Surface tension

In short, the small bubble inflates the large one, because the total surface is reduced in that way.

The soap bubbles are spheres with a surface area $A$ and volume $V$ with $\begin{aligned} A &= 4 \pi r^2 \\ V &= \frac{4}{3} \pi r^3 \end{aligned}$ An infinitesimal change of radius $dr$ results in a corresponding change of the surface and the volume $\begin{aligned} dA &= \frac{dA}{dr} dr = 8 \pi r dr \\ dV &= \frac{dV}{dr} dr = 4 \pi r^2 dr = A dr \end{aligned}$ When air flows from one bubble to the other, the total volume remains the same at first, so that $\dot V_1 = -\dot V_2 = C = \text{const}$ for constant flow. (We neglect the pressure change in the bubbles and the resulting expansion or compression by the ambient pressure) But the changes in the radii $\frac{dr}{dt} = \frac{1}{A}\frac{dV}{dt} = \pm \frac{C}{4 \pi} \frac{1}{r^2}$ are very different, so that also the surface changes $\frac{dA}{dt} = 8 \pi r \frac{dr}{dt} = \pm 2 C \frac{1}{r}$ scale proportional to the inverse radius. Therefore, the large bubble is subjected to a smaller surface change than the small one. Since forming the surface between soap solution and air costs an certain energy per area, the so-called surface tension $\sigma = \frac{dW}{dA}$ , the system tends to minimize its surface area. (Therefore, the bubbles take on the shape of a sphere in the first place.) If the small bubble inflates the big one, its surface is reduced by an large amount, whereas the area of the large bubble increases by a small amount. Therefore, the total surface is reduced: $\frac{dA_\text{tot}}{dt} = \frac{dA_1}{dt} + \frac{dA_2}{dt} = 2 C \left[ \frac{1}{r_1} - \frac{1}{r_2} \right] < 0 \text{ for } C > 0$ The optimal state is achieved when both bubbles unite into a single large one.

In addition, we can also calculate the internal pressure of the soap bubbles. The corresponding work due the surface change results to $dW = 2 \sigma dA = 16 \pi \sigma r dr$ with the surface tension $\sigma = \frac{dW}{dA}$ . (The factor 2 is due to the fact that both surfaces on the outside and inside of the bubble must be considered.) On the other hand, the volume change is connected with a mechanical work against the air pressure difference $\Delta p$ between inside and outside.of the bubble $dW = \Delta p dV = 4 \pi \Delta p r^2 dr$ Since both energies has to be equal, the overpressure inside the bubble results to $\Delta p = \frac{4 \sigma}{r}$ Therefore, the small bubble has a much larger internal pressure. Since air flow from high to low pressure, the small bubble inflates the big one. (This formula for the pressure explains also, why the inflating of a balloon in the beginning is difficult and later easier.)