# Let $G$ be a planar (simple) graph with $m (≥2)$ edges and $f$ faces. Prove that $f ≤ \frac{2}{3} m$.

I am a first year math student, and I'm studying some exercises for my graph theory exam. This exercise I didn't understand very well:

Let $G$ be a planar (simple) graph with $m (≥2)$ edges and $f$ faces. Prove that $f ≤ \frac{2}{3} m$.

If I look at examples this make sense, but I find it hard to prove it. In algebra or analysis I'm used to prove something from other theorems, but here it seems like I need to prove it really out of nothing.

This is an example I can find where $3f=2m$:

And now, if I add more lines, I can't get examples that disprove it, but I find it hard to reason why this is true. Any hints ?

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@Kasper By facets you probably mean faces. Are you aware of Euler's Formula? Moreover, if the graph has at least 2 edges, what is the minimum number of vertices it can have? –  user39280 May 11 '13 at 10:14
@dado The Euler formula is introduced exactly after this exerise. So yes, I'm aware of it, but I think I need to prove this without that theorem. –  Kasper May 11 '13 at 10:18

Put a pebble on each side of each edge. Each face will have at least $3$ pebbles (why?), so number of pebbles is at least $3$ times number of faces, but also each edge has $2$ pebbles, so number of pebbles is twice number of edges.
I love your explanation. So you need at least three lines to enclose a face. So that is at least 3 pebbles. For the face that is not enclosed you have two options: Or there is a three like graph, and as $m ≥2$ you will get at least 4 pebbles. Or there is no three, and then you have at least 3 pebbles. –  Kasper May 11 '13 at 10:30
Hint: You know that the sum of the degrees of all vertices is $2m,$ and each face consists of at least three vertices.