Question: 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!

Answer: 
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.
If $z =0 $ there is nothing to be done, so suppose $z \neq 0$.

Let $w =\frac{1}{\|z\|} z$. Note that $\|w\| = 1$. Then $|\sum\_{j=1}^{n}z\_{j}w\_{j}| = \frac{1}{\|z\|}|\sum\_{j=1}^{n}z\_{j}^2| = \|z\| \leq 1$.