I'm trying to interpret a verified solution for the following problem.
Show that $v_3+f_3>0$. Here, $v_n$ denotes the number of vertices of a convex polyhedron that meet with $n$ edges, and $f_n$ denotes the number of faces that have $n$ edges.
The solution that I'm reading uses the property that the sum of angles (between two consecutive edges) around any vertex is less than $2\pi$, and it proceeds as follows.
Let $F,V,E$ be # of faces, vertices, and edges of a convex polyhedron. And, assume that $v_3+f_3=0$.
As we already know that the sum of angles around a vertex must be less than $2\pi$, we get a following inequality: $\sum \text{angles} < 2\pi V$.
But, $\sum \text{angles} = \sum (n-2)f_n \pi$ because the sum of angles of an $n$-gon is $(n-2)\pi$.
i.e. $V>\sum (n-2)f_n$. (This is where I'm having problem.)
Note that $E=\frac{1}{2}\sum nf_n$.
So, $2 = V + F - E > \sum (\frac{n}{2}-1)f_n$.
It is clear to see that $v_3+f_3 \geq 0$, but assuming $v_3+f_3=0$ gives us a contradiction because when $n \geq 4$, $2>\sum f_n$, which is not possible.
The problematic part: $V>\sum (n-2)f_n$
First, I thought this was a typo, and I thought it was supposed to be $2V>\sum (n-2)f_n$ as we can cancel $\pi$ out on both sides of the previous inequality. However, then this inequality only gives me that $2>0$, which doesn't imply anything.
$$2V>\sum (n-2)f_n \\ \Rightarrow V>\sum(\frac{n}{2}-1)f_n \\ \Rightarrow 2=V+F-E>0$$
So, now I'm trying to observe for any weird convex polyhedron so that I can find a counterexample of $V>\sum (n-2)f_n$, but the inequality seems right.
I am wondering where did this inequality come from, and if possible, I would like to know details behind here that are probably omitted because of the obvious reasons that I'm missing here.
Any help would be greatly appreciated.
Unfortunately, you’re right, the proof is just wrong. You don’t even need a weird convex polyhedron; the inequality
$$ V\gt\sum(n-2)f_n $$
is violated for the dodecahedron,
$$ V = 20 \lt 36 = (5-2)\cdot12\;, $$
and for the icosahedron,
$$ V = 12 \lt 20 = (3-2)\cdot20\;. $$
If I had to guess how a correct proof might work, I’d suspect one might be able to use duality (since $v_3+f_3\gt0$ means that either the polyhedron or its dual has at least one triangle). If I find a proof I’ll update the answer.
Update:
Indeed, duality allows for a nice straightforward symmetric proof. If $v_3=f_3=0$, then $V\gt\frac12\sum(n-2)f_n\ge\frac12\sum(4-2)f_n=F$, and by duality analogously $F\gt V$, a contradiction.