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Math Help - An Algebra question!

  1. #1
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    An Algebra question!

    Hey guys, I am quite new to this forum, and just encountered this seemingly simple problem but still was not successful in solving it.

    If,
    An Algebra question!-gif1.latex.gif

    then prove that,

    An Algebra question!-gif2.latex.gif

    where x,y and z are real numbers.

    I tried everything I could from basic identities to substituting x=a+y and x=b+z which one of my friend suggested but no luck yet.
    The 'prove that' part seems obvious but I cannot make out how to write a proof.

    I would appreciate if you guys can help me out with this.
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  2. #2
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    Re: An Algebra question!

    Hey sam300.

    As a suggestion, you could re-write x,y,z as x* y* z* where x* = x + a, y* = y + b and z* = z + c and then simplify in terms of a, b, and c on one side. If you can prove that they have to be zero then you are done.

    Also note that when you have squared terms, you know that things have to be positive.

    A starting suggestion though would be firstly to get rid of the denominator on both sides by cross multiplying terms between the left and right hand sides.

    In addition to this if you need more tools, take a look at the properties of norms if you have to. I don't think you will need to but norms are used with questions like this.
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  3. #3
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    Re: An Algebra question!

    Quote Originally Posted by sam300 View Post
    Hey guys, I am quite new to this forum, and just encountered this seemingly simple problem but still was not successful in solving it.

    If,
    Click image for larger version. 

Name:	gif1.latex.gif 
Views:	3 
Size:	994 Bytes 
ID:	28547

    then prove that,

    Click image for larger version. 

Name:	gif2.latex.gif 
Views:	57 
Size:	265 Bytes 
ID:	28548

    where x,y and z are real numbers.

    I tried everything I could from basic identities to substituting x=a+y and x=b+z which one of my friend suggested but no luck yet.
    The 'prove that' part seems obvious but I cannot make out how to write a proof.

    I would appreciate if you guys can help me out with this.
    Let a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}. Your equation becomes a^2+b^2+c^2=a+b+c and the question reduces to proving a=b=c=1

    This should be a much easier problem to work with. See if you make any headway here.

    Some hints: Show (a-b)^2+(b-c)^2+(a-c)^2=0. Use the fact that ab+bc+ac=a+b+c.
    Last edited by Gusbob; June 7th 2013 at 10:38 PM.
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  4. #4
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    Re: An Algebra question!

    chiro,

    I did not quite get you, are you saying I should get rid of denominators and substitute x as x'-a (or x'+a?)? How can I put x'=x+a?
    I am sorry, I'm not acquainted with the properties of norms as you are saying.


    Gusbob,

    I tried your approach but no success yet,
    at most I got,
    An Algebra question!-png.latex.png

    OR by using (a+b+c)^2 identity:
    An Algebra question!-png2.latex.png

    But I'm stuck at that. Thanks for your replies anyway, please do reply if you have any idea what to do next.

    PS. I have no idea why the latex is appearing so small.
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  5. #5
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    Re: An Algebra question!

    Quote Originally Posted by sam300 View Post
    chiro,

    I did not quite get you, are you saying I should get rid of denominators and substitute x as x'-a (or x'+a?)? How can I put x'=x+a?
    I am sorry, I'm not acquainted with the properties of norms as you are saying.


    Gusbob,

    I tried your approach but no success yet,
    at most I got,
    Click image for larger version. 

Name:	png.latex.png 
Views:	2 
Size:	880 Bytes 
ID:	28550

    OR by using (a+b+c)^2 identity:
    Click image for larger version. 

Name:	png2.latex.png 
Views:	4 
Size:	992 Bytes 
ID:	28551

    But I'm stuck at that. Thanks for your replies anyway, please do reply if you have any idea what to do next.

    PS. I have no idea why the latex is appearing so small.
    Quick sketch of proof: (a-b)^2=a^2-2ab+b^2 and similarly for (b-c)^2 and (c-a)^2. Adding the three together, one gets

    (a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2)-2(ab+bc+ca)

    Since ab+bc+ca=a+b+c=a^2+b^2+c^2, the right hand side is 0. As squares of real numbers are non-negative, it must be that a-b=b-c=c-a=0. That is, a=b=c. Finally, the equation a^2+b^2+c^2=a+b+c means 3a^2=a^2, so a=0 or a=1. We know the former case is impossible as it requires x=0, so we conclude that a=b=c=1
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  6. #6
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    Re: An Algebra question!

    Well you are correct, and I had been trying to arrive at same proof.


    But how did you assume that ab+bc+ca=a+b+c?
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  7. #7
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    Re: An Algebra question!

    Quote Originally Posted by sam300 View Post
    Well you are correct, and I had been trying to arrive at same proof.


    But how did you assume that ab+bc+ca=a+b+c?
    Sorry I made a mistake here. I will fix it tomorrow morning.
    Last edited by Gusbob; June 8th 2013 at 09:17 AM.
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  8. #8
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    Re: An Algebra question!

    Bump, no one can do this?
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  9. #9
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    Re: An Algebra question!

    An Algebra question!-10-jun-13.png
    Thanks from topsquark
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  10. #10
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    Re: An Algebra question!

    Quote Originally Posted by ibdutt View Post
    Click image for larger version. 

Name:	10 Jun 13.png 
Views:	11 
Size:	20.4 KB 
ID:	28589
    I have already tried that method, but to say that x=y=z we have to have squares on LHS.
    Your answer also has cubes and expressions like (x-y) which may make the term negative
    Thus when some terms are negative and some positive, their sum can be zero.
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  11. #11
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    Re: An Algebra question!

    the thinking is good. i would like to mention that on the LHS we have sum squares of rationals and so that will always be positive.
    On the RHS we have the sum of the same rationals so the sign of x,y and z has to be the same.
    I hope that is oK.

    Quote Originally Posted by sam300 View Post
    I have already tried that method, but to say that x=y=z we have to have squares on LHS.
    Your answer also has cubes and expressions like (x-y) which may make the term negative
    Thus when some terms are negative and some positive, their sum can be zero.
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  12. #12
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    Re: An Algebra question!

    The stated problem is equivalent to a^2 + b^2 + c^2 = a + b + c and abc=1

    We have (a-1)^2 + (b-1)^2 + (c-1)^2 + (a + b + c) = 3 which shows that a + b + c <= 3

    If a,b,c are >0 then we also have a + b + c >= 3 (Arithmetic average is greater than the geometric average)
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  13. #13
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    Re: An Algebra question!

    Quote Originally Posted by Idea View Post
    The stated problem is equivalent to a^2 + b^2 + c^2 = a + b + c and abc=1

    We have (a-1)^2 + (b-1)^2 + (c-1)^2 + (a + b + c) = 3 which shows that a + b + c <= 3

    If a,b,c are >0 then we also have a + b + c >= 3 (Arithmetic average is greater than the geometric average)
    You are a genius dude, but there is one little important flaw.
    How did you assume a,b,c>0, they can be -ve also, can't they?
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  14. #14
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    Re: An Algebra question!

    And due to your arithmetic mean and geometric mean hint,
    I found a simpler way,
    As, abc=1
    Each of a,b,c is -1 or +1,
    So we have three cases,
    1) All three are +1
    2) All three are -1
    3) Any two are -1 and one is +1

    Try each case in,
    (a+b+c)/3>=(abc)^(1/3) OR (a+b+c)/3>= 1 (As abc=1)

    And you will get the answer, just the flaw is we don't know if a,b,c>0.
    And therefore this is not a perfect proof. But it is a proof nonetheless.
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  15. #15
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    Re: An Algebra question!

    abc=1 does not imply a,b,c is -1 or +1 as these are real numbers, e.g. a=1/3, b=2, c=3/2 , abc=1
    As for the other question, a,b,c>0 is not an assumption. It follows from the given.
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