0
$\begingroup$

Suppose $x >0, y \in (0,1), y(1+x) \in (0, 1)$, how can I prove that

$(x+y) [\ln (x+y) - \ln y] +(1-x-y) [\ln (1-x-y) - \ln (1-y)] \geq y [x - (1+x) \ln (1+x)]?$

I have no clue about how to start. It is a part that will help solving another problem. $y$ and $y(1+x)$ can be thought as the parameters of two Bernoulli distributions. Thanks in advance!

  • 0
    Thanks! The question is motivated under probability background. However, the question itself is purely real analysis.2012-09-20

1 Answers 1

1

The inequality you want to prove is equivalent to: $\frac{1}{y}[(x+y) [\ln (x+y) - \ln y] +(1-x-y) [\ln (1-x-y) - \ln (1-y)]] \geq x - (1+x) \ln (1+x)?$

So why not studying $\varphi_x(y) = \frac{1}{y}[(x+y) [\ln (x+y) - \ln y] +(1-x-y) [\ln (1-x-y) - \ln (1-y)]]$

and minimizing it on $(0,1)$?

  • 0
    @steveO yes! Just edited the answer accordingly. I am too lazy to study the function myself but that shouldn't be so hard...2012-09-20