Well, $x$, $y$ and $z$ are the lengths of the segments connecting the vertices of a right angled triangle with side lengths $\frac{5}{2}, 6\textrm{ and }\frac{13}{2}$ to its Fermat - Toricelli point. Now, the area of the right - triangled triangle is equal to $\frac{15}{2}$, but it is also equal to the sum of the areas of the three triangles formed by two vertices and the Fermat point, which is $\frac{\sqrt{3}}{4}(xy + yz + zx)$ and hence $xy + yz + zx = 10 \sqrt{3}$. Correct?

Arne wrote: Well, $x$, $y$ and $z$ are the lengths of the segments connecting the vertices of a right angled triangle with side lengths $\frac{5}{2}, 6\textrm{ and }\frac{13}{2}$ to its Fermat - Toricelli point. why is this??

Take some point $O$ and points $X$, $Y$, $Z$ such that $OX = x$, $OY = y$ and $OZ = z$, where $\angle{XOY} = \angle{YOZ} = \angle{ZOX} = 120^{\circ}$. By the cosine rule, the lengths of the sides of triangle $XYZ$ are $\sqrt{x^2 + xy + y^2}$, $\sqrt{y^2 + yz + z^2}$ and $\sqrt{z^2 + zx + x^2}$. And it's obvious that $O$ is the Fermat point of $XYZ$. Clear enough?

Hmm... I don't know that point. Is there an elementary solution?

Well, you don't actually need to know the name of some point. Just take SOME point $O$ and points $X$, $Y$, $Z$ such that $OX = x$ etcetera, and $\angle{XOY} = \angle{YOZ} = \angle{ZOX} = 120^{\circ}$. Then $XYZ$ is right angled etcetera. You see? You don't need to know ANYTHING about the Fermat point. And, believe me, this is the shortest solution you are going to get

I wonder, we can associate the given condition with Area?