Derivation of Symmetry Property of Metric Spaces I am given the following modified triangle inequality property of metric spaces, where for any $x_1$, $x_2$, $x_3 \in X$, we have $d(x_1, x_2) \le d(x_1, x_3)+d(x_2, x_3)$.  I am tasked to show that the symmetry property follows from the use of this and the positive definite property.
I am having difficulty moving beyond the definition of Euclidean distance in thinking about this problem.  Can someone offer a suggestion to get started?
 A: Let $x_3=x_1$ in the formula above. This gives $d(x_1,x_2) \le d(x_1,x_1) + d(x_2,x_1) = d(x_2,x_1)$.
Exchanging the roles of $x_1,x_2$ gives $d(x_2,x_1) \le d(x_1,x_2)$.
Hence $d(x_2,x_1) = d(x_1,x_2)$.
A: Although the accepted answer is very intuitive, I think it contains a mistake. Exchanging the roles doesn't lead to the obtained(desired) right hand side $d(x_1, x_2)$. For instance:
$d(x_2, x_1) \leq d(x_2, x_3) + d(x_1, x_3)$. Using the fact that $x_1 = x_3$ we have: $d(x_2, x_1) \leq d(x_2, x_1) + d(x_1, x_1)$. Since $d(x_1, x_1) = 0$ (non-degeneracy property) then we concluded that $d(x_2, x_1) \leq d(x_2, x_1)$. As can be observed, the RHS differs from the desired one.
In my opinion, deriving the symmetry property from the non-negative and triangular inequality properties is not possible in the general case (that is, without any information about the underlying distance function. Perhaps, defining better what you meant with "positive definite property" could help here). Aside from the fact that the symmetry property is an axiom, theory states that an space without this axiom is known as quasi-metric space, and no proof exits about that every quasi-metric space can be transformmed into a metric space. Note that the last is a consequence of deriving symmetry property from the other ones only.
