Voltage loss on DC circuits
You have come across the reason why high voltages are used in the home (230V, 110V), even higher are used to transmit power over longer distances (10kV and higher). This is also the reason many cars are moving away from 12V power to 48V, and lorries/trucks use 24V rather than 12V.

As you mention, Power is the current times the voltage, P=IV. If you remember Ohm's law: V=IR, you can put the two equations together: P=Ix(IxR), P=I^2R. Looking at the wires, they have a fixed resistance, other than increasing the diameter of the wirse(expensive and additional weight) or using some very fancy high end superconductor (very very expensive and most are still just not practical for a whole plethora of reasons), the only thing we can change is the current. To get the same power, if you half the current, you need to double the voltage, which isn't too hard to do. But with half the current, the power lost due to resistance is reduced by a factor of 4.

So, to answer your question: how is it solved in real life, we increase the voltage.

Other ideas you may want to consider for your application:
Reduce wire length
Increase wire diameter (reduce wire resistance)
Check any connectors, which often have higher resistance than the wires

As mentioned in the comments, a "star" configuration could also improve matters. The reason for this is pretty obvious, as using a "star" (each load being powered by its own wire) decreases the load per wire. But you are having to spend more money and space on wire, which may not be suitable for your application.

If those aren't useful for whatever reason, increase input voltage.

Increasing input voltage may mean you need to drop the voltage back down at the point of load to make things work. But that can be done with a simple buck converter. There are lots of things to consider when designing the power connection system, but that is beyond what I can do in a quick answer on here.