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Following a problem I found on mathstack, 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 $n$?

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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.

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  • $\begingroup$ Oh,Nice,Thank you very much! for $n=2$ you have easy to prove? $\endgroup$ – function sug May 11 '16 at 6:47
  • $\begingroup$ Nice,I conjecture $|1+x|+|1+y|+|1+z|+|1+xyz|\ge |x|+|y|+|z|?$.maybe is also hold.and for $n=4,5$ and so on maybe is hold too $\endgroup$ – function sug May 12 '16 at 4:30
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    $\begingroup$ this is already false, take $x=y=z=-1$ $\endgroup$ – Fedor Petrov May 12 '16 at 6:18
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Add Edit In deed,this problem in 2012 by A Catalin Tigaeru have prove it,But He methods is very ugly.

For $n=2$ it seem can also following prove it

\begin{align*}(|1+z_{i}|+|1+z_{j}|+|1+z_{i}z_{j}|)^2&=(|1+z_{i}|+|1+\overline{z_{j}}|+|1+z_{i}z_{j}|)^2\\ &\ge (|z_{i}-\overline{z_{j}}|+|1+z_{i}z_{j}|)^2\ge |z_{i}-\overline{z_{j}}|^2+|1+z_{i}z_{j}|^2\\ &=|z_{i}|^2+|z_{j}|^2+|z_{i}z_{j}|^2+1-2\Re{(z_{i}{z_{j}})}+2\Re{(z_{i}z_{j})}\\ &\ge |z_{i}|^2+|z_{j}|^2+2|z_{i}z_{j}|\\ &= (|z_{i}|+|z_{j}|)^2 \end{align*}

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