Results 1 to 11 of 11

Math Help - Prove an identity

  1. #1
    Member
    Joined
    Aug 2008
    Posts
    91

    Prove an identity

    Hello,
    show that if a^2+b^2+c^2=ab+ac+bc, then a=b=c.

    I tried to roll stuff up using binomial expansions, eg. for (a+b+c)^2, bt it only yields some obvious things like a=a. Any ideas?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2010
    From
    Staten Island, NY
    Posts
    451
    Thanks
    2
    Try (a-b)^2+(a-c)^2+(b-c)^2

    Edit: Oh never mind - I think that's just 0.

    Edit 2: Lol. Oh yeah - that's why it works.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2010
    From
    Tétouan/Morocco
    Posts
    44
    Assuming that a,b and c are positif real numbers .
    Using the Arithmetic mean-Geometric mean inequality we get a˛+b˛>=2ab also a˛+c˛>=2ac and b˛+c˛>=2bc summing we get a˛+b˛+c˛>=ab+ac+bc and equality holds in this inequality when a=b=c !

    You can also use the Cauchy-Schwartz inequality and in the same way you get equality holds when a=b=c
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Dec 2010
    Posts
    470
    I believe something similar was posted before:
    http://www.mathhelpforum.com/math-he...ht=equilateral
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by atreyyu View Post
    Hello,
    show that if a^2+b^2+c^2=ab+ac+bc, then a=b=c.

    I tried to roll stuff up using binomial expansions, eg. for (a+b+c)^2, bt it only yields some obvious things like a=a. Any ideas?

    Thanks.
    a^2+b^2+c^2-ab-ac-bc=0\Rightarrow\ 2a^2+2b^2+2c^2-2ab-2ac-2bc=0

    \Rightarrow\ a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2=0

    (a-b)^2+(b-c)^2+(a-c)^2=0

    but all three terms on the left are non-negative and sum to zero.
    As there are no negative terms, they are all zero.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Dec 2010
    From
    Tétouan/Morocco
    Posts
    44
    Quote Originally Posted by snowtea View Post
    I believe something similar was posted before:
    http://www.mathhelpforum.com/math-he...ht=equilateral
    It's exactly the same problem with geometric interpertation (with some complex geometry from "Also sprach Zarathustra" )
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Aug 2008
    Posts
    91
    Thanks for all, the replies, but all your solutions make use of the fact that a, b, c are positive. There is no word of it in the original wording I have got, and without that I don't think you can sum inequalities together or draw a concluion Archie Meade has done... maybe there is a problem with the phrasing of the question and a, b, c should indeed be positive?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Nov 2010
    From
    Staten Island, NY
    Posts
    451
    Thanks
    2
    We all haven't assumed that a,b,c are positive. We have used the fact that x^2 is nonnegative for all real  x.

    For example (a-b)^2 is always positive or zero regardless of what a and b are.
    Last edited by DrSteve; January 9th 2011 at 10:56 AM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Yes,

    by "all three terms", I meant the terms (a-b)^2,\;\; (b-c)^2,\;\; (a-c)^2,
    not the terms a, b, c.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Dec 2010
    From
    Tétouan/Morocco
    Posts
    44
    Well , i guess you have the solution now ! And all the solutions are with a,b and c real numbers ! (i made a mistake by assuming they were positive , but a˛+b˛>=2ab is true for all a and b real numbers )
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Aug 2008
    Posts
    91
    That's correct. Thank you all a lot.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove this identity
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: November 21st 2010, 04:56 PM
  2. Prove the identity
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: September 7th 2010, 06:34 AM
  3. prove the identity
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: April 25th 2010, 04:22 PM
  4. one more identity to prove
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: January 19th 2010, 06:16 AM
  5. Prove the Identity
    Posted in the Trigonometry Forum
    Replies: 10
    Last Post: February 25th 2008, 05:29 PM

Search Tags


/mathhelpforum @mathhelpforum