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!

To have a chance of success, one must choose a family $(w_j)_j$ adapted to the input $(z_j)_j$. If $z_j=0$ for every $j$, the result holds. Otherwise, try $w_j=c^{-1}\bar z_j$ with $c^2=\sum\limits_{k=1}^n|z_k|^2$.
