Find minimum value of $|z+1+i|+|z-1-i|+|z-2|+|z-3|$ If $z \in \mathbb{C}$  Find minimum value of $|z+1+i|+|z-1-i|+|z-2|+|z-3|$
All that i know is $$|z_1+z_2| \le |z_1|+|z_2|$$ and i am not sure which two complex numbers i combine and use the above inequality. any geometrical approach?
 A: My previous answer turned out to be wrong as I misread the input. 
Introducing the points as $A = (1,1)$, $B=(-1,-1)$, $C = (2,0)$ and $D = (3,0),$ we are supposed to find the minimum value of $$f(P) = PA+PB+PC+PD.$$
This is known as the Fermat-Torricelli problem in general and its unique solution is called the Geometric Median. In general, there is no closed form solution to this, but in the case of $4$ points it is known to be solvable. 
To be explicit, the point that minimizes $f(P)$ is the intersection of two diagonals formed by $A,B,C,D$, if they formed a convex quadrilateral. 
But our points form a non-convex quadrilateral with $C$ being inside the triangle formed by $A,B,D$. In this case, it is known that $f(P)$ attains its minimum at $P=C$, the point that is strictly inside the convex hull of $A,B,C,D.$ 
You can find a proof of the above in this paper.
A: The following contour plot suggests that the minimum is taken at $z_*:=2$. The minimum value would then be 
$$f(2)=1+\sqrt{2}+\sqrt{10}\doteq5.57649\ .$$

In order to prove that $z_*=2$ indeed gives the minimum we argue as follows: The function $f$ (considered as function $(x,y)\mapsto f(x+iy)$) is a sum of four convex functions, hence convex, and thus can have at most one isolated local minimum, which is then the global minimum. It is therefore sufficient to prove that $z_*$ is an isolated local minimum of $f$.
Consider the function
$$g(x,y):=\sqrt{(x-1)^2+(y-1)^2}+\sqrt{(x+1)^2+(y+1)^2}+\sqrt{(x-3)^2+y^2}\ ,$$
which is $f$ with the term $|z-2|$ removed. Compute
$$a:=g_x(2,0)=-1+{1\over\sqrt{2}}+{3\over\sqrt{10}},\qquad b:=g_y(2,0)={1\over\sqrt{10}}-{1\over\sqrt{2}}\ .$$
It follows that
$$\bigl|\nabla g(2,0)\bigr|=\sqrt{a^2+b^2}\doteq0.763444<1\ .\tag{1}$$
Now at $z_*=2$ the directional derivative of $z\mapsto |z-2|$ is $\ +1$ in all directions. Together with $(1)$ we may therefore conclude that $f(z)=g(z)+|z-2|$ takes an isolated local minimum at $z_*$.
