# triangle inequality proof

by chocolatelover
Tags: inequality, proof, triangle
 P: 239 1. The problem statement, all variables and given/known data Prove that |x+y|
 P: 1,635 since -|x|-(|x|+|y|)
 P: 1,635 Or we could prove it this way also: to prove it it is also sufficient to prove that $$|x+y|^{2}\leq (|x|+|y|)^{2} ???$$ so we have $$|x+y|^{2}=(x+y)^{2}=x^{2}+2xy+y^{2}=|x|^{2}+2xy+|y|^{2}<|x|^{2}+2|x||y| +|y|^{2}=(|x|+|y|)^{2}$$ hence we are done.
P: 239

## triangle inequality proof

Thank you very much

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P: 1
 Quote by sutupidmath -(|x|+|y|)
Why do absolute value properties yield to |x+y|<|x|+|y| from -(|x|+|y|)<x+y<|x|+|y|? Could someone please explain this further?
 HW Helper P: 3,353 I don't think its valid in general to add two inequalities like that. But if you accept up to there, then to see the final step is just seeing that if -A < a < A, then we can say |a| < A.
 HW Helper P: 3,353 Correction and sincere apologies to sutupidmath, you CAN add inequalities as long as they are pointing the same same direction which in this case in true. And that goes to make an extremely simple proof!
 P: 15 I think vrdfx wasn't asking how -|x| -(|x|+|y|)

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