overdrive voltage v.s. speed

the GBW of the opamp is gm/C1, and the fT of a MOS is gm/Cgs.
we can represent both as parameters of Vov (overdrive voltage) and L (channel length)

it is said for high-speed, larger Vov is preferred in lots of literature.

such as
https://cmosedu.com/cmos1/email/email2.htm
https://cmosedu.com/jbaker/papers/talks/IEEE_EDS_WMED_2005.pdf
https://www.ece.tamu.edu/~spalermo/ecen474/lecture07_ee474_gmid.pdf

what i am confused about is the gm has three different equations.

1. gm=KP*(W/L)*Vov
2. gm=sqrt(2KP*W*I/L)
3. gm=2I/Vov

it seems the conclusion of the larger Vov, the higher the bandwidth derived from the 1st equation, as we increasing Vov, if the MOS still in saturation, the bias current will also be increased, no doubt that the gm is increased, in turn, higher speed.

However, for opamp design, when we use diff-pair input, the DC bias current through the device is always defined as the tail current source is fixed, so that eq.1 can't be used, we have to use eq. 2 or eq. 3, so that the statement of increasing Vov, increasing speed falls apart.

do you think my understanding correct?

Thanks!

In differential amplifiers, the gm is defined in terms of the current through the device i.e. equation 2 is used. When the current is fixed, Vov is no in our control and is fixed by the current itself. So only way to increase the gm and effectively the bandwidth of an amplifier would be to increase the current.