I have solve this following
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 $$\sum\limits_{i=1}^n{\left| {{z_i}}\right|}\leqslant n$$.
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)
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 $$\sum\limits_{i=1}^n{\left| {{w_i}}\right|}\leqslant n$$.
