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Prove that |A-B| >= |A| - |B|

Note that A and B are vectors, and that |A| means magnitude of vector A.

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I have already proved this using a cheap trick, but I want to prove it using geometry instead. My question is about what cases I would need. Meaning that since there is a triangle formed (if I put A, B, and A-B together) what would I need to do in order to provide a sufficient geometric proof.

I was thinking of having A - B form a triangle and having a triangle with an acute angle between A and B, and then having a triangle with an obtuse angle between A and B, and then having A and B form a straight line. And of course showing on each of these that |A-B| >= |A| - |B|

Would that be sufficient?

Thanks.