We prove a stronger result $ \sum\limits_{i = 1}^{n}{\left|{{z_{i}}}\right|} ^2\leqslant n $ and statement follows from an application of Cauchy-Schwartz then . $ \sum\limits_{i = 1}^{n}{\left|{{z_{i}}}\right|}^2 \\ =\sum\limits_{i = 1}^{n} \left ( xy_i + x_i y -x_i y_i \right )\left ( \frac{\bar{x}}{y_i} + \frac{\bar{y}}{x_i} - \frac{1}{x_i y_i} \right ) \\ =\sum\limits_{i = 1}^{n} \left ( x \bar{x} +x \bar{y} \frac{y_i}{x_i} - \frac{x}{x_i} + y \bar{y} +y \bar{x} \frac{x_i}{y_i} - \frac{y}{y_i} -\bar{x}x_i -\bar{y}y_i +1 \right ) \\ =n|x|^2 + \sum\limits_{i = 1}^{n}\left ( x \bar{y} \frac{y_i}{x_i} \right ) - n |x|^2 + n|y|^2 + \sum\limits_{i = 1}^{n} y \bar{x} \frac{x_i}{y_i} - n|y|^2 -n(|x|^2 + |y|^2 )+n \\ =\sum\limits_{i = 1}^{n} \left (x \bar{y} \frac{y_i}{x_i} +y \bar{x} \frac{x_i}{y_i} \right ) \- -n(|x|^2 + |y|^2 )+n $ By triangle inequality we have : $\left |\sum\limits_{i = 1}^{n} \left (x \bar{y} \frac{y_i}{x_i} +y \bar{x} \frac{x_i}{y_i} \right ) \right | \leq |x\bar{y}|\left ( \sum\limits_{i = 1}^{n}|\frac{y_i}{x_i}| \right ) + |y\bar{x}|\left ( \sum\limits_{i = 1}^{n}|\frac{x_i}{y_i}| \right ) = 2n|xy|$ $\implies \sum\limits_{i = 1}^{n}{\left|{{z_{i}}}\right|}^2 = \sum\limits_{i = 1}^{n} \left (x \bar{y} \frac{y_i}{x_i} +y \bar{x} \frac{x_i}{y_i} \right ) \- -n(|x|^2 + |y|^2 )+n \leq n - n (|x| -|y|)^2 \leq n$

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.