Optimal Bounds for Neuman Mean Using Arithmetic and Centroidal ...

Hindawi Publishing Corporation Journal of Function Spaces Volume 2016, Article ID 5131907, 7 pages http://dx.doi.org/10.1155/2016/5131907

Research Article Optimal Bounds for Neuman Mean Using Arithmetic and Centroidal Means Ying-Qing Song,1 Wei-Mao Qian,2 and Yu-Ming Chu1 1

School of Mathematics and Computation Sciences, Hunan City University, Yiyang 413000, China School of Distance Education, Huzhou Broadcast and TV University, Huzhou 313000, China

2

Correspondence should be addressed to Yu-Ming Chu; [email protected] Received 13 September 2015; Accepted 11 January 2016 Academic Editor: Henryk Hudzik Copyright © 2016 Ying-Qing Song et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We present the best possible parameters 𝛼1 , 𝛼2 , 𝛽1 , 𝛽2 ∈ R and 𝛼3 , 𝛽3 ∈ (1/2, 1) such that the double inequalities 𝛼1 𝐴(𝑎, 𝑏) + (1 − 𝛼1 )𝐶(𝑎, 𝑏) < 𝑁𝑄𝐴 (𝑎, 𝑏) < 𝛽1 𝐴(𝑎, 𝑏) + (1 − 𝛽1 )𝐶(𝑎, 𝑏), 𝐴𝛼2 (𝑎, 𝑏)𝐶1−𝛼2 (𝑎, 𝑏) < 𝑁𝑄𝐴 (𝑎, 𝑏) < 𝐴𝛽2 (𝑎, 𝑏)𝐶1−𝛽2 (𝑎, 𝑏), and 𝐶[𝛼3 𝑎 + (1 − 𝛼3 )𝑏, 𝛼3 𝑏 + (1 − 𝛼3 )𝑎] < 𝑁𝑄𝐴 (𝑎, 𝑏) < 𝐶[𝛽3 𝑎 + (1 − 𝛽3 )𝑏, 𝛽3 𝑏 + (1 − 𝛽3 )𝑎] hold for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏 and give several sharp inequalities involving the hyperbolic and inverse hyperbolic functions. Here, 𝑁(𝑎, 𝑏), 𝐴(𝑎, 𝑏), 𝑄(𝑎, 𝑏), and 𝐶(𝑎, 𝑏) are, respectively, the Neuman, arithmetic, quadratic, and centroidal means of 𝑎 and 𝑏, and 𝑁𝑄𝐴 (𝑎, 𝑏) = 𝑁[𝑄(𝑎, 𝑏), 𝐴(𝑎, 𝑏)].

1. Introduction Let 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏. Then, the Schwab-Borchardt mean SB(𝑎, 𝑏) [1–7] of 𝑎 and 𝑏 is given by √𝑏2 − 𝑎2 { , 𝑎 < 𝑏, { { { { arccos (𝑎/𝑏) SB (𝑎, 𝑏) = { { { √𝑎2 − 𝑏2 { { , 𝑎 > 𝑏, { cosh−1 (𝑎/𝑏)

inverse hyperbolic tangent function, 𝐺(𝑎, 𝑏) = √𝑎𝑏, 𝐴(𝑎, 𝑏) = (𝑎 + 𝑏)/2, and 𝑄(𝑎, 𝑏) = √(𝑎2 + 𝑏2 )/2 are, respectively, the geometric, arithmetic, and quadratic means of 𝑎 and 𝑏, and the inequalities 𝐻 (𝑎, 𝑏) < 𝐺 (𝑎, 𝑏) < 𝐿 (𝑎, 𝑏) < 𝑃 (𝑎, 𝑏) < 𝐴 (𝑎, 𝑏)

(1)

where cosh−1 (𝑥) = log(𝑥 + √𝑥2 − 1) is the inverse hyperbolic cosine functions. It is well known that the Schwab-Borchardt mean SB(𝑎, 𝑏) is strictly increasing in both 𝑎 and 𝑏, nonsymmetric, and homogeneous of degree 1 with respect to 𝑎 and 𝑏. Many symmetric bivariate means are special cases of the SchwabBorchardt mean. For example, 𝑃(𝑎, 𝑏) = (𝑎 − 𝑏)/[2arcsin((𝑎 − 𝑏)/(𝑎 + 𝑏))] = SB[𝐺(𝑎, 𝑏), 𝐴(𝑎, 𝑏)] is the first Seiffert mean, 𝑇(𝑎, 𝑏) = (𝑎 − 𝑏)/[2arctan((𝑎 − 𝑏)/(𝑎 + 𝑏))] = SB[𝐴(𝑎, 𝑏), 𝑄(𝑎, 𝑏)] is the second Seiffert mean, 𝑀(𝑎, 𝑏) = (𝑎 − 𝑏)/[2sinh−1 ((𝑎 − 𝑏)/(𝑎 + 𝑏))] = SB[𝑄(𝑎, 𝑏), 𝐴(𝑎, 𝑏)] is the Neuman-S´andor mean, and 𝐿(𝑎, 𝑏) = (𝑎 − 𝑏)/[2tanh−1 ((𝑎 − 𝑏)/(𝑎 + 𝑏))] = SB[𝐴(𝑎, 𝑏), 𝐺(𝑎, 𝑏)] is the logarithmic mean, where sinh−1 (𝑥) = log(𝑥 + √1 + 𝑥2 ) is the inverse hyperbolic sine function, tanh−1 (𝑥) = log((1 + 𝑥)/(1 − 𝑥))/2 is the

< 𝑀 (𝑎, 𝑏) < 𝑇 (𝑎, 𝑏) < 𝑄 (𝑎, 𝑏)

(2)

hold for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏 (see [1]), where 𝐻(𝑎, 𝑏) = 2𝑎𝑏/(𝑎 + 𝑏) is the harmonic mean of 𝑎 and 𝑏. Recently, the Schwab-Borchardt mean and its generated means have been the subject intensive research. In particular, many remarkable inequalities for these means can be found in the literature [1–5, 8–16]. Liu and Meng [17] proved that the double inequality 𝛼𝐶 (𝑎, 𝑏) + (1 − 𝛼) 𝐻 (𝑎, 𝑏) < 𝑇 (𝑎, 𝑏) < 𝛽𝐶 (𝑎, 𝑏) + (1 − 𝛽) 𝐻 (𝑎, 𝑏)

(3)

holds for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏 if and only if 𝛼 ≤ 3/𝜋 and 𝛽 ≥ 1, where 𝐶 (𝑎, 𝑏) =

2 (𝑎2 + 𝑎𝑏 + 𝑏2 ) 3 (𝑎 + 𝑏)

is the centroidal mean of 𝑎 and 𝑏.

(4)

2

Journal of Function Spaces

Very recently, Neuman [18] introduced the Neuman mean 𝑁 (𝑎, 𝑏) =

1 𝑏2 ], [𝑎 + 2 SB (𝑎, 𝑏)

(5)

hold for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏? The main purpose of this paper is to answer these questions.

2. Lemmas

presented the explicit formula for 𝑁𝑄𝐴(𝑎, 𝑏) as

In order to prove our main results, we need several lemmas, which we present in this section.

𝑁𝑄𝐴 (𝑎, 𝑏) ≜ 𝑁 [𝑄 (𝑎, 𝑏) , 𝐴 (𝑎, 𝑏)]

Lemma 1 (see [19–21]). For 𝑎, 𝑏 ∈ R, let 𝑓, 𝑔 : [𝑎, 𝑏] → R be continuous on [𝑎, 𝑏] and be differentiable on (𝑎, 𝑏), and let 𝑔󸀠 (𝑥) ≠ 0 on (𝑎, 𝑏). If 𝑓󸀠 (𝑥)/𝑔󸀠 (𝑥) is increasing (decreasing) on (𝑎, 𝑏), then so are

1 sinh−1 (V) = 𝐴 (𝑎, 𝑏) [√1 + V2 + ], 2 V

(6)

𝑓 (𝑥) − 𝑓 (𝑎) , 𝑔 (𝑥) − 𝑔 (𝑎)

and proved that the double inequality 𝑀 (𝑎, 𝑏) < 𝑁𝑄𝐴 (𝑎, 𝑏) < 𝑇 (𝑎, 𝑏)

(7)

holds for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏, where V = (𝑎 − 𝑏)/(𝑎 + 𝑏). From (2), (3), and (7), we clearly see that the double inequality 𝐴 (𝑎, 𝑏) < 𝑁𝑄𝐴 (𝑎, 𝑏) < 𝐶 (𝑎, 𝑏)

(8)

holds for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏. Let 𝐽(𝑥) = 𝐶[𝑥𝑎 + (1 − 𝑥)𝑏, 𝑥𝑏 + (1 − 𝑥)𝑎]. Then, it is not difficult to verify that the function 𝐽(𝑥) is continuous and strictly increasing on [1/2, 1] for fixed 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏. Note that 1 𝐽 ( ) = 𝐴 (𝑎, 𝑏) < 𝑁𝑄𝐴 (𝑎, 𝑏) < 𝐶 (𝑎, 𝑏) = 𝐽 (1) 2

(9)

for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏. Motivated by inequalities (8) and (9), it is natural to ask what are best possible parameters 𝛼1 , 𝛼2 , 𝛽1 , 𝛽2 ∈ R and 𝛼3 , 𝛽3 ∈ (1/2, 1) such that the double inequalities 𝛼1 𝐴 (𝑎, 𝑏) + (1 − 𝛼1 ) 𝐶 (𝑎, 𝑏) < 𝑁𝑄𝐴 (𝑎, 𝑏)

(10) < 𝐴 (𝑎, 𝑏) 𝐶

(𝑎, 𝑏) ,

𝐶 [𝛼3 𝑎 + (1 − 𝛼3 ) 𝑏, 𝛼3 𝑏 + (1 − 𝛼3 ) 𝑎] < 𝑁𝑄𝐴 (𝑎, 𝑏) < 𝐶 [𝛽3 𝑎 + (1 − 𝛽3 ) 𝑏, 𝛽3 𝑏 + (1 − 𝛽3 ) 𝑎]

𝑓 (𝑥) =

If 𝑓󸀠 (𝑥)/𝑔󸀠 (𝑥) is strictly monotone, then the monotonicity in the conclusion is also strict. 𝑛 Lemma 2 (see [22–24]). Let 𝑓(𝑥) = ∑∞ 𝑛=0 𝑎𝑛 𝑥 and 𝑔(𝑥) = ∞ 𝑛 ∑𝑛=0 𝑏𝑛 𝑥 be two real power series converging on R with 𝑏𝑛 > 0 for all 𝑛 ∈ N. Then, the following statements are true:

(1) If the nonconstant sequence {𝑎𝑛 /𝑏𝑛 } is increasing (decreasing) for all 𝑛 ∈ N, then the function 𝑥 󳨃→ 𝑓(𝑥)/𝑔(𝑥) is strictly increasing (decreasing) on (0, ∞). (2) If there exists 𝑚 ≥ 1 such that the nonconstant sequence {𝑎𝑛 /𝑏𝑛 } is increasing (decreasing) for 0 ≤ 𝑛 ≤ 𝑚 and decreasing (increasing) for 𝑛 ≥ 𝑚, then there exists 𝑥0 ∈ (0, ∞) such that the function 𝑥 󳨃→ 𝑓(𝑥)/𝑔(𝑥) is strictly increasing (decreasing) on (0, 𝑥0 ) and strictly decreasing (increasing) on (𝑥0 , ∞).

𝑓 (𝑥)

𝐴𝛼2 (𝑎, 𝑏) 𝐶1−𝛼2 (𝑎, 𝑏) < 𝑁𝑄𝐴 (𝑎, 𝑏) 1−𝛽2

(11)

Lemma 3. The function

< 𝛽1 𝐴 (𝑎, 𝑏) + (1 − 𝛽1 ) 𝐶 (𝑎, 𝑏) ,

𝛽2

𝑓 (𝑥) − 𝑓 (𝑏) . 𝑔 (𝑥) − 𝑔 (𝑏)

=

6𝑥 + 6𝑥 cosh (2𝑥) + (1/2) sinh (4𝑥) − 7 sinh (2𝑥) (12) −4𝑥 + 4𝑥 cosh (2𝑥) + sinh (4𝑥) − 2 sinh (2𝑥)

is strictly increasing from (0, log(1 + √2)) onto (1/2, [3 log(1 + √2) − √2]/[log(1 + √2) + √2]). Proof. Making use of the power series formulas sinh(𝑥) = 2𝑛+1 2𝑛 ∑∞ /(2𝑛 + 1)! and cosh(𝑥) = ∑∞ 𝑛=0 𝑥 𝑛=0 𝑥 /(2𝑛)!, we have

∞ 2𝑛 2𝑛 2𝑛+1 2𝑛+1 6𝑥 + 6𝑥 ∑∞ / (2𝑛 + 1)!) 𝑥2𝑛+1 − 7 ∑∞ / (2𝑛 + 1)!) 𝑥2𝑛+1 𝑛=0 (2 / (2𝑛)!) 𝑥 + (1/2) ∑𝑛=0 (4 𝑛=0 (2 ∞ 2𝑛 2𝑛 2𝑛+1 / (2𝑛 + 1)!) 𝑥2𝑛+1 − 2 ∑∞ (22𝑛+1 / (2𝑛 + 1)!) 𝑥2𝑛+1 −4𝑥 + 4𝑥 ∑∞ 𝑛=0 (2 / (2𝑛)!) 𝑥 + ∑𝑛=0 (4 𝑛=0

(13) =

2𝑛+4 ∑∞ (22𝑛+1 + 3𝑛 + 1) / (2𝑛 + 3)!) 𝑥2𝑛 𝑛=0 (2 . 2𝑛+5 (22𝑛+1 + 𝑛 + 1) / (2𝑛 + 3)!) 𝑥2𝑛 ∑∞ 𝑛=0 (2

Journal of Function Spaces

3

Let

Let ∞

𝑐𝑛 = ∑

22𝑛+4 (22𝑛+1 + 3𝑛 + 1) (2𝑛 + 3)!

𝑛=0 ∞

𝑑𝑛 = ∑

2𝑛+5

2

2𝑛+1

(2

+ 𝑛 + 1)

(2𝑛 + 3)!

𝑛=0

𝑒𝑛 = 1 + (1 − 3𝑛) 22𝑛+1 .

,

Then, simple computations lead to

(14)

𝑒0 = 3 > 0,

.

𝑒𝑛 < 1 − 22𝑛+2 < 0

Then, we clearly see that 𝑑𝑛 > 0

(15)

𝑐𝑛+1 𝑐 1 + (1 − 3𝑛) 22𝑛+1 − 𝑛 = 2𝑛+1 . 𝑑𝑛+1 𝑑𝑛 (2 + 𝑛 + 1) (22𝑛+3 + 𝑛 + 2)

(16)

6 − 8 cosh (2𝑥) + 12𝑥 sinh (2𝑥) + 2 cosh (4𝑥) −4𝑥 + 4𝑥 cosh (2𝑥) + sinh (4𝑥) − 2 sinh (2𝑥) −

𝑓󸀠 [log (1 + √2)] =

[6𝑥 + 6𝑥 cosh (2𝑥) + (1/2) sinh (4𝑥) − 7 sinh (2𝑥)] [−4 + 8𝑥 sinh (2𝑥) + 4 cosh (4𝑥)] , [−4𝑥 + 4𝑥 cosh (2𝑥) + sinh (4𝑥) − 2 sinh (2𝑥)]2 6 − 24 + 24√2 log (1 + √2) + 34

−4 log (1 + √2) + 12 log (1 + √2) + 12√2 − 4√2 −

=

[−4 log (1 + √2) + 12 log (1 + √2) + 12√2 − 4√2]

10√2 − 12 log (1 + √2) − 3√2 log2 (1 + √2) 2

[√2 + log (1 + √2)]

2

.

Therefore, Lemma 3 follows from (21) and the monotonicity of 𝑓(𝑥) on the interval (0, log(1 + √2)).

10√2 − 12 log (1 + √2) − 3√2 log2 (1 + √2)

(20)

= 0.2698 ⋅ ⋅ ⋅ > 0, 𝑐0 1 = , 𝑑0 2

𝑓 [log (1 + √2)] =

(19)

[6 log (1 + √2) + 18 log (1 + √2) + 6√2 − 14√2] [−4 + 16√2 log (1 + √2) + 68]

Note that

𝑓 (0+ ) =

Lemma 4. The function sinh (𝑥) [2 − cosh (𝑥)] − 𝑥 (22) sinh3 (𝑥) is strictly increasing from (0, log(1 + √2)) onto (−1/3, 2 − √2 − log(1 + √2)). 𝑔 (𝑥) =

3 log (1 + √2) − √2 log (1 + √2) + √2

(21) .

It follows from (19) and (20) together with the piecewise monotonicity of 𝑓(𝑥) that log(1+√2) < 𝑥0 and 𝑓(𝑥) is strictly increasing on (0, log(1 + √2)).

Proof. Making use of the power series formulas sinh(𝑥) = 2𝑛+1 2𝑛 /(2𝑛 + 1)! and cosh(𝑥) = ∑∞ ∑∞ 𝑛=0 𝑥 𝑛=0 𝑥 /(2𝑛+)! together 3 with the identity 4 sinh (𝑥) = sinh(3𝑥) − 3 sinh(𝑥), one has

∞ 2𝑛+1 2𝑛+1 − (1/2) ∑∞ / (2𝑛 + 1)!) 𝑥2𝑛+1 − 𝑥 2 sinh (𝑥) − (1/2) sinh (2𝑥) − 𝑥 2 ∑𝑛=0 (1/ (2𝑛 + 1)!) 𝑥 𝑛=0 (2 = 𝑔 (𝑥) = ∞ ∞ (1/4) [sinh (3𝑥) − 3 sinh (𝑥)] (1/4) [∑𝑛=0 (32𝑛+1 / (2𝑛 + 1)!) 𝑥2𝑛+1 − 3 ∑𝑛=0 (1/ (2𝑛 + 1)!) 𝑥2𝑛+1 ]

=

(18)

for all 𝑛 ≥ 1. It follows from (16)–(18) that the nonconstant sequence {𝑐𝑛 /𝑑𝑛 } is increasing for 0 ≤ 𝑛 ≤ 1 and decreasing for 𝑛 ≥ 1. Then, from (13)–(15) and Lemma 2(2) we know that there exists 𝑥0 ∈ (0, ∞) such that 𝑓(𝑥) is strictly increasing on (0, 𝑥0 ) and strictly decreasing on (𝑥0 , ∞). Differentiating 𝑓(𝑥) gives

for all 𝑛 ≥ 0 and

𝑓󸀠 (𝑥) =

(17)

2𝑛+2 ∑∞ ) / (2𝑛 + 3)!) 𝑥2𝑛 𝑛=0 ((2 − 2

2𝑛+3 − 3) / (4 × (2𝑛 + 3)!)) 𝑥2𝑛 ∑∞ 𝑛=0 ((3

.

(23)

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Journal of Function Spaces

Let 𝑢𝑛 =

From (6) and (28), we clearly see that inequality (27) is equivalent to

2 − 22𝑛+2 , (2𝑛 + 3)!

(24)

32𝑛+3 − 3 . V𝑛 = 4 × (2𝑛 + 3)!

𝛽1
0, V𝑛+1 V𝑛 (32𝑛+4 − 1) (32𝑛+2 − 1)

2V2

(25) =

V𝑛 > 0, for all 𝑛 ≥ 0. Note that 𝑔 (0+ ) =

2V2 − 3 (√1 + V2 + (sinh−1 (V)) /V) + 6

sinh (3𝑥) − 3 sinh (2𝑥) + 9 sinh (𝑥) − 6𝑥 sinh (3𝑥) − 3 sinh (𝑥)

=1+

𝑢0 1 =− , V0 3

(26)

𝑔 [log (1 + √2)] = 2 − √2 − log (1 + √2) .

3 [sinh (𝑥) (2 − cosh (𝑥)) − 𝑥] . 2 sinh3 (𝑥)

Therefore, Theorem 5 follows easily from (29) and (30) together with Lemma 4. Theorem 6. The double inequality

Therefore, Lemma 4 follows easily from (23)–(26) and Lemma 2(1).

𝐴𝛼2 (𝑎, 𝑏) 𝐶1−𝛼2 (𝑎, 𝑏) < 𝑁𝑄𝐴 (𝑎, 𝑏) < 𝐴𝛽2 (𝑎, 𝑏) 𝐶1−𝛽2 (𝑎, 𝑏)

3. Main Results Theorem 5. The double inequality 𝛼1 𝐴 (𝑎, 𝑏) + (1 − 𝛼1 ) 𝐶 (𝑎, 𝑏) < 𝑁𝑄𝐴 (𝑎, 𝑏) < 𝛽1 𝐴 (𝑎, 𝑏) + (1 − 𝛽1 ) 𝐶 (𝑎, 𝑏)

(27)

holds for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏 if and only if 𝛼1 ≥ 4 − 3[√2 + log(1 + √2)]/2 = 0.5566 ⋅ ⋅ ⋅ and 𝛽1 ≤ 1/2. Proof. Since all the means 𝐴(𝑎, 𝑏), 𝑁𝑄𝐴 (𝑎, 𝑏), and 𝐶(𝑎, 𝑏) are symmetric and homogeneous of degree 1, without loss of generality, we assume that 𝑎 > 𝑏. Let V = (𝑎 − 𝑏)/(𝑎 + 𝑏) ∈ (0, 1). Then, (4) leads to 1 𝐶 (𝑎, 𝑏) = (1 + V2 ) 𝐴 (𝑎, 𝑏) . 3

(28)

(30)

(31)

holds for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏 if and only if 𝛼2 ≥ 1−[log(√2+ log(1 + √2)) − log 2]/[2 log 2 − log 3] = 0.5208 ⋅ ⋅ ⋅ and 𝛽2 ≤ 1/2. Proof. Without loss of generality, we assume that 𝑎 > 𝑏. Let V = (𝑎 − 𝑏)/(𝑎 + 𝑏) ∈ (0, 1). Then, from (6) and (28) we clearly see that inequality (31) is equivalent to 𝛽2
𝑏 > 0 and 𝑡 = (log 𝑎 − log 𝑏)/2 > 0. Then, we clearly see that 𝑎−𝑏 = tanh (𝑡) , 𝑎+𝑏

𝑔󸀠 (0+ ) 𝑐 1 𝐺 (0 ) = 1󸀠 + = 0 = , 𝑔2 (0 ) 𝑑0 2 +

𝐺 [log (1 + √2)] =1−

2 1/2 √ 1 + ( 𝑎 − 𝑏 ) = cosh (2𝑡) , 𝑎+𝑏 cosh (𝑡)

(39)

log [√2 + log (1 + √2)] − log 2 2 log 2 − log 3

where 𝑐𝑛 and 𝑑𝑛 are defined by (14). Therefore, Theorem 6 follows easily from (32)–(34) and (39) together with the monotonicity of 𝐺(𝑥) on the interval (0, log(1 + √2)). Theorem 7. Let 𝛼3 , 𝛽3 ∈ (1/2, 1). Then, the double inequality

𝐶 (𝑎, 𝑏) =

4 cosh2 (𝑡) − 1 𝐴 (𝑎, 𝑏) , 3 cosh2 (𝑡)

𝐶 [𝛼𝑎 + (1 − 𝛼) 𝑏, 𝛼𝑏 + (1 − 𝛼) 𝑎] =

2 (𝛼2 − 𝛼 + 1) cosh (2𝑡) + (1 + 2𝛼 − 2𝛼2 ) 3 cosh2 (𝑡)

(40)

⋅ 𝐴 (𝑎, 𝑏) ,

holds for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏 if and only if 𝛼3 ≤ 1/2 +

𝑁𝑄𝐴 (𝑎, 𝑏) =

√6[√2 + log(1 + √2)] − 12/4 = 0.8329 ⋅ ⋅ ⋅ and 𝛽3 ≥ 1/2 + √2/4 = 0.8535 ⋅ ⋅ ⋅ .

⋅ 𝐴 (𝑎, 𝑏) .

< 𝐶 [𝛽3 𝑎 + (1 − 𝛽3 ) 𝑏, 𝛽3 𝑏 + (1 − 𝛽3 ) 𝑎]

Proof. Without loss of generality, we assume that 𝑎 > 𝑏. Let V = (𝑎 − 𝑏)/(𝑎 + 𝑏) ∈ (0, 1) and 𝑝 ∈ (1/2, 1). Then, (4) and (6) lead to 1 𝐶 [𝑝𝑎 + (1 − 𝑝) 𝑏, 𝑝𝑏 + (1 − 𝑝) 𝑎] − 𝑁𝑄𝐴 (𝑎, 𝑏) = 3 1 2 ⋅ 𝐴 (𝑎, 𝑏) [3 + (2𝑝 − 1) V2 ] − 𝐴 (𝑎, 𝑏) 2 ⋅ [√1 + V2 + (sinh−1 (V)) /V] = 𝐴 (𝑎, 𝑏) ⋅[

V (2 − √1 + V2 ) − sinh−1 (V) 2V3

+

1 2 (2𝑝 − 1) ] V2 . 3

(43)

and (4) and (6) lead to

,

𝐶 [𝛼3 𝑎 + (1 − 𝛼3 ) 𝑏, 𝛼3 𝑏 + (1 − 𝛼3 ) 𝑎] < 𝑁𝑄𝐴 (𝑎, 𝑏)

(42)

(44)

1 cosh1/2 (2𝑡) sinh−1 (tanh (𝑡)) + ] [ 2 cosh (𝑡) tanh (𝑡)

From (44) and Theorems 5–7, we get Corollary 8 immediately. Corollary 8. Let 𝛼1 , 𝛼2 , 𝛽1 , 𝛽2 ∈ R and 𝛼3 , 𝛽3 ∈ (1/2, 1). Then, the double inequalities 8 − 2𝛼1 2 (1 − 𝛼1 ) − 3 3 cosh2 (𝑡)

(41)