I have this maximum,
$$ \max_{a\leq \frac{b}{1-db}}\frac{1}{1+da}(-a\log(a)-(1-a)\log(1-a)) $$
where $b\rightarrow 0$, $d=\frac{n}{b}$ and $n\in(0,1)$ is a real constant.
Yes, I know that since $a\leq \frac{b}{1-n}\rightarrow 0$ you can prove that te hole expression is going to zero, but I want to know How fast is this going to zero and how this is affected by $d$.
Here is what I've done,
first I tried to simplify the problem, so I studied the case where $b$ is still going to $0$ but $d$ is a constant. And the maximum is (obviously) archived just taking $a$ to be the higher bound, so:
\begin{eqnarray} \nonumber &max&_{a\leq \frac{b}{1-db}}\frac{1}{1+da}(-a\log(a)-(1-a)\log(1-a))\\ \nonumber &=&\frac{1}{1+d\frac{b}{1-db}}(-\frac{b}{1-db}\log(\frac{b}{1-db})-(1-\frac{b}{1-db})\log(1-\frac{b}{1-db}))\\ \nonumber &=& (1-db)\log(1-db)-b\log(b)-(1-db-b)\log(1-db-b)\\ \nonumber &\approx& -db-b\log(b)+db+b+o(b)\\ \nonumber &=&-b\log(b)+b+o(b) \end{eqnarray}
And so, the value of $d$ does not affect the decreasing in the first or second order term...
With the complete problem (where $d=\frac{n}{b}$ and $b$ is going to $0$), I started for doing some plots of specific cases, and it seems that the evaluation of the higher bound is again the maximum in some cases but not always... but of course this is just to have an idea, and it is not a proof...
For the proof I tried to optimize the expression evaluated in some $xb$ (because the value of $a$ I think will depend on $b$) but I didn't get any interesting.
Any help will be appreciated!
Calling
$$ f(a) = \frac{(a-1) \log (1-a)-a \log (a)}{\frac{a n}{b}+1} $$
The lagrangian
$$ L(a,\lambda,\epsilon) = f(a)+\lambda \left(a (1-n)-b-e^2\right) $$
gives the stationary points
$$ \begin{array}{rcl} (b+n) \log (1-a) b-\lambda (n-1) (b+a n)^2 & = & 0\\ a-b-a n -\epsilon^2& = & 0\\ \epsilon \lambda & = & 0 \end{array} $$
getting as solution
$$ a = \frac{b}{1-n},\lambda = g(b,n), \epsilon = 0 $$
$\epsilon = 0\rightarrow a-b-a n=0$ and the maximum/minimum is
$$ (b+n-1) \log \left(\frac{b}{n-1}+1\right)-b \log \left(-\frac{b}{n-1}\right) $$
As we can observe this stationary value is dependent on $n$
This stationary point should be qualified regarding the sign of $f''(a)$ or
$$ \frac{(n-1)^3 \left(2 n (b+n-1) \left(n \log \left(\frac{b}{n-1}+1\right)+2 b \tanh ^{-1}\left(\frac{2 b}{n-1}+1\right)\right)-b\right)}{b^2 (b+n-1)} $$
I hope this helps.