If $z_1,z_2,z_3,z_4\in C $ satisfy $z_1+z_2+z_3+z_4=0$ and $|z_1|^2+|z_2|^2+|z_3|^2+|z_4|^2=1$ then what will be the least value of $|z_1-z_2|^2+|z_1-z_4|^2+|z_2-z_3|^2+|z_3-z_4|^2$?
What approach should I follow?Expanding the equation does not yield anything.Please do not give the answer; what I want is an approach to these kind of problems.I have been stuck here for long?
We will show that
$$|z_1-z_2|^2+|z_2-z_3|^2+|z_3-z_4|^2+|z_4-z_1|^2=2+|z_1+z_3|^2+|z_2+z_4|^2$$
The latter is clearly at least 2, and equality is reached whenever $z_1=-z_3$ and $z_2=-z_4$.
Firstly, note that when you expand $|z_1-z_2|^2$, you get
\begin{align*} (z_1-z_2)(\overline{z_1}-\overline{z_2})&=z_1\overline{z_1}-z_1\overline{z_2}-z_2\overline{z_1}+z_2\overline{z_2} \\ &= |z_1|^2-z_1\overline{z_2}-z_2\overline{z_1}+|z_2|^2 \\ \end{align*}
Therefore:
\begin{align*} &\phantom{=}|z_1-z_2|^2+|z_2-z_3|^2+|z_3-z_4|^2+|z_4-z_1|^2 \\ &=2(|z_1|^2+|z_2|^2+|z_3|^2+|z_4|^2)-z_1(\overline{z_2}+\overline{z_4})-z_2(\overline{z_1}+\overline{z_4})-z_3(\overline{z_2}+\overline{z_4})-z_4(\overline{z_1}+\overline{z_3}) \\ &= 2-z_1(\overline{z_2}+\overline{z_4})-z_2(\overline{z_1}+\overline{z_4})-z_3(\overline{z_2}+\overline{z_4})-z_4(\overline{z_1}+\overline{z_3}) \end{align*}
It remains to show the following
$$-z_1(\overline{z_2}+\overline{z_4})-z_2(\overline{z_1}+\overline{z_4})-z_3(\overline{z_2}+\overline{z_4})-z_4(\overline{z_1}+\overline{z_3})=|z_1+z_3|^2+|z_2+z_4|^2 $$
If you expand the right hand side similarly to before, and collect terms, it turns out this equality is equivalent to the following:
$$(z_1+z_2+z_3+z_4)(\overline{z_1}+\overline{z_2}+\overline{z_3}+\overline{z_4})=0$$
which is of course true, since $z_1+z_2+z_3+z_4=0$.
This proves that
\begin{align*}|z_1-z_2|^2+|z_2-z_3|^2+|z_3-z_4|^2+|z_4-z_1|^2&=2+|z_1+z_3|^2+|z_2+z_4|^2\\&\geq2\end{align*}
It can be checked that the conditions are met, and equality holds, when $z_1=z_2=\frac{1}{2}$, $z_3=z_4=-\frac{1}{2}$. In fact, equality will hold whenever $$z_1+z_3=z_2+z_4=0.$$