Explain a surprisingly simple optimization result

(Sigh. Of course simply writing down the question made me think enough about it to see the solution, but not before I'd already posted the question. Move along, nothing to see here).

Somehow it had escaped me that the $\frac{c_A}{p_0}$ term is there in both inequalities. It is simply the time it will take for improvement $A$ to be finished. In the same language $\frac{c_A}{p_A}$ is the time it will take for improvement $A$ to produce enough to pay for its own production cost after it is finished.

That makes perfect intuitive sense.

I'm not sure whether looking for physical meaning makes much sense here because swapping of A and C in the ABC sequence is governed by a completely different law, which does also depend on the B parameters. An interesting problem, by the way! I have no idea how to do the general case...

An example:

Put $c_1=4/3, p_1=1, p_0=p_2=p_3=c_2=c_3=2$. This gives $c_1(\frac 1{p_0}+\frac 1{p_1})=c_2(\frac 1{p_0}+\frac 1{p_2})=c_3(\frac 1{p_0}+\frac 1{p_3})=2$, which means that no particular choice of the first step has any advantage according to the "heuristic strategy", and if we decrease $c_1$ just a tiny bit, it becomes the choice to make.

However, every order starting with $c_1$ currently gives $\frac 46+\frac 23+\frac 25=\frac{26}{15}=\frac 53+\frac 1{15}$ and every order ending with $c_1$ gives $\frac 22+\frac 24+\frac 4{18}=\frac{31}{18}=\frac 53+\frac 1{18}$, which is strictly shorter time, so it remains shorter under small perturbations.