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Math Help - Proof inequality

  1. #1
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    Proof inequality

    For all real numbers a, b, c, prove that:

    a(a - b) + b(b - c) + c(c - a) >= 0.



    Thanks!
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  2. #2
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    Quote Originally Posted by Ideasman View Post
    For all real numbers a, b, c, prove that:

    a(a - b) + b(b - c) + c(c - a) >= 0.
    Open paranthesis,
    a^2 - ab + b^2 - bc + c^2 - ac \geq 0
    Iff,
    a^2+b^2+c^2 \geq ab+bc+ac
    Note, by Cauchy-Swartz
    a^2+b^2+c^2 \geq |ab+bc+ac|
    But,
    |ab+bc+ac|\geq ab+bc+ac
    So,
    a^2+b^2+c^2 \geq ab+bc+ac
    And the inequality is true.
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  3. #3
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    You'd have to prove the cauchy-shwartz thm wouldn't you?

    I don't see how

    a^2 + b^2 + c^2 >= |ab + bc + ac|
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    You'd have to prove the cauchy-shwartz thm wouldn't you?

    I don't see how

    a^2 + b^2 + c^2 >= |ab + bc + ac|
    It's a well-known theorem, so i don't suppose you would have to prove it. However, if it makes you feel better, you shouldn't have much trouble finding a proof on the internet, or maybe TPH would like to give you one
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  5. #5
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    Quote Originally Posted by Ideasman View Post
    You'd have to prove the cauchy-shwartz thm wouldn't you?

    I don't see how

    a^2 + b^2 + c^2 >= |ab + bc + ac|
    We know that,
    \ \left(\sum_{k=1}^n a_k^2\right) \cdot \left( \sum_{k=1}^n b_k^2 \right)  \geq \left( \sum_{k=1}^n a_kb_k \right)^2

    Let, a_1=a , \ a_2 = b , \ a_3 = c
    And, b_1 = b, \ b_2 = c, \ b_3 = a
    With n=3

    Then,
    (a^2+b^2+c^2)^2=(a^2+b^2+c^2)(b^2+c^2+a^2) \geq (ab+bc+ac)^2

    Take square roots, and note that \sqrt{(ab+bc+ac)^2} = |ab+bc+ac|

    So,
    a^2+b^2+c^2 \geq |ab+bc+ac|
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