To allow continued operation without damage, every segment on an LCD panel must have an average of zero volts DC across it. Further, if one or more segments is active and one or more rows is active, there will be four kinds of segments. Assume the active row is driven with voltage R and the inactive rows with voltage I. The active segments are driven with voltage A and the inactive ones with voltage B. There will be four kinds of segments, driven with the following voltages
A-R: Active segment on active row
B-R: Inactive segment on active row
A-I: Active segment on inactive row
B-I: Inactive segment on inactive row
Ideally, one would like to have A-R be a nice big voltage while the other three differential voltages are zero, but that is mathematically impossible. Since the amount to which a display segment is darkened is a function of the RMS voltage (which doesn't care about polarity), having A be one rail, R and B be the opposite rail, and I be mid-rail (swap the R/B and A each frame to avoid DC offset), will mean that |A-R| will be full voltage, |B-R| will be zero, and both |A-I| and |B-I| will be half voltage. Note that |A-I| and |B-I| are equal--that means that the darkness of a segment on an inactive row will not be affected by whether the segment on the active row is active or not; if you see LCDs with "streaks" on the columns, it generally means that for some reason those two voltages are not quite equal.
If one has a 3:1 multiplex display with the above drive method, then active segments will spend 1/3 of the time with full voltage, and 2/3 of the time with 1/2 voltage. Their RMS drive will thus be sqrt(1/3 + 2/3*1/4) = sqrt(1/2). Inactive segments will spend 1/3 of the time with zero voltage and 2/3 with 1/2 voltage, so their RMS drive will be sqrt(2/3*1/4) = sqrt(1/6). The ratio of RMS voltage will thus be sqrt(3):1.
One may improve things slightly if on half the frames (reverse polarity on the other half) one sets R to be VDD, I to be 1/3 VDD, A to be zero, and B to be 2/3 VDD. In that case, |A-R| will be VDD, |B-R| will be 1/3 VDD, and both |A-I| and |B-I| will also be 1/3 VDD. Active segments will spend 1/3 of the time with full voltage and 2/3 with 1/3 voltage, for an RMS drive of sqrt(1/3 + 2/3*1/9) = sqrt(11/27). Inactive segments will spend their entire time at 1/3 voltage, for an rms drive of sqrt(1/9) [i.e. sqrt(3/27)]. The ratio of RMS voltage will thus be sqrt(11/3):1.