Planar graph $G$ that does not contain any cycle of length $\leq k-1$

Question

Let $G=(V,E)$ be a planar graph that does not contain any cycle of length $\le k-1$ where $3\le k \le |V|$.

Prove the inequality $(k-2)|E|\le k|V|-2k$

My attempt so far. If the graph has has $f$ faces/regions than it follows that $(k-2)f\le2m$ where $m$ denotes the number of edges.

Applying this to the Euler formula: $n+f=m+2$. I don't get the desired inequality.

What am I missing ? Would appreciate any help

Answer 1

You should consider the inequality $k|F|\leq 2|E|$. Then $$|E|=|V|+|F|-2\leq |V|+\frac{2|E|}{k}-2\implies (k-2)|E|\le k|V|-2k.$$