Solving the recursive equation $T(n)=T(k)+T(n-k-1)+O(n)$
Consider the recurrence
$$
T(n+1) = \max_{0 \leq k \leq n} T(k) + T(n-k) + n+1, \qquad T(0) = 0,
$$
which is one formalization of the worst-case running time of quicksort. Let us show by induction that the maximum is attained at $k = 0$ (or $k = n$). We will show this while at the same time showing that $T(n) = \frac{n(n+1)}{2}$.
The base case, $n=0$, is clear. Now suppose that $T(m) = \frac{m(m+1)}{2}$ for all $m \leq n$, and consider $T(n+1)$. The function $T(k)$ is convex on the interval $[0,n]$ (we extend it to real values using the same formula), and so the maximum of $T(k) + T(n-k)$ is attained at $k=0$ (or $k=n$). Therefore
$$
T(n+1) = T(0) + T(n) + n+1 = \frac{n(n+1)}{2} + n+1 = \frac{(n+1)(n+2)}{2}.
$$
Instead of arguing using convexity, we can also argue directly:
$$
T(k) + T(n-k) = \frac{k(k+1) + (n-k)(n-k+1)}{2} =
\frac{n(n+1)}{2}-k(n-k),
$$
which is clearly maximized (uniquely) when $k = 0$ or $k = n$.