Use the triangle inequality to show that $|a|+|b| \leq |a+b|+|a-b|$
Hint: Note that
$2|a|=|a+b+a-b|\leq|a+b|+|a-b|$
$2|b|=|b+a+b-a|\leq|b+a|+|b-a|=|a+b|+|a-b|$
then
$$2|a|+2|b|\leq 2(|a+b|+|a-b|)$$ then $$|a|+|b|\leq |a+b|+|a-b|$$
An inelegant solution uses cases:
If $a \ge 0$ and $b \ge 0$ then $|a| + |b| = a+b = |a + b| \le |a+b| + |a-b|$.
If $a \ge 0$ and $b < 0$ then $|a| + |b| = a - b = |a-b| \le |a + b| + |a-b|$.
If $a < 0$ and $b \ge 0$ then $|a| + |b| = -a + b = |b-a| = |a-b| \le |a + b| + |a-b|$.
If $a< 0$ and $b < 0$ then $|a| + |b| = -a - b = |-a -b | = |a+b| \le |a+b| + |a-b|$.