Let $z_i=\frac{y_i}n\sum_{j=1}^nx_j+\frac{x_i}n\sum_{j=1}^ny_j-x_iy_i$ with $|x_i|=|y_i|=1$. Prove that $\sum_{i=1}^n |z_i| \leq n$.

Question

Suppose that, for $i=1,2,\ldots,n$, complex numbers $x_i$ and $y_i$ satisfy $|x_i|=|y_i|=1$ . Let $$x=\frac 1 n \sum_{i=1}^n x_i\text{ and } y=\frac 1 n \sum_{i=1}^n y_i\,.$$ Define $$z_i=xy_i+yx_i-x_iy_i\text{ for }i=1,2,\ldots,n\,.$$

Prove that $$\sum_{i=1}^n |z_i| \leq n\,.$$

Answer 1

From $$z_i=xy_i+x_iy-x_iy_i\,,$$ we get $$\bar{z}_i=\bar{x}\bar{y}_i+\bar{x}_i\bar{y}-\bar{x}_i\bar{y}_i$$ for $i=1,2,\ldots,n$. Thus, $$\begin{align}|z_i|^2&=z_i\bar{z}_i\\&=x\bar{x}y_i\bar{y}_i+x\bar{x}_iy_i\bar{y}-x\bar{x}_iy_i\bar{y}_i+x_i\bar{x}y\bar{y}_i+x_i\bar{x}_iy\bar{y}-x_i\bar{x}_iy\bar{y}_i\\&\phantom{aaaaa}-x_i\bar{x}y_i\bar{y}_i-x_i\bar{x}_iy_i\bar{y}+x_i\bar{x}_iy_i\bar{y}_i\,.\end{align}$$ As $x_i\bar{x}_i=|x_i|^2=1$, $y_i\bar{y}_i=|y_i|^2=1$, $x\bar{x}=|x|^2$, and $y\bar{y}=|y|^2$, we get $$|z_i|^2=1+|x|^2-x\bar{x}_i-\bar{x}x_i+|y|^2-y\bar{y}_i-\bar{y}y_i+x\bar{y}\bar{x}_iy_i+\bar{x}yx_i\bar{y}_i\,.$$ Take the sum over $i=1,2,\ldots,n$ to get $$\begin{align}\sum_{i=1}^n\,|z_i|^2&=n+n\,|x|^2-x\,\sum_{i=1}^n\,\bar{x}_i-\bar{x}\,\sum_{i=1}^n\,x_i\\&\phantom{aaaaa}+n\,|y|^2-y\,\sum_{i=1}^n\,\bar{y}_i-\bar{y}\,\sum_{i=1}^n\,y_i+x\bar{y}\,\sum_{i=1}^n\,\bar{x}_iy_i+\bar{x}y\,\sum_{i=1}^n\,x_i\bar{y}_i\,.\end{align}$$ Because $\sum\limits_{i=1}^n\,x_i=n\,x$ and $\sum\limits_{i=1}^n\,y_i=n\,y$, we obtain $$\begin{align}\sum_{i=1}^n\,|z_i|^2&=n+n|x|^2-x(n\,\bar{x})-\bar{x}(n\,x)+n|y|^2-y(n\,\bar{y})+\bar{y}(n\,y)\\ &\phantom{aaaaa}+x\bar{y}\,\sum_{i=1}^n\,\bar{x}_iy_i+\bar{x}y\,\sum_{i=1}^n\,x_i\bar{y}_i \\&=n-n\,|x|^2-n\,|y|^2+x\bar{y}\,\sum_{i=1}^n\,\bar{x}_iy_i+\bar{x}y\,\sum_{i=1}^n\,x_i\bar{y}_i \\&\leq n-n\,|x|^2-n\,|y|^2+|x|\,|\bar{y}|\,\left|\sum_{i=1}^n\,\bar{x}_iy_i\right|+|\bar{x}|\,|y|\,\left|\sum_{i=1}^n\,x_i\bar{y}_i\right| \\&=n-n|x|^2-n|y|^2+2\,|x|\,|y|\,\left|\sum_{i=1}^n\,\bar{x}_iy_i\right|\,.\end{align}$$ Using the Cauchy-Schwarz Inequality, $$\left|\sum_{i=1}^n\,\bar{x}_iy_i\right|\leq\sqrt{ \sum_{i=1}^n\,|\bar{x}_i|^2}\sqrt{\sum_{i=1}^n\,|y_i|^2}=\sqrt{n}\sqrt{n}=n\,.$$ Thus, $$\sum_{i=1}^n\,|z_i|^2\leq n-n|x|^2-n|y|^2+2n\,|x|\,|y|=n-n\big(|x|-|y|\big)^2\leq n\,.$$ That is, $$\sum_{i=1}^n\,|z_i|^2\leq n$$ with equality iff $\sum\limits_{i=1}^n\,x_i=\sum\limits_{i=1}^n\,y_i=0$ or there exists a complex number $u$ such that $u=1$ for which $$y_i=ux_i\text{ for all }i=1,2,\ldots,n\,.$$

Consequently, the Cauchy-Schwarz Inequality or the Power-Mean Inequality implies $$\sum_{i=1}^n\,|z_i|\leq \sqrt{n\,\sum_{i=1}^n\,|z_i|^2}\leq \sqrt{n\cdot n}=n\,.$$ The equality holds if and only if

• $\sum\limits_{i=1}^n\,x_i=\sum\limits_{i=1}^n\,y_i=0$, or
• $x_1=x_2=\ldots=x_n$ and $y_1=y_2=\ldots=y_n$, or
• $n$ is even, there exists a complex number $u$ such that $|u|=1$ for which $y_i=ux_i$ for each $i=1,2,\ldots,n$, and there exists a subset $S$ of $\{1,2,\ldots,n\}$ of size $\dfrac{n}{2}$ such that $$x_i=x_j\text{ for all }i,j\in S\text{ and } x_i=x_j\text{ for all }i,j\in\{1,2,\ldots,n\}\setminus S\,.$$