Find a neccessary and sufficient condition for $G$ to be Hamiltonian a) prove that $K_{r,2r,3r}$ is Hamiltonian for every postive integer $r$
b) prove that $K_{r,2r,3r+1}$ is Hamiltonian for no postive integer $r$
c) Let $G=K_{n_1 , n_2, \dots ,n_k}$ be the complete k-partite graph of order $n \geq 3$, where $n_1 \leq n_2 \leq \dots \leq n_k$. Find a neccessary and sufficient condition for $G$ to be Hamiltonian

Dirac's Theorem: If $G$ is a graph of order $n \geq 3$ such that $deg(v) \geq \frac{n}{2}$ for each $v \in V(G)$ then $G$ is Hamiltonian
Theorem 3.16: If $G$ is Hamiltonian and $S$ is vertex cut in $G$ then $k(G-S) \leq |S|$

Here is what I got so far
a) Let $r$ be a positive number. Let $X,Y,Z$ be disjoint vertex set of  $K_{r,2r,3r}$ that have order $r,2r,3r$ respectively. Since $K_{r,2r,3r}$is a 3-partite graph, $K_{r,2r,3r}$ has order $6r$. Note that every vertex in $X,Y,Z$ degree $5r,4r,3r$ respectively so for each $v \in V(K_{r,2r,3r})$, $deg(v) \geq 3r = \frac{n}{2}$. By Dirac's theorem, $K_{r,2r,3r}$ is Hamiltonian for every positive integer $r$
b) Let $r$ be a positive number. Let $X,Y,Z$ be disjoint vertex set of  $K_{r,2r,3r+1}$ that have order $r,2r,3r+1$ respectively. Since $K_{r,2r,3r+1}$is a 3-partite graph, $K_{r,2r,3r+1}$ have the vertex cut $S = X \cup Y$, so $|S| = 3r$, but $k(G-S)=3r+1 >|S|$. By the contrapositive of theorem 3.16, we can conclude that for any positive number $r$, $K_{r,2r,3r+1}$ isn't Hamiltonian.
c) the book says $G$ is Hamiltonian iff $n_k \leq \sum_{i=1} ^{k-1} n_i$. But I can't see how they got this.
 A: a) Correct!
b) Correct!
c) The idea for this is to generalize your idea for Parts (a) and (b). But maybe instead of starting from that formula, just think about how you could apply Dirac's Theorem to $K_{n_1,\ldots,n_k}$. In Part (a), the idea you used is that vertices in $Z$ have the smallest degree amongst all vertices, but that since these degrees are at least half the total number of vertices, you were able to conclude that the graph is Hamiltonian. 
Now, let $X_1,\ldots,X_k$ be the parts of $K_{n_1,\ldots,n_k}$ where $|X_i|=n_i$ and suppose you want to apply Dirac's Theorem. To do this, you need to know that no part uses more than half the vertices (otherwise any vertex in that big part has degree strictly less than $\frac{n}{2}$ and you couldn't apply Dirac's Theorem). 
Since $X_k$ is the largest part, this means that you want to assume $n_k\leq \frac{n}{2}=\frac{n_1+\cdots +n_k}{2}$. That inequality is equivalent to the one in your book.
On the other hand, if any part did have more than half the vertices, you can apply a similar argument as in Part (b) to conclude non-Hamiltonicity. 
So in my opinion, your book doesn't really give the most illuminating answer in this case. I think it would be better stated as "$K_{n_1,\ldots,n_k}$ is Hamiltonian if and only if no part contains more than half the vertices."
