Planar graph, number of faces, minimum vertex degree 3 In a finite connected planar graph $G$, such that $\delta(G)\geq 3$, show that at least $2$ faces have at most $5$ edges. I have tried writing out degree sum from Handshaking lemma and then substituting in Euler's formula but I am stuck there.
 A: Using the fact $\delta(G)\geq 3$, one could prove each edge is part of a cycle, and therefore bound exactly two faces. 
Denote by $e,v,f$ the number of edges, vertices and faces in $G$ respectively.  
Intuition
Since the proof is quite technical, I'll try to sum up the proposition I use:


*

*$(f-1)\leq \frac{e}{3}$

*$2e\geq 3v$ which implies $v-\frac{2e}{3}\leq 0$

*Euler formula $v-e+f=2$
Proof
Now assume by contradiction the statement is false, and therefore at least $f-1$ of the faces have at least $6$ edges.
This implies:
$$2e=\sum\limits_{i=1}^{f} \text{#edges if the face } f_i\geq
6(f-1)\implies \\ 
2e\geq 6(f-1)\implies (f-1)\leq \frac{e}{3}$$
Using Euler formula $v-e+f = 2$, we get $$v-e+f\leq v-e+\frac{e}{3} +1=v-\frac{2e}{3} + 1 =2 $$ 
Since
$$2e = \sum\limits_{u\in V(G)}deg(u) \geq 3v\implies \frac{2e}{3}\geq v$$
and finaly we get a contradiction:
$$v-\frac{2e}{3}\leq 0 \implies v-\frac{2e}{3} + 1 \leq 1$$
and because $$v-e+f \leq v-\frac{2e}{3} + 1 $$
we get
$$2 = v-e+f \leq 1 $$
which is a contradiction.
