Proving that $\sum_{j=1}^{n}|z_{j}|^{2}\le 1$ when $|\sum_{j=1}^{n}z_{j}w_{j}| \le 1$ for all $\sum_{j=1}^{n}|w_{j}|^{2}\le 1$ The problem is like this:


Fix $n$ a positive integer. 

Suppose that $z_{1},\cdots,z_{n} \in \mathbb C$ are complex numbers satisfying $|\sum_{j=1}^{n}z_{j}w_{j}| \le 1$ for all $w_{1},\cdots,w_{n} \in \mathbb C$ such that $\sum_{j=1}^{n}|w_{j}|^{2}\le 1$. 

Prove that $\sum_{j=1}^{n}|z_{j}|^{2}\le 1$. 


For this problem, I so far have that $|z_{i}|^{2}\le 1$ for all $i$ by plugging $(0, \cdots,0,1,0,\cdots,0)$ for $w=(w_{1},\cdots,w_{n} )$
Also, by plugging $(1/\sqrt{n},\cdots,1/\sqrt{n})$ for $w=(w_{1},\cdots,w_{n} )$
 we could have $|z_{1}+\cdots+z_{n}|\le \sqrt{n}$
I wish we can conclude that $|z_{i}|\le 1/\sqrt{n}$ for each $i$.
Am I in the right direction?
Any comment would be grateful!

Let $w_j:=\frac{\bar z_j}{\sqrt{\sum_{k=1}^n|z_k|^2}}$ (when $\sum_j|z_j|^2\neq 0$). Then $|w_j|^2=\frac{|z_j|^2}{\sum_{k=1}^n|z_k|^2}$ and we get $\sqrt{\sum_{j=1}^n|z_j|^2}\leq 1$, hence the result.
