proof of arithmetic-geometric-harmonic means inequality


Let M be max{x1,x2,x3,,xn} and let m be min{x1,x2,x3,,xn}.

Then

M=M+M+M++Mnx1+x2+x3++xnn
m=nnm=n1m+1m+1m++1mn1x1+1x2+1x3++1xn

where all the summations have n terms. So we have proved in this way the two inequalitiesMathworldPlanetmath at the extremes.

Now we shall prove the inequality between arithmetic meanMathworldPlanetmath and geometric meanMathworldPlanetmath.

1 Case n=2

We do first the case n=2.

(x1-x2)2 0
x1-2x1x2+x2 0
x1+x2 2x1x2
x1+x22 x1x2

2 Case n=2k

Now we prove the inequality for any power of 2 (that is, n=2k for some integer k) by using mathematical induction.

x1+x2++x2k+x2k+1++x2k+12k+1
= (x1+x2++x2k2k)+(x2k+1+x2k+2++x2k+12k)2

and using the case n=2 on the last expression we can state the following inequality

x1+x2++x2k+x2k+1++x2k+12k+1
(x1+x2++x2k2k)(x2k+1+x2k+2++x2k+12k)
x1x2x2k2kx2k+1x2k+2x2k+12k

where the last inequality was obtained by applying the induction hypothesis with n=2k. Finally, we see that the last expression is equal to x1x2x3x2k+12k+1 and so we have proved the truth of the inequality when the number of terms is a power of two.

3 Inequality for n numbers implies inequality for n-1

Finally, we prove that if the inequality holds for any n, it must also hold for n-1, and this propositionPlanetmathPlanetmathPlanetmath, combined with the preceding proof for powers of 2, is enough to prove the inequality for any positive integer.

Suppose that

x1+x2++xnnx1x2xnn

is known for a given value of n (we just proved that it is true for powers of two, as example). Then we can replace xn with the average of the first n-1 numbers. So

x1+x2++xn-1+(x1+x2++xn-1n-1)n
= (n-1)x1+(n-1)x2++(n-1)xn-1+x1+x2++xn-1n(n-1)
= nx1+nx2++nxn-1n(n-1)
= x1+x2++xn-1(n-1)

On the other hand

x1x2xn-1(x1+x2++xn-1n-1)n
= x1x2xn-1nx1+x2++xn-1n-1n

which, by hypothesisMathworldPlanetmathPlanetmath (the inequality holding for n numbers) and the observations made above, leads to:

(x1+x2++xn-1n-1)n(x1x2xn)(x1+x2++xn-1n-1)

and so

(x1+x2++xn-1n-1)n-1x1x2xn

from where we get that

x1+x2++xn-1n-1x1x2xnn-1.

So far we have proved the inequality between the arithmetic mean and the geometric mean. The geometric-harmonic inequality is easier. Let ti be 1/xi.

From

t1+t2++tnnt1t2t3tnn

we obtain

1x1+1x2+1x3++1xnn1x11x21x31xnn

and therefore

x1x2x3xnnn1x1+1x2+1x3++1xn

and so, our proof is completed.

Title proof of arithmetic-geometric-harmonic means inequality
Canonical name ProofOfArithmeticgeometricharmonicMeansInequality
Date of creation 2013-03-22 12:41:25
Last modified on 2013-03-22 12:41:25
Owner drini (3)
Last modified by drini (3)
Numerical id 6
Author drini (3)
Entry type Proof
Classification msc 26D15
Related topic ArithmeticMean
Related topic GeometricMean
Related topic HarmonicMean
Related topic GeneralMeansInequality
Related topic WeightedPowerMean
Related topic PowerMean