if $G$ is a connected planar graph with $|V| = v$ and $|E| =e$ and each cycle in the graph is of at least length $k$

Question

if $G$ is a connected planar graph with $|V| = v$ and $|E| =e$ and each cycle in the graph is of at least length $k$, Prove that $e \leq { \big( \frac{k}{k-2} \big)} {(v -2)}$.

I was thinking to use euler's formula $e = f + v -2$ to prove this, But first I have to argue that $kf \leq 2e$ and then from here I will out this into euler's formula to get $$ke = kv + kf - 2k \implies ke = kv + kf - 2k \leq kv + 2e -2k $$

So now we have $$ ke \leq kv + 2e -2k \implies ke -2e \leq k(v -2) \implies e(k-2) \leq k(v-2) \implies e \leq \big( \frac{k}{k-2} \big) (v -2)$$.

But How would I argue in the first place that $kf \leq 2e$ to be able to proceed with my argument ?

Answer 1

The answer is the proof of Theorem 6.10 in:

http://compalg.inf.elte.hu/~tony/Oktatas/TDK/FINAL/Chap%206.PDF

As you guess at first we must prove: $kf\leq 2e$

Note: girth is the length of the smallest cycle