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Let $p_1>p_2$ and $n_1>n_2$ be positive numbers. I want to show that,

$$ \frac{\log \left(\frac{p_1}{n_1}+1\right)}{\log \left(\frac{p_2}{n_2}+1\right)}\leq \frac{\log \left(\frac{p_1}{c+n_1}+1\right)}{\log \left(\frac{p_2}{c+n_2}+1\right)} $$

where $c$ is a positive number. The simplest solution I can imagine of is to define,

$$ f(u)=\frac{\log \left(\frac{p_1}{n_1+u}+1\right)}{\log \left(\frac{p_2}{n_2+u}+1\right)} $$

and show $f(u)$ is an increasing function over $u \ge 0$. We have,

$$ \frac{\partial{f}}{\partial{u}}=\frac{\frac{p_2 \log \left(\frac{p_1}{n_1+u}+1\right)}{\left(n_2+u\right) \left(n_2+p_2+u\right)}-\frac{p_1 \log \left(\frac{p_2}{n_2+u}+1\right)}{\left(n_1+u\right) \left(n_1+p_1+u\right)}}{\log ^2\left(\frac{p_2}{n_2+u}+1\right)} $$

Since, the denominator is obviously positive, we need to show the numerator is also positive. But I cannot do that. Any help on the issue is appreciated. Any other (simpler) way to prove the first inequality is also welcomed.

EDIT: In addition to the above assumptions, we have $u \le p_2$. So we need to show the inequality over $0 < u \le p_2$.

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I am not sure, but couldn't you just do away with all the logs and have it not affect the inequality? –  soandos Jun 14 '11 at 6:57
You mean $\frac{\log(f(x))}{\log(g(x))}$ is increasing iff $\frac{f(x)}{g(x)}$ is increasing? –  Mohsen Jun 14 '11 at 7:03
@soandos: I guess we need some additional conditions like $f(x)g(x)>0$ and $f(x) \le g(x)$. The second one does not necessarily hold. But it doesn't mean that the above inequality does not hold. –  Mohsen Jun 14 '11 at 7:31
No hint after 45 views! It doesn't look that hard :) –  Mohsen Jun 14 '11 at 9:44

1 Answer 1

up vote 3 down vote accepted

I continue where you left off in your analysis. So the object is to prove $$ \frac{p_2}{(n_2 + u) (n_2 + p_2 + u)} \log \left( \frac{p_1}{n_1 + u} + 1 \right) \geq \frac{p_1}{(n_1 + u) (n_1 + p_1 + u)} \log \left( \frac{p_2}{n_2 + u} + 1 \right) $$ which can be written as $$ \frac{(n_1 + u)(n_1 + p_1 + u)}{p_1} \log \left( \frac{p_1}{n_1 + u} + 1 \right) \geq \frac{(n_2 + u)(n_2 + p_2 + u)}{p_2} \log \left( \frac{p_2}{n_2 + u} + 1 \right). $$ This means, that the function $$ g(p, v) = \frac{v (p+v)}{p} \log \left( \frac{p}{v} + 1 \right) $$ should be increasing in $p$ and $v$.

The partial derivatives of $g$ are given by $$ \frac{\partial g}{\partial v} = \frac{(2 v+ p) \log \left(\frac{p}{v} + 1 \right) - p}{p} $$ and $$ \frac{\partial g}{\partial p} = \frac{v \left(p - v \log \left( \frac{p}{v} + 1 \right) \right)}{p^2} $$

Now, $\frac{\partial g}{\partial v} \geq 0$ follows from $\log (1+t) \geq \frac{t}{2+t}$, and $\frac{\partial g}{\partial p} \geq 0$ follows from $\log (1+t) \leq t$.

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Thanks Michael, you're awesome :) –  Mohsen Jun 14 '11 at 18:48

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