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Homework Help: Algebra: proving an inequality

  1. Jul 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Let
    f(x) = (1/2+x) log [(1+x)/x] + (3/2-x) log [(1-x)/(2-x)]
    where log is natural logarithm and 0 < x <= 1/2. Show that f(x) >= 0 for all x.


    2. Relevant equations
    The only inequality that I can think of is the log sum inequality:
    http://en.wikipedia.org/wiki/Log_sum_inequality


    3. The attempt at a solution
    I try to plot f (see figure), and it seems that the statement is correct. In fact, the figure suggests that f is decreasing and equals 0 at x = 1/2. However, I can't figure out an analytical proof of this.
     

    Attached Files:

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  2. jcsd
  3. Jul 14, 2010 #2
    I haven't done this out, but, if you find f(1/2) = 0 and that f(x) is decreasing on (0,1/2], what does that tell you about f(x) on (0,1/2]? How would you show that f(1/2) = 0 and f(x) is decreasing on (0,1/2]?

    EDIT: On second thought, it may not be completely straightforward to show that f(x) is decreasing on (0,1/2]. If so, let us know what you've tried.
     
  4. Jul 14, 2010 #3
    Certainly if I can show that f'(x) <=0 then the inequality is proved. However, it seems to me that showing f' being negative is even more difficult than showing f being nonnegative directly.

     
  5. Jul 15, 2010 #4
    It turns out that the function f can be simplified as
    [tex]
    f(x) = d\left(\frac{1+x}{2} \parallel \frac{x}{2}\right) - d\left(\frac{x}{2} \parallel \frac{1+x}{2}\right)
    [/tex]
    where
    [tex]
    d(p \parallel q) = p \log \frac{p}{q} + (1 - p) \log \frac{1-p}{1-q}
    [/tex]
    is the KL divergence between two Bernoulli random variables with respective expectations p and q. So all I need to do now is to show that the difference of these two KL divergences is nonnegative when x <= 1/2. Any idea how to do that?
     
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