Summing fractions the wrong way

$\dfrac{a}{b} < \dfrac{c}{d}$ and $a,b>0$

$\Rightarrow \boxed{ ad < bc }$ [Multiplying with the positive expression $bd$ on both sides].

Therefore, $ad < bc \Rightarrow ad + ab < ab + bc \Rightarrow a(b+d) < b(a+c)$

$\Rightarrow \boxed{ \dfrac{a}{b} < \dfrac{a+c}{b+d} }$ [Dividing by the positive expression $(b+d)b$ on both sides]

Similarly, we also have $ad < bc \Rightarrow ad + cd < bc + cd \Rightarrow \boxed{ \dfrac{a+c}{b+d} < \dfrac{c}{d} }$ .

We now finally have $\boxed{ \dfrac{a}{b} < \dfrac{a+c}{b+d} < \dfrac{c}{d} }$

Imagine for a second that you had $a=c$ and $b=d$ . This would mean that $\frac{a+c}{b+d}=\frac{2c}{2d}=\frac{c}{d}$ . Now think about increasing $b$ . This would clearly make $\frac{a}{b}<\frac{c}{d}$ , but it would also make the denominator of $\frac{a+c}{b+d}$ bigger, meaning that $\frac{a+c}{b+d}$ must now be less than $\frac{c}{d}$ .

The same argument can be made by making $c$ bigger. $\frac{a+c}{b+d}=\frac{2a}{2b}=\frac{a}{b}$ . Making $c$ bigger gives us $\frac{a}{b}<\frac{c}{d}$ and makes the numerator of $\frac{a+c}{b+d}$ bigger, meaning that we now have the identity $\frac{a}{b}<\frac{a+c}{b+d}$ .

Given a/b<c/d, then:

ad<bc ...(1)

Let us prove that a/b < (a+c)/(b+d) < c/d, by considering both inequalities separately:

To show that a/b < (a+c)/(b+d);

a(b+d)<b(a+c)
ad<bc ...(2)

To show that (a+c)/(b+d) < c/d;

(a+c)d<c(b+d)
ad<bc ...(3)

Hence, a/b < (a+c)/(b+d) < c/d must be true, and it is the only option that is true.