Prove that $||a|-|b||\leq |a-b|$ for all real numbers

Question

I was thinking divide it into $a\geq b$ and $a<b$, but then I realized I need to include situations when they are greater than zero and less than zero, will that too comlicated? Is there a simpler way to show this?

Work backwards. Undo the absolute value bars on the LHS and consider the two resulting inequalities. Can you see how to proceed now? — Sri-Amirthan Theivendran, Feb 07 '16 at 23:05
@FoobazJohn You mean like this $-||a|-|b||\leq |a|-|b|\leq ||a|-|b||$ and $|a|-|b|\leq |a-b|$? But this can only show $||a|-|b||=|a-b|$ — mathmathmath, Feb 07 '16 at 23:09
Anyway, considering the various cases works. This is actually a very natural thing to try. — Giuseppe Negro, Feb 07 '16 at 23:12

score 5 · Accepted Answer · answered Feb 07 '16 at 23:07

5

With a little trick one has $$|a|= |a-b+b| \leq |a-b| + |b|.$$

Hence $$|a|-|b| \leq |a-b|.$$ With the same trick applied to $|b|$ you find $$|b|-|a| \leq |a-b|.$$ Therefore $$||a|-|b||\leq |a-b|.$$

answered Feb 07 '16 at 23:07

C. Dubussy

9,120

Prove that $||a|-|b||\leq |a-b|$ for all real numbers

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