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Math Help - Algebraic inequality

  1. #1
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    Algebraic inequality

    If a, b, c are real numbers and a + b + c = 0, prove that -1/2 \le ab + bc + ac \le 1.

    EDIT: Damned typo! The upper bound should be 1 instead of -1!
    Last edited by mathwizard; June 17th 2008 at 02:03 PM.
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  2. #2
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    Quote Originally Posted by mathwizard View Post
    If a, b, c are real numbers and a + b + c = 0, prove that -1/2 \le ab + bc + ac \le -1.
    a + b + c = 0 \Rightarrow a^2 + b^2 + c^2 = -2(ab+bc+ca)

    WE have to prove -1/2 \le ab + bc + ac \le -1 \Leftrightarrow 1 \ge -2(ab + bc + ac) \ge 2 \Leftrightarrow 1 \ge a^2 + b^2 + c^2 \ge 2

    But this is false, because we choose a= 1.03, b = -0.47, c = -0.56, a+b+c = 0, but a^2 + b^2 + c^2 = 1.594 < 2
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  3. #3
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    Hello,

    Quote Originally Posted by Isomorphism View Post
    a + b + c = 0 \Rightarrow a^2 + b^2 + c^2 = -2(ab+bc+ca)

    WE have to prove -1/2 \le ab + bc + ac \le -1 \Leftrightarrow 1 \ge -2(ab + bc + ac) \ge 2 \Leftrightarrow 1 \ge a^2 + b^2 + c^2 \ge 2

    But this is false, because we choose a= 1.03, b = -0.47, c = -0.56, a+b+c = 0, but a^2 + b^2 + c^2 = 1.594 < 2
    This is false because you can't have a number which is both superior to 2 and inferior to 1

    Like in #1, you can't have something that is both greater than -1/2 and smaller than -1.
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