# proof of Chebyshev’s inequality

Let $x_{1},x_{2},\ldots,x_{n}$ and $y_{1},y_{2},\ldots,y_{n}$ be real numbers such that $x_{1}\leq x_{2}\leq\cdots\leq x_{n}$. Write the product $(x_{1}+x_{2}+\cdots+x_{n})(y_{1}+y_{2}+\cdots+y_{n})$ as

 $\displaystyle(x_{1}y_{1}+x_{2}y_{2}+\cdots+x_{n}y_{n})$ (1) $\displaystyle+$ $\displaystyle(x_{1}y_{2}+x_{2}y_{3}+\cdots+x_{n-1}y_{n}+x_{n}y_{1})$ $\displaystyle+$ $\displaystyle(x_{1}y_{3}+x_{2}y_{4}+\cdots+x_{n-2}y_{n}+x_{n-1}y_{1}+x_{n}y_{2})$ $\displaystyle+$ $\displaystyle\cdots$ $\displaystyle+$ $\displaystyle(x_{1}y_{n}+x_{2}y_{1}+x_{3}y_{2}+\cdots+x_{n}y_{n-1}).$
• If $y_{1}\leq y_{2}\leq\cdots\leq y_{n}$, each of the $n$ terms in parentheses is less than or equal to $x_{1}y_{1}+x_{2}y_{2}+\cdots+x_{n}y_{n}$, according to the rearrangement inequality. From this, it follows that

 $(x_{1}+x_{2}+\cdots+x_{n})(y_{1}+y_{2}+\cdots+y_{n})\leq n(x_{1}y_{1}+x_{2}y_{% 2}+\cdots+x_{n}y_{n})$

or (dividing by $n^{2}$)

 $\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)\left(\frac{y_{1}+y_{2}+\cdots+% y_{n}}{n}\right)\leq\frac{x_{1}y_{1}+x_{2}y_{2}+\cdots+x_{n}y_{n}}{n}.$
• If $y_{1}\geq y_{2}\geq\cdots\geq y_{n}$, the same reasoning gives

 $\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)\left(\frac{y_{1}+y_{2}+\cdots+% y_{n}}{n}\right)\geq\frac{x_{1}y_{1}+x_{2}y_{2}+\cdots+x_{n}y_{n}}{n}.$

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 (1) can only be equal to $x_{1}y_{1}+x_{2}y_{2}+\cdots+x_{n}y_{n}$ 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 $x_{1}y_{1}+x_{2}y_{2}+\cdots+x_{n}y_{n}$ 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 $y_{1}=y_{2}=\cdots=y_{n}$.

Title proof of Chebyshev’s inequality ProofOfChebyshevsInequality 2013-03-22 13:08:38 2013-03-22 13:08:38 pbruin (1001) pbruin (1001) 4 pbruin (1001) Proof msc 26D15