Size of outerplanar graph If $G$ is an outerplanar graph of order $n \geq 2$ and size $m$, show that $m \leq 2n -3$

[I can show the result for Hamiltonian outerplanar graphs, and I think its posible to extend the result, but my proof is very inelegant and tedious. Any ideas?]
We can assume that the graph $G$ is 2-connected, otherwise, for two adjacent blocks, let $v$ be their common vertex, in a planar embedding, we can see that we can add an edge $\{w_1,w_2\}$, where $w_1,w_2\in\Gamma(v)$, such that $G$ is still an outerplanar graph, but one block less.

Then, we assume that all inner faces are triangles. Hence, with these conditions, $2m=n+3(n-2)$, that is $m=2n-3$, as an $n$-gon is triangulated by $n-2$ triangles.

Note that what we are doing is just adding edges to make it Hamiltonian, and then triangulate it. Hence, back to the original graph, we have $m\leq 2n-3$.
We can assume that the graph $G$ is 2-connected, otherwise, for two adjacent blocks, let $v$ be their common vertex, in a planar embedding, we can see that we can add an edge $\{w_1,w_2\}$, where $w_1,w_2\in\Gamma(v)$, such that $G$ is still an outerplanar graph, but one block less.

Then, we assume that all inner faces are triangles. Hence, with these conditions, $2m=n+3(n-2)$, that is $m=2n-3$, as an $n$-gon is triangulated by $n-2$ triangles.

Note that what we are doing is just adding edges to make it Hamiltonian, and then triangulate it. Hence, back to the original graph, we have $m\leq 2n-3$.