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volume 4, issue 3, article 59, 2003.

Received 21 January, 2003;

accepted 22 July, 2003.

Communicated by:P. Cerone

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Journal of Inequalities in Pure and Applied Mathematics

ON SOME SPECTRAL RESULTS RELATING TO THE RELATIVE VALUES OF MEANS

C.E.M. PEARCE

School of Applied Mathematics, Adelaide University,

Adelaide SA 5005, Australia.

EMail:cpearce@maths.adelaide.edu.au

c

2000Victoria University ISSN (electronic): 1443-5756 008-03

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On Some Spectral Results Relating to the Relative Values

of Means C.E.M. Pearce

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J. Ineq. Pure and Appl. Math. 4(3) Art. 59, 2003

http://jipam.vu.edu.au

Abstract

In the case of two positive numbers, the geometric mean is closer to the harmonic than to the arithmetic mean. We derive some spectral results relating to corresponding properties with more than two positive numbers.

2000 Mathematics Subject Classification:26E60, 26D15 Key words: AGH inequality, Means, Relative values, Spectrum.

1. Introduction

LetA,G,Hdenote respectively the arithmetic, geometric and harmonic means ofnpositive real numbersx₁, . . . , x_n, which are not all equal. It is well–known thatH < G < A. Scott [3] has shown in the casen = 2thatGis closer toH than toA, so that

(1.1) A−G

A−H > 1 2.

He showed by a counterexample that this need not be the case whenn >2.

Subsequently Lord [1] and Pearce and Peˇcariˇc [2] addressed the question of the behaviour of the quotient

f_n(x₁, . . . , x_n) := A−G A−H

in the case of general n. Several generalisations and extensions of (1.1) were obtained. The following are pertinent to the present article.

Since

f_n(ax₁, . . . , ax_n) =f(x₁, . . . , x_n)

fora > 0, it suffices to consider the values taken byf_nwhenx= (x₁, . . . , x_n) lies on the intersection

K:=

(

x∈Rⁿ : x_i ≥0for 1≤i≤n and

n

X

i=1

x²_i = 1 )

of the nonnegative orthant and the surface of the unit hypersphere. The function fn is clearly well–defined and continuous on the interior of K except ate :=

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√1

n, . . . ,^√¹_n

, where it is undefined since A, G and H all coincide. In fact this singularity is removable. It is shown in [1] that defining f_n(x) = 1 for boundary points ofK(where some but not all valuesx_i vanish) andf_n(e) = ¹₂ makesfncontinuous on the whole ofK. SinceKis compact,fnpossesses and realises an infimumα_n. Further, the range off_nconstitutes the interval[α_n,1], the sequence(α_n)^∞₂ is strictly decreasing to limit zero andα_n > ¹_n forn ≥ 3.

The seminal paper of Scott givesα2 = ¹₂.

In this article we continue the development of [1] and [2] and derive some striking structural results, principally as follows. In Section2, Theorem2.1, we show that ifxis such thatf_n(x) =α_n, then{x₁, . . . , x_n}contains precisely two distinct values. In Section3, Theorem3.3, we show that iff_n(x) = α_n, then the smaller of the two distinct components of xmust occur with multiplicity one.

We conclude in Section 4 by giving characterisations ofα_n and some related infima arising naturally in our analysis.

We postpone consideration of asymptotics to a subsequent article.

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2. The Dichotomy Theorem

Theorem 2.1. Forn >2, any set{x₁, . . . , x_n}for whichf_n(x) =α_n contains precisely two distinct values.

Proof. First suppose that S₁ := Pn

i=1x_i andS_n := Qn

i=1x_i are fixed. Subject to these constraints, the mimimum off_ncorrespond to an extremum ofPn

i=1 1 xi

and satisfies

∂L

∂x_i = 0 for i= 1, . . . , n, whereLdenotes the Lagrangian

L:=

n

X

i=1

1 x_i −λ

n

X

i=1

x_i−S₁

!

−µ

n

Y

i=1

x_i−S_n

! .

Then ∂L

∂xi

=− 1

x²_i −λ−µY

j6=i

x_j = 0 (i= 1, . . . , n), that is,

1

x_i +λx_i+µS_n = 0 (i= 1, . . . , n).

Hence eachx_i must be equal to one of the two solutions of the quadratic λx²+µS_nx+ 1 = 0.

For a minimum, these solutions must be distinct, sincef_n(e) = ¹₂ whileα_n < ¹₂ forn≥3.

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Forj = 1,2, . . . , nand fixedn >2, define

Vj ={x: {x1, . . . , xn}contains preciselyj distinct values}, V_j^∗ ={f_n(x) : x∈ V_j}

and

δj = infV_j^∗.

An immediate implication of Theorem2.1is the following result.

Corollary 2.2. We have

δ2 =δ3 =· · ·=δn =αn. Forj >1, the setV_j^∗contains its infimum only forj = 2.

Proof. If 1 ≤ j ≤ n −1, any element of V_j can be approximated arbitrarily closely by elements of Vj+1, but not conversely. Since K is compact and fn

continuous, we must therefore have thatδ_j+1 ≤δ_j. Thus δn ≤δn−1 ≤ . . . ≤δ2 ≤δ1 = 1

2. On the other hand, by Theorem2.1

δ₂ =α_n = inf{f_n(x)}= min{δ₁, δ₂, . . . , δ_n}.

The first part of the corollary follows.

The second part follows by invoking Theorem2.1again.

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3. Comparison Results

In the remaining sections of the paper we examine more closely the central case when {x1, . . . , xn} contains only two distinct values, that isx ∈ V2. We may assume without loss of generality an ordering

x₁ ≤x₂ ≤ · · · ≤x_n.

We decompose

V₂ =

n−1

[

k=1

U_k,

where

U_k ={x:x₁ =x₂ =· · ·=x_k < x_k+1 =· · ·=x_n} (1≤k < n).

Forx ∈ U_k we have for the k equal points denoted by xand the rest by y that

A−G A−H =

k

nx+ 1− ^k_n

y−x^k/ny^1−k/n

k

nx+ 1−_n^k y−n

.k x +^n−k_y

.

If we setβ =k/nandu=x/y, this gives

f_n(x) = βu+ 1−β−u^β βu+ 1−β−1 _β

u + 1−β withβ ∈₁

n,_n², . . . ,ⁿ⁻¹_n and0< u < 1.

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This may be rearranged as

(3.1) f_n(x) = 1− u

(u−1)²g(u, β), where

(3.2) g(u, β) = u^β −1

β + u^−(1−β)−1 1−β .

We shall find it convenient to have alternative sets of variables and functions.

Setv =u^1/n. Then forx∈ U_kwe put

h_k(n, v) = f_n(x) and φ_k(v) =g

u,k n

.

Proposition 3.1. For fixedn ≥3andv ∈(0,1), the sequence(hk(n, v))ⁿ⁻¹_k=1 is strictly increasing.

Proof. By virtue of the representation (3.1), (3.2), it suffices to prove that the sequence(φ_k(v))ⁿ⁻¹_k=1 is strictly decreasing. To show thatφ_k(v) > φ_k+1(v), we need to establish the inequality

v^k−1

k + v^−(n−k)−1

n−k > v^k+1−1

k+ 1 + v^{−(n−k−1)}−1 n−k−1 , which on multiplication byv^n−k becomes

Θ(v)<0,

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whereΘis the polynomial (3.3) Θ(v) = vⁿ⁺¹

k+ 1 − vⁿ

k +v^n−k 1

k + 1

n−k − 1

k+ 1 − 1

n−k−1

+ v

n−k−1− 1 n−k . Sincen+ 1> n > n−k >1>0, (3.3) expressesΘin descending powers of v. The coefficients taken in sequence have exactly three changes in sign, regardless of whether the expression in brackets is positive, negative or zero.

Hence by Descartes’ rule of signs the polynomial equation

(3.4) Θ(w) = 0

has at most three positive solutions.

Now by elementary algebra we have that

Θ(1) = Θ⁰(1) = Θ⁰⁰(1) = 0,

so thatw = 1is a triple zero ofΘ(w). HenceΘ(w)has no zeros on (0,1)and therefore must have constant sign on(0,1). BecauseΘ(0) < 0, we thus have Θ(w)<0throughout(0,1)and we are done.

For1≤k < n, put

U_k^∗ ={f_n(x) : x∈ U_k} and

ε_k= inf U_k^∗.

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Lemma 3.2. For eachn≥3we have

ε_k











< 1

2 for 1≤k < n 2

= 1

2 for n

2 ≤k ≤n−1.

Proof. Since v = 1 gives f_n = ¹₂ and v = 0 gives f_n = 1, a necessary and sufficient condition thatεk < ¹₂ is that there should existv ∈(0,1)for which

nvⁿ (1−vⁿ)²

v^k−1

k + v^−(n−k)−1 n−k

> 1 2 or

(3.5) Ω(v)<0,

where

Ω(v) =v²ⁿ− 2n

k v^n+k+ 2vⁿ n

k + n

n−k −1

−v^k 2n n−k + 1

=v²ⁿ− 2n

k v^n+k+ 2vⁿ n

k + k

n−k

−v^k 2n n−k + 1.

The polynomialΩhas four changes of sign in its coefficients, and so has at most four positive zeros. We may verify readily that

(3.6) Ω(1) = Ω⁰(1) = Ω⁰⁰(1) = 0,

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while

(3.7) Ω⁰⁰⁰(1) = 2n²(n−2k).

If ⁿ₂ < k ≤ n −1, then Ω(v) has a triple zero at v = 1 and so can have at most one zero on (0,1). Since Ω(0) > 0, condition (3.5) can thus be satisfied if and only if there is such a zero, in which caseΩ(1−∆) < 0for all ∆ > 0 sufficiently small. But by Taylor’s theorem

Ω(1−∆) = Ω(1)−∆Ω⁰(1) + ∆²

2! Ω⁰⁰(1)− ∆³

3! Ω⁰⁰⁰(1) + 0(∆⁴)

≈ −∆³

3 n²(n−2k), (3.8)

which is positive.

Hence we must have ε_k ≥ ¹₂. But since e can be approximated arbitrarily closely by elements of U_k by lettingv → 1, we must have ε_k ≤ f_n(e) = ¹₂. Thusε_k = ¹₂.

Ifk = ⁿ₂, then Ω(v)has exactly four positive zeros, all at v = 1, soΩhas constant sign on (0,1). Since Ω(0) > 0, we thus have Ω(v) > 0 on (0,1).

Arguing as in the previous paragraph, we derive again thatε_k = ¹₂.

Finally, ifk < ⁿ₂, we have by (3.8) thatΩ(1−∆)<0for∆>0sufficiently small, so that condition (3.5) is satisfied. This completes the proof.

Theorem 3.3. The sequence(εk)1≤k<ⁿ₂ is strictly increasing.

Proof. The desired result is equivalent to (ξ_k)_1≤k<ⁿ

2 being strictly decreasing, where

ξ_k= sup

u∈(0,1)

u

(1−u)²φ_k(u) = 1−ε_k.

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By Proposition3.1,

u

(1−u)²φ_k(u)> u

(1−u)²φ_k+1(u) for eachu∈(0,1), so that

ξ_k ≥ξ_k+1 for 1≤k≤n−1.

Further,ξ_k is realised for some choice ofu, foru = u_k, say, and arguing as in Lemma3.2we must haveuk ∈(0,1)for1≤k < ⁿ₂.

To show the inequalities are strict, suppose if possible that equality holds for some value ofk, so that

(3.9) u_k

(1−uk)²φ_k(u_k) = u_k+1

(1−uk+1)²φ_k+1(u_k+1).

By Proposition3.1, u_k+1

(1−u_k+1)²φ_k(u_k+1)> u_k+1

(1−u_k+1)²φ_k+1(u_k+1), so that by (3.9)

u_k+1

(1−uk+1)²φ_k(u_k+1)> u_k

(1−uk)²φ_k(u_k) = ξ_k, contradicting the definition ofξ_k.

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4. Characterisation of ε

_k

In the previous section we saw that for1≤k < ⁿ₂ the supremumξ_k is realised for some u = uk ∈ (0,1). We now consider the determination of uk. For convenience we again employv_k=u^1/n_k .

Theorem 4.1. (i) For1 ≤ k < ⁿ₂,v = v_k is the unique solution on(0,1)of the equation

(4.1) Φk(v) = 0,

where

Φ_k(v) = (vⁿ−1)

vⁿn+k

k −v^n−k n

k + n

n−k

+ k

n−k

−2nvⁿ vⁿ

k −v^n−k 1

k + 1

n−k

+ 1

n−k

.

(ii) Ifv ∈(0,1), thenv < v_korv > v_kaccording asΦ_k(v)<0orΦ_k(v)>0.

Proof. Sincef_nachieves a minimum atv =v_k ∈(0,1), we have that d

dv

nvⁿ (vⁿ−1)² ·

v^k−1

k + v^−(n−k)−1 n−k

= 0

forv =vk, this value ofv corresponding to a local maximum of the differenti-

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ated expression. The left–hand side is the quotient of n(vⁿ−1)²

v^n+k−1n+k

k −vⁿ⁻¹ n

k + n

n−k

+v^k−1 k n−k

−2n²(vⁿ−1)²vⁿ⁻¹ v^n+k

k −vⁿ 1

k + 1

n−k

+ v^k n−k

by(vⁿ−1)⁴. Removing this denominator and the factorn(vⁿ−1)v^k−1from the numerator gives that v = v_ksatisfies (4.1). Statement (i) will therefore follow if it can be shown that (4.1) has a unique solution on(0,1). Uniqueness gives that the differentiated expression has positive gradient forv < v_k and negative gradient for v > v_k. Statement (ii) will then follow, since the term cancelled is negative.

It therefore remains only to show thatΦ_k(v)has a unique zero on(0,1). This we do as follows. The polynomialΦ_k(v)may be written in descending powers ofvas

−v²ⁿn−k

k +v^2n−k n

k + n

n−k

−vⁿ

2n−k

n−k + n+k k

+v^n−k n

k + n

n−k

− k n−k , the coefficients of which exhibit four changes of sign. Hence by Descartes’ rule of signs,Φ_k(v)has at most four positive zeros.

By elementary algebra,

(4.2) Φk(1) = Φ⁰_k(1) = Φ⁰⁰_k(1) = 0, Φ⁰⁰⁰_k(1) =n²(2k−n),

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so thatΦ_k(v)has a triple zero atv = 1. HenceΦ_k(v)has at most one zero on (0,1).

NowΦ_k(v)<0and for∆>0small Φ_k(1−∆) =−∆³

3!Φ⁰⁰⁰_k(1) + 0(∆⁴)>0,

by Taylor’s theorem and (4.2). HenceΦ_k(v)has a zero on(0,1)and this must be unique.

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References

[1] N. LORD, More on the relative location of means II, Math. Gaz., 85 (2001), 114–116.

[2] C.E.M. PEARCEANDJ. PE ˇCARI ´C, More on the relative location of means I, Math. Gaz., 85 (2001), 112–114.

[3] J.A. SCOTT, On the theorem of means: is GnearerAorH?, Math. Gaz., 82 (1998), 104.