Suppose that $G$ is a planar triangulation (also the outer face has to be triangle) on $n \geq 4$ vertices. Let X be a subset of vertices of $G$ such that $|X| \leq n-3$. Let $F(X)$ be the set of all faces that are incident to at least one vertex in $X$. How would you this lower bound:
$|F(X)| > 2|X|$?
My idea was that if $X$ is an independent set, then since $G$ must be $3$-connected, it follows that the degree of every vertex of $G$ is at least $3$, so it has at least $3$ incident triangles and $|F(X)| \geq 3|X|$.
However, if $X$ is not independent, then I would like to proceed by contracting and edge $e = \{uv\}$ such that $u, v \in X$. Such an edge can be contracted if every triangle containing $e$ is also a face. Here, I am stuck.
Do you have any ideas?
The total number of triangles is $2n-4$, by Euler's formula and double-counting. So, you want to prove that the number of triangles not incident to $X$ is at most $2(n-|X|)-4$. An easy way is as follows: Let $Y$ be the complement of $X$. Triangles not incident to $X$ have their vertices contained in $Y$, so there are at most $2|Y|-4$ of them (you can always complete the edges contained in $Y$ to a triangulation of $Y$).