + b + c| <--[Factor out negative.]
≥ −|a + b| − |c| + |a + b + c| <--[By Ineq. 1, where x = a and y = b.]
= −(|a + b| + |c|) + |a + b + c| <--[Factor out negative.]
≥ −|a + b + c| + |a + b + c| <--[By Ineq. 1, where x = a + b and y = c.]
= 0, as desired.