Results 1 to 4 of 4

Math Help - Pre-algebra: inequality

  1. #1
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Pre-algebra: inequality

    Suppose x,y,z are any positive real numbers such that x+y+z=xyz. Prove that: (x-1)(y-1)(z-1) \leq 6\sqrt{3}-10.

    Source: JIPAM
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2

    Exclamation

    Quote Originally Posted by NonCommAlg View Post
    Suppose x,y,z are any positive real numbers such that x+y+z=xyz. Prove that: (x-1)(y-1)(z-1) \leq 6\sqrt{3}-10.

    Source: JIPAM
    We are looking for \max_{x>0,y>0,z>0}f_{0}(x,y,z), where f_{0}(x,y,z):=(x-1)(y-1)(z-1) under the restriction g(x,y,z):=x+y+z-xyz=0.
    Hence, we set f_{1}(x,y):=f_{0}(x,y,-(x+y)/(1-xy)), therefore it suffices to find \max_{x>0,y>0}f_{1}(x,y).
    Considering the symmetricity, and it suffices to examine the partial derivative of f_{1} with respect to x.
    It follows that the critical points are x_{0}=1 and x_{0}=2y/(y^{2}-1).
    Clearly, f_{1}(1,y)=0.
    On the other hand, let f_{2}(y):=f_{1}(2y/(y^{2}-1),y).
    We find that the critical point for f_{2} is y_{0}=\sqrt{3}, and f_{2}(\sqrt{3})=(\sqrt{3}-1)^{3}=6\sqrt{3}-10, which is the max value of f_{1} (at (\sqrt{3},\sqrt{3})) at the same time because the other critical point (1,y) makes f_{1} attain 0.
    Therefore, the given inequality holds.

    Not. I know this is a long solution and not so nice. :S
    Last edited by bkarpuz; March 31st 2009 at 03:23 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by bkarpuz View Post
    We are looking for \max_{x>0,y>0,z>0}f_{0}(x,y,z), where f_{0}(x,y,z):=(x-1)(y-1)(z-1) under the restriction g(x,y,z):=x+y+z-xyz=0.
    Hence, we set f_{1}(x,y):=f_{0}(x,y,-(x+y)/(1-xy)), therefore it suffices to find \max_{x>0,y>0}f_{1}(x,y).
    Considering the symmetricity, and it suffices to examine the partial derivative of f_{1} with respect to x.
    It follows that the critical points are x_{0}=1 and x_{0}=2y/(y^{2}-1).
    Clearly, f_{1}(1,y)=0.
    On the other hand, let f_{2}(y):=f_{1}(2y/(y^{2}-1),y).
    We find that the critical point for f_{2} is y_{0}=\sqrt{3}, and f_{2}(\sqrt{3})=(\sqrt{3}-1)^{3}=6\sqrt{3}-10, which is the max value of f_{1} (at (\sqrt{3},\sqrt{3})) at the same time because the other critical point (1,y) makes f_{1} attain 0.
    Therefore, the given inequality holds.

    Not. I know this is a long solution and not so nice. :S
    well, the question is supposed to be a Pre-algebra (not calculus) problem. that's what the title of my post says.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2

    Unhappy

    Quote Originally Posted by NonCommAlg View Post
    well, the question is supposed to be a Pre-algebra (not calculus) problem. that's what the title of my post says.
    I know, as I know in JIPAM there are so mant inequalities using convex functions, I guess u wish to see a solution in that direction. But I have no idea since I am not focused on this subject.
    I just wanted to share my long and poor solution.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. algebra inequality
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 21st 2010, 05:25 AM
  2. algebra inequality help
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 29th 2009, 01:14 PM
  3. Help with True/False algebra inequality
    Posted in the Algebra Forum
    Replies: 9
    Last Post: September 10th 2009, 05:11 PM
  4. Linear Algebra. Ptolemy's inequality in R^n
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 3rd 2009, 01:44 PM
  5. algebra inequality help
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 15th 2007, 10:17 AM

Search Tags


/mathhelpforum @mathhelpforum