Prove that |v+u| is less than or equal to |v| + |u| wher v and u are complex numbers.
My attempt:
I though I would substitute v = a + bi and u = x + yi to the equation |v+u| > |v| + |u| and try to find a contradiction.
I got to 2abxy > (ay)^2 + (bx)^2 but this doesn't seem to be obviously false. Is my approach correct or is there a better way to prove this?
Your approach *will* work, you just quit a little early. Write:
2abxy = (ay)(bx) + (ay)(bx), so that we have:
(ay)(bx) + (ay)(bx) > (ay)(ay) + (bx)(bx), or:
(ay)(bx - ay) + (ay - bx)(bx) > 0 and, re-arranging:
(-ay)(ay - bx) + (bx)(ay - bx) > 0, that is:
(bx - ay)(ay - bx) = -(ay - bx)^{2} > 0, which is clearly false, since the negative of a square is always less than or equal to 0.