Prove $|a+b|+|a-b| \geq |a|+|b|$

$2|a|=|a+b+a-b|\leq |a+b|+|a-b|$

$2|b|=|b+a+b-a|\leq |b+a|+|b-a|=|a+b|+|a-b|$

Add these inequalities and then halve.

Hint: If $a$ and $b$ have different signs, then their magnitudes are adding up in the same direction in $|a-b|$. Otherwise, i.e. if they have the same signs, this is happening in $|a+b|$. And the other term on the LHS ($|a+b|$ in the first case and $|a-b|$ in the second) is always positive. It is all in magnitudes when working with absolute values.

Denote $$ {x=a+b \\ y=a-b} $$ Then, using the triangle inequality, $$|a|=\dfrac{1}{2}(|x+y|)\leqslant\dfrac{1}{2}(|x|+|y|) \\ |b|=\dfrac{1}{2}(|x-y|)\leqslant\dfrac{1}{2}(|x|+|y|), $$ hence $$ |a|+|b|\leqslant |x|+|y|=|a+b|+|a-b|. $$