Derivative of a 'weighted average' of decreasing fractions I'm having some trouble showing the following statement (which intuitively seems to hold): 
Suppose I have a series of fractions indexed by $i$ , each of them a function
of $N:f_{i}\left(  N\right)  =\frac{A_{i}\left(  N\right)  }{B_{i}\left(
N\right)  }$. Assume that:


*

*(a) $f_{i}\left(  N\right)  $ is increasing in $N$ 

*(b) $B_{i}\left(  N\right)  $ is decreasing in $N$

*(c)  $0<A_{i}<1, 0<B_{i}<1  $. 


Now consider the following 'weighted average':
$AV\left(  N\right)  =\frac{\sum\limits_{i}p_{i}A_{i}\left(  N\right)  }{\sum\limits_{i}p_{i}B_{i}\left(  N\right)  }$ where $\sum p_{i}=1$.
Q: Are the above conditions sufficient to guarantee that $AV\left(  N\right)  $
is increasing in $N$ as well?
Below you will find what I have been able to show. Any help or insights would be greatly appreciated!

(a) implies that $\frac{d A_{i}}{d N}B_{i}-A_{i}\frac{d
B_{i}}{d N}>0$
Now note that $\frac{dAV\left(  N\right)  }{dN}>0$  if
$\left(  \sum\limits_{i}p_{i}\frac{dA_{i}\left(  N\right)  }{dN}\right)
\left(  \sum\limits_{j}p_{j}B_{j}\left(  N\right)  \right)  -\left(
\sum\limits_{i}p_{i}A_{i}\left(  N\right)  \right)  \left(  \sum
\limits_{j}p_{j}\frac{dB_{j}\left(  N\right)  }{dN}\right)  >0$
$\sum\limits_{i}\sum\limits_{j}\left(  p_{i}p_{j}\left[  \frac{dA_{i}\left(
N\right)  }{dN}B_{j}\left(  N\right)  -A_{i}\left(  N\right)  \frac
{dB_{j}\left(  N\right)  }{dN}\right]  \right)  >0$
$\sum\limits_{i}\left(  \left(  p_{i}\right)  ^{2}\left[  \frac{dA_{i}\left(
N\right)  }{dN}B_{i}\left(  N\right)  -A_{i}\left(  N\right)  \frac
{dB_{i}\left(  N\right)  }{dN}\right]  \right)$
$+\sum\limits_{i}%
\sum\limits_{j\neq i}\left(  p_{i}p_{j}\left[  \frac{dA_{i}\left(  N\right)
}{dN}B_{j}\left(  N\right)  -A_{i}\left(  N\right)  \frac{dB_{j}\left(
N\right)  }{dN}\right]  \right)  >0$
$\sum\limits_{i}\left(  \left(  p_{i}\right)  ^{2}\left[  \frac{dA_{i}\left(
N\right)  }{dN}B_{i}\left(  N\right)  -A_{i}\left(  N\right)  \frac
{dB_{i}\left(  N\right)  }{dN}\right]  \right)$
$+\sum\limits_{i}%
\sum\limits_{j\neq i}\left[  \left(  p_{i}p_{j}\right)  \frac{dA_{i}\left(
N\right)  }{dN}B_{j}\left(  N\right)  \right]  -\left(  p_{i}p_{j}\right)
\sum\limits_{i}\sum\limits_{j\neq i}\left[  A_{i}\left(  N\right)
\frac{dB_{j}\left(  N\right)  }{dN}\right]  >0$
The first sum is positive (follows from (a)) and the third sum is negative (follows from (b)). But the second sum does is not necessarily positive.
 A: No, AV is not necessarily increasing in $N$ under the stated assumptions. Proof by counterexample:
Consider the case where there are only two fractions which are equally weighted:
$$AV\left(  N\right)  =\frac{0.5A_{1}\left(  N\right)  +0.5A_{2}\left(
N\right)  }{0.5B_{1}\left(  N\right)  +0.5B_{2}\left(  N\right)  }=\frac
{A_{1}\left(  N\right)  +A_{2}\left(  N\right)  }{B_{1}\left(  N\right)
+B_{2}\left(  N\right)  }$$
We now want to know if $AV\left(  N^{\prime}\right)  >AV\left(  N\right)
$ (where $N^{\prime}>N$). 
Consider the following values for $A$ and $B$:
$$A_{1}\left(  N\right)  =\frac{5}{8},A_{2}\left(  N\right)  =\frac{1}{2}$$
$$A_{1}\left(  N^{\prime}\right)  =\frac{5}{16},A_{2}\left(  N^{\prime}\right)
=\frac{2}{5}$$
$$B_{1}\left(  N\right)  =\frac{1}{2},B_{2}\left(  N\right)  =\frac{5}{8}$$
$$B_{1}\left(  N^{\prime}\right)  =\frac{1}{4},B_{2}\left(  N^{\prime}\right)
=\frac{1}{2}$$
Which satisfies assumptions (b)-(c) strictly, and (a) weakly.We then have that:
$$AV\left(  N\right)  =\frac{\frac{5}{8}+\frac{1}{2}}{\frac{1}{2}+\frac{5}{8}
}=1$$
$$AV\left(  N^{\prime}\right)  =\frac{\frac{5}{16}+\frac{2}{5}}{\frac{1}
{4}+\frac{1}{2}}=\frac{19}{20}<1$$
So that $AV\left(  N\right)  $ is decreasing in $N$.
