$\sum_{k=1}^{n}|1+z_{k}|+\frac{1}{n-1}\sum\limits_{1\le iLet $n\ge 2$is a integer,$z_{1},z_{2},\cdots,z_{n}$ are $n$ complex numbers
Prove that
$$\color{crimson}{\sum_{k=1}^{n}|1+z_{k}|+\dfrac{1}{n-1}\sum_{1\le i<j\le n}|1+z_{i}z_{j}|\ge\sum_{k=1}^{n}|z_{k}|}$$
Proof for $n=2$:
$$
\DeclareMathOperator{\Re}{Re}
| 1+x  |+ | 1+y |+  | 1+xy  |- | x  |- | y  |=
\frac{(| 1+x  |+ | 1+y|+| 1+xy  |)^{2}-(| x |+|y |)^{2}}{| 1+x|+ | 1+y|+|1+xy|+|x|+|y|}=\frac{A}{| 1+x|+|1+y |+ |1+xy |+ |x  |+ |y|}$$where
$$A=(  | 1+x  |+ | 1+y  |+ | 1+xy  |)^{2}- (  | x  |+ | y  |  )^{2}
 \\=
  | 1+x  |^{2}+ | 1+y  |^{2}+  | 1+xy |^{2}+2  | 1+x  | | 1+y|+2  | 1+y  | | 1+xy|+2  | 1+xy  | | 1+x|- | x  |^{2}- | y  |^{2}-2 | x  | | y  |
 \\=1+ | x  |^{2}+2\Re(x)+1+ | y  |^{2}+2\Re(y)+ 1+ |xy |^{2}+2\Re(xy)+2  | 1+x  | | 1+y|+2  | 1+y  | | 1+xy|+2  | 1+xy  | | 1+x|- | x  |^{2}- | y  |^{2}-2 | x  | | y  |
 \\=2\Re ( 1+x  ) ( 1+y  )+2  | 1+x  | | 1+y|+ ( 1- | xy  |  )^{2}+ 2  | 1+y  | | 1+xy|+2  | 1+xy  | | 1+x|
 \\ \geq2  | 1+y  | | 1+xy|+2  | 1+xy  | | 1+x|
\\ \Longrightarrow  | 1+x  |+ | 1+y |+  | 1+xy  |\geq  | x  |+ | y  |
$$
Is it true for a general $n$?
 A: For $z, w \in \Bbb C$ we have (and this is what you already did)
$$
 \DeclareMathOperator{\Re}{Re}
 \bigl( \lvert 1+z \rvert + \lvert 1+w \rvert  + \lvert 1+zw \rvert  \bigr)^2 \\
 = 1 + \lvert z \rvert ^2 + 2 \Re z + 1 + \lvert w \rvert ^2 + 2 \Re w + 1 + \lvert zw \rvert ^2 + 2 \Re (zw) \\
  + 2 \lvert 1+z \rvert  \lvert 1+w \rvert  + 2 \bigl(\lvert 1+z \rvert  + \lvert 1+w \rvert \bigr) \lvert 1+zw \rvert  \\
 = 2 \Re \bigl((1+z)(1+w) \bigr) + 2 \lvert 1+z \rvert  \lvert 1+w \rvert  \\
  + \bigl(\lvert z \rvert  + \lvert w \rvert  \bigr)^2 
   + 2 \bigl(\lvert 1+z \rvert  + \lvert 1+w \rvert \bigr) \lvert 1+zw \rvert  \\
 \ge \bigl(\lvert z \rvert  + \lvert w \rvert  \bigr)^2 
   + 2 \bigl(\lvert 1+z \rvert  + \lvert 1+w \rvert \bigr) \lvert 1+zw \rvert  \\
  \ge \bigl(\lvert z \rvert  + \lvert w \rvert  \bigr)^2 
$$
and therefore
$$
\lvert 1+z \rvert  + \lvert 1+w \rvert  + \lvert 1+zw \rvert  \ge \lvert z \rvert  + \lvert w \rvert  \, ,
$$
which proves the assertion for $n = 2$.
The general case $n \ge 2$ now follows from the case $n=2$:
If $z_1, \ldots z_n \in \Bbb C$ then for $1 \le j < k \le n$
$$
\lvert 1+z_j \rvert  + \lvert 1+z_k \rvert  + \lvert 1+z_j z_k \rvert  \ge \lvert z_j \rvert  + \lvert z_k \rvert  \, .
$$
Adding all these inequalities gives 
$$
 (n-1) \sum_{j=1}^n \lvert 1+z_j \rvert  + \sum_{1 \le j < k \le n} \lvert 1+z_j z_k \rvert 
 \ge (n-1) \sum_{j=1}^n \lvert z_j \rvert 
$$
as desired.
