Find the upper bound for $\chi(G)$ using theorem 8.20 Given following graph 

a) Find the upper bound for $\chi(G)$ using theorem 8.20
b)What is $\chi(G)$?

Theorem 8.20: For every graph $G$, $\chi(G) \leq 1+\max\{\delta(H)\}$, where maximum is taken over all subgraph $H$ of $G$

I know that $\chi(G)=5$ so part b) is done. For part a), I'm not quite sure about $\max\{\delta(H)\}$ part, so I need to find the minimum degree in all subgraph $H$ and pick the biggest one? Is $\max\{\delta(H)\}=5$? Do I need to prove it?
 A: As JMoravitz points out, there are only two vertices of degree at least six, so $\max_{H \subseteq G} \delta(H) \leq 5$. By deleting the edge that connects these two vertices of degree six, we obtain a $5$-regular subgraph $H'$ so that $\max_{H \subseteq G} \delta(H) \geq \delta(H') = 5$. Hence, using the given theorem, it follows that:
$$
\chi(G) \leq 1 + \max_{H \subseteq G} \delta(H)
= 1 + 5
= 6
$$
Now to obtain the lower bound $\chi(G) \geq 5$, we argue by contradiction. Suppose instead that $G$ is $4$-colourable. Then without loss of generality, we can colour the four vertices of the $4$-clique subgraph with four distinct colours as follows:

But by symmetry, $G$ has another $4$-clique, two of which are coloured blue and purple. This forces the remaining two vertices of this reflected $4$-clique to be coloured as follows:

But it is now impossible to properly colour the topmost vertex, as it is adjacent to four vertices with four different colours. To prove the tighter upper bound $\chi(G) \leq 5$, we use the following proper $5$-colouring:

Thus, $\chi(G) = 5$, as desired.
