Inequality assurance Suppose we have $$0<a_{1},a_{2},b_{1},b_{2},c_{1},c_{2},d_{1},d_{2}<1$$ and we know that $$\frac{a_{1}}{b_{1}} > \frac{c_{1}}{d_{1}},$$
$$\frac{a_{2}}{b_{2}} > \frac{c_{2}}{d_{2}}$$
What else we need to prove in order to be sure that:
$$\frac{a_{1}+a_{2}}{b_{1}+b_{2}} > \frac{c_{1}+c_{2}}{d_{1}+d_{2}}?$$
This is just an example. The number of variables might be more albeit obeying the same conditions as mentioned. 
In general, I want to prove that:
$$\frac{\sum_{i=1}^{n}a_{i}}{\sum_{i=1}^{n}b_{i}} > \frac{\sum_{i=1}^{n}c_{i}}{\sum_{i=1}^{n}d_{i}}?$$
where $$\frac{a_{i}}{b_{i}} > \frac{c_{i}}{d_{i}},$$
$$\forall 1 \le i \le n, \;\;\; n \in \mathbb{N}^{+}$$

EDIT1 (bounty will be awarded for this part)
Alternatively, we can state the problem as follows: 
The (least) necessary conditions we need in order to guarantee that:
$$\dfrac{\sum_{i=1}^{n} \alpha_{i} x_{i}}{\sum_{i=1}^{n}x_{i}}
>
\dfrac{\sum_{i=1}^{n} \beta_{i} y_{i}}{\sum_{i=1}^{n}y_{i}},$$
where we know that:
$$\alpha_{i} > \beta_{i} > 1,$$
$$\bigg( \sum_{i=1}^{n}x_{i} + \sum_{i=1}^{n}y_{i} \bigg) \in (0,1).$$

EDIT2 (as for the answer to which the bounty is awarded)
Although the condition suggested by @quasi is a very strong one, I awarded him/er the bounty since his/er answer was the cleanest. My problem remains unsolved, though!
 A: If the variables $a_1,b_1,c_1,d_1,a_2,b_2,c_2,d_2$ are scaled by a positive factor, the values of the fractions being compared will still be the same, hence the condition
$$0 < a_1,b_1,c_1,d_1,a_2,b_2,c_2,d_2 < 1$$
can be replaced by the simpler condition
$$a_1,b_1,c_1,d_1,a_2,b_2,c_2,d_2 > 0$$
To see that the conditions
$$a_1,b_1,c_1,d_1,a_2,b_2,c_2,d_2 > 0$$
$$\frac{a_{1}}{b_{1}} > \frac{c_{1}}{d_{1}},$$
$$\frac{a_{2}}{b_{2}} > \frac{c_{2}}{d_{2}}$$
are not sufficient to force
$$\frac{a_{1}+a_{2}}{b_{1}+b_{2}} > \frac{c_{1}+c_{2}}{d_{1}+d_{2}}$$
let $a_1,b_1,c_1,d_1,a_2,b_2,c_2,d_2$ be given by
$$
a_1 = 1,\;\; b_1 = 2,\;\; c_1 = 3,\;\;d_1 = 7,\;\; 
a_2 = 1,\;\; b_2 = 5,\;\; c_2 = 1,\;\;d_2 = 6
$$
Then
$$\frac{a_1}{b_1} = \frac{1}{2} > \frac{3}{7} = \frac{c_1}{d_1}$$ 
$$\frac{a_2}{b_2} = \frac{1}{5} > \frac{1}{6} = \frac{c_2}{d_2}$$
$$\text{but}$$
$$
\frac{a_1 + a_2}{b_1 + b_2} 
= \frac{1 + 1}{2 + 5} 
= \frac{2}{7}
< \frac{4}{13}
= \frac{3 + 1}{7 + 6}
= \frac{c_1 + c_2}{d_1 + d_2} 
$$

However, for positive real numbers $x_1,...,x_n$ and $y_1,...,y_n$, we always have
$$
\min\left\{\frac{x_i}{y_i}\right\}
\le \frac{\sum_{i=1}^{n}x_{i}}{\sum_{i=1}^{n}y_{i}}
\le \max \left\{\frac{x_i}{y_i}\right\}
$$
To justify the above claim, let
$$m = \min\left\{\frac{x_i}{y_i}\right\}$$
$$M = \max\left\{\frac{x_i}{y_i}\right\}$$
Then
$$m 
\;\;\le\;\; 
\frac{x_i}{y_i}
\;\;\le\;\;
M\;\;\text{ for all }i$$
$$
\implies\;
my_i 
\;\;\le\;\; 
x_i 
\;\;\le\;\; 
My_i
\;\;\text{ for all }i
\qquad\;\;\;\;
$$
$$
\implies\; m\sum_{i=1}^{n}y_i 
 \;\;\le\;\;
\sum_{i=1}^{n}x_i 
 \;\;\le\;\;
M\sum_{i=1}^{n}y_i
\qquad\qquad\qquad
$$
$$
\implies\; m  \;\;\le\;\;  \frac{\sum_{i=1}^{n}x_{i}}{\sum_{i=1}^{n}y_{i}}  \;\;\le\;\;  M
\qquad\qquad\qquad
$$
$$
\implies\; 
\min\left\{\frac{x_i}{y_i}\right\}
\;\;\le\;\;
\frac{\sum_{i=1}^{n}x_{i}}{\sum_{i=1}^{n}y_{i}}
\;\;\le\;\;
\max \left\{\frac{x_i}{y_i}\right\}
\qquad\qquad\qquad
$$
as claimed.

Thus, for the question at hand, if 
$a_1,...,a_n$ and $b_1,...,b_n$ 
are positive real numbers, the condition
$$\min\left\{\frac{a_i}{b_i}\right\} > \max\left\{\frac{c_i}{d_i}\right\}$$
would be sufficient$\,-\,$even without the other conditions, to force 
$$\frac{\sum_{i=1}^{n}a_{i}}{\sum_{i=1}^{n}b_{i}} > \frac{\sum_{i=1}^{n}c_{i}}{\sum_{i=1}^{n}d_{i}}$$
since in that case,
$$
\frac{\sum_{i=1}^{n}a_{i}}{\sum_{i=1}^{n}b_{i}}
\ge
\min\left\{\frac{a_i}{b_i}\right\} > \max\left\{\frac{c_i}{d_i}\right\}
\ge
\frac{\sum_{i=1}^{n}c_{i}}{\sum_{i=1}^{n}d_{i}}
$$
A: Take the case where $n=2$ then you want to show:$$
\frac{\alpha_1x_1+\alpha_2x_2}{x_1+x_2} \gt \frac{\beta_1y_1+\beta_2y_2}{y_1+y_2}
$$ which is equivalent to $$
\alpha_1x_1y_1+\alpha_2x_2y_1+\alpha_1x_1y_2+\alpha_2x_2y_2 \gt \beta_1y_1x_1+\beta_2y_2x_1+\beta_1y_1x_2+\beta_2y_2x_2 \tag{0}
$$ given that $\alpha_1 > \beta_1$ and $\alpha_2>\beta_2$ and matching terms from each side $$
\alpha_1x_1y_1 > \beta_1y_1x_1 \tag{1}
$$$$
\alpha_2x_2y_1 \;?\; \beta_1y_1x_2 \tag{2}
$$$$
\alpha_1x_1y_2 \;?\; \beta_2y_2x_1 \tag{3}
$$$$
\alpha_2x_2y_2 > \beta_2y_2x_2 \tag{4}
$$ (I'm using the symbol $?$ to mean unknown inequality)
As you can see the two unknown inequalities prevent a guarantee of the result you want. Specifically, if the LHS of (2) and (3) is sufficiently smaller than the RHS.
There are a couple of ways to "satisfy" the inequality (0):


*

*Require $\alpha_2 > \beta_1$ and $\alpha_1 > \beta_2$

*Require $\alpha_1x_1y_1 - \beta_1y_1x_1 + \alpha_2x_2y_2 - \beta_2y_2x_2 > \beta_1y_1x_2 - \alpha_2x_2y_1 + \beta_2y_2x_1 - \alpha_1x_1y_2$


Extrapolating to the full problem, #1 means $min(\alpha_i) > max(\beta_j)$, which sounds like a stronger condition than what you want. On the other hand #2 is almost as bad as just requiring a precondition that is the result you want to guarantee. Not quite as bad, since the actual condition is: $$ \sum_{i=1}^n{(\alpha_i-\beta_i)x_iy_i} > \sum_{i=1}^n\sum_{j=1, j \neq i}^n{(\beta_j-\alpha_i)x_iy_j} \tag{5}
$$
This is the minimal condition (if relying on your input data can be specified as part of your conditions).
