On a particular kind of simplicial complex with maximum facet size of $3$. Let $\Delta$ be an abstract simplicial complex  on $n$ vertices such that $\max \{|F| : F$ is a face of $\Delta \}=3$. 
Let $f_2$ be the number of faces of size (cardinality) $3$ and $f_1$ be the number of faces of size (cardinality) $2$. 
Also assume that the Stanley-Reisner ring $k[\Delta]$ is Cohen-Macaulay for every field $k$ of characteristic $0$. 
My question is: If $3f_2=2f_1$, then is it necessarily true that $3n-6\ge f_1$ ? 
NOTE: Some examples of simplicial complexes with $3f_2=2f_1$ are pseudo-manifolds. 
 A: No. In fact, for pseudomanifolds (without boundary) the opposite inequality holds.  If $\Delta$ is a pseudomanifold of dimension $n$, then $$H_n(\Delta,\mathbb{Z}) \cong \begin{cases} \mathbb{Z} & \text{if $\Delta$ is orientable} \\ 0 & \text{if $\Delta$ is not orientable.} \end{cases}$$  If particular, if $n=2$, then $\chi(\Delta) \le 2$. Then
$3\chi(\Delta)=3f_0-3f_1+3f_2=3f_0-3f_1+2f_1=3f_0-f_1 \le 6$. Thus $3f_0-6 \le f_1$.
If we further assume $\Delta$ is Cohen-Macaulay (over some field), then $\chi(\Delta)=2$ if $\Delta$ is orientable, and $\chi(\Delta)=1$ if $\Delta$ is not orientable. The same argument as above then shows that $f_1=3f_0-6$ in the orientable case, and $f_1=3f_0-3$ in the non-orientable case. 
For a concrete example of the orientable case, take a triangulation of a $2$-dimensional sphere, and for the non-orientable case, take a triangulation of the real projective plane $\mathbb{R}P^2$. The latter of these is not Cohen-Macaulay over every field, but is Cohen-Macaulay over fields of characteristic $0$ (or any field of characteristic not $2$).  
