ProofOfChebyshevsInequality2 2002-11-10 14:02:37 2002-11-10 14:02:37 Proof Chebyshev's inequality 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Let $x_1,x_2,\ldots,x_n$ and $y_1,y_2,\ldots,y_n$ be real numbers such that $x_1\le x_2\le\cdots\le x_n$. Write the product $(x_1+x_2+\cdots+x_n)(y_1+y_2+\cdots+y_n)$ as \begin{eqnarray} \nonumber &&(x_1y_1+x_2y_2+\cdots+x_ny_n)\\ \nonumber &+&(x_1y_2+x_2y_3+\cdots+x_{n-1}y_n+x_ny_1)\\ \nonumber &+&(x_1y_3+x_2y_4+\cdots+x_{n-2}y_n+x_{n-1}y_1+x_ny_2)\\ \nonumber &+&\cdots\\ &+&(x_1y_n+x_2y_1+x_3y_2+\cdots+x_ny_{n-1}). \label{expanded} \end{eqnarray} \begin{itemize} \item If $y_1\le y_2\le\cdots\le y_n$, each of the $n$ terms in parentheses is less than or equal to $x_1y_1+x_2y_2+\cdots+x_ny_n$, according to the rearrangement inequality. From this, it follows that $$ (x_1+x_2+\cdots+x_n)(y_1+y_2+\cdots+y_n)\le n(x_1y_1+x_2y_2+\cdots+x_ny_n) $$ or (dividing by $n^2$) $$ \left(\frac{x_1+x_2+\cdots+x_n}{n}\right) \left(\frac{y_1+y_2+\cdots+y_n}{n}\right)\le \frac{x_1y_1+x_2y_2+\cdots+x_ny_n}{n}. $$ \item If $y_1\ge y_2\ge\cdots\ge y_n$, the same reasoning gives $$ \left(\frac{x_1+x_2+\cdots+x_n}{n}\right) \left(\frac{y_1+y_2+\cdots+y_n}{n}\right)\ge \frac{x_1y_1+x_2y_2+\cdots+x_ny_n}{n}. $$ \end{itemize} It is clear that equality holds if $x_1=x_2=\cdots=x_n$ or $y_1=y_2=\cdots=y_n$. To see that this condition is also necessary, suppose that not all $y_i$'s are equal, so that $y_1\neq y_n$. Then the second term in parentheses of (\ref{expanded}) can only be equal to $x_1y_1+x_2y_2+\cdots+x_ny_n$ if $x_{n-1}=x_n$, the third term only if $x_{n-2}=x_{n-1}$, and so on, until the last term which can only be equal to $x_1y_1+x_2y_2+\cdots+x_ny_n$ if $x_1=x_2$. This implies that $x_1=x_2=\cdots=x_n$. Therefore, Chebyshev's inequality is an equality if and only if $x_1=x_2=\cdots=x_n$ or $y_1=y_2=\cdots=y_n$.