Suppose I have eight positive numbers $a_i,b_i,c_i,d_i$ for $i=1,2$ satisfying $$\frac{a_i}{b_i}\le \frac{c_i}{d_i}\hspace{1in}(1)$$ I'm looking for additional conditions that will ensure \begin{align}\frac{a_1+a_2}{b_1+b_2}\le\frac{c_1+c_2}{d_1+d_2} \hspace{1in}(*)\end{align}
I found something nice but perhaps not tight enough. I'm hoping to use any slack in equations (1) to tighten (2) and (3) below.
Claim: For (*) to hold, it is sufficient that $$\frac{a_1}{b_1}\le\frac{a_2}{b_2}\hspace{1in}(2)$$ and $$\frac{d_1}{b_1}\le\frac{d_2}{b_2}\hspace{1in}(3)$$
Pf: Start by subtracting the two inequalities in (1),
$$\frac{a_2}{b_2}-\frac{c_1}{d_1} \le \frac{c_2}{d_2}-\frac{a_1}{b_1}$$
Since the right side is positive (due to combining (1) and (2)), we may multiply by $d_1b_2\le b_1d_2$ (assumed true),
\begin{align*} a_2d_1-c_1b_2\le c_2b_1-a_1d_2\\ \end{align*} Add in $a_1d_1\le c_1b_1$ and $a_2d_2\le c_2b_2$ then factor, \begin{align*} a_1d_1+a_2d_2+d_1a_2+a_1d_2\le& b_1c_2+c_1b_2+c_1b_1+c_2b_2\\ (a_1+a_2)(d_1+d_2) \le & (b_1+b_2)(c_1+c_2)\\ \frac{a_1+a_2}{b_1+b_2}\le &\frac{c_1+c_2}{d_1+d_2} \end{align*}
Note, I did not require $a_i,c_i>0$.