a problem in complex-variable inequality Let $n\ge2$ be a given positive integer, and $z_{1},z_{2},\cdots,z_{n}\in \mathbb{C}$,such
$$|z_{1}|^2+|z_{2}|^2+\cdots+|z_{n}|^2\ge n.$$
Prove or disprove
$$f_{n}=\sum_{j=1}^{n}\left|\sum_{I\subseteq \{1,2,3,\cdots,n\},|I|=j}\prod_{k\in I}z_{k}\right|^2\ge 1$$
In the particular case when $n=2$, it can be proved that
\begin{align*}f_{2}&=|z_{1}+z_{2}|^2+|z_{1}z_{2}|^2=(|z_{1}|^2+|z_{2}|^2)+|z_{1}z_{2}|^2+2\Re(z_{1}\overline{z_{2}})\\
&\ge |z_{1}|^2+|z_{2}|^2+|z_{1}z_{2}|^2-2|z_{1}z_{2}|\\
&=(|z_{1}|^2+|z_{2}|^2-1)+(|z_{1}z_{2}|-1)^2\\
&\ge(|z_{1}|^2+|z_{2}|^2-1)\\
&\ge 1  
\end{align*}
 A: For a polynomial $Q(x)=\sum_i q_ix^i$, define $N(Q)=\sum_i|q_i|^2$. We need to show that $N(R)\geq 2$, where $R(x)=\prod_i (x-z_i)$.
For a polynomial $Q(x)=\sum_{i=0}^k q_ix^i$, define $Q^*(x)=\sum_{i=0}^k\overline{q_i}x^{k-i}=x^k\overline{Q(\bar x^{-1})}$. Here is the lemma which I definitely saw somewhere.
Lemma 1. $N(FG)=N(FG^*)$.
Proof. This can be shown directly, by expanding the brackets. However, a more neat way is to apply the discrete Fourier transform, assigning to a polynomial $Q$ of degree $<n$ the collection of values $Q(\zeta_i)$, where $\zeta_1,\dots,\zeta_n$ are the $n$th degree roots of unity. The Plancherel identity claims
$$
  N(Q)=\frac1n\sum_{i=1}^n|Q(\zeta_i)|^2.
$$
On the other hamd, we have
$$
  |F(\zeta_i)G^*(\zeta_i)|=\left|F(\zeta_i)\overline{G(\bar\zeta_i^{-1})}\right|
  =|F(\zeta_i)G(\zeta_i)|,
$$
hence $N(FG)=N(FG^*)$. $\Box$
Back to the problem, let $|z_1|\leq|z_2|\leq\dots\leq|z_k|\leq 1\leq |z_{k+1}|\leq\dots\leq |z_n|$. Then, instead of $R$, we may consider the polynomial
$$
  S(x)=\prod_{i=1}^k(1-\bar z_i x)\prod_{i=k+1}^n(x-z_i),
$$
as Lemma 1 claims $N(R)=N(S)$. In the polynomial $S(x)$, we are interested only in the coefficients of $x^n$ and $x^0$, whose absolute values are $|\prod_{i=1}^kz_i|$ and $|\prod_{i=k+1}^nz_i|$: the next Lemma shows they suffice.
Lemma 2. Assume that $a_1,\dots,a_k\leq 1$ and $a_{k+1},\dots,a_n\geq 1$ are nonnegative real numbers such that $\sum_ia_i^2=n$. Then $(a_1\dots a_k)^2+(a_{k+1}\dots a_n)^2\geq 2$.
Proof. Induction on the number of indices $i$ such that $a_i\neq 1$. If all the $a_i$ equal $1$, the claim is trivial. Also it is trivial if $a_i=1$ for all $i\leq k$.
Assume that $a_1<1$; then, without loss of generality, $a_n>1$. Replace $a_1$ and $a_n$ with $b_1\leq b_n$ such that $b_1^2+b_n^2=a_1^2+a_n^2$, and one of the $b_i$ equals $1$ (then $a_1<b_1\leq b_n<a_n$). Denote $B_1=\prod_{i=2}^k a_i$ and $B_n=\prod_{i=k+1}^{n-1} a_i$; the inductive hypothesis yields
$$
  b_1^2B_1^2+b_n^2B_n^2\geq 2.
$$
On the other hand, we have
$$
  \left(a_1B_1\right)^2+\left(a_nB_n\right)^2
  =(b_1B_1)^2+(b_nB_n)^2+(a_1^2-b_1^2)B_1^2+(a_n^2-b_n^2)B_n^2\\
  =(b_1B_1)^2+(b_nB_n)^2+(a_1^2-b_1^2)(B_1^2-B_n^2)
  \geq (b_1B_1)^2+(b_nB_n)^2\geq 2,
$$
as desired. $\Box$
