Results 1 to 7 of 7

Math Help - inequality problem

  1. #1
    shosho
    Guest

    inequality problem

    prove that

    a^4 + b^4 + c^2 greater than or equal to a^2bc + b^2ac + c^2ab

    a,b,c are all real numbers

    thanx for any help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by shosho View Post
    prove that

    a^4 + b^4 + c^2 greater than or equal to a^2bc + b^2ac + c^2ab

    a,b,c are all real numbers

    thanx for any help
    I think you means, a,b,c>0:
    a^4+b^4+c^4 \geq a^2bc+b^2ac+c^2ab
    Because most of these inequalities are cyclic.

    Now, by the Cauchy-Swartz Inequality:
    a^4+b^4+c^4 = (a^2)^2+(b^2)^2+(c^2)^2 \geq a^2b^2+b^2c^2+a^2c^2

    By the Cauchy-Swartz Inequality again:
    a^2b^2+b^2c^2+a^2c^2 = (ab)^2+(bc)^2+(ac)^2 \geq (ab)(ac)+(bc)(ab)+(ac)(bc)= a^2bc+b^2ac+c^2ab

    Thus, (by the transitive inequality )
    a^4+b^4+c^4 \geq a^2bc+b^2ac+c^2ab
    And only equality when a=b=c.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    shosho
    Guest
    thanx for the help

    but im not sure about it...
    i dont really understand the theorem thing you used.

    could you please explain in another way

    it would be much appreciated
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by shosho View Post
    thanx for the help

    but im not sure about it...
    i dont really understand the theorem thing you used.

    could you please explain in another way

    it would be much appreciated
    Tell us what course this is part off, or what you were covering imeadiatly
    before the question was set.

    RonL
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker View Post
    I think you means, a,b,c>0:
    a^4+b^4+c^4 \geq a^2bc+b^2ac+c^2ab
    Because most of these inequalities are cyclic.

    Now, by the Cauchy-Swartz Inequality:
    a^4+b^4+c^4 = (a^2)^2+(b^2)^2+(c^2)^2 \geq a^2b^2+b^2c^2+a^2c^2

    By the Cauchy-Swartz Inequality again:
    a^2b^2+b^2c^2+a^2c^2 = (ab)^2+(bc)^2+(ac)^2 \geq (ab)(ac)+(bc)(ab)+(ac)(bc)= a^2bc+b^2ac+c^2ab

    Thus, (by the transitive inequality )
    a^4+b^4+c^4 \geq a^2bc+b^2ac+c^2ab
    And only equality when a=b=c.
    The restriction a,b,c>0 is not needed.

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    shosho
    Guest
    the course i was previously doiing was inequality. it wasnt all that of a difficult section of inequality. nothing specific but the question was given as an extension type. i seem to be terribly stuck
    please help .. this is urgent

    Thank you
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by shosho View Post
    could you please explain in another way
    How about this way?

    We need to use the inequality:
    x^2+y^2+z^2 \geq xy+yz+xz
    For all real numbers x,y,z>0.

    Begin by noticing that, by the AM-GM inequality,
    x^3+y^3+z^3 \geq 3xyz
    Thus,
    x^3+y^3+z^3 - 3xyz \geq 0
    Factor,
    (x+y+z)(x^2+y^2+z^2 - xy - yz - xz) \geq 0
    Since (x+y+z)>0 we can cancel to obtain,
    x^2+y^2+z^2 \geq xy+yz+xz

    Which is the Cauchy-Swartz inequality.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inequality Problem
    Posted in the Algebra Forum
    Replies: 6
    Last Post: February 22nd 2011, 03:09 AM
  2. Inequality Problem
    Posted in the Pre-Calculus Forum
    Replies: 11
    Last Post: October 22nd 2010, 11:56 AM
  3. Inequality problem.
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 3rd 2009, 03:06 PM
  4. Another inequality problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 3rd 2009, 09:26 AM
  5. Inequality problem
    Posted in the Algebra Forum
    Replies: 6
    Last Post: May 22nd 2009, 11:04 PM

/mathhelpforum @mathhelpforum