Proving a complex inequality using Cauchy Schwartz inequality

Question

Let $z_1,z_2,...,z_n \in \Bbb C , M \in \Bbb R^* $such as $\sum_{k=1}^nz_k =0 $ and $\sum_{k=1}^n\lvert z_k\rvert^2\leq M$

show that $$\forall k \in \{1,...,n\},\lvert z_k \rvert\leq\sqrt{\frac{n-1}{n}M}$$

I think that I have to use this inequality $$(\sum_{k=0}^na_kb_k)^2\leq(\sum_{k=0}^na_k)^2(\sum_{k=0}^nb_k)^2$$

But I have no idea how to start

Answer 1

$|z_k|\le \sqrt{\frac{n-1}{n}M}$ $\iff n|z_k|^2\le (n-1)M$
Cauchy-Schwartz inequality yields
$(\sum_{i\neq k}|z_i| )^2\le (n-1)(\sum_{i\neq k}|z_i|^2)\le (n-1)(M-|z_k|^2)$
But $|z_k|=|\sum_{i\neq k}z_i|\le \sum_{i\neq k}|z_i|$ so
$|z_k|^2\le (\sum_{i\neq k}|z_i| )^2\le (n-1)(M-|z_k|^2)$
$\iff n|z_k|^2\le (n-1)M$

Answer 2

Assume that $|z_1| > \sqrt{\frac{n-1}{n} M}$. Then your second condition yields that $$ \sum_{k=2}^n |z_k|^2 < M / n $$ On the other hand the first condition implies $$ \left| \sum_{k=2}^n z_k \right|^2 > \frac{n-1}{n} M $$ However, applying Cauchy-Schwarz to the LHS with $a_k = z_k$ and $b_k = 1$ gives you $$ \left| \sum_{k=2}^n z_k \right|^2 \leq \sum_{k=2}^n |z_k|^2 (n-1) $$ Putting everything together, we get $M (n-1)/n > M (n-1)/n$, which is a contradiction.