Prove that a specific inequality holds Let $n \in \mathbb{N}$. Let $z_1, \ldots, z_n$ and $w_1, \ldots, w_n$ be complex numbers such that
$$ \sum_{j = 1}^n |w_j|^2 \leq 1 $$ 
and
$$ \left| \sum_{j = 1}^n z_j w_j \right| \leq 1 $$
Show that
$$ \sum_{j = 1}^n |z_j|^2 \leq 1$$ 
The reverse is easy, but I have no clue as to why this is true.

EDIT The question should be:

Let $n \in \mathbb{N}$. Let $z_1, \ldots, z_n$ such that
$$ \left| \sum_{j = 1}^n z_j w_j \right| \leq 1 $$
for all complex numbers $w_1, \ldots, w_n$ with
$$ \sum_{j = 1}^n |w_j|^2 \leq 1 $$ 
Show that
$$ \sum_{j = 1}^n |z_j|^2 \leq 1$$
There is no reason why this should be true. It certainly isn't.

Take for instance $w_1 = 0.1$, $w_2 = 0.01$, $z_1 = 1$ and $z_2 = 10$.

Edited:

For each $1 \le j \le n$, assume that $z_j \neq 0$ and let:

$$w_j = \frac{|z_j|^2}{z_j \sqrt{\sum_{i=1}^n |z_i|^2}}$$

Then,

$$\sum_{j=1}^n |w_j|^2 = \sum_{j=1}^n \frac{|z_j|^2}{\sum_{i=1}^n |z_j|^2} = 1 $$

Hence, 

$$\left| \sum_{j=1}^n z_j w_j \right| \le 1$$

But,

$$\left| \sum_{j=1}^n w_j z_j \right| = \sum_{j=1}^n \frac{|z_j|^2}{\sqrt{\sum_{i=1}^n |z_i|^2}} = \sqrt{\sum_{j=1}^n |z_j|^2}$$

Therefore,

$$\sum_{j=1}^n |z_j|^2 \le 1$$

One doesn't need to take into consideration the case where one of the $z_j$'s is equal to $0$ because it just amounts to reducing $n$ to a smaller number.
The Cauchy-Schwarz inequality states that
$$\left| \sum_{j=1}^n z_j w_j \right|^2 \leq \left( \sum_{j=1}^n |z_j|^2 \right) \left( \sum_{j=1}^n |w_j|^2 \right)$$
with equality if and only if $z = (z_1, \ldots, z_n)$ and $w = (w_1, \ldots, w_n)$ are parallel. You only know that one of the factors to the right is less than zero. If you could make the inequality into an equality for some choice of $w$, then this would do it. Therefore, try to pick $w$ parallel to $z$, but remember that $w$ must have length less than one.