How can I show, that every algorithm based on pairwise comparisons for sorting a total ordered set consisting of $n$ elements requires in the $\Omega(n \log n)$ comparisons in the average case.
The hint I was already given is to show that for all $x \in (0,1)$ the inequality $$x \log_2(x) + (1-2) \log_2(1-x) \geq -1 $$ holds, but I have no clue how to begin.
Suppose that we have $n$ elements that need to be sorted and we can only do pairwise comparisons. There are $n!$ ways to arrange these elements, and if we assume the elements are unique, then there is only one way to sort them.
Let's think of each pairwise comparison as a yes/no question. Each pairwise comparison rules out some set of the $n!$ permutations that cannot be the correctly sorted order. Before answering yes or no, there is a set of possible permutations assuming the answer is yes and there is a set of possible permutations assuming the answer is no. These sets are disjoint because the elements being compared are unique, and have only one correct ordering.
If the "yes" set is bigger than the "no" set or vice versa, then we can define our ordering so that it answers the pairwise comparison that gives us the bigger set. Therefore, our algorithm can at best choose a comparison that makes the "yes" set and "no" sets equal. Upon the answer, our choice of permutations is cut in half.
This implies that the best comparison based algorithm runs in $\Omega(\log_2(n!))$ time, because there are $n!$ permutations and each comparison eliminates at best half of the choices in worst-case time. Notice that $$ \log_2(n!) = \sum_{i=1}^n \log_2(i) \le \sum_{i = \lfloor n/2 \rfloor}^n \log_2(i) \le \lceil n/2 \rceil \log_2(\lfloor n/2 \rfloor) \le \frac{n \log(n-1)}{2} - 1 \in \Omega(n\log(n))$$