Results 1 to 3 of 3

Math Help - TA’s Challenge Problem #3

  1. #1
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1

    TA’s Challenge Problem #3

    If x,y,z are non-negative real numbers, show that

    (x+y+z)^2+xy+yz+zx\ \geqslant\ 2\sqrt{xy}(x+y)+2\sqrt{yz}(y+z)+2\sqrt{zx}(z+x)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by TheAbstractionist View Post
    If x,y,z are non-negative real numbers, show that


    (x+y+z)^2+xy+yz+zx\ \geqslant\ 2\sqrt{xy}(x+y)+2\sqrt{yz}(y+z)+2\sqrt{zx}(z+x)

    we have (x+y-2\sqrt{xy})^2 \geq 0, which gives us: x^2 + y^2+6xy \geq 4\sqrt{xy}(x+y). similarly: y^2 + z^2+6yz \geq 4\sqrt{yz}(y+z) and x^2 + z^2+6xz \geq 4\sqrt{xz}(x+z).

    add these three inequalities together and then divide by 2 to get the result.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    Very good!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Challenge Problem
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: July 12th 2010, 03:07 PM
  2. Challenge problem.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 21st 2010, 06:55 AM
  3. TA’s Challenge Problem #7
    Posted in the Math Challenge Problems Forum
    Replies: 5
    Last Post: August 7th 2009, 09:39 AM
  4. TA’s Challenge Problem #5
    Posted in the Math Challenge Problems Forum
    Replies: 1
    Last Post: July 14th 2009, 07:44 PM
  5. TA’s Challenge Problem #2
    Posted in the Math Challenge Problems Forum
    Replies: 7
    Last Post: June 9th 2009, 03:35 AM

Search Tags


/mathhelpforum @mathhelpforum