Algebra: proving an inequality

[vers]

Homework Statement

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.

Homework Equations

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

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.

 

[Tedjn]

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.

 

[vers]

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.

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]?

 

[vers]

It turns out that the function f can be simplified as
$$f(x) = d\left(\frac{1+x}{2} \parallel \frac{x}{2}\right) - d\left(\frac{x}{2} \parallel \frac{1+x}{2}\right)$$
where
$$d(p \parallel q) = p \log \frac{p}{q} + (1 - p) \log \frac{1-p}{1-q}$$
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?