TITLE: (2016 China team selection Test) with a complex inequality
QUESTION [7 upvotes]: Find the minimum possible value of the positive real number $A$ such that
$$|z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}|^2+|z_{1}z_{2}z_{3}|^2<A\,,$$
where $z_{1}$, $z_{2}$, and $z_{3}$ are complex numbers satisfying the conditions below: $$z_{1}+z_{2}+z_{3}=0$$ and
$$|z_{i}|<1\text{ for all }i=1,2,3\,.$$

REPLY [2 votes]: Hint:  Using $z_1+z_2+z_3 =0$, with notation $a_i = |z_i| < 1$, show
$$\left|\sum z_1z_2\right|^2=\sum a_1^4-\sum a_1^2a_2^2$$
Hence $LHS < \sum a_1^2-\sum a_1^2a_2^2+(\prod a_1)^2=1-\prod(1-a_1^2)< 1$
It is obvious $LHS \to 1^-$ as $z_3=0, |z_1|=|z_2|\to 1^-$, so that $A=1$ is minimal.

Note: $\sum$ and $ \prod$ represent cyclic sums and products.