proof of arithmetic-geometric means inequality

A short geometrical proof can be given for the case n=2 of the arithmetic-geometric means inequality.

Let a and b be two non negative numbers. Draw the line AB such that AP has length a, and PB has length b, as in the following picture, and draw a semicircle with diameterMathworldPlanetmathPlanetmath AB. Let O be the center of the circle.


Now raise perpendicularsPlanetmathPlanetmathPlanetmath PQ and OT to AB. Notice that OT is a radius, and so


Also notice that PQOT for any point P, and equality is obtained only when P=O, that is, when a=b.


Notice also that PQ is a height over the hypotenuseMathworldPlanetmath on right triangleMathworldPlanetmath AQB. We have then triangle similaritiesMathworldPlanetmath AQBAPQQPB, and thus


which implies PQ=APPB=ab. Since PQOT, we conclude


This special case can also be proved using rearrangement inequality. Let a,b non negative numbers, and assume ab. Let x1=a,x2=b, and then x1x2. Now suppose y1 and y2 are such that one of them is x1 and the other is x2. Rearrangement inequality states that x1y1+x2y2 is maximum when y1y2 and x1x2. So, we have


and substituting back a,b gives


where it follows the desired result.

One more proof can be given as follows. Let x=a,y=b. Then (x-y)20, and equality holds only when x=y. Then, x2-2xy+y20 becomes


and substituting back a,b gives the desired result as in the previous proof.

Title proof of arithmetic-geometric means inequality
Canonical name ProofOfArithmeticgeometricMeansInequality
Date of creation 2013-03-22 14:49:14
Last modified on 2013-03-22 14:49:14
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 10
Author mathcam (2727)
Entry type Proof
Classification msc 26D15