# Chebyshev Inequality

### Statement

There at least two different mathematical statements that go under the name of Chebyshev Inequality (Pafnuty Chebyshev (1821-1894), a famous Russian mathematician.) One in the realm of Probability Theory is of fundamental importance, the other - the subject of the present page - is an extremely useful and powerful tool for solving olympiad-kind inequalities.

$\{x_i\}$ and $\{y_i\}$ $(i=1,\ldots,n)$ be two monotone sequences. If sequences $\{x_i\}$ and $\{y_i\}$ are similarly ordered then

$\displaystyle n\sum_{i=1}^{n}x_iy_{i}\ge\sum_{i=1}^{n}x_i\sum_{i=1}^{n}y_i.$

If sequences $\{x_i\}$ and $\{y_i\}$ are oppositely ordered then

$\displaystyle n\sum_{i=1}^{n}x_iy_{i}\le\sum_{i=1}^{n}x_i\sum_{i=1}^{n}y_i.$

The equality is achieved when either of the sequences is constant.

Note that the inequalities can be put into a more symmetric form, as a statement of the averages, e.g.,

$\displaystyle \frac{1}{n}\sum_{i=1}^{n}x_iy_{i}\ge\left(\frac{1}{n}\sum_{i=1}^{n}x_i\right)\left(\frac{1}{n}\sum_{i=1}^{n}y_i\right),$

for similarly ordered sequences.

### Proof 1

Assume, the sequences are similarly ordered. The second cases follows from this by changing the signs of all the terms in one of the sequences. With a reference to the Rearrangement Inequality,

$ \displaystyle \sum_{i=1}^{n}x_iy_{i}\ge x_1y_1+x_2y_2+\ldots+x_ny_n\\ \displaystyle \sum_{i=1}^{n}x_iy_{i}\ge x_1y_2+x_2y_3+\ldots+x_ny_1\\ \displaystyle \sum_{i=1}^{n}x_iy_{i}\ge x_1y_3+x_2y_4+\ldots+x_ny_2\\ \displaystyle \sum_{i=1}^{n}x_iy_{i}\ge x_1y_4+x_2y_5+\ldots+x_ny_3\\ \cdots \\ \displaystyle \sum_{i=1}^{n}x_iy_{i}\ge x_1y_n+x_2y_1+\ldots+x_ny_{n-1}\\ $

What remains is to sum all these up.

### Proof 2

$\displaystyle\begin{align} n\sum_{i=1}^{n}x_iy_{i} - \sum_{i=1}^{n}x_i\sum_{j=1}y_i &= \sum_{j=1}^{n}\sum_{i=1}^{n}x_iy_{i} - \sum_{i=1}^{n}x_i\sum_{j=1}^{n}y_j\\ &= \sum_{i=1}^{n}\sum_{j=1}^{n}x_jy_{j} - \sum_{i=1}^{n}\sum_{j=1}^{n}x_iy_j\\ &= \sum_{j=1}^{n}\sum_{i=1}^{n}x_iy_{i} - \sum_{j=1}^{n}\sum_{i=1}^{n}x_jy_i\\ &= \frac{1}{2}\sum_{j=1}^{n}\sum_{i=1}^{n}(x_iy_{i}+x_jy_{j} -x_jy_i-x_iy_j)\\ &= \frac{1}{2}\sum_{j=1}^{n}\sum_{i=1}^{n}(x_i-x_j)(y_{i}-y_j)\\ &\ge 0. \end{align}$

Since both sequences are monotone, the inequality becomes equality when all the products vanish. This happens when $(x_1-x_n)(y_1-y_n)$ vanishes, i.e., when at least one of the sequences is constant.

### Example 1

Let $x_1,x_2,\ldots,x_n$ be positive numbers. Then

$\displaystyle x_{1}^{n+1}+x_{2}^{n+1}+\ldots+x_{n}^{n+1}\ge x_1x_2\ldots x_n(x_1+x_2+\ldots+x_n).$

We may consider the sequence $\{x_i\}$ to be ordered and note that the sequences $\{x_i\}$ and $\displaystyle\{x_i^{n}\}\;$ are ordered similarly. Then by the Chebyshev inequality,

$\displaystyle\begin{align} \frac{1}{n}\sum_{i=1}^{n}x_{i}^{n+1} &\ge \frac{1}{n}\sum_{i=1}^{n}x_{i}^{n}\frac{1}{n}\sum_{i=1}^{n}x_i\\ &\ge\sqrt[n]{x_{1}^{n}x_{2}^{n}\cdots x_{n}^{n}}\sum_{i=1}^{n}x_i\\ &= x_1x_2\ldots x_n\cdot\frac{1}{n}\cdot (x_1+x_2+\ldots+x_n). \end{align}$

This is equivalent to the required inequality.

**Note:** the above can be shown directly with the Rearrangement Inequality.

### Example 2 (Nesbitt's inequality)

For positive $a,b,c,$

$\displaystyle\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}.$

Set $s=a+b+c\;$ and rewrite the inequality as

$\displaystyle\frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c}\ge\frac{3}{2}.$

Assuming $a,b,c$ ordered, the terms $s-a,s-b,s-c\;$ are oppositely ordered. Thus by the Chebyshev inequality

$\displaystyle \left(\frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c}\right)\left[(s-a)+(s-b)+(s-c)\right]\ge 3(a+b+c).$

Since $(s-a)+(s-b)+(s-c)=2s,\;$ this is equivalent to Nesbitt's inequality.

### References

- Kin-Yin Li, Chebyshev Inequality, Mathematical Excalibur, Volume 4, Number 3, January, 1999 - March, 1999
- Yue Kwok Choy, Chebyshev Inequality
- Samin Riasat, Basics of Olympiad Inequalities
- IMOMath, Chebyshev Inequality. Chebyshev’s Inequality
- G. H. Hardy, J. E. Littlewood, G. Polya,
*Inequalities*, Cambridge University Press (2^{nd}edition) - A. Engel, Problem-Solving Strategies, Springer Verlag, 1998 1988.
- Xu Jiagu,
*Lecture Notes on Mathematical Olympiad Courses*, v 8, (For senior section, v 2), World Scientific, 2012 - P. Zeitz, The Art and Craft of Problem Solving, John Wiley & Sons, 1999

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