For any triangulation of a compact surface, show that
$$v \leq f$$
where $v,e,f$ is vertices, edges and triangle respectively.
Since each triangle has 3 edges and each edge is the common edge of exactly two triangles, we get $3f=2e$ and $v-e+f= \chi$. Hence $v=\dfrac{1}{2}f+\chi$ but i can't continue to indicate $f \geq v$