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 A⁢B such that A⁢P has length a, and P⁢B has length b, as in the following picture, and draw a semicircle with diameterMathworldPlanetmathPlanetmath A⁢B. Let O be the center of the circle.

QOABPab

Now raise perpendicularsPlanetmathPlanetmathPlanetmath P⁢Q and O⁢T to A⁢B. Notice that O⁢T is a radius, and so

O⁢T=A⁢B2=a+b2

Also notice that P⁢Q≤O⁢T for any point P, and equality is obtained only when P=O, that is, when a=b.

QOABPT

Notice also that P⁢Q is a height over the hypotenuseMathworldPlanetmath on right triangleMathworldPlanetmath △⁢A⁢Q⁢B. We have then triangle similaritiesMathworldPlanetmath △⁢A⁢Q⁢B∼△⁢A⁢P⁢Q∼△⁢Q⁢P⁢B, and thus

A⁢PP⁢Q=P⁢QP⁢B

which implies P⁢Q=A⁢P⋅P⁢B=a⁢b. Since P⁢Q≤O⁢T, we conclude

a⁢b≤a+b2.

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

x1⁢x2+x2⁢x1≤x12+x22

and substituting back a,b gives

2⁢a⁢b≤(a)2+(b)2=a+b

where it follows the desired result.

One more proof can be given as follows. Let x=a,y=b. Then (x-y)2≥0, and equality holds only when x=y. Then, x2-2⁢x⁢y+y2≥0 becomes

x2+y2≥2⁢x⁢y

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