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Math Help - Proof q.

  1. #1
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    Proof q.

    a, b, c are three positive numbers, no two of which are equal. Prove that
    a^2 - ab + b^2 > ab
    a^3 + b^3 > ab (a + b)
    2(a^3 + b^3 + c^3) > ab (a + b) + bc (b + c) + ca (c + a)
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  2. #2
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    Quote Originally Posted by Detanon View Post
    a, b, c are three positive numbers, no two of which are equal. Prove that
    a^2 - ab + b^2 > ab
    This is equivalent to (a-b)^2 > 0
    a^3 + b^3 > ab (a + b)
    Multiply the preceding inequality by a+b
    2(a^3 + b^3 + c^3) > ab (a + b) + bc (b + c) + ca (c + a)
    From the preceding,
    a^3 + b^3 > ab \; (a + b)
    a^3 + c^3 > ac \; (a + c)
    b^3 + c^3 > bc \; (b + c)
    (Just substituting different variables)
    Now add.
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  3. #3
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    Why is <br />
(a-b)^2 > 0<br />
??
    Is it because its a perfect square or something so its greater than zero?

    I dont really get the second steps either.. what did you do with a^3 + b^3 > ab (a + b)
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  4. #4
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    Quote Originally Posted by Detanon View Post
    Why is <br />
(a-b)^2 > 0<br />
??
    Is it because its a perfect square or something so its greater than zero?
    The square of any non-zero number is positive. We know a - b \neq 0 because we were told a \neq b.

    I dont really get the second steps either.. what did you do with a^3 + b^3 > ab (a + b)
    Multiply a^2 -ab +b^2 by a+b. What do you get?
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  5. #5
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    a^3 + b^3? But why do you multiply a^2 -ab +b^2 by a+b. Can you show all the steps please.
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  6. #6
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    Quote Originally Posted by Detanon View Post
    a^3 + b^3? But why do you multiply a^2 -ab +b^2 by a+b. Can you show all the steps please.
    I have listed all the steps.

    I suggest you go back and re-read my first post.
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