<s>[INST] Fermat-Torricelli minimum distance The Fermat - Torricelli point minimizes sum of distances $S$ taken from vertices of a triangle of sides $a,b,c. $  Find $S$ in terms of $a,b,c$.
Am trying to set up problem with a Lagrange multiplier or partial derivatives for extremization but it seems tedious even with a CAS.
Although it is supposed to be known from the earliest Greek times, it is not seen (by me) in these modern times.
 [/INST] Let


*

*$a, b, c$ be the sides of $\triangle ABC$ whose angles are smaller than $120^\circ$. 

*$P$ be the Fermat-Torricelli point for $\triangle ABC$. 

*$\alpha = |AP|$, $\beta = |BP|$, $\gamma = |CP|$ and $S = \alpha+\beta+\gamma$.

*$\mathcal{A}$ be the area of $\triangle ABC$.


It is known that for such a triangle, $P$ is lying in its interior and
$$\angle APB = \angle BPC = \angle CPA = 120^\circ$$
Using these, we can express the sides and area of triangle as
$$\begin{cases}
a^2 = \beta^2 + \beta\gamma + \gamma^2\\
b^2 = \gamma^2 + \gamma\alpha + \alpha^2\\
c^2 = \alpha^2 + \alpha\beta + \beta^2
\end{cases}
\quad\text{and}\quad
\mathcal{A} = \frac{\sqrt{3}}{4}(\alpha\beta + \beta\gamma+\gamma\alpha)
$$
Summing the three equations from left and apply the equation from right, we find
$$a^2+b^2+c^2 = 2(\alpha^2 + \beta^2 + \gamma^2) + \frac{4}{\sqrt{3}}\mathcal{A}$$
As a result,
$$\begin{align}
S^2 &= (\alpha+\beta+\gamma)^2 
= \alpha^2 + \beta^2 + \gamma^2 + \frac{8}{\sqrt{3}}\mathcal{A}
= \frac12\left(a^2 + b^2 + c^2 - \frac{4}{\sqrt{3}}\mathcal{A}\right) + 
\frac{8}{\sqrt{3}}\mathcal{A}\\
&= \frac12\left(a^2 + b^2 + c^2\right) + 2\sqrt{3}\mathcal{A}\\
&= \frac12\left(a^2+b^2+c^2 + \sqrt{3(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}\,\right)
\end{align}
$$
An expression equivalent to what Quang Hoang obtained in another answer.
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