For all x,y in R, |x+y| <= |x| + |y| Solution This is a different proof :- if po

Question

For all x,y in R, |x+y| <= |x| + |y|

Explanation / Answer

This is a different proof :- if possible, let |x+y| > |x| + |y|. Then as lhs and rhs are positive, (|x+y|)^2 > (|x| + |y|)^2 (x+y)^2 > |x|^2 + |y|^2 + 2|x||y|. x^2 + y^2 + 2xy > |x|^2 + |y|^2 + 2|x||y|. now x^2 and |x|^2 are equal also, y^2 and |y|^2. Hence 2xy > 2|x||y|. This cannot happen because, |x| > x and |y|>y and rhs is always positive while lhs takes positive and negative values and max value being |2xy| = 2|x||y|. Hence contradiction. So the inequality holds. Message me if you have any doubts. BTW The first proof is also correct, its a valid way of proving it.

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