Proving that $|a+b| + |a-b| \geq |a| + |b|$ for the absolute value function

Question

I'm trying to prove the property $|a+b| + |a-b| \geq |a| + |b|$ of the absolute value function. I already know the triangle inequality $|x+y| \leq |x| + |y|$ and that with this I have $$|x| = |y + (x-y)| \leq |y| + |x-y|$$ so that $$|x|-|y| \leq |x-y|$$ and analogous $$|y|-|x| \leq |y-x|.$$ Choosing $|x| = |-y + (x+y)|$ instead gives me $$|x|-|y| \leq |x+y|$$ and $$|y|-|x| \leq |x+y|.$$ But how can I proceed?

Answer 1

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

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

$$ 2(|a+b|+|a-b|)\ge2|a|+2|b|$$

$$ |a+b|+|b-a|\ge|a|+|b|$$