How is this a legitimate application of the handshake lemma

Question

Let $G$ be a graph such that $E(G)$ can be partitioned into $E1,E2$ such that $G/E1$ and $G/E2$ are planar, show that the chromatic (vertex) number of $G$ is $\le 12.$

My professor said:

By induction on $|V(G)|$, the base case is clear because if we have less than or equal to 12 vertices, then clearly dont need as many colours.

Now assume $n=|V(G)| \ge 13$

then be Euler

$|E(G/E1)| \le 3n-6$ and $|E(G/E2)| \le 3n-6 $

so $|E(G)| \le 6n -12$

So far im completely following, what he said next is where I am not understanding.

He then said, it thus follows from the handshaking lemma that

$\sum_{v \in V(G)} deg(v) \lt 12n$

How?

I thought the handshake lemma tells us

$\sum_{v \in V(G)} deg(v)= 2|E(G)| $

so wouldnt we have

$\sum_{v \in V(G)} deg(v)=12n-24$ ?

Where am I missing something? How is this a valid application of handshake lemma?

Thanks

Answer 1

No, you are not missing anything. We have that $\sum\limits_{v\in V(G)}deg(v)= 2|E|\leq 2(6n-12)=12n-24<12n$.

Which is what you wanted.