Recurrence relation of quicksort depending on its pivot
When the 4th smallest element is always chosen as a pivot then the recurrence relation is 
$$T(n) = T(n/4)+T(3n/4) + \mathcal{O}(n).$$

If we look at the recursion tree we will see that the left branch has $\log_4 n$ depth and the right has $\log_{4/3} n$ depth. At each step, until the leftmost branch terminates, the sum of levels is equal to $cn \in \mathcal{O}(n)$, for the remaining the levels the sum $\leq cn$. Therefore in the worst case calculation, if we assume all have depth $\log_{4/3} n$ and have $cn$ cost then the cost is;

$$c n \log_{4/3} \in \mathcal{O}(n\log n)$$