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 $k+1$

Since $G$ has only vertices of degrees $k+1$, $k+2$ and $k+3$ you can split all of the vertices into 3 sets: $X$, $Y$ and $Z$ where $X$ is set of vertices with degree $k+1$, $Y$ with $k+2$ and $Z$ with $k+3$.
Now, since you already know the answers for cases:
$1)$ $|X|<k+3$ and $|Y|<k+1 \Rightarrow  |Z|\geq k+2$
$2)$ $|X|<k+3$ and $|Z|<k+2 \Rightarrow  |Y|\geq k+1$
we are interested in case 3, which is
$3)$ $|Y|<k+1$ and $|Z|<k+2 \Rightarrow  |X|\geq k+3$
Proof for that case:
Suppose $k$ is odd. Then $|Y|$ can't equal $k$ since it is odd number and if it was equal $k$ it would contradict to the theorem that states that there is even number of vertices of odd degree so since $|Y|\leq k-1$ and $|Z| \leq k+1$, $|X| \geq k+3$.
Now suppose $k$ is even. Then $|Z|$ can't equal $k+1$ for the same reason as above (that means contradiction to the theorem that states that there is even number of vertices of odd degree), so $|Z|\leq k$ and then, since $|Y|\leq k$, $|X|$ must be equal or greater $k+3$.
For the cases
$4)$ $|X|<k+3$
$5)$ $|Y|<k+1$
$6)$ $|Z|<k+2$
the reasoning is very similar. Look at case $4)$.
If $|X|\leq k+2$ then we have at least $2k+1$ vertices for $Y$ and $Z$.
When $k$ is odd, $|Y|$ must be even and thus $|Y|\neq k$. If $|Y|<k$ then $|Z|\geq k+2$. If $|Y|>k$ then it is done, since $|Y|\geq k+1$.
When $k$ is even, $|Z|$ must be even. If $|Z|=k+2$ it is done. If not, then $|Z|\leq k$ or $|Z|\geq k+4$. In the second case it is done. If $|Z|\leq k$ then $|Y|\geq k+1$. 
