Minimum value of $|z+1|+|z-1|+|z-i|$ How to find the minimum value of $|z+1|+|z-1|+|z-i|$.
I have tried geometrically etc but failed.

The post has received good and simple answers based on triangles.
However, if you really want to see the calculus based problem, let me start using Dr. Sonnhard Graubner's approach to the solution. Writing $$f(a,b)=\sqrt{(a+1)^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}$$ Computing the partial derivatives lead to the equations $$\frac{df(a,b)}{da}=\frac{a}{\sqrt{a^2+(b-1)^2}}+\frac{a-1}{\sqrt{(a-1)^2+b^2}}+\frac{a+1}{\sqrt{(a+1)^2+
   b^2}}=0$$ $$\frac{df(a,b)}{db}=\frac{b-1}{\sqrt{a^2+(b-1)^2}}+\frac{b}{\sqrt{(a-1)^2+b^2}}+\frac{b}{\sqrt{(a+1)^2+b^
   2}}=0$$ which are extremely complex to solve except if you know (or are able to identify) that one solution corresponds to $a=0$. In such a case, the first partial is effectively equal to $0$ and the second one becomes $$\frac{df(0,b)}{db}=\frac{2 b}{\sqrt{b^2+1}}+\frac{b-1}{\sqrt{(b-1)^2}}=\frac{2 b}{\sqrt{b^2+1}}\pm 1=0$$ for which the solutions are $\mp \frac{1}{\sqrt{3}}$. To these solutions correspond $$f(0,-\frac{1}{\sqrt{3}})=1+\frac{5}{\sqrt{3}}$$ $$f(0,\frac{1}{\sqrt{3}})=1+\sqrt{3}$$ and then the result ($a=0,b=\frac{1}{\sqrt{3}},f=1+\sqrt{3}$).
You can notice that, for $b=\pm \frac{1}{\sqrt{3}}$, $$\frac{d^2f(0,b)}{db^2}=\frac{3 \sqrt{3}}{4} \gt 0$$
Please, notice that the problem with $|z+\alpha|+|z+\beta|+|z-i|$ will be extremely difficult to solve as soon as $\alpha+\beta \neq 0$.
