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Math Help - Prove that |v+u| is less than or equal to |v| + |u|

  1. #1
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    Prove that |v+u| is less than or equal to |v| + |u|

    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?
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    Re: Prove that |v+u| is less than or equal to |v| + |u|

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  3. #3
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    Re: Prove that |v+u| is less than or equal to |v| + |u|

    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.
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