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 $k+1$

Let $a,b,c$ denote the number of vertices of degree $k+1,k+2,k+3$ respectively.
Then we have $a+b+c=3k+3$
If $a\geq k+3$ or $c\geq k+2$ then we are done.
Now if $a< k+3$ and $c< k+2$ then we have to show that $b\geq k+1$.
Thus we have $a\leq k+2$ and $c\leq k+1$. Now we consider two cases.
Case $1$: k is odd.
Putting the values of $a$ and $c$ in $a+b+c=3k+3$ we have $b\geq k$. Now if $b=k$, then the number of vertices containing odd($k+2$) degree is odd. So, $b\neq k$. Thus $b\geq k+1$.
Case $2$: k is even.
Now $a+c\leq 2k+3$. Now if $a+c=2k+3$, then the number of vertices containing odd($k+1$ and $k+3$) degree is odd($2k+3$). So, $a+c\neq 2k+3$. Therefore $a+c\leq 2k+2$. Putting this value in the equation $a+b+c=3k+3$ we have $b\geq k+1$. Hence proved.
