Let $G$ be a connected graph with $n$ vertices and let $d$ be the maximum degree over all $v\in V(G)$. Prove that $G$ has two vertex-disjoing connected subgraphs containing at least $\lceil\frac{n-1}{d}\rceil$ vertices each.
Let $G$ be a tree with $n$ vertices and led $d$ be the maximum degree over all $v\in V(G)$. Prove that there exists an edge $e\in E(G)$ such that removing $e$ results in two trees with at least $\lceil\frac{n-2}{d}\rceil$ vertices each.
These problems were in a set of notes that first proved the following result:
Let $G$ be a tree and $v\in V(G)$ be any vertex. Let $N(v)=\{v_1,v_2,\dots,v_t\}$ (the neighbors of $v$ in $G$), let $e_i$ denote the edge joining $vv_i$, and let $T_i$ denote the subtree containing $v_i$ after the removal of edge $e_i$. Define $f(v)=\max_{i=1}^t|V(T_i)|$ and let $d$ denote the maximum degree over all $v\in V(G)$. Then there exists a vertex $v$ such that: $$\frac{1}{d}(n-1)\leq f(v)\leq\frac{d-1}{d}(n-1)$$
So here is some intuition (using the above result):
For the first problem, take the vertex $v$ such that $f(v)\approx\lceil\frac{1}{d}(n-1)\rceil$. Let $v_i$ be the neighbor of $v$ such that $|T_i|$ is maximal, and consider $V\setminus T_i$ (deleting $T_i$ from $V$). Then we can use the above result again, this time we get the lower bound on $f(v)$ to be $$f(v)\geq \frac{(n-1)-\lceil\frac{1}{d}(n-1)\rceil}{d}$$ (note that $d$ might decrease by $1$, or something), which is probably close to but less than or equal to $\lceil\frac{1}{d}(n-1)\rceil$, so maybe a slight fix will improve this bound (note that this is not a rigorous as I do not examine the bounds with care, but it is based off of my intuition).
The second problem is similar to the first; do what is described above to find which edge to delete.
It would be nice if someone could prove these two results rigorously using my observations and the result above about $f(v)$, but I would like to see any ideas/alternate solutions.