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.

Here's the way I see this geometrically in $\mathbb{C}$: let's say we have two complex numbers $a, b$ and consider the parallelogram formed by $0, a, a + b, b$. The midpoints of the diagonals coincide at the point $\frac{a + b}{2}$. These diagonals cut the parallelogram into four triangles, on each of which we can perform the triangle inequality. We get the following inequalities:
\begin{align*}
|a - 0| &\le \left| a - \frac{a + b}{2} \right| + \left| \frac{a + b}{2} - 0 \right| \\
|(a + b) - a| &\le \left| (a + b) - \frac{a + b}{2} \right| + \left| \frac{a + b}{2} - a \right| \\
|b - (a + b)| &\le \left| b - \frac{a + b}{2} \right| + \left| \frac{a + b}{2} - (a + b) \right| \\
|0 - b| &\le \left| a - \frac{a + b}{2} \right| + \left| \frac{a + b}{2} - 0 \right|
\end{align*}
Simplifying the above inequalities and summing them up yields the desired inequality.
