I am trying to find $T(n)=O(f(n))$, where $$T(n)\leq n^2+n\left[T(n-m)+T(m-1)\right],$$ where $m\in\{1,2,\ldots,n\}$.
Is it possible to find $f(n)$ such that $T(n)=O(f(n))$?
I started to fix $m=n/2$ (which I assume the worst scenario, isn't it?). I found $$\begin{align}T(n)&\leq n^2+\frac{n^3}{2}+\frac{n^4}{8}+\frac{n^5}{64}+\frac{n^4}{4}T(n/16),\\ &\;\vdots\\& \leq n^2+T(1)\sum_{i=1}^{(\lg n)-2}\frac{n^{i+2}}{\sqrt{2^{i(i+1)}}}.\end{align}$$