A $(3k+3)$-node graph with degrees $k+1,\dots,k+3$ has $k+3$ degree-$(k+1)$ nodes or $k+1$ degree-$(k+2)$ nodes or $k+2$ degree-$(k+3)$ nodes? 
Graph $G$ has order $n=3k+3$ for some positive integer $k$ . Every vertex of $G$ has degree $k+1$, $k+2$, or $k+3$. Prove that $G$ has at least $k+3$ vertices of degree $k+1$ or at least $k+1$ vertices of degree $k+2$ or at least $k+2$ vertices of degree $k+3$

I used the proof by cases.
In case 1: $k$ is odd I can show that $G$ has at least $k+1$ vertices of degree $k+2$. Otherwise it will contradict the theorem that say every graph have even number of odd vertices.
Same reason, in case 2: $k$ is odd I can show that at least $k+2$ vertices of degree $k+3$
I just can't figure out how to show $G$ has at least $k+3$ vertices of degree $k+1$

It seems like this final case has nothing to do with the actual value of each degree.
You're really after showing it has at least $k+3$ vertices of degree $k+1$ if it doesn't have at least $k+1$ vertices of degree $k+2$ and at least $k+2$ vertices of degree $k+3$. 
Let $a,b,c$ denote the number of vertices with $k+1,k+2,k+3$ edges respectively. Assuming this is a simple graph (is it?), we have that $a+b+c=3k+3$. We are assuming $b<k+1$ and $c<k+2$. It follows that $a+k+1+k+2>3k+3$. So $a>k$, which means $a\geq k+1$. 
