Why are unbalanced partitions worse than balanced partitions in Quicksort?
for balanced partitions the work actually remains constant on all the steps


That's just false. Have another look at the algorithm: each recursive call gets an input that is properly smaller than the current one.

You also seem to be missing that, for the total cost, you need to add up all the steps. Just saying, "the inputs get smaller", doesn't tell you anything besides that the algorithm terminates.

In the worst case, you get the solution of a recurrence similar to

$\qquad\begin{align*}
 T(1) &amp;= 0 \\
 T(i) &amp;= T(i-1) + i
\end{align*}$

which solves to some $T \in \Theta(n^2)$.

In the average case, you get the solution of a recurrence similar to

$\qquad\begin{align*}
 T(1) &amp;= 0 \\
 T(i) &amp;= 2T(i/2) + i
\end{align*}$

which solves to some $T \in \Theta(n \log n)$.



To stress my point from above, consider

$\qquad\begin{align*}
 T(1) &amp;= 0 \\
 T(i) &amp;= T(i/2) + i
\end{align*}$

which solves to some $T \in \Theta(n)$.

Or

$\qquad\begin{align*}
 T(1) &amp;= 0 \\
 T(i) &amp;= T(i/2) + 1
\end{align*}$

which solves to some $T \in \Theta(\log n)$.

You see that the specifics of the recurrence matter.