Modulus of difference is bigger than modulus of difference of modulus

Question

Simple question:

Can anyone provide a formal proof for: $$|a-b| \geq ||a|-|b||.$$

Thanks.

Answer 1

It's the reverse triangle inequality. $$ |a|=|b+(a-b)|\leq|b|+|a-b|. $$ So $$\tag1 |a|-|b|\leq|a-b|. $$ Now repeat with the roles reversed to get $$\tag2 |b|-|a|\leq|b-a|=|a-b|. $$ Now combine $(1)$ and $(2)$ to get $$ |\,|a|-|b|\,|\leq|a-b|. $$

Answer 2

Since both sides are nonnegative, we can check that the inequality holds for the squares: $$ |a-b|^2\ge\bigl||a|-|b|\bigr|^2 $$ becomes $$ a^2-2ab+b^2\ge a^2-2|a|\,|b|+b^2 $$ that's equivalent to $$ ab\le |ab| $$ which is true.