If F is the Fermat point of triangle ABC, then what is the algebraic formula that gives minimum sum distance, i.e., FA+FB+FC=? Fermat point (F) of a triangle is at the least distance from triangle vertices. If one of the vertex angle is greater than 120 degrees, then F is at that vertex, and minimum distance is equal to sum of the two short sides of triangle. Otherwise, F lies within triangle ABC and the sides AB/BC/CA subtend equal angle of 120 degrees at F. In this case, what is the algebraic formula for finding the minimum distance (in terms of triangle lengths a,b,c)?

If $ABC$ is a triangle with every angle being $\leq 120^\circ$ and $F$ is its Fermat-Torricelli point, we may introduce the points $A',B',C'$ given by the vertices of the equilateral triangles externally built on the sides of $ABC$. Then we have that $(A,F,A'),(B,F,B'),(C,F,C')$ are collinear and $$ FA+FB+FC = AA' = BB' = CC'.\tag{1}$$ We may compute the (squared) length of $CC'$ through the cosine theorem: $$\begin{eqnarray*} CC'^2 &=& a^2+c^2-2ac\cos\left(B+60^\circ\right) \\\ &=&a^2+c^2-ac\cos(B)+\sqrt{3}\,ac\sin B\\\&=&a^2+c^2-\frac{a^2+c^2-b^2}{2}+2\sqrt{3}\Delta\\\&=&\color{red}{\frac{a^2+b^2+c^2}{2}+2\sqrt{3}\Delta}.\tag{2}\end{eqnarray*}$$ Here $\Delta=[ABC]$ can be found in terms of $a,b,c$ through Heron's formula.