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.

Very Simple Trick: We have that 
\begin{split} (|a|-|b|)^2 +2|a^2-b^2| \ge 0&\Longleftrightarrow & a^2+b^2 -2|a||b|+ 2|a +b||a-b| \ge 0\\
&\Longleftrightarrow & a^2+b^2 + 2|a +b||a-b|\ge 2|a||b|\\
&\Longleftrightarrow&  \color{red}{2a^2+2b^2} + 2|a +b||a-b|\ge  \color{red}{a^2+b^2}+2|a||b|\\
&\Longleftrightarrow&  (|a +b|+|a-b|)^2 \ge  (|a|+|b|)^2\\
&\Longleftrightarrow&  |a +b|+|a-b| \ge  |a|+|b|\end{split}
Given that $$\color{red}{ (|a +b|+|a-b|)^2 = 2a^2+2b^2 + 2|a +b||a-b|}$$
