Recall that the girth of a graph is the length of one of its shortest cycles; if the graph has no cycles, we define its girth as being infinite. Prove that if $G$ is a planar graph of girth at least $g$ that is not a forest, then $e(G)\leq g (v(G) − 2)$.
To prove that a planar graph $G$ of girth at least $g$ that is not a forest satisfies $e(G) \leq g(v(G) - 2)$, we will use the fact that a planar graph follows Euler's formula for connected graphs, which relates the number of vertices $v$, edges $e$, and faces $f$ by the equation $v - e + f = 2$.
Given that $G$ is not a forest, it has at least one cycle. The girth $g$ of $G$ is the length of its shortest cycle, so $g$ is at least $3$ because the shortest possible cycle in a graph is a triangle.
Every face in a planar graph of girth $g$ is bounded by at least $g$ edges. However, since each edge may be counted as the boundary for up to two faces (except for the edges on the outer face, which are counted once), the following inequality holds for the number of edges $e$ with respect to the number of faces $f$: $$g\cdot f\leq 2e.$$ Now, let's rewrite the inequality to solve for $f$: $$f\leq\frac{2e}{g}.$$ Since we are given Euler's formula: $$v - e + f = 2.$$
Substitute the inequality for $f$ into Euler's formula: $$v-e+\frac{2e}{g}\geq2.$$
Multiply through by $g$ to clear the fraction: $$g\cdot v-g\cdot e+2e\geq2g.$$
Rearrange the inequality to solve for $e$: $$g\cdot e-2e\leq g\cdot v-2g.$$
Factor out $e$ from the left side: $$e(g-2)\leq g(v-2).$$
Assuming $g\neq 2$, which must be true for a planar graph of girth at least $g$ (since $g$ is the length of the shortest cycle and must be at least $3$, we can divide both sides of the inequality by $g-2$ to get: $$e\leq g\cdot\frac{v-2}{g-2}.$$
Since $g>2$, we know that $g-2<g$. Therefore, we can state that: $$e\leq g\cdot\frac{v - 2}{g - 2}\leq g(v - 2).$$
Thus, for a planar graph $G$ with girth at least $g$ that is not a forest, it holds that: $$e(G)\leq g(v(G)-2).$$
Is this proof valid or is there a flaw somewhere?