proof of general means inequality


Let w1, w2, …, wn be positive real numbers such that w1+w2++wn=1. For any real number r0, the weighted power mean of degree r of n positive real numbers x1, x2, …, xn (with respect to the weights w1, …, wn) is defined as

Mwr(x1,x2,,xn)=(w1x1r+w2x2r++wnxnr)1/r.

The definition is extended to the case r=0 by taking the limit r0; this yields the weighted geometric mean

Mw0(x1,x2,,xn)=x1w1x2w2xnwn

(see derivation of zeroth weighted power mean). We will prove the weighted power means inequalityMathworldPlanetmath, which states that for any two real numbers r<s, the weighted power means of orders r and s of n positive real numbers x1, x2, …, xn satisfy the inequality

Mwr(x1,x2,,xn)Mws(x1,x2,,xn)

with equality if and only if all the xi are equal.

First, let us suppose that r and s are nonzero. We distinguish three cases for the signs of r and s: r<s<0, r<0<s, and 0<r<s. Let us consider the last case, i.e. assume r and s are both positive; the others are similar. We write t=sr and yi=xir for 1in; this implies yit=xis. Consider the function

f:(0,) (0,)
x xt.

Since t>1, the second derivative of f satisfies f′′(x)=t(t-1)xt-2>0 for all x>0, so f is a strictly convex function. Therefore, according to Jensen’s inequality,

(w1y1+w2y2++wnyn)t = f(w1y1+w2y2++wnyn)
w1f(y1)+w2f(y2)++wnf(yn)
= w1y1t+w2y2t++wnynt,

with equality if and only if y1=y2==yn. By substituting t=sr and yi=xir back into this inequality, we get

(w1x1r+w2x2r++wnxnr)s/rw1x1s+w2x2s++wnxns

with equality if and only if x1=x2==xn. Since s is positive, the function xx1/s is strictly increasing, so raising both sides to the power 1/s preserves the inequality:

(w1x1r+w2x2r++wnxnr)1/r(w1x1s+w2x2s++wnxns)1/s,

which is the inequality we had to prove. Equality holds if and only if all the xi are equal.

If r=0, the inequality is still correct: Mw0 is defined as limr0Mwr, and since MwrMws for all r<s with r0, the same holds for the limit r0. The same argumentMathworldPlanetmath shows that the inequality also holds for s=0, i.e. that MwrMw0 for all r<0. We conclude that for all real numbers r and s such that r<s,

Mwr(x1,x2,,xn)Mws(x1,x2,,xn).
Title proof of general means inequality
Canonical name ProofOfGeneralMeansInequality
Date of creation 2013-03-22 13:10:26
Last modified on 2013-03-22 13:10:26
Owner pbruin (1001)
Last modified by pbruin (1001)
Numerical id 5
Author pbruin (1001)
Entry type Proof
Classification msc 26D15
Related topic ArithmeticMean
Related topic GeometricMean
Related topic HarmonicMean
Related topic RootMeanSquare3
Related topic PowerMean
Related topic WeightedPowerMean
Related topic ArithmeticGeometricMeansInequality
Related topic JensensInequality