The statement of the problem: Let $n \in \mathbb N $ \ {0} and $z_1,z_2, … z_n \in \mathbb C $. Prove that $$\sum_{i=1}^n |z_i||z-z_i| \ge \sum_{i=1}^n |z_i|^2$$ holds for any $z \in \mathbb C $ $\iff$
$z_1+z_2+…z_n=0$
My approach : I tried to prove the inequality using Cauchy's inequality and the modulus inequality, but I reached some ambiguous results, from which it does not appear that their sum would be 0. I think that the solution can also be a geometric one given the condition that z1 +z2+… zn = 0 it turns out that the vertices of the polygon with these affixes is regular, but I'm not sure.
Any and all proofs will be helpful. Thanks a lot!
Best Answer
Here's a proof using Triangle Inequality and Cauchy-Schwarz
(i.) when $\sum_{k=1}^n z_k = 0$
$\sum_{k=1}^n \vert z_k\vert^2= \Big \vert \sum_{k=1}^n \overline z_k \cdot z_k - \overline z_k\cdot z\Big\vert = \Big \vert \sum_{k=1}^n \overline z_k (z_k-z)\Big\vert$
$\leq \sum_{k=1}^n \Big \vert \overline z_k (z_k-z)\Big\vert=\sum_{k=1}^n \Big \vert z_k\Big \vert \cdot \Big \vert z-z_k\Big\vert$
by triangle inequality
(ii.) now suppose the inequality remains true when $\sum_{k=1}^n z_k =\lambda\neq 0$, then Cauchy-Schwarz tells us
$\sum_{k=1}^n \vert z_k\vert^2\leq \sum_{k=1}^n \Big \vert z_k\Big \vert \cdot \Big \vert z-z_k\Big\vert\leq \Big(\sum_{k=1}^n \vert z_k\vert^2\Big)^\frac{1}{2} \cdot \Big(\sum_{k=1}^n \Big \vert z-z_k\Big\vert^2\Big)^\frac{1}{2}$
$\implies\sum_{k=1}^n \vert z_k\vert^2\leq \sum_{k=1}^n \Big(z -\overline z_k\big)\overline{\Big(z - \overline z_k\Big)}= n \vert z\vert^2 + \Big(\sum_{k=1}^n \vert z_k\vert^2\Big)- 2\cdot \text{Re}\Big(z \sum_{k=1}^n z_k\Big)$
$\implies 2\cdot \text{Re}\Big(z\cdot \lambda\Big)\leq n \vert z\vert^2 $
which is obviously false, e.g. set $z:=\frac{\overline \lambda}{2n}$