Prove a statement about searching in binary search tree
Let $h$ be longest path from root $T$ to leaves. We must read $h+1$ nodes. So searching for root of $T$ we have cost $1$, for child of root we have cost $2$ , and etc. it's sufficient to sum over cost at each nodes on $h$. As a result we have
\begin{gather*} 
1+2+3...+h=O(h^2).
\end{gather*}
Now as mentioned in question,
$O(h^2)$ must be $O(n\log n)$. More formally, $O(h^2)=O(n\log n)$ that we can conclude that $h=O(\sqrt{n\log n})$.
For example consider a binary search tree with $n$ nodes such that root have sub tree with height $\sqrt{n\log n}$, and right sub tree of root have height $\log (n-\sqrt{n\log n}).$