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AG mean Oscar G. Villareal vol. 9, iss. 3, art. 78, 2008

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## THE ARITHMETIC-ALGEBRAIC MEAN INEQUALITY VIA SYMMETRIC MEAN

OSCAR G. VILLAREAL

Department of Mathematics University of California, Irvine

340 Rowland Hall, Irvine, CA 92697-3875, USA EMail:ovillare@math.uci.edu

URL:http://math.uci.edu/∼oscar

Accepted: 06 August, 2008

Communicated by: P.S. Bullen 2000 AMS Sub. Class.: Primary 26D15

Key words: Arithmetic mean, Geometric mean, Symmetric mean, Inequality

Abstract: We give two proofs of the arithmetic-algebraic mean inequality by giving a char- acterization of symmetric means.

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### Contents

1 Introduction 3

2 First Proof of Theorem 1.1 4

3 Second Proof of Theorem 1.1 7

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### 1.Introduction

Let (a1, . . . , an) ∈ Rn be an n-tuple of positive real numbers. The inequality of arithmetic-algebraic means states that

n

a1a2· · ·an≤ a1+· · ·+an

n .

The left-hand side of the inequality is called the geometric mean and the right-hand side the arithmetic mean. We will refer to this inequality asAGnto specify the size of then-tuple. This inequality has been known in one form or another since antiquity and numerous proofs have been given over the centuries. Bullen’s book [1], for example, gives over seventy proofs. We give two proofs based on a characterization of symmetric means as the smallest among the means constructed by homogeneous symmetric polynomials. The main result is

Theorem 1.1. Let(a1, . . . , an) ∈ Rnbe ann-tuple of positive real numbers,f(x1, . . . , xn)be a homogenous symmetric polynomial of degree k, 1 ≤ k ≤ n, having positive coefficients, and letsk(x1, . . . , xn)be thek-th elementary symmetric poly- nomial. Then

sk(a1, . . . , an)

n k

≤ f(a1, . . . , an) f(1, . . . ,1) . There is equality if and only if thea0is are all equal.

Note that nk

= sk(1, . . . ,1). Similarly we note that if the coefficents of f are all equal to one, thenf(1, . . . ,1)is the number of monomials comprisingf. Thus it is reasonable to think of f(af(1,...,1)1,...,an) as a mean forf as in the theorem. The theorem implies the arithmetic-algebraic mean inequality by takingk =n,f(x1, . . . , xn) = (x1+· · ·+xn)nso thatf(1, . . . ,1) = nn, and then takingn-th roots.

We shall give two proofs of Theorem1.1. The first depends on Muirhead’s The- orem. The second provesAGnand Theorem1.1in one induction step.

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### 2.First Proof of Theorem1.1

For any functionf(x1, . . . , xn), the symmetric groupSnacts on thexk’s, and so we

set X

!f(x1, . . . , xn) = X

σ∈Sn

f(xσ(1), . . . , xσ(n)).

In particular, for ann-tuple of nonnegative real numbersα= (α1, α2, . . . , αn), when f(x1, . . . , xn) =xα =xα11xα22· · ·xαnn,

we set

[α] = 1 n!

X!xα11xα22· · ·xαnn.

Note that [1,0, . . . ,0] is the arithmetic mean while [1n,n1, . . . ,n1] is the geometric mean.

Let α = (α1, α2, . . . , αn), β = (β1, β2, . . . , βn)be two n-tuples of nonnegative real numbers. Muirhead’s theorem gives conditions under which an inequality exists of the form

[α] = 1 n!

X!xα11xα22· · ·xαnn ≤[β] = 1 n!

X!xβ11xβ22· · ·xβnn

valid for all positvexi’s. To do this we first note that[α]is invariant under permu- tations of theαi’s and so we introduce an eqivalence relation as follows. We write α≤βif some permutation of the coordinates ofαandβsatisfies

α12+· · ·+αn12+. . .+βn, α1 ≥α2 ≥ · · · ≥αnandβ1 ≥β2 ≥. . .≥βn,

α12+· · ·+αk ≤β12+. . .+βkfork= 1,2, . . . , n.

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Theorem 2.1. The inequality [α] = 1

n!

X!xα11xα22· · ·xαnn ≤[β] = 1 n!

X!xβ11xβ22· · ·xβnn

is valid for all positvexi’s if and only ifα ≤ β. There is equality only whenα =β or thexi’s are all equal.

We refer to [2] for the proof of this theorem and further discussion. Before giving the first proof of Theorem1.1we need a lemma.

Lemma 2.2. Let (a1j, . . . , anjj) ∈ Rnj for j = 1, . . . , m, and let c1, . . . , cm be positive real numbers. Supposea≤ a1j+···+an nj j

j for eachj. Then

a≤ c1(a11+· · ·+an11) +c2(a12+· · ·+an22) +· · ·+cm(a1m+· · ·+anmm)

c1n1+c2n2+· · ·+cmnm .

There is equality if and only if the original inequalities are all equalities.

Proof. For eachj we rewritea ≤ a1j+···+an nj j

j as nja ≤ a1j +· · ·+anjj. We then multiply bycj to obtaincjnja ≤ cj(a1j +· · ·+anjj). We now add over all j to obtain

(c1n1+c2n2+· · ·+cmnm)a

≤c1(a11+· · ·+an11) +c2(a12+· · ·+an22) +· · ·+cm(a1m+· · ·+anmm).

By dividing by the coefficient ofa we get the lemma. Note that if at least one of the original inequalities is strict, then the argument shows the final inequality is also strict.

Proof of Theorem1.1. Letf(x1, . . . , xn)be a homogenous symmetric polynomial of degreek with positive coefficients. The monomials off break up into orbits under

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the action of the symmetric groupSnand so we may writef =c1f1+· · ·+cmfm, cj > 0 where eachfj is a homogenous polynomial with all non-zero coefficients equal to one and for whichSnacts transitively. In view of Lemma2.2, for the proof of Theorem1.1we may assumef(x1, . . . , xn)itself is a homogenous polynomial of degreekwith all non-zero coefficients equal to one and for whichSnacts transitively.

For such an f, it follows that there exists an α such thatf(x1, . . . , xn) = t[α], where t = f(1,1, . . . ,1) is the number of monomials comprising f. We note that sk(x1, . . . , xn) = nk

[1,1, . . . ,1,0, . . . ,0] with k 1’s and n − k 0’s. Since [1,1, . . . ,1,0, . . . ,0]≤α, Theorem2.1gives the result.

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### 3.Second Proof of Theorem1.1

The inequality of arithmetic-geometric means can be stated in polynomial form in two ways. By takingn-th powers we get

a1· · ·an

a1+· · ·+an n

n

.

Alternately, if we letai =Ani we get

A1· · ·An ≤ An1 +· · ·+Ann

n .

We will refer to these equivalent inequalities also asAGn.

Letf(x1, . . . , xn)be a homogenous symmetric polynomial. The monomials off break up into orbits under the action of the symmetric groupSnand so we may write f = c1f1 +· · ·+cmfm, cj ∈ R where each fj is a homogenous polynomial with all non-zero coefficients equal to one and for whichSn acts transitively. In view of Lemma2.2, for the proof of Theorem 1.1 we may assume f(x1, . . . , xn)itself is a homogenous polynomial with all non-zero coefficients equal to one and for which Snacts transitively.

Proposition 3.1. AssumeAG2,. . . ,AGn−1. Letf(x1, . . . , xn)be a homogenous sym- metric polynomial of degreek, 1 ≤ k ≤ n, with all non-zero coefficients equal to one and for whichSnacts transitively. Assumef(x1, . . . , xn)6=xn1+· · ·+xnn. Then the conclusion of Theorem1.1holds.

Proof. The polynomialf(x1, . . . , xn)has a monomial of the formx`11x`22· · ·x`sswhere k = degf =`1+· · ·+`sand0< `j < n. ByAG`j for eachj we have

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`1x1x2· · ·x`1 ≤x`11 +· · ·+x``1

1,

`2x`1+1x`1+2· · ·x`1+`2 ≤x``2

1+1+· · ·+x``2

1+`2, ...

`sx`1+···+`s−1+1x`1+···+`s−1+2· · ·x`1+···+`s ≤x``s1+···+`s−1+1+· · ·+x``s1+···+`s.

Sincek = degf =`1+· · ·+`s, we multiply the inequalities to obtain (3.1) `1· · ·`sx1· · ·xk≤(x`11 +· · ·+x``1

1)· · ·(x``s

1+···+`s−1+1+· · ·+x``s

1+···+`s).

Inequality (3.1) now yields (3.2) X

!`1· · ·`sx1· · ·xk

≤X

! (x`11 +· · ·+x``11)· · ·(x``s1+···+`s−1+1+· · ·+x``s1+···+`s).

SinceP

!x1· · ·xkconsists ofn!monomials with coefficient one, we get X!x1· · ·xk = n!

n k

sk(x1, . . . , xn).

Similarly since(x`11+· · ·+x``1

1)· · ·(x``s

1+···+`s−1+1+· · ·+x``s

1+···+`s)consists of`1· · ·`s monomials with coefficient one, it follows thatP

! ((x`11+· · ·+x``1

1)· · ·(x``s

1+···+`s−1+1+

· · ·+x``s

1+···+`s)consists of`1· · ·`sn!monomials with coefficient one. Thus we have X! ((x`11+· · ·+x``1

1)· · ·(x``s

1+···+`s−1+1+· · ·+x``s

1+···+`s) = `1· · ·`sn!

t f(x1, . . . , xn),

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wheret = f(1, . . . ,1)is the number of monomials of f. Plugging this into (3.2), then we see that if thexi’s are not all equal then at least one permutation of (3.1) is a strict inequality and hence inequality (3.2) is also strict.

By the previous proposition and the discussion preceding it, in order to prove Theorem1.1, it suffices to proveAGnfor alln≥2.

Theorem 3.2. AGnis true for alln≥2.

Proof. The proof is by induction onn. The casen = 2is standard. For x, y ∈ R, x, y > 0we have (√

x−√

y)2 ≥ 0with equality if and only ifx = y. Expanding we get.

x−2√

xy+y≥0, x+y≥2√

xy, x+y

2 ≥√ xy.

We now assumeAG2, . . . , AGn and we proveAGn+1. To this end, it suffices to show that

x1· · ·xn+1

x1+· · ·+xn+1 n+ 1

n+1

.

Now, byAG2 andAGnwe have for eachk, q

xkn

x1· · ·xk−1xk+1· · ·xn+1 ≤ xk+ √n

x1· · ·xk−1xk+1· · ·xn+1

2

≤ x1+· · ·+nxk+· · ·+xn+1 2n

= s+ (n−1)xk

2n .

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Where we have sets =x1 +· · ·+xn+1. Multiplying these inequalities overk, we get

x1· · ·xn+1

n+1

Y

k=1

s+ (n−1)xk 2n

= 1

(2n)n+1

n+1

Y

k=1

(s+ (n−1)xk).

Multiplying through by(2n)n+1 and expanding we get, (3.3) (2n)n+1x1· · ·xn+1

n+1

X

k=0

(n−1)ksk(x1, . . . , xn+1)sn+1−k.

We now use Proposition3.1and the discussion preceding it to conclude sk(x1, . . . , xn+1)≤

n+ 1 k

sk (n+ 1)k for0< k < n+ 1. Plugging this into (3.3), we get,

(3.4) (2n)n+1x1· · ·xn+1

n

X

k=0

n+ 1 k

n−1 n+ 1

k

sn+1+ (n−1)n+1sn+1(x1, . . . , xn+1).

Moving

(n−1)n+1sn+1(x1, . . . , xn+1) = (n−1)n+1x1· · ·xn+1

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to the other side, we get (2n)n+1−(n−1)n+1

x1· · ·xn+1

n

X

k=0

n+ 1 k

n−1 n+ 1

k

sn+1

=

"n+1 X

k=0

n+ 1 k

n−1 n+ 1

k

n−1 n+ 1

n+1# sn+1

=

"

n−1 n+ 1 + 1

n+1

n−1 n+ 1

n+1# sn+1

=

"

2n n+ 1

n+1

n−1 n+ 1

n+1# sn+1

= (2n)n+1−(n−1)n+1 sn+1 (n+ 1)n+1.

Cancelling((2n)n+1−(n−1))n+1, we get (3.5) x1· · ·xn+1

x1+· · ·+xn+1 n+ 1

n+1

,

as desired. We note that if the xk’s are distinct, then by Proposition 3.1, the in- equalites used in equation (3.4) are strict. It follows that in this case inequality (3.5) is also strict.

To recap our argument, Lemma 2.2 reduces the proof of Theorem 1.1 to the case wheref(x1, . . . , xn)is a homogenous polynomial with all non-zero coefficients equal to one, for whichSnacts transitively. Proposition3.1further reduces the proof to theAGn. Finally, the proof ofAGnis achieved in Theorem3.2.

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### References

[1] P.S. BULLEN, Handbook of Means and their Inequalities, Kluwer Acad. Press, Dordrecht, 2003.

[2] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cambridge Univ. Press, 1964.

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