An O(n log sqrt(n)) sorting algorithm? Skiena TADM 2nd Question 4-28
Azzo is correct.  It's very simple:

$n\log \sqrt{n} = n\log n^{\frac{1}{2}} = \frac{1}{2}n\log n = \Omega (n\log n)$

Therefore Skiena's $O(n\log\sqrt{n})$ sorting algorithm does not violate the lower bound on comparison sorts.