I have solve this following 



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> Question: Complex numbers ${x\_i},{y\_i}$ satisfy $\left| {{x\_i}} \right| = \left| {{y\_i}} \right| = 1$ for $i=1,2,\ldots ,n$. Let $x=\frac{1}{n}\sum\limits\_{i=1}^n{{x\_i}}$, $y=\frac{1}{n}\sum\limits\_{i=1}^n{{y\_i}}$ and $z\_i=x{y\_i}+y{x\_i}-{x\_i}{y\_i}$. Prove that $$\sum\limits\_{i=1}^n{\left| {{z\_i}}\right|}\leqslant n$$.
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**Proof:** Note that $|z\_i| = |(x-x\_i)(y-y\_i) - xy| \leq |x-x\_i| |y-y\_i| + |x| |y|$. Hence by Cauchy-Schwarz Inequality,
\begin{align\*} \sum |z\_i| &\leq \sum |x-x\_i| |y-y\_i| + n|x| |y| \\ &\leq \sqrt{\sum |x-x\_i|^2}\sqrt{\sum |y-y\_i|^2} + n|x| |y|.\end{align\*}Now we have
\begin{align\*}\sum |x-x\_i|^2 &= \sum \left(|x|^2 + |x\_i|^2 + 2{\rm Re}(x\overline{x\_i})\right)\\
&= n|x|^2 + n - 2{\rm Re}(x\sum\overline{x\_i})\\
&= n|x|^2 + n - 2n{\rm Re}(x\overline{x})\\
&= n(1-|x|^2),
\end{align\*}the last equality following from ${\rm Re} (x\overline{x}) = x\overline{x} = |x|^2$.


Thus
$$\sqrt{\sum |x-x\_i|^2}\sqrt{\sum |y-y\_i|^2} + n|x| |y| = n\left(\sqrt{1-|x|^2}\sqrt{1-|y|^2} + |x| |y|\right) \leq n,$$


where the last inequality is also proven by Cauchy-Schwarz, and we are done.


Now I conjecture and try prove(Now I can't prove it)



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> **Question** Complex numbers ${x\_i},{y\_i},z\_{i}$ satisfy $\left| {{x\_i}} \right| = \left| {{y\_i}} \right| = \left|z\_{i}\right|=1$ for $i=1,2,\ldots ,n$. Let $x=\frac{1}{n}\sum\limits\_{i=1}^n{{x\_i}}$, $y=\frac{1}{n}\sum\limits\_{i=1}^n{{y\_i}},z=\dfrac{1}{n}\sum\_{i=1}^{n}z\_{i}$ and $w\_i=xy{z\_i}+yz{x\_i}+xzy\_{i}-2{x\_i}{y\_i}z\_{i}$. Prove that $$\sum\limits\_{i=1}^n{\left| {{w\_i}}\right|}\leqslant n$$.
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maybe this conjecture also hold to this complex inequality
The following seems to be a counterexample:


Let $n=2$, $x\_1 = 1 = - y\_1$, $x\_2= -1 = - y\_2$, $z\_1=z\_2 =1$.


Then $x=y=0$, $z=1$. So $w\_i = -2 x\_i y\_i z\_i = 2$ for $i=1,2$.


Finally, $w\_1 + w\_2 = 4$.


