Given $G = (V,E)$, a planar, connected graph with cycles, Prove: $|E| \leq \frac{s}{s-2}(|V|-2)$. $s$ is the length of smallest cycle 
Given $G = (V,E)$, a planar, connected graph with cycles, where the smallest simple cycle is of length $s$. Prove: $|E| \leq \frac{s}{s-2}(|V|-2)$.

The first thing I thought about was Euler's Formula where $v - e + f = 2$. But I really could not connect $v$, $e$ or $f$ to the fact that we have a cycle with minimum length $s$.
Any direction will be appreciated, thanks!
 A: This is a question of Grimaldi, Discrete and Combinatorial Mathematics, chapter 11 section 4 exercise 23,a.
You may read the last few pages of this section, if you still do not see I can give you the hint.
Good luck
The responce for the last comment:
Each edge is in the boundary of at most 2 faces. Then if you consider the sum of lengths of all faces in G, or regions, the total lenght is always less equal to 2$e$. And obviously, since $s$ is the smallest lenght, the sum of lengths is greater and equal to $sr$. Finally we have this: $2e \geq \sum deg(r_i) \geq sr$ where $deg(r_i)$ is the number of edges in the boundary of face. 
A: Lets use euler's theorem for this n-e+f=2 where:
n-vertices, e-edges and f-faces
Let $d_1,d_2,...,d_f$ where each $d_i$ is the number of edges in face $i$ of our grph. A cycle causes us to have a face, and according to this our smallest cycle is of size s $\Rightarrow |d_i| \geq s$
$d_1+d_2+...+d_f = 2e \Rightarrow s \times f\leq 2e\Rightarrow $
(f=2-n+e)
$ 2e\geq s(2-n+e)\Rightarrow 2e\geq 2s-sn+se\Rightarrow s(n-2) \geq e(s-2)$
Now just divide and you'll get the desired result.
