View of Bilateral inequalities for means

Rev. Anal. Num´er. Th´eor. Approx., vol. 42 (2013) no. 2, pp. 94–102 ictp.acad.ro/jnaat

BILATERAL INEQUALITIES FOR MEANS

MIRA-CRISTIANA ANISIU^∗and VALERIU ANISIU^†

Abstract. Let (M1, M2, M3) be three means in two variables chosen fromH, G,L,I,A,Q,S,C so that

M1(a, b)< M2(a, b)< M3(a, b), 0< a < b.

We consider the problem of findingα, β∈Rfor which

αM1(a, b) + (1−α)M3(a, b)< M2(a, b)< βM1(a, b) + (1−β)M3(a, b).

We solve the problem for the triplets (G, L, A), (G, A, Q), (G, A, C), (G, Q, C), (A, Q, C), (A, S, C), (A, Q, S) and (L, A, C). The Symbolic Algebra Program Maple is used to determine the range where some parameters can vary, or to find the minimal polynomial for an algebraic number.

MSC 2000. 26D15, 26E60; 26-04

Keywords. Two-variable means, weighted arithmetic mean, inequalities, sym- bolic computer algebra.

1. INTRODUCTION

We remind the definitions of the classical means, namely, for 0< a < b

• thearithmetic,geometric and harmonic ones A= ^a+b₂ , G=√

ab, H = _a+b^2ab, as well as

• theH¨older and theanti-harmonic meanQ= a²+b²

2

1/2

, C = ^a_a+b^2+b2;

• theP´olya & Szeg˝o logarithmic mean, theexponential (oridentric), and theweighted geometric mean

L= _ln^b−a_b−lna, I = ¹_e

b^b a^a

1/(b−a)

, S =

a^ab^b1/(a+b)

.

References on means and inequalities between them can be found in [5].

At first, the following inequalities between means were established (1) H < G < L < I < A < Q < S < C,

∗“T. Popoviciu” Institute of Numerical Analysis, Romanian Academy, P.O. Box 68, 400110 Cluj-Napoca, Romania, e-mail: [email protected].

†“Babe¸s-Bolyai” University, Faculty of Mathematics and Computer Science, 1 Kog˘alniceanu St., 400084 Cluj-Napoca, Romania, e-mail: [email protected].

followed by relations between some means and the arithmetic means of two others ([7], [3])

(2) L < ^G+A₂ , ^G+Q₂ < A < ^G+C₂ < Q < ^A+C₂ < S.

A more difficult problem is to obtain results of the type (2) for weighted arithmetic means and to determine the maximal interval for the parameter for which the inequalities hold.

We mention here an inequality proved by Alzer and Qiu for the meansG, I and A.

Theorem 1. [1] The double inequality

(3) αA(a, b) + (1−α)G(a, b)< I(a, b)< βA(a, b) + (1−β)G(a, b)

holds true for all positive real numbers a 6= b, if and only if α ≤ 2/3 and β ≥2/e.

Results of this type continued to appear, recent ones are given in [9] for (H, L, A) and (H, I, A), and in [4] for (G, L, C).

LetM1,M2,M3 be three means out of the eight listed in (1) so that (4) M1(a, b)< M2(a, b)< M3(a, b).

We consider the problem of findingα, β∈R for which (5) αM₁(a, b) + (1−α)M₃(a, b)< M₂(a, b) and

(6) M2(a, b)< βM1(a, b) + (1−β)M3(a, b).

The inequalities (5) and (6) are equivalent to

(7) α > ^M_M^{3(a,b)−M2(a,b)}

3(a,b)−M1(a,b), respectively

(8) β < ^M_M^{3(a,b)−M2(a,b)}

3(a,b)−M1(a,b).

Basically, denoting byt=b/a, t >1, the problem reduces to find inff and supf, where

(9) f(t) = ^M_M^{3(1,t)−M2(1,t)}

3(1,t)−M1(1,t).

The function f is obviously bounded, 0 ≤ f(t) ≤ 1. If supf is attained at some t ∈ (1,∞), then α ∈ (supf,∞); otherwise α ∈ [supf,∞). Similarly, β ∈(−∞,inff) if inff is attained in (1,∞), and β ∈(−∞,inff] otherwise.

Symbolic Algebra Programs can be of great help to determine the range where the parameters can vary. Maple was used in [3] to find the interval forα in Theorem 9 below. We also use it to simplify the polynomials in the proof of Theorem 5 and to obtain the optimal value β₀ ofβ.

Starting from the means listed in (1), we can formulate ⁸₃

= 56 bilateral inequalities of the type (3). We shall choose seven of them, for which one of

(5) and (6) was already proved in [3], and we shall find the possible values of the parameter for the remaining one. Then, for L < A < C we find the optimal intervals forα andβ in order that both inequalities (5) and (6) hold.

To this aim Maple is again very useful.

2. BILATERAL INEQUALITIES

We consider means in two variables, but we prefer to use a simpler (and shorter) notation.

Let us denote for 0 < a < b, t = b/a, t > 1. It it obvious, due to the homogeneity, that, if M(a, b) is any mean from (1), it suffices to prove the inequalities forM(1, t). We shall write from now onM(t) instead ofM(1, t).

Theorem 2. The double inequality

αG(t) + (1−α)A(t)< L(t)< βG(t) + (1−β)A(t), ∀t >1, holds if and only if α≥1 and β ≤2/3.

Proof. We denote, fort >1,

(10) f₁(t) = ^A(t)−L(t)_A(t)−G(t) = ^{(t+1) ln}_(t+1−2^{t−2(t−1)√}

t) lnt .

Let us suppose that the first inequality in the theorem holds. From limt→∞f1(t) = 1 it follows obviously that α ≥ 1. Conversely, if α ≥ 1 it suffices to have

A(t)−L(t) A(t)−G(t) <1,

which is true because L(t) > G(t). We evaluate f1(t)−2/3, where 2/3 = limt→1f₁(t) and show that it is positive. The denominator is obviously posi- tive; we substituteu=√

tin the numerator and obtain f(u) = (u²+ 4u+ 1) lnu−3u²+ 3.

We have f(1) =f⁰(1) =f⁰⁰(1) = 0 andf⁰⁰⁰(u) = 2(u−1)²/u³ >0 foru > 1, hencef₁(t)>2/3 fort >1. It follows thatL(t)< βG(t) + (1−β)A(t), ∀t >1

if and only if β≤2/3.

Theorem 3. The double inequality

αG(t) + (1−α)Q(t)< A(t)< βG(t) + (1−β)Q(t), ∀t >1, holds if and only if α≥1/2 and β≤1−√

2/2.

Proof. Let us consider fort >1, the function (11) f₂(t) = ^Q(t)−A(t)_Q(t)−G(t) = 1−

√ t²+1+√

√ 2t 2(√

t+1)². We have f2(t) < 1/2, since √

t²+ 1 +√

2t > √

2/2 (√

t+ 1)² ⇔ √

t²+ 1 >

√2/2 (t+ 1) ⇔(t−1)² >0. Since limt→1f₂(t) = 1/2, it follows thatαG(t) + (1−α)Q(t)< A(t), ∀t >1 if and only if α≥1/2.

Let us suppose that A(t) < βG(t) + (1 − β)Q(t), ∀t > 1. Since limt→∞f₂(t) = 1 −√

2/2, it follows that β ≤ 1 − √

2/2. Conversely, if

β ≤ 1−√

2/2, it suffices to prove that f2(t) > 1−√

2/2. This is equiva-

lent with _√

t²+1+√ 2t

(^√t+1)² <1, i. e. √

t²+ 1 +√

2t < √ t+ 12

and this is true because √

t²+ 1 +√ 2t <

t+ 1 +√

2t < √ t+ 12

.

Theorem 4. The double inequality

αG(t) + (1−α)C(t)< A(t)< βG(t) + (1−β)C(t), ∀t >1, holds if and only if α≥2/3 and β≤1/2.

Proof. Fort >1 we define

(12) f₃(t) = _C(t)−G(t)^C(t)−A(t) = ^(t−1)2

2(t²+1−√ t(t+1)).

Since limt→1f₃(t) = 2/3, fromαG(t) + (1−α)C(t)< A(t), ∀t >1 it follows that α ≥ 2/3. If α ≥ 2/3, it is true that f3(t) < α, because f3(t) < 2/3 is

equivalent with _√

t t+√

t+1 < ¹₃, or (√

t−1)²>0.

Similarly, it follows thatf₃(t)>1/2,since f₃(t)−¹₂ =

√t(√ t−1)² 2(t²+1−√

t(t+1)) =

√t 2(t+√

t+1) >0.

The infimum off₃ on (1,∞) is precisely 1/2, because lim_t→∞f₃(t) = 1/2 Theorem 5. The double inequality

αG(t) + (1−α)C(t)< Q(t)< βG(t) + (1−β)C(t), ∀t >1, holds if and only if α ≥1−√

2/2 and β < β0, where β0 u0.3471574308... is the unique positive root of the polynomial

9x⁴−26x³+ 22x²−2x−1.

Proof. We have to find, fort >1,the extreme values of (13) f4(t) = ^C(t)−Q(t)_C(t)−G(t) = ^2t2−

√ 2√

t²+1−√ 2t√

t²+1+2 2(t²−t³²−√

t+1) .

Denoting byu=√

t, we compute the derivative of h(u) =f4(u²) and we obtain

h⁰(u) = ^(u+1)

√

2(u⁴+1)(^{−u4−4u2−1})^{+u6+2u5+3u4+3u2+2u+1}

√

2(u⁴+1)(u−1)³(u²+u+1)² . So, the roots of the derivative satisfy the algebraic equation

2(u⁴+ 1) u⁴+ 4u²+ 12

= u⁶+ 2u⁵+ 3u⁴+ 3u²+ 2u+ 12

.

After the simplification of a quartic polynomial whose roots are not in the interval (1,∞), we obtain the equation

(14) u⁸−8u⁵−10u⁴−8u³+ 1 = 0,

which has a unique root u₀ in the interval (1,∞). This can be easily proved by using the Sturm sequence. Thenu0 will be the unique root ofh⁰ in (1,∞).

Now h⁰(2)>0, h⁰(3)<0,so 2< u₀ <3 and h is strictly increasing in the interval (1, u0) and strictly decreasing in the interval (u0,∞). We also have limu→1h(u) = 1/3, limu→∞h(u) = 1−√

2/2 and therefore inff₄ = infh = 1−√

2/2, supf4= suph=h(u0) =β0.

Since h(u₀) is an algebraic number, we can easily find its minimal polyno- mial by performing the following commands inMaple:

>theta:=RootOf(u^8-8*u^5-10*u^4-8*u^3+1, u):

> M:= g(theta):

> sqrfree(evala(Norm(convert(Z-M,RootOf))),Z)[2][1][1];

9x⁴−26x³+ 22x²−2x−1

Notice thatMaple is of course able to express the maximumh(u₀) in terms of radicals by executing the command:

> select(u-> is(u>0),[solve(9*x^4-26*x^3+22*x^2-2*x-1,Explicit)]);

but the resulting expression is cumbersome and we will not print it here.

Theorem 6. The double inequality

αA(t) + (1−α)C(t)< Q(t)< βA(t) + (1−β)C(t), ∀t >1, holds if and only if α≥2−√

2 and β ≤1/2.

Proof. Let us consider, fort >1 (15) f₅(t) = ^C(t)−Q(t)_C(t)−A(t) = ^2(t

2+1)−(t+1)√

2(t²+1)

(t−1)² .

From limt→∞f₅(t) = 2−√

2 it follows thatαA(t)+(1−α)C(t)< Q(t), ∀t >

1 impliesα≥2−√

2. Now ifα≥2−√

2 we have to prove thatf5(t)<2−√ 2, which can be written as p

2(t²+ 1) (t+ 1) > −√

2t² + 4 + 2√ 2

t−√ 2. If

−√

2t²+ 4 + 2√ 2

t−√

2≤0, the inequality holds. Otherwise, squaring both sides it reduces to 4 3 + 2√

2

t(t−1)² >0.

We obtain also

f5(t)−¹₂ = ^3(t

2+1)+2t−2(t+1)√

2(t²+1) 2(t−1)² >0,

because (3(t²+ 1) + 2t)² −8(t+ 1)²(t²+ 1) > 0 ⇔ (t−1)⁴ > 0. We have limt→1f₅(t) = 1/2, hence this is the infimum of f₅ on (1,∞) and the second

part of the theorem is also true.

Lemma 7. [3] For t >1, the following inequality holds

(16) t

t

t+1 > t−lnt.

Proof. The inequality (16) is equivalent to

t

t+1lnt >ln(t−lnt).

We consider the function

k(t) = ln(t−lnt)−^t−1_t lnt, t >1, with

k⁰(t) = ^{(lnt−1) ln}_t2(t−lnt)^t.

It has limt→1k(t) = 0, limt→∞k(t) = 0 and a minimum at t0 =e. It follows that k(t)<0 on (1,∞), hence ((t−1)/t) lnt >ln(t−lnt). It follows that

t

t+1lnt > ^t−1_t lnt >ln(t−lnt).

Theorem 8. The double inequality

αA(t) + (1−α)C(t)< S(t)< βA(t) + (1−β)C(t), ∀t >1, holds if and only if α≥1/2 and β≤0.

Proof. We define

(17) f₆(t) = _C(t)−A(t)^C(t)−S(t) = 2^{t2+1−(t+1)t}

t t+1 (t−1)² . We have

f₆(t)−¹₂ = ^3(t2+1)+2t−4(t+1)t t t+1

2(t−1)² = ^2(t+1)_(t−1)2 ·g(t), where

g(t) = ^3(t_4(t+1)^2+1)+2t−t

t t+1. Then

g⁰(t) =− ^g1(t)

4(t+1)², where

(18) g1(t) = 4t

t

t+1(t+ 1 + lnt) + 1−3t²−6t.

Using the fact that S > Q, i. e. t^t/(t+1) > p

(t²+ 1)/2, we obtain that g₁(t)>p

2(t²+ 1)g₂(t),where

g₂(t) = 2(t+ 1 + lnt)−(3t²+ 6t−1)/p

2(t²+ 1).

The derivative ofg₂ is

g₂⁰(t) = ^(t+1)

√

2(t²+1) 3

−(^3t4+7t2+6t)

t(t²+1)√

2(t²+1) .

In order to establish its sign we consider the polynomial P(t) = (t+ 1)²p

2(t²+ 1)6

− 3t⁴+ 7t²+ 6t2

=t⁶(t−1)² _t⁸6 +³²_t5 +⁵²_t4 +³⁶_t3 +¹⁹_t2 +¹⁴_t −1 .

The expression from the last parenthesis is obviously decreasing for t ≥ 1 and it is positive for t= 10. It follows that it is positive on (1,10), hence on this interval P is also positive. Therefore g⁰₂(t)>0, g₂(t)> g₂(1) = 0, so g₁ is positive too for 1< t <10.

Let us consider now that t≥10. Using (16) in (18) we obtain that g1(t)>

g₃(t), where

g₃(t) =t²−2t+ 2−(2 lnt+ 1)². For

g₄(t) =p

t²−2t+ 2−2 lnt−1, the sign ofg₄⁰ is given byt²−t−2√

t²−2t+ 2; but (t²−t)²−4(t²−2t+ 2) = (t−10)⁴+ 38(t−10)³+ 537(t−10)²+ 3348(t−10) + 7772>0 for t≥10. It follows thatg3(t)≥g3(10) = 3.45... >0, hence g1 is positive fort≥10 too.

In conclusion, g1(t) > 0 on (1,∞), therefore g⁰(t) < 0 on (1,∞). The function g being decreasing, g(t) < g(1) = 0 for t > 1 and f₆(t) < 1/2 for t >1.

The second part of the theorem follows from limt→∞f₆(t) = 0 andf₆(t)>0,

∀t >1.

Theorem 9. The double inequality

αA(t) + (1−α)S(t)< Q(t)< βA(t) + (1−β)S(t), ∀t >1 holds if and only if α≥2−√

2 and β ≤0.

Proof. We shall prove that the first inequality holds forα= 2−√

2 (hence a fortiori forα ≥2−√

2).

Let us denote (19)

H(t, α) =Q(t)−αA(t)−(1−α)S(t) = ¹₂p

2 + 2t²−¹₂α(1 +t)−(1−α)t

t 1+t. and

h₁(t) = (√

2 + 1)H(t,2−√ 2),

whereHis given in (19). We have to prove that h₁(t)>0 fort >1. It follows that

h1(t) = ⁽

√2+1)√

2(t²+1)

2 −

√ 2(t+1)

2 −t

t t+1.

We put in the inequality (1 +x)^q<1 +qx,which holds for x >0, 0< q <1, x=t−1 and q=t/(t+ 1). It follows that

t

t

t+1 < ^t_t+1²⁺¹,

and

h₁(t)> ⁽¹⁺

√2) 2(t+1)

(t+ 1)p

2 + 2t²−√

2(t²+ 2(√

2−1)t+ 1) .

Let us denote the positive expressions h2(t) = (t+ 1)p

2 + 2t², h3(t) =√

2(t²+ 2(√

2−1)t+ 1);

it follows easily that h²₂(t)−h²₃(t) = 4t(t−1)², therefore h₁(t)>0.

The second part of the theorem is obvious, since f₇(t) = ^S(t)−Q(t)_S(t)−A(t)

satisfiesf7(t)>0,∀t >1 and limt→1f7(t) = 0.

Theorem 10. The double inequality

αL(t) + (1−α)C(t)< A(t)< βL(t) + (1−β)C(t), ∀t >1, holds if and only if α≥3/4 and β≤1/2.

Proof. We have to find the extreme values of f8(t) = ^C(t)−A(t)_C(t)−L(t) fort >1, wheref₈ is given by

f8(t) = ¹₂ ·_lnt+t^(t−1)2lnt−t^2lnt2+1. We obtain

f₈⁰(t) =−_2t(lnt+t^(t−1)h_2lnt−t^3(t)₂₊₁₎₂, where

h4(t) =t³−2 t²lnt2

+ 2t²lnt−t²−2tlnt−2t(lnt)²−t+ 1.

Now, h4(1) =h⁰₄(1) =h⁰⁰₄(1) =h⁰⁰⁰₄(1) = 0 and

h⁽⁴⁾₄ (t) = ^{8(t−1) ln}_t3 ^t >0 for t >1.

Thereforeh4>0 in (1,∞) andf₈⁰ <0 in (1,∞). The functionf8being strictly decreasing on (1,∞), inff₈ = limt→∞f₈(t) = 1/2 and supf₈ = limt→1f₈(t) =

3/4.

3. FINAL REMARKS

From the eight means considered in this paper, two enter the class of Gini means [6] defined fora, b >0,u, v∈R,

G_u,v(a, b) =





 a^u+b^u

a^v+b^v

_u−v¹

, u6=v

exp

a^uloga+b^ulogb a^u+b^u

, u=v

namely S = G1,1, C = G2,1; two belong to the class of Stolarsky means [8]

defined fora, b >0,a6=b, u, v∈R,

E_r,s(a, b) =

















 s

r a^r−b^r a^s−b^s

_r−s¹

, rs(r−s)6= 0

exp

−¹_r +^arlog_a^a−br−b^rrlogb

, r=s6= 0 1

s a^s−b^s loga−logb

¹_s

, r= 0, s6= 0

√

ab, r=s6= 0,

namely L = E_1,0, I = E_1,1. The other four are in both classes, namely H =G−1,0 =E−2,−1,G=G_0,0 =E_0,0,A=G_1,0=E_2,1 andQ=G_2,0 =E_4,2. As it was shown in [2], the families of Gini means Gu,v and Stolarsky means E_r,s have in common only the power means. So even if general results will be proved for these two classes of means, not all the inequalities from this paper will be consequences; for example, in the last theorem Lis a Stolarsky mean, while C is a Gini one.

REFERENCES

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[5] P. S. Bullen,Handbook of Means and Their Inequalities, Series: Mathematics and Its Applications, vol. 560, 2nd ed., Kluwer Academic Publishers Group, Dordrecht, 2003.

[6] C. Gini,Di una formula comprensiva delle medie, Metron,13(1938), pp. 3–22.

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Received by the editors: October 24, 2013.