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Math Help - Prove positive real numbers a,b,c.

  1. #1
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    Prove positive real numbers a,b,c.

    Prove that for any three distinct positive real numbers a, b and c:
    ((a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3
    ----------------------------------------------------------- > 8abc
    (a-b)^3 + (b-c)^3 + (c-a)^3

    (where dotted line shows it is a fraction)
    I have no idea what to do here, help!!
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Re: Prove positive real numbers a,b,c.

    Quote Originally Posted by pikachu26134 View Post
    Prove that for any three distinct positive real numbers a, b and c:
    ((a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3
    ----------------------------------------------------------- > 8abc
    (a-b)^3 + (b-c)^3 + (c-a)^3

    (where dotted line shows it is a fraction)
    I have no idea what to do here, help!!
    there's an interesting property that if x+y+z=0 then x^3+y^3+z^3=3xyz.
    In the numerator take x=a^2-b^2,y=b^2-c^2,z=c^2-a^2
    in the denominator take x=a-b,y=b-c,z=c-a.
    then the LHS is equal to (a+b)(b+c)(c+a). now try to prove that for positive a,b,c, (a+b)(b+c)(c+a)>8abc. Its easy. Use AM>GM.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Prove positive real numbers a,b,c.

    Quote Originally Posted by abhishekkgp View Post
    there's an interesting property that if x+y+z=0 then x^3+y^3+z^3=3xyz.
    In the numerator take x=a^2-b^2,y=b^2-c^2,z=c^2-a^2
    in the denominator take x=a-b,y=b-c,z=c-a.
    then the LHS is equal to (a+b)(b+c)(c+a). now try to prove that for positive a,b,c, (a+b)(b+c)(c+a)>8abc. Its easy. Use AM>GM.

    First you take x=a^2-b^2 and then x=a-b... It makes no sense!
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    Senior Member abhishekkgp's Avatar
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    Re: Prove positive real numbers a,b,c.

    Quote Originally Posted by Also sprach Zarathustra View Post
    First you take x=a^2-b^2 and then x=a-b... It makes no sense!
    In order to use the formula x^3+y^3+z^3=3xyz if  x+y+z=0 you have to appropriately set x,y \, and \, z. we get
    (a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3= 3(a^2-b^2)(b^2-c^2)(c^2-a^2)
    (a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)
    now do you understand what i was trying to say??
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  5. #5
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    Re: Prove positive real numbers a,b,c.

    Im a little confused, are you trying to say that (a+b)/2 (b+c)/2 (c+a)/2 >(or equal) to square root ab * square root bc *square root ca?
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  6. #6
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    Re: Prove positive real numbers a,b,c.

    I'm still confused, i dont get what you mean!!
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  7. #7
    Senior Member abhishekkgp's Avatar
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    Re: Prove positive real numbers a,b,c.

    Quote Originally Posted by pikachu26134 View Post
    I'm still confused, i dont get what you mean!!
    okay then. for the time being forget about the question you have posted. Try to prove that if x+y+z=0 then  x^3+y^3+z^3=3xyz. Post the proof of this in your reply and i will tell you next step. This way you will understand what i was trying to say.
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  8. #8
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    Re: Prove positive real numbers a,b,c.

    Wait, first of all i dont know how to prove that. But, wouldn't (coming back to my question) it end up (a+b)/2 x (b+c)/2 x (c+a)/2 = sqr ab x sqr bc x sqr ca ?? (using AM>GM inequality) I need to get this done by tomorrow so yeah.
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  9. #9
    Senior Member abhishekkgp's Avatar
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    Re: Prove positive real numbers a,b,c.

    Quote Originally Posted by pikachu26134 View Post
    Wait, first of all i dont know how to prove that. But, wouldn't (coming back to my question) it end up (a+b)/2 x (b+c)/2 x (c+a)/2 = sqr ab x sqr bc x sqr ca ?? (using AM>GM inequality) I need to get this done by tomorrow so yeah.
    it can be proved by cubing both sides or by theory of equations. Suppose its been proved.
    then look at the numerator. we have (a^2-b^2)+(b^2-c^2)+(c^2-a^2)=0 so we get (a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3=3(a^2-b^2)(b^2-c^2)(c^2-a^2)

    if you agree with this then i can continue. also tell me why you agree with this in case you do.
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  10. #10
    Senior Member abhishekkgp's Avatar
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    Re: Prove positive real numbers a,b,c.

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  11. #11
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    Re: Prove positive real numbers a,b,c.

    Yes i do agree with this, because you simply cubed both sides, and as x=(a^2-b^2) and y=....etc, you've just shown an expanded version of this. Hence it is true.
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  12. #12
    Senior Member abhishekkgp's Avatar
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    Re: Prove positive real numbers a,b,c.

    Quote Originally Posted by pikachu26134 View Post
    Yes i do agree with this, because you simply cubed both sides, and as x=(a^2-b^2) and y=....etc, you've just shown an expanded version of this. Hence it is true.
    now look at the denominator.
    we have (a-b)+(b-c)+(c-a)=0 so we have (a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a).
    agreed? why, why not?
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  13. #13
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    Re: Prove positive real numbers a,b,c.

    Same rule as before, x^3 + y^3 + z^3 = (x^2+y^2+z^2-xy-xz-yz) +3xyz.... simplifies down to x+y+z = 3xyz. So yes, i do agree.
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  14. #14
    Senior Member abhishekkgp's Avatar
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    Re: Prove positive real numbers a,b,c.

    Quote Originally Posted by pikachu26134 View Post
    Same rule as before, x^3 + y^3 + z^3 = (x^2+y^2+z^2-xy-xz-yz) +3xyz.... simplifies down to x+y+z = 3xyz. So yes, i do agree.
    so [(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3]/[(a-b)^3+(b-c)^3+(c-a)^3]= \text{??} when simplified using what we have just discussed.
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  15. #15
    Senior Member abhishekkgp's Avatar
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    Re: Prove positive real numbers a,b,c.

    Quote Originally Posted by pikachu26134 View Post
    Same rule as before, x^3 + y^3 + z^3 = (x^2+y^2+z^2-xy-xz-yz) +3xyz.... simplifies down to x+y+z = 3xyz. no!! So yes, i do agree.
    theres a mistake i pointed out in red.
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