Proving that: $ | a + b | + |a-b| \ge|a| + |b|$ I am trying to prove this for nearly an hour now:

$$
\tag{$\forall a,b \in \mathbb{R}$}| a + b |  + |a-b| \ge|a| + |b|
$$

I'm lost, could you guys give me a tip from where to start, or maybe show a good resource for beginners in proofs ? 
Thanks in advance.

To prove
$$
| a + b |  + |a-b| \ge|a| + |b|
$$
Square both the sides. This does not change inequality. We have
$$
| a + b |^2  + |a-b|^2 + 2|a+b||a-b| \ge|a|^2 + |b|^2 + 2|a||b|
$$
$$
(|a|^2 + |b|^2 +2|a||b|cos\theta) + (|a|^2 + |b|^2 -2|a||b|cos\theta) + 2|a+b||a-b| \ge|a|^2 + |b|^2 + 2|a||b|
$$ where $\theta$ is angle between a and b
$$
2|a|^2 + 2|b|^2  + 2|a+b||a-b| \ge|a|^2 + |b|^2 + 2|a||b|
$$ 
$$
|a|^2 + |b|^2  + 2|a+b||a-b| \ge 2|a||b|
$$ 
$$
|a|^2 + |b|^2  + 2|a+b||a-b| - 2|a||b| \ge 0 
$$ 
$$
(|a|-|b|)^2  + 2|a+b||a-b|\ge 0 
$$ 
So on left hand side we have both terms which are always greater than 0, hence this inequality always holds
Equality exists when a=b
QED
