On two new means of two arguments III Текст научной статьи по специальности «Математика»

116

Probl. Anal. Issues Anal. Vol. 7(25), No. 1, 2018, pp. 116-133

DOI: 10.15393/j3.art.2018.4590

UDC 517.16

J. SÄNDÜR AND B. A. BHAYO ON TWO NEW MEANS OF TWO ARGUMENTS III

Abstract. In this paper we establish two sided inequalities for the following two new means

X = X (a,b) = AeG/P-1, Y = Y (a,b) = CeL/A-1,

where A, G, L and P are the arithmetic, geometric, logarithmic, and Seiffert means, respectively. As an application, we refine many other well known inequalities involving the identric mean I and the logarithmic mean L.

Key words: inequalities, means of two arguments, identric mean, logarithmic mean

2010 Mathematical Subject Classification: 26D05, 26D15, 26D99

1. Introduction. The study of the inequalities involving the classical means such as arithmetic mean A, geometric mean G, identric mean I and logarithmic mean L have been of the extensive interest for several authors, e.g., see [2, 3, 9, 11, 21, 22, 30, 31, 32, 40].

In 2011, Sandor [27] introduced a new mean X(a,b) for two positive real numbers a and b, defined by

X = X(a, b) = AeG/P-1, where A = A(a, b) = (a + b)/2, G = G(a, b) = \fab, and

a — b

P = P(a, b) =-(-—), a = b, P(a, a) = a,

i a b \

2 arcsin -r

\a + b

are the arithmetic mean, geometric mean, and Seiffert mean [38], respectively.

© Petrozavodsk State University, 2018

For p £ R and a,b > 0 with a = b, we define the pth power mean Mp(a, b) and the pth power-type Heronian mean Hp(a, b) by

{( ap + bP )1/p Mp = Mp(a, b) = M 2 ) , P = ° I \J~ab, p = 0,

and

'ap + (ab)p/2 + bp )1/p

, p = 0, Vab, p = 0,

Hp = Hp(a, b) = M 3 J

respectively.

The present paper contains essentially results on the X mean, in particular, several inequalities involving the X mean and the refinements of the following double inequalities are established. For all a, b > 0 with a = b

Mp <X <Mq (1)

holds if and only if p < 1/3 and q > log(2)/(1 + log(2)) « 0.4094, and

Ha <X<Hp (2)

holds if and only if a < 1/2 and p > log(3)/(1 + log(2)) « 0.6488.

In the same paper, Sandor [27] introduced another mean Y(a, b) for two positive real a and b by

Y = Y(a, b) = GeL/A-1,

where

ab

L = L(a, b) = -—^--—-—, a = b, L(a, a) = a,

log(a) - log(y)

is a logarithmic mean. For two positive real numbers a and b, the identric mean and harmonic mean are defined by

1 ( aa \1/(a-b) I = I (a, b) = — ( ~bb J , a = b, I (a, a) = a,

and

H = H(a, b) = 2ab/(a + b),

respectively. For the sharp inequalities of logarithmic and identric means, see ([25, 18]). See also [23], [24], [36]. In 2012, the X mean appeared in [27]. In 2014, X and Y means were published in the journal of Notes on Number Theory and Discrete Mathematics [29]. For the historical background and the generalization of these means we refer the reader to, e.g. [3, 9, 17, 21, 22, 28, 30, 31, 32, 33, 34, 40]. Connections of these means and the trigonometric or hyperbolic inequalities are given in [5, 27, 29, 32].

In [29], Sandor proved inequalities for X and Y means in terms of other classical means as well as their relationship. Let us recall some of the results for easy reference.

Theorem 1. [29] For a,b > 0 with a = b, the following inequalities

1) g<ag<x< p-g<p •

2) H<LG<Y<-A^ < G, J A 2A - L

L2 L • eG/L-1 PX

3) 1 <1G< G < AG,

__ G2 LG G(A + L)

4) H<T<A<1A-T <Y

hold.

In [5] a series expansion of X and Y was presented. Theorem 2. [5] For a,b > 0 with a = b, the following inequalities

1) 1 (G + H) < Y< 2(G + H), e2

2) G2I <IY <IG< L2,

„. G - Y Y + G 3G + H ,

3) A-L < "A" < ^^ < ^

4) L < <X<L(X,A) <P< <I,

33

5)2 (1 - A) < * (A) < (A)2

are true.

Chu et al. [10] and Zhou et al. [41] proved the double inequalities (1) and (2), respectively.

This paper is organized as follows: in Section 1, we give the introduction. Section 2 consists of main results and remarks. In Section 3, some connections of X, Y and other means are given with trigonometric and hyperbolic functions. Some lemmas are also proved in this section which will be used in the proof of the main result. Section 4 deals with the proof of the main result and corollaries. In the computations we have used also the Mathematica software (see e.g.[26]).

2. Main result and motivation. Making contribution to the topic, we refine some previous results appeared in the literature [1, 2, 5, 10, 41, 29] as well as establish new results involving the X mean.

Theorem 3. For a,b > 0

aG +(1 - a)A < X < fG + (1 - f)A, (3)

with the best possible constants a = 2/3 ~ 0.6667 and f = (e — 1)/e ~ « 0.6321, and

A + G — ai P < X < A + G — fi P, (4)

with the best possible constants a1 = 1 and f1 = n(e — 1)/(2e) ~ 0.9929. Remark. In [29, Theorem 2.7], Sandor proved that for a = b,

X < A

and

Y <G

1 A ^ G

~e + 1 " ~e)p

Mi - !) A

e V e A

(5)

(6)

As A/P > 1, the right side of (3) gives a slight improvement to (5). From (6), as clearly G ■ L/A < A, we get a similar inequality. The second inequality in (4) could be a counterpart of the inequality L + G — A <Y studied in [5, Theorem 20].

H. Alzer [1] proved the following inequalities:

1 < (A + G)/(L + I) < e/2,

(7)

where the constants 1 and 2/e are the best possible ones. The following result improves among others the right side of (7).

Theorem 4. For a = b

(A + G)/e <X <Mq < (L + I )/2 < (A + G)/2, (8)

where q = log(2)/(1 + log(2)) « 0.4094 is the best possible constant. Remark. Particularly, (8) implies that

X < (L + I)/2, (9)

which is new. Since L < X < I (see Theorems 1 and 2), X is less than the arithmetic mean of L and I. In fact, by left side of (1), and by L < M1/3 (see [25], [16]), and L < I < M2/3 (see [25]; see also [30], for other references), we get also

L < M1/3 <X <Mq < (L + I )/2 <I < M2/3. (10)

Theorem 5. For a = b

A + G - P<X<P2/A< (A + G)/2. (11)

Remark. The right hand side of (11) offers another refinement to X <

< (A + G)/2. An improvement of P2 > XA appears in [29, Theorem 2.9]:

P2 > (A2((A + G)/2)4)1/3 > AX, so (11) could be further refined. For the following inequalities

L < < A + G - P<X< VPX < A+G < (12)

32

P + X n 2 A + G r

< — <P<—T~ <L

one can see that the first inequality is Carlson's inequality, while the second written in the form P < (2A + G)/3 is due to Sandor [33]. The third inequality is Theorem 2.10 in [29], while the fourth, written as PX <

< ((A + G)/2)2 is Theorem 2.11 of [29]. The inequality (P + X)/2 < P follows by X < P, while the last two inequalities are due to Saandor ([33, 31]).

Theorem 6. For a = b

Mi/2 < (P + X)/2 <Mk, (13)

where k = (5 log 2 + 2)/(6(log 2+1)) « 0.5380. Remark. One has

r 2G + A v L + I A + G P + X „ 2A + G r

L < --- < X < ^ < —-— < —-— < P < --- < I

3 2 2 2 3

(14)

and

A + G

v—G < VPX < —+—. (15)

2

Inequalities (15) show that VPX lies between the geometric and arithmetic means of A and G, while (12) shows among others that (A + G)/2 lies between the geometric and arithmetic means of P and X.

Theorem 7. The following inequalities

Mp < M1/3 < (2G + A)/3 < X, for p < 1/3, (16)

Ha < H1/2 < (2G + A)/3 < X, for a < 1/2, (17)

hold.

Theorem 8. For a = b

(AX)1/a2 <P< (AX)1/(1+ft)

with the best possible constants a2 = 2 and f2 = log(n/2)/log(2e/n) ~ « 0.8234.

3. Preliminaries and lemmas. We use the following result by Biernacki and KrzyZ [8] in studying the monotonicity of certain power series.

Lemma 1. Let A(x) = ^Lo anxn and C(x) = xn be two

real power series converging on the interval (—R, R), 0 < R < rn. If the sequence {an/cn} is increasing (decreasing) and cn > 0 for all n, then the function A(x)/C(x) is also increasing (decreasing) on (0, R).

For \x\ < n, the following power series expansions

xcotx = 1 - ^ -=—\B2n\x2n, (18)

2

1 (2n)

1 ^ o2n

COt * = ^ -£ T^T B2n l*2™-1' (19)

n=1 1 ^ 22n

cothx = 1 + V — \B2n\x2n-1, (20)

x (2n)!

n=1

can be found in [13, 1.3.1.4 (2)—(3)]. Here B2n are the even-indexed Bernoulli numbers (see [12, p. 231]). We get the following expansions directly from (19) and (20)

1 1 ^ 22

= -(cotx)' = + £ ^\B2n\(2n - 1)x2n-2, (21)

(sinx)2 x2—^ (2n)!

\ y n = 1

1 1 ^ 22n

= -(cothx)' = -2 - £ — (2n - 1)\B2n\x2n-2. (22)

(sinhx)2 x2 (2n)!

n=1

For the following expansion formula

x ^ 22n_2

— = \B2n\x2n (23)

sin x ^ (2n)!

n=1

see [15].

For easy reference we recall the following lemma from [5, 6].

Lemma 2. For x = arcsin((a - b)/(a + b)) and y = (1/2) log(a/b), with a > b > 0, one has

P sin(x) G H 2 X xcot(x)-1

— = ——, — = cos(x), — = cos(x)2, — = excot(x) 1, A x A W'A W'A

L = sinh(y) L = tanh(y) H = 1 Y = etanh(y)/y-1 G y , a y , G cosh(y), G

=A -1 lo<G)=A -1.

Remark. It is well known that many inequalities involving the means can be obtained from the classical inequalities for trigonometric functions. For example, the following inequality

e(x/tanh(x) —1)/2 < sinh(x) , x > 0,

x

recently appeared in [7, Theorem 1.6], is equivalent to

sinh(x) > ^x/ tanh(x) — 1 x (24)

x sinh(x)

By Lemma 2, this can be written as

L I G = I

G > G ^ L = L,

or

L >VIG. (25)

The inequality (25) was proved by Alzer [3].

The following trigonometric inequalities (see [7, Theorem 1.5]) imply an other double inequality for Seiffert mean P,

'exp(2(tan^-1))<^exp((logIHdb-l)) -6(26,

The second mean inequality in (26) was also pointed out by Saandor (see [29, Theorem 2.12]). For various related trigonometric and hyperbolic inequalities, see also [14], [19].

Lemma 3. [4, Theorem 2] For < a < b < rn, let f,g : [a, b] ^ R be continuous on [a, b], and differentiable on (a,b). Let g (x) = 0 on (a,b). If f (x)/g (x) is increasing (decreasing) on (a,b), then so are

f (x) — f (a) and f (x) — f (b) g(x) — g(a) g(x) — g(b)"

If f (x) /g (x) is strictly monotone, then the monotonicity in the conclusion is also strict.

Lemma 4. The following function

log(x/ sin(x))

h(x) =

log(e1_x/tan(x) sin(x)/x)

is strictly decreasing from (0, n/2) onto (f2,1), where f2 = log(n/2)/log(2e/n) « 0.8234. In particular, for x E (0, n/2) we have

(e1-x/tan(x) sin(x)\132 < x <ie1-x/ tan(x) sin(x)\ (27)

\ x J sin(x) \ x J

Proof. Let

hi (x) log(x/ sin(x))

h(x) = —

h2 (x) l0g(e1-x/ tan(x) sin(x)/x)

for x E (0, n/2). Differentiating with respect to x we get

hi(x) 1 — x/ tan(x) A1 (x) h'2(x) (x/ sin(x))2 — 1 B1 (x)

Using the expansion formula we have

^ 22n2n ^

A1(x) = Y.2222n B2n\x2n = £ anx2n

n=1 n=1

and

^ 22n2n ^

B1(x) = Z ^\B2n\(2n — 1)x2n = £ bnx2n.

n=1 n=1

Let cn = an/bn = 1/(2n — 1), which is the decreasing in n E N. Thus, by Lemma 1 h1 (x)/h'2(x) is strictly decreasing in x E (0, n/2). In turn, this implies by Lemma 3 that h(x) is strictly decreasing in x E (0, n/2). Applying L'Hopital rule, we get limx^0 h(x) = 1 and limx^^/2 h(x) = f2. This completes the proof. □

Remark. It is observed that the inequalities in (27) coincide with the trigonometric inequalities given in (26). Here Lemma 4 gives a new and an optimal proof for these inequalities.

Lemma 5. The following function

1 _ e^/ tan(x)-1

f (x) =

1 — cos(x)

is strictly decreasing from (0, n/2) onto ((e — 1)/e, 2/3) where (e — 1)/e ~ 0.6321. In particular, for x E (0, n/2), we have

1 tan(x) 1

< -— <

log(1 + (e — 1) cos(x)) x 1 + log((1 + 2 cos(x))/3)'

Proof. Write f(x) = f1 (x)/f2(x), where f1(x) = 1 — ex/tan(x) —1 and f2(x) = 1 — cos(x) for all x E (0n/2). Clearly, f1 (x) = 0 = f2 (x). Differentiating with respect to x, we get

f1 (x) _ ex/tan(x) 1 / x cos(x) f2 (x) sin(x)3 \sin(x)2 sin(x)

/ x cos(x)\ . , ,

--r_TT = f3(x)-

\sm(x)2 sin(x) J

Again

ex) tan(x) —1

f3 (x) =--: (x3 (c(x) - 2) ,

sin(x)3

( ) / cos(x) + x \ \sin(x) sin(x)2/

where

-(-) = _T_

sin(x) sin(x)2 In order to show that f < 0, it is enough to prove that

c(x) > 2,

which is equivalent to

sin(x) x + sin(x) cos(x) x 2 sin(x)

Applying the Cusa-Huygens [20] inequality

sin(x) cos(x) + 2 x3

we get

cos(x) + 2 x + sin(x) cos(x)

3 2sin(x) '

which is equivalent to (cos(x) — 1)2 > 0. Thus f' > 0, clearly f1 /f2 is strictly decreasing in x E (0, n/2). By Lemma 3, we conclude that the function f (x) is strictly decreasing in x E (0,n/2). The limiting values follow easily. This completes the proof of the lemma. □

Lemma 6. The following function

sin(x)

f4(x) =

x (cos(x) — excot(x)-1 + 1)

is strictly increasing from (0, n/2) onto (1, c), where c = 2e/(n(e — 1)) ~ ~ 1.0071. In particular, for x E (0, n/2) we have

1 + cos(x) — ex/tan(x)-1 < < c(1 + cos(x) — ex/tan(x)-1).

x

Proof. Differentiating with respect to x we get

e(x — sin(x)) (ecos(x) — (x + sin(x))ex cot(x) csc(x) + e) f 4 (x) — 2 •

x2 (e cos(x) — ex cot(x) + e)

Let f5(x) = log ((x + sin(x))ex cot(x)/ sin(x)) — log(e cos(x) + e) for x E (0, n/2). Differentiation yields

2 — x (cot(x) + x csc2(x))

f5(x) = , • / \ ,

x + sin(x)

which is negative by the proof of Lemma 5, and limx^0 f5 (x) = 0. This implies that f4(x) > 0, and f4(x) is strictly increasing. The limiting values follow easily. This implies the proof. □

Lemma 7. For a = b, one has

M1/3 < (2G + A)/3. (28)

Proof. Let G = G(a,b), etc. Divide both sides with b and put a/b = x. Then inequality (28) becomes the following:

3

ix1'3 + 1\ 3

i 2 j < 4(x + 4\fx +1). (29)

Let x = t6, where t > 1. Then raising both sides of (29) to 3th power, after elementary transformations we get,

t6 — 9t4 + 16t3 — 9t2 + 1 > 0,

which can be written as (t — 1)4(t2 + 4t +1) > 0, so it is true. Thus (29) and (28) are proved. □

Since L < M1/3, by (28) we get a new proof , as well as a refinement of Carlson's inequality L < (2G + A)/3.

Lemma 8. The inequality

H1/2 < (2G + A)/3 (30)

holds for a = b.

Proof. By definition of Ha one has

H1/2 = ((Va + (ab)1/4 + Vb)/3)2 = (V2(A + G) + V—)2 /9,

by remarking that \fa + Vb = \J2(A + G). Therefore, (2) can be written equivalently as

(2(A + G) + 2y/2G(A + G) + G)/9 < (2G + A)/3. (31)

Now, it is immediate that (31) becomes, after elementary computations

A + 3G> 2^2G(A + G), (32)

or by raising both sides to the 2th power:

A2 + 6AG + 9G2 > 8AG + 8G2,

which becomes (A — G)2 > 0, true. Thus (32) and (31) are proved, and (30) follows. □

4. Proof of main result. Proof of Theorem 3. By Lemma 5

e— 1 1 — 1/e1—x/tan(x) 2 - < --- < —

e cos(x)/e1—x/tan(x) — 1/e1—x/ tan(x) 3

Now we get the proof of (3) by utilizing the Lemma 2. The proof of (4) follows easily from Lemmas 2 and 5. □

Proof of Theorem 4. The second inequality of (8) is the right hand side of (1). In [2], Alzer and Qiu proved the third inequality of (8). The last inequality is the left side of (7). By [10] and [2], q is the best possible constant in both sides.

Now let us prove the first inequality of (8). By Lemma 2 this becomes equivalent to 1 + cos(x) < ex cot(x), or

log(1 + cos(x)) <x cot(x), x e (0, n/2). (33)

Now, by the classical inequality log(1 + t) < t (t > 0), applied to t = cos(x), we get log(1 + cos(x)) < cos(x). Now cos(x) < x cot(x) = = x cos(x)/sin(x) is true by sin(x) < x. The proof of (33) follows. □ One has the following relation, in analogy with relation (7) of Theorem 2 for the mean Y:

Corollary. The inequality (A + G)/e < X < (A + G)/2 holds. The constants e and 2 are the best possible ones.

The inequalities (A + G)/e < X and (2G + A)/3 < X are not comparable.

Proof of Theorem 5. The second inequality of (11) appeared in [27] in the form P2 > AX. The last inequality follows by P < (2A + G)/3. Indeed, one has ((2A + G)/3)2 < A(A + G)/2 becomes 2G2 < A2 + AG, and this is true by G < A. □

Proof of Theorem 6. By [29, Theorem 2.10], one has P + X > A + G, and remarking that (A + G)/2 = M1/2, the left side of (13) follows. For the right hand side of (13), we will use P < Mt with t = 2/3 (see [33]), and X < Mq ([10]), where q = (log2)/(log2 + 1). On the other hand the function f (t) = Mt is known to be strictly log-concave for t > 0 (see [35]). Particularly, this implies that f (t) is strictly concave. Thus

(Mt + Mq)/2 < M(t+q)/2. As (t + q)/2 = k « 0.5380, the result follows. □

Corollary. One has the following two sets of inequalities:

1) PX > PL > AG,

2) IL > PL > AG.

Proof. The first inequality of (1) follows by X > L, while the second appears in [33]. The first inequality of (2) follows by I > P, while the second one is the same as the second one in (1). □

Remark. Particularly in Corollary , (2) improves Alzer's inequality IL > > AG. Inequality (1) improves PX > AG, which appears in [29].

Corollary. One has

1) X > A(P + G)/(3P - G) > (2G + A)/3 > L,

2) P21A >X> (P + G)/2.

Proof. The first two inequalities of (1) appear in [29, Theorem 2.5 and Remark 2.3]. The second inequality of (2) follows by the first inequality of (1) and the remark that A/(3P - G) > 1/2 , since this is P < (2A + G)/3; while the first one is P2 > AX ([27]). □

Remark. Since it is known that P > (2/n)A (due to Seiffert, see [33]). By X > (P + G)/2 we get the inequality X > [(2/n)A + G]/2, which is not comparable with (A + G)/e < X.

Proof of Theorem 7. The first inequality of (16) follows, since the function f (t) = Mt is known to be strictly increasing. The second inequality follows by (28), while the third one can be found in Theorem 2.

It is known that Hp is an increasing function of p. Therefore, the proof of (17) follows by (30). □

Corollary. For a,b > 0 with a = b, we have

iL<lg< 1+H - G <*>

Proof. The first inequality is due to Alzer [3], while the second inequality follows from the fact that the function

1 _ eXl tanh(x) —1

1 " 1 - coshW :(0- ^ ^ (0'

is strictly decreasing. The proof of the monotonicity of the function is the analogue to the proof of Lemma 5. □

The right hand side of (34) may be written as L + I < G + A (by H = G2/A), and this is due to Alzer (see [2, 30] for history of early results).

Proof of Theorem 8. The proof follows easily from Lemma 4. □

In [37] (see also [39]), Seiffert proved that

2 A<P n

for all a, b > 0 with a = 0. As a counterpart of the above result we give the following inequalities.

Corollary. The following inequalities

1 n

-A<—P<X <P e 2e

hold true for a, b > 0 with a = b.

Proof. The first inequality follows from (35). For the proof of the second inequality we write by Lemma 2

X xex/ tan(x)-1

f5(x) = x = —— = f5(x)

P sin(x) for x E (0, n/2). Differentiation gives

ex/tan(x)-1 / x2 \

-1 — ~7^2 < 0.

sin(x) V sin(x)2 J

Hence the function f5 is strictly decreasing in x, with

lim f5(x) = 1 and lim f5(x) = n/(2e) « 0.5779.

This implies the proof. □

We finish this paper by giving the following open problem and a conjecture.

Open problem. What are the best positive constants a and b, such that

Ma < (P + X)/2 <Mb. Conjecture. For a = b, one has PX > IL.

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Received November 21, 2017. In revised form, February 28, 2018. Accepted February 28, 2018. Published online March 31, 2018.

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