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Proving this algebra question!

  1. Jun 9, 2013 #1
    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,

    [tex]\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}[/tex]

    then prove that,

    [tex]x=y=z[/tex]

    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 have been thinking and trying different ways to solve this for past 3 hours.
    I would appreciate if you guys can help me out with this.
     
    Last edited: Jun 9, 2013
  2. jcsd
  3. Jun 9, 2013 #2

    Integral

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    Try rearranging to look like this:

    [tex]\frac{x^2}{y^2}-\frac{x}{y}+\frac{y^2}{z^2}-\frac{y}{z}+\frac{z^2}{x^2}-\frac{z}{x}=0 [/tex]

    Now think about how this could be true.

    This is a homework style question, should have been posted in the homework forum. I am moving it there. You need to show us some effort.
     
  4. Jun 9, 2013 #3

    Hmm, if we pair the terms like this,
    [tex](\frac{x^2}{y^2}-\frac{x}{y})+(\frac{y^2}{z^2}-\frac{y}{z})+(\frac{z^2}{x^2}-\frac{z}{x})=0 [/tex]
    We can see that [tex]\frac{x^2}{y^2}≥\frac{x}{y}[/tex]
    even if x or y are negative
    that implies,
    [tex]\frac{x^2}{y^2}-\frac{x}{y}[/tex]
    will either be positive or zero.
    But if either of the pairs is positive then sum cannot be zero.
    Therefore, [tex]\frac{x^2}{y^2}-\frac{x}{y}=0[/tex]
    which further implies [tex]\frac{x^2}{y^2}=\frac{x}{y}[/tex]
    which will give [tex]x=y[/tex] and eventually [tex]x=y=z[/tex]

    Is this correct or I am going wrong somewhere?
     
  5. Jun 9, 2013 #4

    Office_Shredder

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    x=1, y=2?
     
  6. Jun 9, 2013 #5
    Oh, yea, then how else do I solve that?
     
    Last edited: Jun 10, 2013
  7. Jun 9, 2013 #6

    Ray Vickson

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    You have an equation of the form ##t_1 + t_2 + t_3 = 0,## where
    [tex] t_1 = \frac{x^2}{y^2} - \frac{x}{y}, \:
    t_2 = \frac{y^2}{z^2} - \frac{y}{z}, \:
    t_3 = \frac{z^2}{x^2} - \frac{z}{x}[/tex]
    There are several cases:
    (1) All three t_i equal zero.
    (2) One of the t_i = 0, one is > 0 and one is < 0.
    (3) Two of the t_i are > 0 and one is < 0 (or two are < 0 and one is > 0).
    Look at what happens in each case.
     
  8. Jun 9, 2013 #7

    CAF123

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    Is it a correct observation to note that ##t_1, t_2, t_3## are roots of a cubic polynomial with a zero x^2 coefficient? (not that I can see how(or even if) this helps be it the case).
     
  9. Jun 9, 2013 #8
    I don't think that will get me anything, case (2) and (3) can be zero sometimes or sometimes it may not.
    I don't see how we can prove that in case (2) and (3) ##t_1 + t_2 + t_3 ≠ 0,##
     
  10. Jun 10, 2013 #9

    Ray Vickson

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    What are you talking about? You are told that t_1 + t_2 + t_3 = 0. So, can you have that and cases (2) or (3) as well (remembering also the very special forms of the t_i)? You need to sit down and work it out carefully; don't just "guess".
     
  11. Jun 10, 2013 #10
    well, I thought I needed to prove t_1 = t_2= t_3 = 0 to get x/y=x^2/y^2 which will get me x=y
    and to get that I need to rule out case (2) and (3) as a possibility.
    But you seem to be saying something else. Okay, let me think.
     
  12. Jun 10, 2013 #11

    Ray Vickson

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    No, no, no. I am not saying that cases (2) or (3) can occur; however, we need to have some good reasons to eliminate them. One way would be to get a contradiction: assume case (2) and find out eventually that this leads to something going wrong. Ditto for case (3).
     
  13. Jun 10, 2013 #12
    Well I tried that and got inequalities like
    [tex]\frac{x}{y}>=\frac{x^2}{y^2}[/tex]

    but simplifying them isn't getting me anywhere. Or maybe I did not get the thing required to find case (2) and (3) invalid.
    Sorry, I'm not good at simplifying inequalities of variables.
     
  14. Jun 11, 2013 #13

    haruspex

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    Just checking... the RHs is definitely correct, yes?
    [tex]\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}[/tex]
    It isn't, perhaps,
    [tex]\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}=\frac{y}{x}+\frac{z}{y}+\frac{x}{z}[/tex]
    ?
    I ask because that seems much easier!
     
  15. Jun 11, 2013 #14
    RHS is correct? ofc it is correct, the first equation is given to be true.
    And I don't see how I could show eqn (2) to be incorrect.
    And even if it is incorrect, how should help in proving?
    Can you explain a bit in detail? I did not quite get you.
     
  16. Jun 11, 2013 #15

    Ray Vickson

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    He is asking you whether you might have copied down the original problem incorrectly, because the one you wrote is harder than the one he wrote.
     
  17. Jun 11, 2013 #16
    Oh, okay, no I 'm sure I wrote down the problem correctly.

    I think I found a another method, but it requires a,b,c>0, can that be proved somehow?
     
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