Results 1 to 4 of 4

Math Help - confusion about inequalities involving (x^2 + y^2) ln(x^2 + y^2)

  1. #1
    Member
    Joined
    Aug 2008
    Posts
    249

    confusion about inequalities involving (x^2 + y^2) ln(x^2 + y^2)

    my book said:
    assume 0< x^2 + 2y^2 < 1 then,
    0 > (x^2 + y^2)ln( x^2 + 2y^2) >/= (x^2 + y^2)ln(2(x^2 + y^2)) = (x^2 + y^2)ln(x^2 + y^2) + (x^2 + y^2)ln(2)

    i understand the first part since the inside of the natural log is less than 1 then the value of ln would be less than 0. and i understand the end which is just the property involving products within a logarithm. what i don't understand is the 2nd inequality. i don't understand how my book knew that adding an additional x^2 inside the logarithm would make ln( x^2 + 2y^2) less than ln(2x^2 + 2y^2). the only constraint we have is that 0< x^2 + 2y^2 < 1 so we know that x^2 is less than 1 but that is all we really know and that fact is too general. i thought the logarithm function is always increasing so making the value inside the ln bigger would make the logarithm bigger and not smaller wouldn't it? is the book wrong or am i missing something? thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,914
    Thanks
    1762
    Awards
    1
    If 0<a<b then \ln(a)<\ln(b).

    How does that apply?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2008
    Posts
    249
    well since x^2 + 2y^2 is less than 2(x^2 + y^2), shouldn't ln(x^2 + 2y^2) < ln(2(x^2+y^2))? but the book has it in the opposite direction for some reason saying ln(x^2 + 2y^2) >/= ln(2(x^2 + y^2))
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    If you're accurately relating what the book said, then I would say that because of Plato's post, you can conclude the book's reasoning is wrong at that point. That is, you claim the book is saying this:

    Assuming 0<x^{2}+2y^{2}<1, then

    0 > (x^2 + y^2)\ln( x^2 + 2y^2) \ge (x^2 + y^2)\ln(2(x^2 + y^2)) = (x^2 + y^2)\ln(x^2 + y^2) + (x^2 + y^2)\ln(2),

    right?

    You're correct. The inequality

    \ln( x^2 + 2y^2) \ge \ln(2(x^2 + y^2))

    doesn't work, even for 0<x^{2}+2y^{2}<1.

    Counterexample: choose x=0.5, y=0.6. Then

    0<x^{2}+2y^{2}=0.97<1. However,

    -0.01858\approx\ln(x^{2}+2y^{2})\not\ge\ln(2(x^{2}+  y^{2}))\approx 0.1213.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving inequalities involving ln
    Posted in the Algebra Forum
    Replies: 5
    Last Post: November 26th 2011, 03:55 PM
  2. Replies: 2
    Last Post: September 19th 2010, 01:45 PM
  3. confusion involving method to compute determinant
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 18th 2010, 01:19 PM
  4. Inequalities Involving Triangles, please?
    Posted in the Geometry Forum
    Replies: 3
    Last Post: October 19th 2009, 03:00 PM
  5. Replies: 1
    Last Post: September 27th 2009, 03:12 AM

Search Tags


/mathhelpforum @mathhelpforum