proof of Chebyshev’s inequality

Let x1,x2,,xn and y1,y2,,yn be real numbers such that x1x2xn. Write the product (x1+x2++xn)(y1+y2++yn) as

(x1y1+x2y2++xnyn) (1)
+ (x1y2+x2y3++xn-1yn+xny1)
+ (x1y3+x2y4++xn-2yn+xn-1y1+xny2)
+ (x1yn+x2y1+x3y2++xnyn-1).
  • If y1y2yn, each of the n terms in parentheses is less than or equal to x1y1+x2y2++xnyn, according to the rearrangement inequality. From this, it follows that


    or (dividing by n2)

  • If y1y2yn, the same reasoning gives


It is clear that equality holds if x1=x2==xn or y1=y2==yn. To see that this condition is also necessary, suppose that not all yi’s are equal, so that y1yn. Then the second term in parentheses of (1) can only be equal to x1y1+x2y2++xnyn if xn-1=xn, the third term only if xn-2=xn-1, and so on, until the last term which can only be equal to x1y1+x2y2++xnyn if x1=x2. This implies that x1=x2==xn. Therefore, Chebyshev’s inequalityMathworldPlanetmath is an equality if and only if x1=x2==xn or y1=y2==yn.

Title proof of Chebyshev’s inequality
Canonical name ProofOfChebyshevsInequality
Date of creation 2013-03-22 13:08:38
Last modified on 2013-03-22 13:08:38
Owner pbruin (1001)
Last modified by pbruin (1001)
Numerical id 4
Author pbruin (1001)
Entry type Proof
Classification msc 26D15