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 $\sum\limits_{i=1}^n{\left| {{z_i}}\right|}\leqslant n$.
Problem
Source: 2012 China TST Quiz 1 Day 1 P1
Tags: inequalities, complex numbers, triangle inequality, algebra proposed, algebra
mahanmath
14.03.2012 15:18
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 :
$\left |\sum\limits_{i = 1}^{n} \left (x \bar{y} \frac{y_i}{x_i} +y \bar{x} \frac{x_i}{y_i} \right ) \right | \leq |x\bar{y}|\left ( \sum\limits_{i = 1}^{n}|\frac{y_i}{x_i}| \right ) + |y\bar{x}|\left ( \sum\limits_{i = 1}^{n}|\frac{x_i}{y_i}| \right ) = 2n|xy|$
$\implies \sum\limits_{i = 1}^{n}{\left|{{z_{i}}}\right|}^2 = \sum\limits_{i = 1}^{n} \left (x \bar{y} \frac{y_i}{x_i} +y \bar{x} \frac{x_i}{y_i} \right ) \- -n(|x|^2 + |y|^2 )+n \leq n - n (|x| -|y|)^2 \leq n$
chronondecay
22.03.2012 06:37
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.
P-H-David-Clarence
06.11.2016 08:43
You are right!