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  "body": "\n## Proof\n\nWe count the pairs of a face and an edge side on its boundary, and then apply Euler's formula.\n\n1. Let $G$ be a finite simple [[connected|0FFAYNMAJZ]] [[planar graph|029JQEK15B]] with $v \\ge 3$ vertices and $e$ edges. Fix a drawing of $G$ in the plane with no crossings, and let $f$ be its number of faces.\n2. The graph $G$ has at least two edges. Indeed, $G$ is connected with at least two vertices, so each vertex lies on an edge. One edge covers only two of the $v \\ge 3$ vertices.\n3. Each edge of the drawing has two sides, and each side borders exactly one face. For a face $F$, let $\\ell(F)$ be the number of edge sides that border $F$. An edge with $F$ on both sides counts twice.\n4. Every edge side borders exactly one face, so $\\sum_F \\ell(F) = 2e$, where the sum runs over all $f$ faces. This is a [[double count|0F8YBHF2W3]] of the edge sides.\n5. Since $G$ is connected and has an edge, the boundary of each face $F$ is traced by a closed [[walk|0VJZM0F9YM]] of length $\\ell(F)$ in $G$.\n6. We claim that $\\ell(F) \\ge 3$ for each face $F$. Since $G$ is connected and has an edge, every face borders some edge side, so $\\ell(F) \\ge 1$. A closed walk of length $1$ would use a loop, and a simple [[graph|0JD5C82QMB]] has no loops.\n7. A simple graph has no two edges with the same ends. So a boundary walk of length $2$ goes along one edge $xy$ and straight back. Then $F$ lies on both sides of $xy$ and touches no other edge. At $x$ and $y$ the walk turns around, which happens only at vertices of degree $1$. Then $G$, being connected, would be the single edge $xy$, contrary to Step 2. This proves the claim.\n8. By Steps 4 and 6, $3f \\le \\sum_F \\ell(F) = 2e$.\n9. By [[Euler's formula|0W1C598HMS]], $v - e + f = 2$, so $f = 2 - v + e$.\n10. Substituting into Step 8 gives $3(2 - v + e) \\le 2e$, that is, $6 - 3v + 3e \\le 2e$.\n11. Hence $e \\le 3v - 6$.\n\n## Idea\n\nEach face is surrounded by at least three edge sides, and each edge has exactly two sides. So there are at most $2e/3$ faces. Euler's formula ties the number of faces to $v$ and $e$, and this turns the bound on faces into a bound on edges.\n",
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