Since we don't know the context in which you confront this problem, here's an elementary geometric approach. (It might not be the most efficient way since you may have other tools to bring to bear on the problem.)
This is a classic problem (due to Fermat) for finding what is called the Fermat point (more generally the geometric median) of the triangle you gave.
As the links outline, since the triangle here involves no angle larger than $120^\circ$, the point you seek is found as follows:
- Construct equilateral triangles exterior to each side of the given triangle.
- Form the line segment from the new vertex to opposite old vertex.
- The Fermat point is the intersection of these three lines.
This diagram summarizes the situation:

Computing the coordinates of the northwest black point is easy with a little trigonometry: \begin{align}x_1&=-1-\sqrt{2}\sin(15^\circ)=-{1\over 2}(1+\sqrt{3}),\\
y_1&=\sqrt{2}\sin(75^\circ)={1\over 2}(1+\sqrt{3}),
\end{align}
the northeast point follows from symmetry, and the south point is simple.
The $x$-coordinate of the Fermat point is zero, and a little algebra reveals the $y$-coordinate is $1/\sqrt{3}$.
Finally, to answer the question that was asked: in order to minimize $$w(z):=|z+1|+|z-1|+|z-i|, \quad z\in \mathbb{C},$$ we take $z=0+{1\over \sqrt 3}i$ which results in $w=1+\sqrt 3$.
Here's a contour plot of $w$ superimposed with items above:
