Question 161
Let $n \in \mathbb{N}^{*}, n \geq 2$. Prove that for any complex numbers $a_{1}, a_{2}, \ldots, a_{n}$ and $b_{1}, b_{2}, \ldots, b_{n}$ the following statements are equivalent:
a) $\sum_{k=1}^{n}\left|z-a_{k}\right|^{2} \leq \sum_{k=1}^{n}\left|z-b_{k}\right|^{2}$ for every $z \in \mathbb{C}$;
b) $\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} b_{k}$ and $\sum_{k=1}^{n}\left|a_{k}\right|^{2} \leq \sum_{k=1}^{n}\left|b_{k}\right|^{2}$.
a) $\sum_{k=1}^{n}\left|z-a_{k}\right|^{2} \leq \sum_{k=1}^{n}\left|z-b_{k}\right|^{2}$ for every $z \in \mathbb{C}$;
b) $\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} b_{k}$ and $\sum_{k=1}^{n}\left|a_{k}\right|^{2} \leq \sum_{k=1}^{n}\left|b_{k}\right|^{2}$.