Let $F$ be a simplicial surface with no boundary. Prove the following inequality:
$$V\geq \frac{1}{2}\left(7+\sqrt{49-24\chi(F) }\right) =\frac{1}{2}\left(7+\sqrt{1+48g }\right).$$ Where $V$ is the number of $0-$simplicies in $F$ and $g$ is the genus of $F$.
My ideas: Write $\chi(F)$ as $V-E+F$.(here the $F$ is for faces) If I assume whether the surface is orientable or not I could use $\chi(F) = 2-2g$ or $\chi(F)=2-g$.
I asked the assistant and he told me I should use the property $3F=2E$. I found out that the surface is orientable but not how that helps me. Further I narrowed it down to the follwing two equations:
$V^2-V\geq 2E$ and $V^2-V\geq 3F$ but that doesn't seem to help me either. Anyone??
We have $V-1 \ge \text{deg}(x)$ for each vertex $x$ of the triangulation. Thus, summing over the vertices yields $$V^2-V= V(V-1)= \sum_{x \text{ vertex}} (V-1)\ge \sum_{x \text{ vertex}} \text{deg}(x) = 2E,$$ where the last equality is the handshaking lemma. Using the equality $3F=2E$ (each edge of the triangulation is exactly the boundary of two faces) we can conclude that the desired inequality holds.