$\textbf{Definition 1:}$ A polytope $P \subseteq \mathbb{R^n}$ is the convex hull of a finite points $x_1,...,x_n \in \mathbb{R}^n$, i.e, $P=\text{conv}(\{x_1,...,x_k\})$. The dimension of P is the dimension of the affine subespace generated by the points $x_1,...,x_k$.
Consider a tetrahedron $T$ in $\mathbb{R}^3$. Obviously $T$ is a polytope. It is easy to show that each pair of vertices of $T$ are conected by an edge .
Now suppose that P is a polytope of dimension $3$ having the property that each pair of its vertices is connected by an edge. Is P a tetrahedron?
Intuitively I think it's true but I can't rigorously prove that.
By absurd suppose that $P$ is not a tetrahedron, it means that $P$ has at least $5$ vertices ($x_1,...,x_m, m\geq 5$) affinely independet. So there exist hiperplanes $H_{ij}=\{x\in R^n: c_{ij}^Tx = \alpha_{ij} \}$, with $c_{ij} \in \mathbb{R}^n$, $c_{ij}\neq 0$, $i < j$ ($i,j=1,2,...,m$ ), such that their intersection with $P$ are edges of $P$, and $P$ is in one side of the semiespaces divided by the hiperplanes:
\begin{equation} c_{ij}^Tx \leq \alpha_{ij}, \ \ \ \ \forall x \in P, \ \forall i<j, \ i,j=1,...,m \tag{1}. \end{equation}
The idea is to use $(1)$ with $x=x_i$, so we get $m$ inequalities for each edge, this is $3\cdot 2^{m-3}m\geq 60$ inequalities. I have a progress of this idea but I'm not be able to get a contradiction and I think there is a simple way to prove that $P$ is a tetrahedron.
Could anyone please help me?
(I am using Ziegler's Lectures on Polytopes)
We assume $P$ is a polytope of dimension three in ${\bf R}^3$, with $n$ vertices. The vertices and edges of $P$ form a planar graph, which let's call $G$. If all the vertices of $P$ are connected, then all the vertices of $G$ are connected, so $G$ is the complete graph on $n$ vertices, denoted $K_n$. It's well-known that $K_n$ is planar for $n\le4$, but for all $n\ge5$, it's nonplanar. Therefore, for all the vertices of $P$ to be connected, we must have $n\le4$.
The transition between convex polyhedron and planar graph is well-presented at Every polyhedral graph is planar - proof