**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

*Received:* 02 August, 2008

*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.

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

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**Contents**

**1** **Introduction** **3**

**2** **First Proof of Theorem 1.1** **4**

**3** **Second Proof of Theorem 1.1** **7**

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

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**1.** **Introduction**

Let (a_{1}, . . . , a_{n}) ∈ R^{n} be an n-tuple of positive real numbers. The inequality of
arithmetic-algebraic means states that

√n

a_{1}a_{2}· · ·a_{n}≤ a_{1}+· · ·+a_{n}

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 asAG_{n}to 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*(a

_{1}, . . . , a

_{n}) ∈ R

^{n}

*be an*n-tuple of positive real numbers,f(x

_{1}, . . . , x

_{n})

*be a homogenous symmetric polynomial of degree*k, 1 ≤ k ≤ n, having

*positive coefficients, and let*s

_{k}(x

_{1}, . . . , x

_{n})

*be the*k-th elementary symmetric poly-

*nomial. Then*

sk(a1, . . . , an)

n k

≤ f(a1, . . . , an)
f(1, . . . ,1) .
*There is equality if and only if the*a^{0}_{i}*s are all equal.*

Note that ^{n}_{k}

= s_{k}(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}^{(a}_{f(1,...,1)}^{1}^{,...,a}^{n}^{)} as a mean forf as in the theorem. The theorem
implies the arithmetic-algebraic mean inequality by takingk =n,f(x_{1}, . . . , x_{n}) =
(x_{1}+· · ·+x_{n})^{n}so thatf(1, . . . ,1) = n^{n}, and then takingn-th roots.

We shall give two proofs of Theorem1.1. The first depends on Muirhead’s The-
orem. The second provesAG_{n}and Theorem1.1in one induction step.

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

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**2.** **First Proof of Theorem** **1.1**

For any functionf(x_{1}, . . . , x_{n}), the symmetric groupS_{n}acts on thex_{k}’s, and so we

set X

!f(x_{1}, . . . , x_{n}) = X

σ∈Sn

f(x_{σ(1)}, . . . , x_{σ(n)}).

In particular, for ann-tuple of nonnegative real numbersα= (α_{1}, α_{2}, . . . , α_{n}), when
f(x_{1}, . . . , x_{n}) =x^{α} =x^{α}_{1}^{1}x^{α}_{2}^{2}· · ·x^{α}_{n}^{n},

we set

[α] = 1 n!

X!x^{α}_{1}^{1}x^{α}_{2}^{2}· · ·x^{α}_{n}^{n}.

Note that [1,0, . . . ,0] is the arithmetic mean while [^{1}_{n},_{n}^{1}, . . . ,_{n}^{1}] 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!x^{α}_{1}^{1}x^{α}_{2}^{2}· · ·x^{α}_{n}^{n} ≤[β] = 1
n!

X!x^{β}_{1}^{1}x^{β}_{2}^{2}· · ·x^{β}_{n}^{n}

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

α1+α2+· · ·+αn=β1+β2+. . .+βn,
α_{1} ≥α_{2} ≥ · · · ≥α_{n}andβ_{1} ≥β_{2} ≥. . .≥β_{n},

α_{1}+α_{2}+· · ·+α_{k} ≤β_{1} +β_{2}+. . .+β_{k}fork= 1,2, . . . , n.

Muirhead’s Theorem states

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

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* Theorem 2.1. The inequality*
[α] = 1

n!

X!x^{α}_{1}^{1}x^{α}_{2}^{2}· · ·x^{α}_{n}^{n} ≤[β] = 1
n!

X!x^{β}_{1}^{1}x^{β}_{2}^{2}· · ·x^{β}_{n}^{n}

*is valid for all positve*x_{i}*’s if and only if*α ≤ β. There is equality only whenα =β
*or the*x_{i}*’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* (a

_{1j}, . . . , a

_{n}

_{j}

_{j}) ∈ R

^{n}

^{j}

*for*j = 1, . . . , m, and let c

_{1}, . . . , c

_{m}

*be*

*positive real numbers. Suppose*a≤

^{a}

^{1j}

^{+···+a}

_{n}

^{nj j}

j *for each*j. Then

a≤ c_{1}(a_{11}+· · ·+a_{n}_{1}_{1}) +c_{2}(a_{12}+· · ·+a_{n}_{2}_{2}) +· · ·+c_{m}(a_{1m}+· · ·+a_{n}_{m}_{m})

c_{1}n_{1}+c_{2}n_{2}+· · ·+c_{m}n_{m} .

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

*Proof. For each*j we rewritea ≤ ^{a}^{1j}^{+···+a}_{n} ^{nj j}

j as n_{j}a ≤ a_{1j} +· · ·+a_{n}_{j}_{j}. We then
multiply byc_{j} to obtainc_{j}n_{j}a ≤ c_{j}(a_{1j} +· · ·+a_{n}_{j}_{j}). We now add over all j to
obtain

(c_{1}n_{1}+c_{2}n_{2}+· · ·+c_{m}n_{m})a

≤c_{1}(a_{11}+· · ·+a_{n}_{1}_{1}) +c_{2}(a_{12}+· · ·+a_{n}_{2}_{2}) +· · ·+c_{m}(a_{1m}+· · ·+a_{n}_{m}_{m}).

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. Let*f(x_{1}, . . . , x_{n})be a homogenous symmetric polynomial of
degreek with positive coefficients. The monomials off break up into orbits under

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

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the action of the symmetric groupS_{n}and so we may writef =c_{1}f_{1}+· · ·+c_{m}f_{m},
c_{j} > 0 where eachf_{j} is a homogenous polynomial with all non-zero coefficients
equal to one and for whichS_{n}acts transitively. In view of Lemma2.2, for the proof
of Theorem1.1we may assumef(x_{1}, . . . , x_{n})itself is a homogenous polynomial of
degreekwith all non-zero coefficients equal to one and for whichS_{n}acts transitively.

For such an f, it follows that there exists an α such thatf(x_{1}, . . . , x_{n}) = t[α],
where t = f(1,1, . . . ,1) is the number of monomials comprising f. We note
that s_{k}(x_{1}, . . . , x_{n}) = ^{n}_{k}

[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.

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

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**3.** **Second Proof of Theorem** **1.1**

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

a1· · ·an≤

a_{1}+· · ·+a_{n}
n

n

.

Alternately, if we leta_{i} =A^{n}_{i} we get

A1· · ·An ≤ A^{n}_{1} +· · ·+A^{n}_{n}

n .

We will refer to these equivalent inequalities also asAGn.

Letf(x_{1}, . . . , x_{n})be a homogenous symmetric polynomial. The monomials off
break up into orbits under the action of the symmetric groupS_{n}and so we may write
f = c_{1}f_{1} +· · ·+c_{m}f_{m}, c_{j} ∈ R where each f_{j} is a homogenous polynomial with
all non-zero coefficients equal to one and for whichS_{n} acts transitively. In view of
Lemma2.2, for the proof of Theorem 1.1 we may assume f(x_{1}, . . . , x_{n})itself is a
homogenous polynomial with all non-zero coefficients equal to one and for which
S_{n}acts transitively.

* Proposition 3.1. Assume*AG

_{2}

*,. . . ,AG*n−1

*. Let*f(x

_{1}, . . . , x

_{n})

*be a homogenous sym-*

*metric polynomial of degree*k, 1 ≤ k ≤ n, with all non-zero coefficients equal to

*one and for which*S

_{n}

*acts transitively. Assume*f(x

_{1}, . . . , x

_{n})6=x

^{n}

_{1}+· · ·+x

^{n}

_{n}

*. Then*

*the conclusion of Theorem1.1holds.*

*Proof. The polynomial*f(x_{1}, . . . , x_{n})has a monomial of the formx^{`}_{1}^{1}x^{`}_{2}^{2}· · ·x^{`}_{s}^{s}where
k = degf =`_{1}+· · ·+`_{s}and0< `_{j} < n. ByAG_{`}_{j} for eachj we have

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

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`_{1}x_{1}x_{2}· · ·x_{`}_{1} ≤x^{`}_{1}^{1} +· · ·+x^{`}_{`}^{1}

1,

`_{2}x_{`}_{1}_{+1}x_{`}_{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^{`}_{`}^{s}_{1}_{+···+`}_{s−1}_{+1}+· · ·+x^{`}_{`}^{s}_{1}_{+···+`}_{s}.

Sincek = degf =`_{1}+· · ·+`_{s}, we multiply the inequalities to obtain
(3.1) `_{1}· · ·`_{s}x_{1}· · ·x_{k}≤(x^{`}_{1}^{1} +· · ·+x^{`}_{`}^{1}

1)· · ·(x^{`}_{`}^{s}

1+···+`s−1+1+· · ·+x^{`}_{`}^{s}

1+···+`s).

Inequality (3.1) now yields (3.2) X

!`_{1}· · ·`_{s}x_{1}· · ·x_{k}

≤X

! (x^{`}_{1}^{1} +· · ·+x^{`}_{`}^{1}_{1})· · ·(x^{`}_{`}^{s}_{1}_{+···+`}_{s−1}_{+1}+· · ·+x^{`}_{`}^{s}_{1}_{+···+`}_{s}).

SinceP

!x_{1}· · ·x_{k}consists ofn!monomials with coefficient one, we get
X!x_{1}· · ·x_{k} = n!

n k

s_{k}(x_{1}, . . . , x_{n}).

Similarly since(x^{`}_{1}^{1}+· · ·+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^{`}_{1}^{1}+· · ·+x^{`}_{`}^{1}

1)· · ·(x^{`}_{`}^{s}

1+···+`s−1+1+

· · ·+x^{`}_{`}^{s}

1+···+`_{s})consists of`_{1}· · ·`_{s}n!monomials with coefficient one. Thus we have
X! ((x^{`}_{1}^{1}+· · ·+x^{`}_{`}^{1}

1)· · ·(x^{`}_{`}^{s}

1+···+`s−1+1+· · ·+x^{`}_{`}^{s}

1+···+`s) = `_{1}· · ·`_{s}n!

t f(x_{1}, . . . , x_{n}),

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

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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 thex_{i}’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 proveAG_{n}for alln≥2.

**Theorem 3.2.** AG_{n}*is true for all*n≥2.

*Proof. The proof is by induction on*n. 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 assumeAG_{2}, . . . , AG_{n} and we proveAG_{n+1}. To this end, it suffices to
show that

x_{1}· · ·x_{n+1} ≤

x_{1}+· · ·+x_{n+1}
n+ 1

n+1

.

Now, byAG_{2} andAG_{n}we have for eachk,
q

x_{k}√^{n}

x_{1}· · ·xk−1x_{k+1}· · ·x_{n+1} ≤ xk+ √^{n}

x1· · ·xk−1xk+1· · ·xn+1

2

≤ x_{1}+· · ·+nx_{k}+· · ·+x_{n+1}
2n

= s+ (n−1)x_{k}

2n .

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

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Where we have sets =x_{1} +· · ·+x_{n+1}. Multiplying these inequalities overk, we
get

x1· · ·xn+1 ≤

n+1

Y

k=1

s+ (n−1)x_{k}
2n

= 1

(2n)^{n+1}

n+1

Y

k=1

(s+ (n−1)x_{k}).

Multiplying through by(2n)^{n+1} and expanding we get,
(3.3) (2n)^{n+1}x_{1}· · ·x_{n+1} ≤

n+1

X

k=0

(n−1)^{k}s_{k}(x_{1}, . . . , x_{n+1})s^{n+1−k}.

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

n+ 1 k

s^{k}
(n+ 1)^{k}
for0< k < n+ 1. Plugging this into (3.3), we get,

(3.4) (2n)^{n+1}x_{1}· · ·x_{n+1}

≤

n

X

k=0

n+ 1 k

n−1 n+ 1

k

s^{n+1}+ (n−1)^{n+1}s_{n+1}(x_{1}, . . . , x_{n+1}).

Moving

(n−1)^{n+1}s_{n+1}(x_{1}, . . . , x_{n+1}) = (n−1)^{n+1}x_{1}· · ·x_{n+1}

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

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

x_{1}· · ·x_{n+1} ≤

n

X

k=0

n+ 1 k

n−1 n+ 1

k

s^{n+1}

=

"_{n+1}
X

k=0

n+ 1 k

n−1 n+ 1

k

−

n−1 n+ 1

n+1#
s^{n+1}

=

"

n−1 n+ 1 + 1

n+1

−

n−1 n+ 1

n+1#
s^{n+1}

=

"

2n n+ 1

n+1

−

n−1 n+ 1

n+1#
s^{n+1}

= (2n)^{n+1}−(n−1)^{n+1} s^{n+1}
(n+ 1)^{n+1}.

Cancelling((2n)^{n+1}−(n−1))^{n+1}, we get
(3.5) x_{1}· · ·x_{n+1} ≤

x_{1}+· · ·+x_{n+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(x_{1}, . . . , x_{n})is a homogenous polynomial with all non-zero coefficients
equal to one, for whichS_{n}acts transitively. Proposition3.1further reduces the proof
to theAG_{n}. Finally, the proof ofAG_{n}is achieved in Theorem3.2.

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

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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.