Planar graph, Minimum degree of a vertex at least 3 and Number of faces of a graph

Question

Lemma: Suppose that a plane simple graph on $n ≥ 4$ vertices with the minimal degree of
a vertex at least $3$ does not have faces of degree $4$ or $5$. Prove that there are at least $4$ faces of
degree $3$ (triangles) in this graph.

I saw that a similar question was asked by Dolva Planar graph, number of faces, minimum vertex degree 3. Where the Handshaking lemma and Euler's formula $v-e+f=2$ were suggested. But I am still stuck.

Any advice or help is welcome. Thanks in Advance!

Answer 1

Hints:

1. $v\leq2e/3$;
2. $3k+6(f-k)\leq2e$, $k\leq3$, then $f\leq e/3+k/2$;
3. $2=v-e+f\leq 2e/3-e+e/3+k/2=k/2\leq3/2$.