What is the minimum value of $$|1+z_1|+|1+z_2|+|1+z_3|+...+|1+z_n|+|1+z_1z_2z_3...z_n|$$ for $n$ even, $z\in\mathbb{C}$?
I'm trying to generalize this question. I tried to apply the method in the answer to that question, but I couldn't get it to work.
On
Fact 1: Let $w_1, w_2, w_3 \in \mathbb{C}$ with $|w_3| \le 1$. Then
$|1 + w_1| + |1 + w_2| + |1 + w_1 w_2 w_3|
\ge |1 + w_3|$.
(The proof is given at the end.)
Fact 2: Let $w_1, w_2, w_3 \in \mathbb{C}$ with $|w_1| \ge 1$ and $|w_2| \ge 1$. Then $|1 + w_1| + |1 + w_2| + |1 + w_1 w_2 w_3|
\ge |1 + w_3|$.
(The proof is given at the end.)
WLOG, assume that $|z_1| \le |z_2| \le \cdots \le |z_n|$.
We claim that, for each $n\ge 4$ even, $$\sum_{k=1}^n |1 + z_k| + |1 + z_1 z_2 \cdots z_n| \ge \sum_{k=1}^{n-2} |1 + z_k| + |1 + z_1z_2 \cdots z_{n-2}|. \tag{1}$$ Indeed, if $|z_{n-1}| \ge 1$, by Fact 2, we have \begin{align*} &|1 + z_{n-1}| + |1 + z_n| + |1 + z_1 z_2 \cdots z_{n-2}\cdot z_{n-1}z_n|\\ \ge\,& |1 + z_1z_2 \cdots z_{n-2}|; \end{align*} and if $|z_{n-1}| < 1$, using $|z_1z_2 \cdots z_{n-2}| < 1$, by Fact 1, we have \begin{align*} &|1 + z_{n-1}| + |1 + z_n| + |1 + z_1 z_2 \cdots z_{n-2}\cdot z_{n-1}z_n|\\ \ge\,& |1 + z_1z_2 \cdots z_{n-2}|. \end{align*} The claim is proved.
Now, repeating the process (1), we have $$\sum_{k=1}^n |1 + z_k| + |1 + z_1 z_2 \cdots z_n| \ge \cdots \ge |1 + z_1| + |1 + z_2| + |1 + z_1z_2| \ge 2.$$
Also, when $z_1 = z_2 = \cdots = z_n = -1$, we have $\sum_{k=1}^n |1 + z_k| + |1 + z_1 z_2 \cdots z_n| = 2$.
Thus, the minimum of $\sum_{k=1}^n |1 + z_k| + |1 + z_1 z_2 \cdots z_n|$ is $2$.
Proof of Fact 1:
Use @Carl Schildkraut's idea in this answer.
If $|w_2| \ge 1$, we have \begin{align*} &|1 + w_1| + |1 + w_2| + |1 + w_1 w_2 w_3|\\ \ge\,& |1 + w_1| + |w_2|\, |1 - w_1w_3|\\ \ge\,& |1 + w_1| + |1 - w_1 w_3|\\ \ge\,& |w_3|\, |1 + w_1| + |1 - w_1w_3|\\ \ge\,& |1 + w_3|. \end{align*}
If $|w_2| < 1$, we have \begin{align*} &|1 + w_1| + |1 + w_2| + |1 + w_1 w_2 w_3|\\ \ge\,& |w_2|\, |1 + w_1| + |1 + w_2| + |1 + w_1 w_2 w_3|\\ \ge\,& |1 - w_1w_2| + |1 + w_1 w_2 w_3|\\ \ge\,& |w_3|\, |1 - w_1w_2| + |1 + w_1 w_2 w_3|\\ \ge\,& |1 + w_3|. \end{align*}
We are done.
Proof of Fact 2:
We have \begin{align*} &|1 + w_1| + |1 + w_2| + |1 + w_1w_2w_3|\\ \ge\,& |1 + w_1| + |w_2|\, |1 - w_1w_3|\\ \ge\,& |1 + w_1| + |1 - w_1w_3|\\ \ge\,& |w_1|\, | 1 + w_3|\\ \ge\,& |1 + w_3|. \end{align*}
We are done.
The answer is still $2$. We use induction on $n$. You've linked the basis case $(n=2)$. Now we make $2$ cases:
Case 1: Modulus of at least two of the complex numbers is $1$ or more. Without loss of generality assume $|z_1|\ge1$ and $|z_2|\ge1$. Then \begin{align} \sum\limits_{k=1}^n |1+z_k|+|1+z_1z_2\cdots z_n|&=\color{green}{\sum_{k=2}^n|1+z_k|}+\color{red}{|1+z_1|+|1+z_1z_2\cdots z_n|}\\ &\ge\color{green}{|1+z_2|+\sum_{k=3}^n|1+z_k|}+\color{red}{|z_1-z_1z_2\cdots z_n|}&&(\text{By triangle inequality})\\ &\ge\color{brown}{|1+z_2|}+\color{purple}{\sum_{k=3}^n|1+z_k|}+\color{red}{|1-z_2z_3\cdots z_n|}&&(\because|z_1|\ge1)\\ &\ge \color{purple}{\sum_{k=3}^n|1+z_k|}+|z_2+z_2z_3\cdots z_n|&&(\text{By triangle inequality})\\ &\ge\color{\purple}{\sum_{k=3}^n|1+z_k|}+|1+z_3z_4\cdots z_n|&&(\because|z_2|\ge1)\\ &\ge2&&(\text{By induction hypothesis})\\ \end{align}
Case 2: Modulus of at most one complex numbers is $1$ or more. Let that be $z_1$ without loss of generality. Then \begin{align} \sum\limits_{k=1}^n |1+z_k|+|1+z_1z_2\cdots z_n|&\ge\color{green}{\sum_{k=2}^n|1+z_k|}+\color{red}{|z_2z_3\cdots z_n+z_1z_2z_3\cdots z_n|+|1+z_1z_2\cdots z_n|}&&(\because |z_2z_3\cdots z_n|\le1)\\ &\ge\color{brown}{|1+z_2|}+\color{purple}{\sum_{k=3}^n|1+z_k|}+\color{red}{|1-z_2z_3\cdots z_n|}&&(\text{By triangle inequality})\\ &\ge \color{purple}{\sum_{k=3}^n|1+z_k|}+\color{brown}{|z_3z_4\cdots z_n+z_2z_3z_4\cdots z_n|}+\color{red}{|1-z_2z_\cdots z_n|}&&(\because |z_3z_4\cdots z_n|\le1)\\ &\ge\color{\purple}{\sum_{k=3}^n|1+z_k|}+|1+z_3z_4\cdots z_n|&&(\text{By triangle inequality})\\\\ &\ge2&&(\text{By induction hypothesis})\\ \end{align}
Cases of equality
First we will closely examine the case $n=2$. Allow me to reproduce the (nice) answer to linked question:
If at least one of the variables (say $z_1$) has modulus $1$ or more, then \begin{align} |1+z_1|+|1+z_2|+|1+z_1z_2| &\ge |1+z_2|+|(1+z_1)-(1+z_1z_2)|\\ &=|1+z_2|+|z_1||1-z_2|\\ &\ge |1+z_2|+|1-z_2|\\ &\ge 2 \end{align} For last inequality, note that equality holds iff $z_2$ lies on the line joining $-1$ and $1$ on the complex plane, i.e. it is a real number with absolute value at most $1$.
The penultimate inequality becomes an equality iff $|z_1|=1$ (or $z_2=1$ which forces $z_1=-1$ anyway).
The first inequality becomes an equality iff $\frac{1+z_1z_2}{1+z_1}$ is non-positive real or $z_1=-1$ (remember $|a+b|=|a|+|b|$ iff $a, b$ and $0$ are collinear with origin on a side, and $|a-b|=|a|+|b|$ iff $a, b$ and $0$ are collinear with origin in middle).
Now $\frac{1+z_1z_2}{1+z_1}=-c\iff 1+z_1z_2=-c-cz_1\iff z_1=-\frac{1+c}{z_2+c}\text{ or }-c=z_2=1$. This implies $z_1$ is real (with absolute value $1$).
Hence, the solution is that one of the variables is $-1$ and other is a real number with absolute value at most $1$.
If both the variables have modulus less than $1$, then the linked answer goes like \begin{align} |1+z_1|+|1+z_2|+|1+z_1z_2| &\geq|z_2||1+z_1|+|1+z_2|+|1+z_1z_2|\\ &\geq|(1+z_2)-(z_2+z_1z_2)|+|1+z_1z_2|\\ &=|1-z_1z_2|+|1+z_1z_2|\geq 2. \end{align} Note that the first inequality is strict, because $|1+z_1|>0$ and $|z_2|<1$, so no solution in this case.
The solution set is thus $\{-1\}\times[-1,1]\cup[-1,1]\times\{-1\}$.
A similar analysis works on my solution. Again, note the second case gives no solution (because third inequality is strict, since $|1+z_2|>0$ and $|z_3z_4\cdots z_n|<1$).
In two steps in the first case, I use $|z_1|\ge1$ and $|z_2|\ge1$, they become equality to get equality throughout. Also, "all the variables are real in the solution set with absolute value at most $1$" can be proved by induction. I've proved basis above.
Note that my approach in case 1 is to remove two variables (having absolute values at least $1$) while retaining the same expression. If we have at least two variables with absolute value strictly less than $1$, we can remove all other variables and then get into case 2 producing no solution. So either all variables are $\pm1$ or one of them has absolute value strictly less than $1$ and others are $\pm1$.
The solution set thus comprises of all permutations of $[-1,1]\times\{-1\}^{n-1}$.