Something sharper holds. There exists a spanning tree with at least $\frac{1}{4}n+1$ leaves. https://artofproblemsolving.com/community/c6h446932

Official solution (google translation). We will step by step construct a tree in the graph ~ $ G $ so that after each step, more than a quarter of all the vertices in this tree were hanging. The number of vertices in the tree will increase with each step. First, consider any vertex ~ $ u $ and all vertices adjacent to it. These the vertices form a tree in which more than a quarter (in fact, even at least two-thirds) of all vertices ~ --- hanging. Suppose that after the next step we have a tree ~ $ F $ with many vertices ~ $ W $, in which more than a quarter of all vertices are hanging. Consider a few cases. 1. Let there be a vertex ~ $ x $ in the tree ~ $ F $ adjacent to at least ~ two peaks not included in the tree. Then we attach these vertices to ~ $ F $. The number of vertices will increase by ~ $ m \geq2 $, and ~ the number of hanging vertices ~ --- by $ m-1 $ (since one of the vertices of the tree is no longer hanging). Therefore, we got a larger tree, in which ~ more than a quarter of all peaks are hanging. 2. Let the tree ~ $ F $ have no vertices adjacent to at least two vertices, not falling into a tree. If not all vertices are included in the tree, then consider the set ~ $ W_1 $, consisting of all not included in the tree ~ $ F $ vertices adjacent to at least ~ one vertex of the tree ~ $ F $. If ~ $ W_1 $ has a vertex ~ $ x $ adjacent to at least two vertices not of of the tree ~ $ F $, then we attach ~ $ x $ to the ~ tree, and ~ to ~ the top ~ $ x $ we attach all vertices adjacent to it that are not included in the tree ~ $ F $. The number of vertices in the tree increased by ~ $ m \geq3 $, and the number of hanging vertices ~ --- by ~ $ {m-2} $. Therefore, we got a larger tree in which more than a quarter of all peaks are hanging. 3. Suppose that neither the tree ~ $ F $ nor the set ~ $ W_1 $ has vertices adjacent though would be with two vertices not included in the tree. If this is not all vertices of the graph ~ $ G $, then we consider the set ~ $ W_2 $, consisting of all not included in the tree ~ $ F $ and the set ~ $ W_1 $ vertices adjacent to at least one vertex from ~ $ W_1 $. Let $ x $ ~ be a vertex from the set ~ $ W_2 $. By construction, the vertex ~ $ x $ is not adjacent from none of the vertices of the tree ~ $ F $ and is adjacent to some vertex $ y \in W_1 $. Then we attach to the tree ~ $ F $ the vertex ~ $ y $, to it the vertex ~ $ x $, and to vertex ~ $ x $ --- all remaining vertices adjacent to it. (Such peaks should at least two, and both of them have not yet entered the tree!) Obviously, the increase quantity hanging peaks in the tree will be at least a quarter of the total increase number of vertices. Consequently, we have a larger tree, in which more quarters of all peaks are hanging. We will do the steps described above while it is possible. Let $ T $ ~ be a tree, for which it is impossible to take the step described above, and in it $ w $ vertices and $ e $ hanging vertices, where $ e> \frac w4 $. By construction, non-hanging tree vertices ~ $ T $ are not adjacent to vertices not in ~ $ T $. Since it’s impossible to \hbox {take} a step, then every hanging vertex of the tree ~ $ T $ adjacent to at most one vertex not in ~ $ T $, and ~ each not the vertex ~ $ x $ included in ~ $ T $ is also adjacent to at most ~ one vertex from ~ $ T $. Therefore, ~ $ x $ must be adjacent to at least two hanging peaks from a tree ~ $ T $. This means that the number not included in the vertex tree does not exceed ~ $ \frac e2 $. By attaching these vertices to the ~ tree ~ $ T $, we get a spanning tree in which exactly ~ $ e $ hanging vertices and ~ less than $ 4e + \frac {e}2 $ vertices. Hence, in this tree, at least $ \frac29 $ of the total number of vertices \medskip \dagger Try modifying the proof of the problem so that it turned out a tree in which more than a quarter of all the vertices are hanging. The estimate ~ $ \frac14 $ is accurate: it is easy to construct a series of examples connected graphs for which the maximum possible number of hanging peaks in the core the tree is arbitrarily close to a quarter of the number of vertices and the degree of all there are at least ~ 3 vertices. \medskip \dagger \! \! \dagger And how many hanging vertices can be provided in the spanning tree of a connected graph, whose vertex degrees are at least ~ $ d $ (where ~ $ d \geq4 $)? It is easy to construct a series of examples of connected graphs with a minimal degree vertices ~ $ d $ for which the maximum possible number of hanging vertices in spanning tree arbitrarily close to ~ $ \frac {d-2} {d + 1} $ of the total the number of vertices of the graph. (For ~ $ d = 3 $, the estimate ~ $ \frac14 $ is just obtained.) Most likely, for any such graph there exists a spanning tree in which ~ the number of hanging vertices is more than ~ $ \frac {d-2} {d + 1} $ of the number of vertices of the graph. At the moment, the author of the problem has (quite difficult in comparison with the problem) proofs of this fact for $ d = 4 $ and ~ $ d = 5 $

Would you give a link to the Russian text?

fagot wrote: ... And how many hanging vertices can be provided in the spanning tree of a connected graph, whose vertex degrees are at least $ d $ (where $ d \geq4 $)?... Most likely, for any such graph there exists a spanning tree in which the number of hanging vertices is more than $ \frac {d-2} {d + 1} $ of the number of vertices of the graph. It was conjectured in 1987 by N. Linial, known as Linial's conjecture. In 1990, Noga Alon proved it's not true for large enough $d$. fagot wrote: At the moment, the author of the problem has (quite difficult in comparison with the problem) proofs of this fact for $ d = 4 $ and $ d = 5 $ As I got it, you are the author of this problem!? The case $d=4$ was proved by Kleitman and West (1991). Grigss and Wu proved it for $d=5$ (1992). For $d=3$ (your problem) it was proved by Linial, Sturtevant (1987), thou in 1981, J. R. Storer found an algorithm for constructing a spanning tree in a cubic graph with at least $n/4$ leaves. The same algorithm works for any graph with minimal degree $3$. For more information see my blog.