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 problem in complex-variable inequality
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. $\Box$


