how to increase input impedance via bootstraping?
I think you're mixing up the circuit's behaviour for DC (constant over time) currents and voltages and AC (varying over time like a sinewave) currents and voltages.

If someone talks about "small signal" then that's AC !

You have to "bias" your circuit so that it can work, that is DC.
Let's see if your circuit is biased properly.
For DC you can pretend all the Capacitors are gone, because Capacitors cannot conduct DC current.
This would leave the + input of the opamp connected to ground via 2 resistors. You know no current flowes into the input of an opamp so the + input is at 0 V DC
Since Cb is out, the - input is directly connected to the output of the opamp so we have a unity gain buffer. This means the output voltage will copy the voltage at the + input, so it will be... 0V DC !
Does it matter at what DC voltage Vin is ? No, because the Capacitor will block the DC whatever it is !

Now we will discuss AC :-)
Let's assume the input signal is a 1 V sinewave (AC) at the input with a frequency that is high enough such that the capacitors have a very low impedance. Let's also assume the opamp is ideal and that it can easily handle this high frequency.

Since the Cap at the input has a very low impedance (for our assumed signal) the signal will also appear at the + input, so 1 V AC at the + input of the opamp.
The opamp is still connected as a unity gain buffer isn't it ?
So at the opamp's output we will also have the same (a copy of) the 1 V sinewave.
Capacitor Cb makes sure this 1 V AC will also appear at the node between the 2 68 k ohm resistors.
Now what will be the AC voltage across R1 ?

There's your bootstrap ! If you didn't have the bootstrapping in place the input impedance (for AC) would be 2 x 68 k ohms = 136 komhs (or 1 x 68 komhs in the top circuit).
But because of the bootstrapping the AC voltage across R1 is zero meaning that no AC current will flow meaning infinitely high input impedance for AC !
So in theory, even if R1 and R2 had a very low value, the input impedance would be very high ! So is the magic of bootstrapping ;-)