Proving that any planar embedding of $G$ has at least $4$ triangular faces

Question

Let $G=(V,E)$ be a connected simple planar graph whose edges can be colored red and blue so that for any vertices $u,v∈V$, there is a unique path connecting $u$ and $v$ whose edges are all red, and a unique path whose edges are all blue.
Prove that any planar embedding of $G$ has at least $4$ triangular faces (i.e., faces with degree $3$). This count may include the outer (exterior)face.

I found that each edge in this graph is contained in a cycle and each face has at least degree $3$. Also I have the formula that $|V| - |E| + |F| = 2$ and $3|F| \leqslant 2|E|$, but I am not sure how to relate this to the number of triangular faces in the graph.

Can you help me solve this problem?

Answer 1

Here's an approach, can you fill in the details?

1. Argue that $E = 2V - 2$.
2. Argue that if $G$ has fewer than $4$ triangular faces, then $2E \geq 4F - 3$
3. Explain why those two are contradictory