> But what lower resistance means is that you have higher power consumption as: p=iv & v=ir -> P=V^2/R So we see low resistance-> higher current. and higher current -> more power.

I think you're forgetting time in this equation. If two otherwise identical MOSFETs have different gate resistances, the lower-resistance MOSFET will charge faster and draw switching current for less time.

Take this all the way back to a simple linear circuit: a 1-ohm, 1-farad RC circuit will have a time constant of 1s, and if supplied with 1V the current will be I(t) = exp(-t). The power lost to heat in the resistor is then I^2R = exp(-2t), and integrated from 0 to infinity the net energy loss is 0.5J.

If instead the circuit has a resistance of 0.5 ohm, the resulting current is I(t) = 2exp(-2t). Power loss is I^2R = (0.5)*(4 exp(-4t)) = 2 exp(-4t), and integrated from 0 to infinity gives a net energy loss of again 0.5J.

In the meantime, the lower-resistance MOSFET is likely to save power from other effects. Most notably, faster switching implies the transistor spends less time in its linear region where it itself acts as a resistor between source and drain.