A question on a graph which is a a "union" of two trees Suppose that $T_1=(V,E_1)$ and $T_2=(V,E_2)$ are two trees and let us consider the graph $G=(V,E_1\cup E_2)$.
If $E_1\cap E_2=\varnothing$, then there exists a vertex with $\deg(x)\leqslant3$. Indeed, otherwise $2|E|\geqslant 4|V|$, that is, $|E|\geqslant 2|V|$ by the fundamental Theorem of Grpah Theory. Now, since $T_1,T_2$ are trees, it follows that $2|V|-2\geqslant 2|V|$, a contradction.
(see also Prove that in the union of two trees there exist a vertex with degree of at most $3$)
Remark: By $(V,E_1\cup E_2)$ I mean to the simple graph.
My question: Is this result true also if we allow $E_1\cap E_2\neq\varnothing$ ? I can't find any conuterexample.
 A: Intuition: If this isn't true, then the $G$ has a fairly large number of edges to give every vertex degree $\geq 4$, which seems unlikely since we are starting with trees, so we should be able to find a contradiction in sums of degrees.
Claim: If $T_i=(V,E_i)$ are trees for $i=1,2$, then $G=(V,E_1\cup E_2)$ has a vertex of degree $\leq 3$.
Proof: Suppose for contradiction that no such vertex exists. So $\deg(x)\geq 4$ for all $x\in V$. Therefore if $x$ is a leaf (that is, a vertex of degree 1) in $T_1$, then  $x$ must have degree at least 3 in $T_2$, so that $x$ has degree at least 4 in $G$.
Let $L_1,L_2$ be the number of leaves in $T_1,T_2$ respectively, and $X_1,X_2$ be the number of vertices of degree $\geq 3$ in $T_1,T_2$ respectively.
So the observation above is saying that we must have $X_1\geq L_2$ and $X_2 \geq L_1$ (i.e. we need at least one vertex of degree $\geq 3$ in $T_2$ for every leaf in $T_1$ and vice verse).
Now summing the degrees in $T_i$ and bounding below:
$$
2|V|-2 = \sum_{x\in V} deg(x) \geq L_i  + 2(|V|-L_i-X_i) + 3X_i
$$
Rearranging gives:
$$
X_i \leq L_i - 2
$$
Combining this with previous inequalities:
$$
L_2 \leq X_1 \leq L_1 - 2 \leq X_2 -2 \leq L_2 - 4
$$
which is a contradiction. Hence we have proved the claim.
A: Actually your reasoning goes.

*

*If $\deg(x)\geq4$ for every vertex $x\in V(G)$, then by the handshake lemma
$$
2|E(G)|=\sum_{x}\deg(x)\geq4n\ \Rightarrow\ |E(G)|\geq2n.
$$
($|V|=n$)


*On the other hand we have $|E(T_1)|=|E(T_2)|=n-1$, so
$$
|E(G)|=|E(T_1)\cup E(T_2)|\leq|E(T_1)|+|E(T_2)|\leq2n-2.
$$
