Sorting by adjacent swaps needs exactly the number of inversions

Effect of a swap. Swap $a_i$ and $a_{i+1}$. The relative order of this pair changes, so the pair is an inversion after the swap iff it was not before. Every other pair keeps the same relative order. So the number of inversions changes by exactly $\pm1$.

Lower bound. The sorted sequence has $0$ inversions. If the sequence has $I$ inversions, any sequence of swaps sorting it must decrease the count from $I$ to $0$ in steps of $\pm1$, so it has at least $I$ swaps.

Upper bound. If $I>0$, the sequence is not sorted, so there is an index $i$ with $a_i>a_{i+1}$. Swapping them decreases $I$ by $1$. Repeating, we sort the sequence in exactly $I$ swaps.

Maximum. The number of inversions is at most the number of pairs, $\binom n2$, with equality for the decreasing sequence.

Example. $5,4,3,2,1$ has $10$ inversions and needs $10$ swaps; $2,3,1$ has $2$ inversions.

Remark. The parity of the number of inversions is the sign of the permutation.

Lesson. A quantity that changes by exactly $\pm1$ per step (a monovariant) gives matching lower and upper bounds.