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 $n$?


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.

  • $\begingroup$ 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
  • 3
    $\begingroup$ this is already false, take $x=y=z=-1$ $\endgroup$ – Fedor Petrov May 12 '16 at 6:18

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*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.