# Triangle inequality proof

by chocolatelover
Tags: inequality, proof, triangle
 P: 1,633 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.