locate a point inside a triangle such that the total distance from the vertices is a maximum

We consider the function $$ f(x,y) = \sum_{j=1}^3 [(x-a_j)^2+(y-b_j)^2] $$ where $(a_j,b_j)$ the vertices of the triangle.

There is only one critical point that is $(\frac{a_1+a_2+a_3}{3},\frac{b_1+b_2+b_3}{3})$ which is a minimum and the centroid point of the triangle.

Hence, the point lied inside or on the triangle with the maximum total distance from the 3 vertices should lie on the boundary of the domain for $(x,y)$ i.e. on the triangle. This follows from the fact that $f: K\rightarrow \mathbb{R}$ where $K$ is the triangle with its interior, is a continuous function on a compact set so it attains a maximum value. Since it doesn't have a local maximum the maximum value is attained at the boundary.

Then as already suggested in another answer posted: let $A,B,C$ the vertices and assume that the point $D=(x,y)$ we are looking for, lies on $AB$. Then the total distance from the 3 vertices, let $d$, is equal to $d=AB+DC$. Moreover, $DC$ is bounded above by the $\max{\{AC,BC\}}$. Hence, $D$ is the vertex $A$ or $B$ if $AC\geq BC$ or $BC\geq AC$ respectively.

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In other words the point that maximizes the distance from the 3 vertices of the triangle is the vertex whose adjacent edges are the two longest.

The sum of the distances does not have a maximum inside the triangle, so the maximum must be on the boundary.

If the maximum point $P$ is on edge $AB$, the sum of distances is $AB$ plus the distance from $P$ to the third vertiex $C$. This is maximized at either $P=A$ or $P=B$. Therefore the maximum sum of distances is achieved at a vertex.

Moreover, the maximum will be at the vertex opposite the shortest edge.