 Solution

Count facial boundary walks with each edge-side counted once. Their total length is $2m$. Every face walk has length at least four: simplicity excludes loops and parallel-edge two-cycles, absence of triangles excludes length three, and a length-two face walk consisting solely of a bridge traversed twice could be the whole boundary only for the excluded connected graph on two vertices. Bridges may occur twice in a longer boundary walk and are still correctly counted. Thus $4f\le 2m$.

The Euler formula for a connected planar graph gives $2=n-m+f\le n-m+m/2=n-m/2$. Rearranging yields