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Simple question:

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

Thanks.

jpugliese
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  • How would you prove this using number-testing? – José Carlos Santos Jan 13 '18 at 22:10
  • José Carlos Santos, sorry, I was mistaken. Given the formulation of my sentence the phrase inside the parenthesis was unnecessary. – jpugliese Jan 13 '18 at 22:11
  • The word modulus here is technically correct but would be better to call it absolute value given that this has been tagged as a real analysis question. Modulus implies you are working with complex numbers. – David Reed Jan 13 '18 at 22:19
  • Thanks for the comment David. I'll make sure I make the distinction next time. – jpugliese Jan 13 '18 at 22:21

2 Answers2

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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 $$ |\,|a|-|b|\,|\leq|a-b|. $$

Martin Argerami
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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.

egreg
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