Why do most seismic inversion methods ignore high frequencies?
Because those frequencies are not present in the seismic data.

Here's an example with some typical numbers:


Seismic sample interval: 4 ms
Therefore Nyquist is 0.5 $\times$ 1/0.004 = 125 Hz
Usually there is a filter on the recorder at 0.8 $\times$ Nyquist = 100 Hz
So the seismic does not contain frequencies beyond 100 Hz


The maximum frequency $f_\text{max}$ of the data determines its resolving power. There are various ways to think about and calculate this; here's one formula (Kallweit &amp; Wood, 1982):

$$\tau_\text{min} = \frac{1}{1.4\,f_\text{max}}$$

So in our example, the thinnest resolvable bed is about 1 / (1.4 $\times$ 100) = 7.1 ms, and it will be thicker than that if there is a lot of noise or the bandwidth is not optimal (the earth attenuates high frequencies more than low ones, so this is an absolute minimum). 

If we assume a P-wave velocity of 2500 m/s, then 7.1 ms two-way time is about 2500 * (7.1/2) = 8.9 m (55 ft). Ordinary wireline logs have a resolution of 0.15 m or 6 in, so there's a substantial discrepancy. I call this 'the integration gap'.

Because the seismic contains no information about geology thinner than this 8.9 m (in this example), not only is it pointless trying to invert for it, but it would be dangerous: anything we thought we'd resolved would be erroneous. This is why it's necessary to filter the logs back to the resolution of the seismic.

Reference

Kallweit, R &amp; L Wood (1982). The limits of resolution of zero-phase wavelets. Geophysics 47 (7), p 1035–1046, DOI:10.1190/1.1441367.