Following a problem I found on [mathstack](https://math.stackexchange.com/questions/1762753/proof-of-an-inequality-in-c-2), with no solution, and no comment, so I think this inequality is not easy, so I post it here (because I think there are more some good math job, maybe someone can solve it).


Let $n\ge 2$ be an 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 $n=2$:


We have (denote $z\_1=x,z\_2=y$):


$$
(| 1+x |+ | 1+y |+ | 1+xy |)^{2}- ( | x |+ | y | )^{2}=
 | 1+x |^{2}+ | 1+y |^{2}+ | 1+xy |^{2}+\\2|1+x| | 1+y|+2 | 1+y | | 1+xy|+2 | 1+xy | | 1+x|- | x |^{2}- | y |^{2}-2 | x | | y |=1+ | x |^{2}+2Re(x)+1+ | y |^{2}+2Re(y)+ 1+ |xy |^{2}+2Re(xy)+2 | 1+x | | 1+y|+2 | 1+y | | 1+xy|+2 | 1+xy | | 1+x|- | x |^{2}- | y |^{2}-2 | x | | y |=2Re ( 1+x ) ( 1+y )+2 | 1+x | | 1+y|+ ( 1- | xy | )^{2}+ 2 | 1+y | | 1+xy|+2 | 1+xy | | 1+x|\geq 0$$
as desired.


Is it true for a general $n$?



Absolute value inequality with complex numbers
If you have it for $n=2$, just sum up over all pairs $(z\_i,z\_j)$ with $i<j$ and divide by $n-1$. 


As for the proof for $n=2$, yours is quite ok for me, and the proof by math110 is especially elegant, but well, here is another approach. We need two easy lemmata:


Lemma 1. For real $t$ and non-negative real $R$ we have $|R-e^{it}|\geq \min(1,R) |1-e^{it}|$. 


Lemma 2. For reals $u,v$ we have $|\cos u-\cos v|\leqslant 2|\sin\frac{u-v}2|=|e^{iu}-e^{iv}|$.


Now denote $x=r\_1e^{it}$, $y=-r\_2e^{-is}$, where $r\_1=|x|,r\_2=|y|$. Then we have 
$$
U:=|1+x|+|1+y|+|1+xy|=|r\_1+e^{it}|+|r\_2-e^{is}|+|r\_1r\_2-e^{i(t-s)}|.
$$
Now some cases. 


1) $r\_1r\_2\geqslant 1$. Then we have $|r\_1+e^{it}|\geqslant Re(r\_1+e^{it})=r\_1+\cos t$, $|r\_2-e^{is}|\geqslant r\_2-\cos s$, $|r\_1r\_2-e^{i(t-s)}|\geqslant |1-e^{i(t-s)}|\geqslant \cos s-\cos t$ by our lemmata. Summing up we get $U\geqslant r\_1+r\_2$ as desired.


2) $r\_1\leqslant 1$, $r\_2\leqslant 1$. We have $|r\_1+e^{it}|=|1+e^{-it} r\_1|\geqslant 1+r\_1\cos t$, $|r\_2-e^{it}|\geqslant 1-r\_2\cos s$, $|r\_1r\_2-e^{i(t-s)}|\geqslant r\_1r\_2|\cos s-\cos t|$, thus it suffices to prove that
$2+r\_1\cos t-r\_2\cos s+r\_1r\_2|\cos s-\cos t|\geqslant r\_1+r\_2$. This is linear in $r\_1,r\_2$, so it suffices to check for $r\_1,r\_2\in\{0,1\}$, where inequality is clear.


3) $r\_1\leqslant 1\leqslant r\_2$ and $r\_1r\_2\leqslant 1$. We get 
$$
U\geqslant 1+r\_1\cos t+r\_2-\cos s+r\_1r\_2|\cos t-\cos s|.
$$
For fixed $r\_2$ inequality $1+r\_1\cos t+r\_2-\cos s+r\_1r\_2|\cos t-\cos s|\geqslant r\_1+r\_2$ is linear in $r\_1$, thus we may prove it for all $r\_1\in [0,1]$ checking for $r\_1=0$ and for $r\_1=1$. Both cases are clear.


