Show that the best case time complexity of Quicksort is $\Omega(n \log n)$
Let us replace $\Theta(n)$ with $n$, for concreteness, and assume a base case of $T(0) = 0$. Let's try to prove inductively that $T(n) \geq cf(n)$, where $f(n) = n\log n$ for all $n$ (where $0\log 0 = 0$).

During the proof, we will need to minimize $f(q) + f(m-q)$ for $0 \leq q \leq m$. Since $f'(n) = \log n + 1$, any minimum point must satisfy
$$
\log q + 1 = \log (m-q) + 1,
$$
that is, $q = m/2$. Since $2f(m/2) = m\log(m/2)$ whereas $f(0) + f(m) = m\log m$, we see that $q = m/2$ is indeed a minimum.

We are now ready to prove inductively that $T(n) \geq cf(n)$. The base case is obvious. For the inductive step, note that for each $q \in \{0,\ldots,n-1\}$,
$$
T(q) + T(n-1-q) \geq c(f(q) + f(n-1-q)) \geq c(n-1) \log((n-1)/2).
$$
Therefore to complete the proof, we need to show that
$$
c(n-1) \log \frac{n-1}{2} + n \geq cn\log n.
$$
This holds trivially for $n=1$, and holds for all $n \geq 2$ when $c = 1/\log 3$.

Altogether, this shows that $T(n) \geq n\log_3 n$.