smc

Assuming that one of the above fractions is indeed always the largest, try plugging in $a=1,b=2,c=3,d=4$, since those are valid values for the variables given the constraints of the problem. The options become: $\text{(A)}\frac{1+2}{3+4}\text{(B)}\frac{1+4}{2+3}\text{(C)}\frac{2+3}{1+4}\text{(D)}\frac{2+4}{1+3}\text{(E)}\frac{3+4}{1+2}$ Simplified, the options are $\frac{3}{7},1,1,\frac{3}{2},\frac{7}{3}$, respectively. Since $\frac{7}{3}$ is the only option that is greater than $2$, the answer is $\boxed{\frac{c+d}{a+b}}$.