I have the following setting:


Let $0 \leq r < 1$ and let $\{z\_i\}\_{i=1}^k$ be $k$ complex numbers such that $|z\_i| \leq r$ for all $i$.
Moreover, $r + \sum\_{i=1}^k 2Re(z\_i) \geq 0$


I am interested to know if the following is true:


$(1-r)\prod\_{i=1}^k |1-z\_i|^2 \leq 1$.


The above stems from trying to bound the determinant of $I-B$ where $B$ is nonnegative. One can think of $z\_i$ being the complex eigenvalues of $B$ and the reason for $|1-z\_i|^2$ is due to complex eigenvalues coming in conjugate pairs for real matrices. Also, $r$ is the spectral radius of the nonnegative $B$, and $r + \sum\_{i=1}^k 2Re(z\_i) \geq 0$ is due to the trace of a nonnegative matrix being nonnegative.


All ideas I have so far yield an upper bound greater than $1$, and I thought the expression was false. However, using Mathematica I could not find a counterexample to this, which makes me think it could be true.
The main ideas I used: triangle inequality, and arithmetic-geometric ineq., but could not get the 1 upper bound or a counterexample.



Proving maximum value of a determinant of $I - B$, where $B$ is nonnegative matrix
This is not true in general. E.g., take any $r\in(0,1)$ and let
$$z\_i:=x+iy,\quad x:=-\frac r{2k},\quad 
y:=r\sqrt{1-\frac{1}{{4k^2}}}$$
for all $i$.
Then all the conditions on $r$ and $z\_i$'s hold, whereas
$$(1-r)\prod\_{i=1}^k|1-z\_i|^2=(1-r)(1+r^2+r/k)^k\to\infty \not\le 1$$
as $k\to\infty$.


Alternatively, let
$$z\_i:=ir$$
for all $i$. Then all the conditions on $r$ and $z\_i$'s hold, whereas
$$(1-r)\prod\_{i=1}^k|1-z\_i|^2=(1-r)(1+r^2)^k\to\infty \not\le 1$$
as $k\to\infty$.


