Quicksort $T(n)_{best}=\Omega(n\log n) $ proof
You mentioned the quicksort recursion:
$$
T_{best}(n) = \min_i [T_{best}(i-1) + T_{best}(n-i) + n].
$$
Note how in this recursion $(i-1) + (n-i) = n-1$. This is because the two subproblems don't involve the pivot. This is the case for any node in the recursion tree: if a node corresponds to an array of length $m$, then its two children correspond to arrays of lengths $m_1,m_2$ where $m_1 + m_2 = m-1$. In particular, if a certain level has $x$ nodes then the sum of lengths at that level $S$ and the sum of lengths at the next level $T$ satisfies $S = T + x$; the $x$ pivots disappear from the recursion.

Denote by $N_i$ the number of nodes at depth $i$, and by $S_i$ the sum of lengths at depth $i$. Assuming we start counting at level 0, we have $S_0 = n$, $S_{i+1} = S_i - N_i$, and $N_i \leq 2^i$. You can prove by induction that
$$
 S_i \geq n - (2^i-1).
$$
In particular, $S_{\log_2 n} \geq 1$ (assuming $n$ is a power of 2), so there are at least $\log_2 n$ levels of recursion. Furthermore,
$$
\sum_{i=0}^{\log_2 n} S_i \geq \sum_{i=0}^{\log_2 n} (n+1-2^i) = (n+1)\log_2 n - (2n-1) = \Omega(n\log n).
$$
Since $\sum_{i=0}^\infty S_i$ is exactly the running time of quicksort (up to constants), we see that quicksort always takes at least $\Omega(n\log n)$.