Here's a problem I've been assigned in my graph theory class:
The degree of every vertex of a graph $G$ of order $2n+1\ge5$ is either $n+1$ or $n+2.$ Prove that $G$ contains at least $n+1$ vertices of degree $n+2$ or $n+2$ vertices of degree $n+1.$
I could use some help getting started with this one. My first thought is to utilise the first theorem of Graph Theory, but doing so hasn't led to anything useful. Thanks.
Proof (By Contradiction): Suppose the graph has less than n+1 vertices of degree n+2 and less than n+2 vertices of degree n+1 (i.e., there are at most (n+1)-1=n vertices of degree n+2, and at most (n+2)-1=n+1 vertices of degree n+1). Then the sum of all degrees is at most n(n+2)+(n+1)(n+1)=n^2+2n+n^2+2n+1=2n^2+4n+1. By the first theorem of graph theory, this is 2m, where m is the size of G. The size of G is unknown, but it can be no greater than if G is a complete graph. The complete graph has size z(z-1)/2, where z is the order; so if G is complete, then it has size {(2n+1)[(2n+1)-1]}/2 = (4n^2+2n)/2. So the sum of all degrees is 2n^2+4n+1 = 2m = 4n^2+2n. But 2n^2+4n+1 > 4n^2+2 for all n≥2; this is a contradiction. Any other such graph that is not complete will have a size less than that of this complete one. Thus they will lead to contradictions as well. Therefore G contains at least n+1 vertices of degree n+2, or at least n+2 vertices of degree n+1. QED.
Is this reasonable?
Let $x_1$ be the number of vertices with degree $n+1$, $x_2$ be the number of vertices with degree $n+2$.
For the sake of contradiction, assume that $x_1 < n+2$ AND $x_2 < n+1$.
Thus $x_1 \leq n+1$, $x_2 \leq n$. We know that $x_1 + x_2 = 2n+1$. This equality only holds when $x_1 = n+1$ AND $x_2 = n$.
Thus the sum of degree across all nodes is
$$ (n+1)(n+1) + (n)(n+2) = 2n^2 + 4n + 1 $$
Which is clearly odd. However, the sum of all degrees in a graph must be even. Contradiction.
Thus $x_1 \geq n+2$ OR $x_2 \geq n+1$