# maybe this conjecture also hold to this complex inequality

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 $$\sum\limits_{i=1}^n{\left| {{w_i}}\right|}\leqslant n$$.

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, $w_1 + w_2 = 4$.
• Thanks,can we get a similar conclusion?How to edit this $w_{i}$ Jul 15 '16 at 11:27