Note that in a tree with n vertices the number of edges is (n-1).
Let us assume that the degree of all the vertices be greater than or equal to 4.
Now, assume that the graph G is partitioned into two trees T1 and T2.
Then T1 and T2 both can have at most n vertices.
So, T1 and T2 both can have at most (n-1) edges.
Since the graph G is partitioned into T1 and T2 we get that the graph G can have at most (n-1) + (n-1)= 2n – 2 edges. …………..(1)
Now, we have assumed that the degree of the vertices of G is greater than or equal to 4.
So, the sum of the degrees is greater than or equal to 4n.
Hence by hand-shaking lemma we get the number of edges is greater than or equal to (4n / 2) = 2n ……………….(2)
So, we get that the statements (1) and (2) are contradictory in nature.
Hence, our assumption is wrong.
Hence, there exist at least one vertex with degree less than 4.