Show the correctness of a logarithmic inequality 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 $0 < u \le p_2$.
 A: 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 $\log (1+t) \leq t$.
