Power lost = current x voltage

You can break this down further to figure this out simply as possible I could go into the dy/dx differentiation but I wont.

Ok first for a set power line the resistance is a fixed amount. So it won't change but we can substitute it for some values in the first equation.

First substitute it for current:

Current = voltage / resistance

So:

Power lost = voltage² / resistance

Next substitute it for voltage:

Voltage = current x resistance

So:

Power lost = current² x resistance

You see the subtle difference there? In the first one the resistance is deviding the square of the voltage so in a large power grid with high resistance a high voltage devided by a high resistance is ok.

In the second one the square of the current is multiplied by the resistance so in a high current with a large grid with large resistance this is a worse situation.

Couple this to the need to carry a given amount of power along the grid you get a situation where having a high voltage low current reduces losses but a high current low voltage increases the losses.