It takes at least n log₂(n) comparisons to sort using comparisons, in the worst case. But for large n, n^0.5 (the square root of n) is much smaller than n log₂(n).

The limit as n goes to infinity of (n log₂(n))/n^0.5 is infinite. So for every constant c there is a suitably large n so that cn^0.5 is less than n log₂(n).