Case when there are more leaves than non leaves in the tree Prove that there are more leaves than non-leaves in the graph that don't have vertices of degree 2.
Ideas: If graph doesn't have vertices of degree 2 this means that vertices of the graph have degree 1 or $\geq$ 3. Vertices with degree 1 are leaves and vertices with degree $\geq$ 3 are non-leaves. Particularly, root has degree 3 (therefore 3 vertices on the level 1), level 2 has 6 vertices, level $i$ has $3*2^{i-1}$ vertices. 
Let's assume there are $n$ level, therefore according to assumption $1+\sum_{i=1}^{n-1} 3*2^{i-1} < 3*2^{n-1}$. 
There are few problems: currently I don't have an idea how to show that the above inequality is true. In addition, do I need to consider particular cases when no all leaves on the same level of the tree, or maybe when no all non leaves have the same degree, however intuitively all these cases just extend number of leaves.
I will appreciate any idea or hint how to show that the assumption is right.

If $D_k$ is the number of vertices of degree $k$ then $\sum k \cdot D_k=2E$ where $E$ is the number of edges. In a tree, $E=V-1$ with $V$ the number of vertices. So if $D_2=0$ you have
$$1D_1+3D_3+4D_4+...=2E=2V-2\\ =2(D_1+D_3+D_4+...)-2.$$
From this,
$$2D_1-2-D_1=(3D_3+4D_4+...)-(2D_3+2D_4+...),$$
$$D_1-2=1D_3+2D_4+3D_5+...,$$ and then
$$D_1>D_3+2D_4+3D_5+... \ge D_3+D_4+D_5+...$$
The sum $D_3+D_4+D_5+...$ on the right here is the number of non leaves, since $D_2=0$, while $D_1$ is the number of leaves, showing more leaves than non leaves.
