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Math Help - Algebra: Inequalities

  1. #1
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    Algebra: Inequalities

    The question states this:

    bc(b+c) + ca(c+a) + ab(a+b) > (or equal to) 6abc

    These type of question are quite hard for me, as i do not know how to approach it. In easier variations, such as a^2+b^2 >= 2ab, it is easily recognizable. But not so with these questions

    If anyone could enlighten me on their process of thought, along with solution, it would be very appreciated.
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  2. #2
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    pickslides's Avatar
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    Are there any restrictions on a,b and c?

    i.e. a,b,c \in \mathbb{Z^+}
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Look at this:

    (a+b)(b+c)(a+c)=bc(b+c) + ca(c+a) + ab(a+b)+2abc


    We can notice that if we prove the following we are done.

    Prove for: a,b,c \in \mathbb{Z^+}<br />

    (a+b)(b+c)(a+c)\geq 8abc


    It easy to prove! Try it!
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  4. #4
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    Ok thxs.

    Just with the solutions Also sprach Zarathustra; did you just know the identity in the first line from experience? And also, to prove the first line, i found a solution, that followed your method. But in the given question
    "bc(b+c) + ca(c+a) + ab(a+b) > (or equal to) 6abc",
    is actually the first part, then followed few questions later by (a+b)(b+c)(a+c) >= 8abc. So im am wondering if there is another way to prove the original question?

    And to pickslides; im not sure if thats the conditions since the question dosent present it. However, think you and Also sprach Zarathustra are right and that would be the conditions of the a,b,c.

    EDIT:
    perhaps using a^2+b^2 >= 2ab as hinted by question.
    Last edited by Lukybear; July 8th 2010 at 11:15 PM.
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Proving of (a+b)(b+c)(a+c)\geq 8abc<br />

    a+b\geq 2\sqrt{ab}

    and:


    b+c\geq 2\sqrt{bc}

    and:

    a+c\geq 2\sqrt{ac}


    Now, by multiplying these three inequalities above, you will get the needed!
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  6. #6
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    O no i ve got that. And i really aprreciate your help.

    But is there a way to prove bc(b+c) + ca(c+a) + ab(a+b) > (or equal to) 6abc without first proving the above?
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Lukybear View Post
    O no i ve got that. And i really aprreciate your help.

    But is there a way to prove bc(b+c) + ca(c+a) + ab(a+b) > (or equal to) 6abc without first proving the above?
    Maybe, with more sophisticated tools...

    BUT, the way how I proved it is the most natural way!

    I simplified the original expression and continue from there...


    Experience is the key word!

    I f you want I'll send you a some question this type...(for practice)
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  8. #8
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    Yes that would be absolutely wonderful.
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