Does triangle inequality imply Euclidean metric? We know that the distances between vertices in a Euclidean space satisfy the triangle inequality, but is the converse true? Specifically, given a complete graph $K_n$ of $n>2$ vertices, along with $\binom{n}{2}$ distances $\{d_{ij}>0,\;\;1\le i < j\le n\}$ that satisfy triangle inequality, can the graph be embedded into a Euclidean space (with an arbitrary finite dimensions)?

No. Consider the situation $V=\{a,b,c,d\}$ with distances
$d(a,d)=2$ and $d(a,b)=d(a,c)=d(b,c)=d(b,d)=d(c,d)=1$.
The triangle inequality is satisfied since left-hand side of $d(x,z)\leq d(x,y)+d(y,z)$ is less or equal $2$ and the right-hand side is greater or equal $2$.
We have $d(a,d)=d(a,b)+d(b,d)$ and $d(a,d)=d(a,c)+d(c,d)$. In Euclidean spaces these imply that $b$ and $c$ are midpoints of $[a,d]$. Therefore $b$ and $c$ must be equal (but they aren't).
