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Optimal bounds for the Neuman-Sándor mean in terms of the first Seiffert and quadratic means

Abstract

In this paper, we find the least value α and the greatest value β such that the double inequality

P α (a,b) Q 1 α (a,b)<M(a,b)< P β (a,b) Q 1 β (a,b)

holds true for all a,b>0 with ab, where P(a,b), M(a,b) and Q(a,b) are the first Seiffert, Neuman-Sándor and quadratic means of a and b, respectively.

MSC:26E60.

1 Introduction

Let u, v and w be the bivariate means such that u(a,b)<w(a,b)<v(a,b) for all a,b>0 with ab. The problems of finding the best possible parameters α and β such that the inequalities αu(a,b)+(1α)v(a,b)<w(a,b)<βu(a,b)+(1β)v(a,b) and u α (a,b) v 1 α (a,b)<w(a,b)< u β (a,b) v 1 β (a,b) hold for all a,b>0 with ab have attracted the interest of many mathematicians.

For a,b>0 with ab, the first Seiffert mean P(a,b) [1], the Neuman-Sándor mean M(a,b) [2], the quadratic mean Q(a,b) are defined by

P ( a , b ) = a b 4 arctan ( a / b ) π , M ( a , b ) = a b 2 sinh 1 ( a b a + b ) , Q ( a , b ) = a 2 + b 2 2 ,
(1.1)

respectively. In here, sinh 1 (x)=log(x+ x 2 + 1 ) is the inverse hyperbolic sine function.

Recently, the means P, M and Q have been the subject of intensive research. In particular, many remarkable inequalities for these means can be found in the literature [314]. The first Seiffert mean P(a,b) can be rewritten as (see [[2], Eq. (2.4)])

P(a,b)= a b 2 arcsin [ ( a b ) / ( a + b ) ] .
(1.2)

Let H(a,b)=2ab/(a+b), L(a,b)=(ba)/(logbloga), A(a,b)=(a+b)/2, T(a,b)=(ab)/[2arctan((ab)/(a+b))] and C(a,b)=( a 2 + b 2 )/(a+b) be the harmonic, logarithmic, arithmetic, second Seiffert and contra-harmonic means of a and b, respectively. Then it is known that the inequalities

H(a,b)<L(a,b)<P(a,b)<A(a,b)<M(a,b)<T(a,b)<Q(a,b)<C(a,b)

hold for all a,b>0 with ab.

Neuman and Sándor [2, 15] proved that the inequalities

π 4 log ( 1 + 2 ) T ( a , b ) < M ( a , b ) < A ( a , b ) log ( 1 + 2 ) , 2 T 2 ( a , b ) Q 2 ( a , b ) < M ( a , b ) < T 2 ( a , b ) Q ( a , b ) , H ( T ( a , b ) , A ( a , b ) ) < M ( a , b ) < L ( A ( a , b ) , Q ( a , b ) ) , T ( a , b ) > H ( M ( a , b ) , Q ( a , b ) ) , M ( a , b ) < A 2 ( a , b ) P ( a , b ) , A 2 / 3 ( a , b ) Q 1 / 3 ( a , b ) < M ( a , b ) < 2 A ( a , b ) + Q ( a , b ) 3 , A ( a , b ) T ( a , b ) < M ( a , b ) < A 2 ( a , b ) + T 2 ( a , b ) , A ( x , y ) A ( 1 x , 1 y ) < M ( x , y ) M ( 1 x , 1 y ) < T ( x , y ) T ( 1 x , 1 y ) , 1 A ( 1 x , 1 y ) 1 A ( x , y ) < 1 M ( 1 x , 1 y ) 1 M ( x , y ) < 1 T ( 1 x , 1 y ) 1 T ( x , y ) , A ( x , y ) A ( 1 x , 1 y ) < M ( x , y ) M ( 1 x , 1 y ) < T ( x , y ) T ( 1 x , 1 y )

hold for all a,b>0 and x,y(0,1/2] with ab and xy.

Li et al. [16] proved that the double inequality L p 0 (a,b)<M(a,b)< L 2 (a,b) holds for all a,b>0 with ab, where L p (a,b)= [ ( b p + 1 a p + 1 ) / ( ( p + 1 ) ( b a ) ) ] 1 / p (p1,0), L 0 (a,b)=1/e ( b b / a a ) 1 / ( b a ) and L 1 (a,b)=(ba)/(logbloga) is the p th generalized logarithmic mean of a and b, and p 0 =1.843 is the unique solution of the equation ( p + 1 ) 1 / p =2log(1+ 2 ).

In [13], Neuman proved that the double inequalities

Q α (a,b) A 1 α (a,b)<M(a,b)< Q β (a,b) A 1 β (a,b)
(1.3)

and

C λ (a,b) A 1 λ (a,b)<M(a,b)< C μ (a,b) A 1 μ (a,b)
(1.4)

hold for all a,b>0 with ab if α1/3, β2[log(2+ 2 )log3]/log2, λ1/6 and μ[log(2+ 2 )log3]/log2.

Jiang and Qi [17, 18] gave the best possible parameters α, β, t 1 and t 2 in (0,1/2) such that the inequalities

Q ( α a + ( 1 α ) b , α b + ( 1 α ) a ) < M ( a , b ) < Q ( β a + ( 1 β ) b , β b + ( 1 β ) a ) , Q t 1 , p ( a , b ) < M ( a , b ) < Q t 2 , p ( a , b )

hold for all a,b>0 with ab and p1/2, where Q t , p (a,b)= C p (ta+(1t)b,tb+(1t)a) A 1 p (a,b).

Inspired by inequalities (1.3) and (1.4), in this paper, we present the optimal upper and lower bounds for the Neuman-Sándor mean M(a,b) in terms of the geometric convex combinations of the first Seiffert mean P(a,b) and the quadratic mean Q(a,b). All numerical computations are carried out using Mathematica software.

2 Lemmas

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

Lemma 2.1 The double inequality

x+ x 3 3 2 x 5 15 < 1 + x 2 sinh 1 (x)<x+ x 3 3 2 x 5 15 + 8 x 7 105
(2.1)

holds for x(0,1).

Proof To show inequality (2.1), it suffices to prove that

ω 1 (x)= 1 + x 2 sinh 1 (x) ( x + x 3 3 2 x 5 15 ) >0
(2.2)

and

ω 2 (x)= 1 + x 2 sinh 1 (x) ( x + x 3 3 2 x 5 15 + 8 x 7 105 ) <0
(2.3)

for x(0,1).

From the expressions of ω 1 (x) and ω 2 (x), we get

ω 1 (0)= ω 2 (0)=0,
(2.4)
ω 1 (x)= x ω 1 ( x ) 1 + x 2 , ω 2 (x)= x ω 2 ( x ) 1 + x 2 ,
(2.5)

where

ω 1 ( x ) = sinh 1 ( x ) ( x 2 x 3 3 ) 1 + x 2 , ω 2 ( x ) = sinh 1 ( x ) ( x 2 x 3 3 + 8 x 5 15 ) 1 + x 2 , ω 1 ( 0 ) = ω 2 ( 0 ) = 0 ,
(2.6)
ω 1 (x)= 8 x 4 3 1 + x 2 >0
(2.7)

and

ω 2 (x)= 16 x 6 5 1 + x 2 <0
(2.8)

for x(0,1).

Therefore, inequality (2.2) follows from (2.4)-(2.7), and inequality (2.3) follows from (2.4)-(2.6) and (2.8). □

Lemma 2.2 The inequality

x 3 1 + x 2 > [ sinh 1 ( x ) ] 3

holds for x(0,1).

Proof Let x(0,1), then from (1.3) we have

M(1+x,1x)> A 2 / 3 (1+x,1x) Q 1 / 3 (1+x,1x).
(2.9)

Therefore, Lemma 2.2 follows from (2.9). □

Lemma 2.3 The inequality

1 x 2 arcsin(x)>x x 3 3 x 5 3
(2.10)

holds for x(0,0.7), and the inequality

1 x 2 arcsin(x)<x x 3 3 2 x 5 15
(2.11)

holds for x(0,1), where arcsin(x) is the inverse sine function.

Proof Let

φ 1 (x)= 1 x 2 arcsin(x)x+ x 3 3 + x 5 3 ,
(2.12)
φ 2 (x)= 1 x 2 arcsin(x)x+ x 3 3 + 2 x 5 15 .
(2.13)

Then simple computations lead to

φ 1 (0)= φ 2 (0)=0,
(2.14)
φ 1 (x)= x φ 1 ( x ) 1 x 2 , φ 2 (x)= x φ 2 ( x ) 1 x 2 ,
(2.15)

where

φ 1 ( x ) = ( x + 5 x 3 3 ) 1 x 2 arcsin ( x ) , φ 2 ( x ) = ( x + 2 x 3 3 ) 1 x 2 arcsin ( x ) .

Note that

φ 1 (0)= φ 2 (0)=0, φ 1 (0.7)=0.1327,
(2.16)
φ 1 (x)= x 2 ( 9 20 x 2 ) 3 1 x 2 ,
(2.17)
φ 2 (x)= 8 x 4 3 1 x 2 <0
(2.18)

for x(0,1).

From (2.17) we clearly see that φ 1 (x) is strictly increasing on (0,3 5 /10] and strictly decreasing on [3 5 /10,0.7). This in conjunction with (2.16) implies that

φ 1 (x)>0
(2.19)

for x(0,0.7).

Therefore, inequality (2.10) follows from (2.12), (2.14), (2.15) and (2.19), and inequality (2.11) follows from (2.12) and (2.14)-(2.16) together with (2.18). □

Lemma 2.4 Let

Φ(x)= 1 1 + x 2 sinh 1 ( x ) 1 x ( 1 + x 2 ) .

Then the inequality

Φ(x)> 2 x 3 34 x 3 45 + 754 x 5 945 x 7
(2.20)

holds for x(0,0.7), and

Φ(x)< 2 x 3 34 x 3 45 + 4 x 5 5
(2.21)

holds for x(0,1).

Proof To show inequalities (2.20) and (2.21), it suffices to prove that

ϕ 1 ( x ) : = x ( 1 + x 2 ) sinh 1 ( x ) [ Φ ( x ) ( 2 x 3 34 x 3 45 + 754 x 5 945 x 7 ) ] = x 1 + x 2 sinh 1 ( x ) x ( 1 + x 2 ) sinh 1 ( x ) ( 2 x 3 34 x 3 45 + 754 x 5 945 x 7 ) > 0
(2.22)

for x(0,0.7), and

ϕ 2 ( x ) : = x ( 1 + x 2 ) sinh 1 ( x ) [ Φ ( x ) ( 2 x 3 34 x 3 45 + 4 x 5 5 ) ] = x 1 + x 2 sinh 1 ( x ) x ( 1 + x 2 ) sinh 1 ( x ) ( 2 x 3 34 x 3 45 + 4 x 5 5 ) < 0
(2.23)

for x(0,1).

From the expressions of ϕ 1 (x) and ϕ 2 (x), one has

ϕ 1 (0)= ϕ 2 (0)=0,
(2.24)
ϕ 1 (x)= x 945 1 + x 2 ϕ 1 (x), ϕ 2 (x)= 2 x 45 1 + x 2 ϕ 2 (x),
(2.25)

where

ϕ 1 ( x ) = x ( 1 , 260 + 84 x 2 40 x 4 + 191 x 6 + 945 x 8 ) ϕ 1 ( x ) = 2 ( 630 168 x 2 + 120 x 4 764 x 6 4 , 725 x 8 ) 1 + x 2 sinh 1 ( x ) ,
(2.26)
ϕ 2 ( x ) = x ( 18 x 6 + x 4 2 x 2 30 ) ϕ 2 ( x ) = + 2 ( 15 4 x 2 + 3 x 4 + 72 x 6 ) 1 + x 2 sinh 1 ( x ) .
(2.27)

Note that

630 168 x 2 + 120 x 4 764 x 6 4 , 725 x 8 > 630 168 × ( 0.7 ) 2 764 × ( 0.7 ) 6 4 , 725 × ( 0.7 ) 8 = 185.4 > 0
(2.28)

for x(0,0.7).

Lemma 2.1 and equations (2.26)-(2.28) lead to

ϕ 1 ( x ) > x ( 1 , 260 + 84 x 2 40 x 4 + 191 x 6 + 945 x 8 ) 2 ( 630 168 x 2 + 120 x 4 764 x 6 4 , 725 x 8 ) ( x + x 3 3 2 x 5 15 + 8 x 7 105 ) = x 7 105 ( 157 , 311 + 1 , 151 , 003 x 2 + 307 , 438 x 4 120 , 076 x 6 + 75 , 600 x 8 ) > 0
(2.29)

for x(0,0.7), and

ϕ 2 ( x ) > x ( 18 x 6 + x 4 2 x 2 30 ) + 2 ( 15 4 x 2 + 3 x 4 + 72 x 6 ) ( x + x 3 3 2 x 5 15 ) = x 5 15 ( 5 + 2 , 476 x 2 + 708 x 4 288 x 6 ) > 0
(2.30)

for x(0,1).

Therefore, inequality (2.22) follows from (2.24), (2.25) and (2.29), and inequality (2.23) follows from (2.24), (2.25) and (2.30). □

Lemma 2.5 Let

ϒ(x)= 1 x ( 1 + x 2 ) 1 1 x 2 arcsin ( x ) .

Then the inequality

ϒ(x)> 4 x 3 + 34 x 3 45 3 x 5 2
(2.31)

holds for x(0,0.7), and

ϒ(x)< 4 x 3 + 34 x 3 45 8 x 5 9
(2.32)

holds for x(0,1).

Proof Let

ϵ 1 ( x ) : = x ( 1 + x 2 ) 1 x 2 arcsin ( x ) [ ϒ ( x ) + ( 4 x 3 34 x 3 45 + 3 x 5 2 ) ] = 1 x 2 arcsin ( x ) x ( 1 + x 2 ) + x ( 1 + x 2 ) 1 x 2 arcsin ( x ) ( 4 x 3 34 x 3 45 + 3 x 5 2 )
(2.33)

and

ϵ 2 ( x ) : = x ( 1 + x 2 ) 1 x 2 arcsin ( x ) [ ϒ ( x ) + ( 4 x 3 34 x 3 45 + 8 x 5 9 ) ] = 1 x 2 arcsin ( x ) x ( 1 + x 2 ) + x ( 1 + x 2 ) 1 x 2 arcsin ( x ) ( 4 x 3 34 x 3 45 + 8 x 5 9 ) .
(2.34)

An easy calculation gives rise to

ϵ 1 (0)= ϵ 2 (0)=0,
(2.35)
ϵ 1 (x)= x 90 ( 1 x 2 ) ϵ 1 (x), ϵ 2 (x)= x 45 ( 1 x 2 ) ϵ 2 (x),
(2.36)

where

ϵ 1 ( x ) = x ( 150 202 x 2 15 x 4 68 x 6 + 135 x 8 ) ϵ 1 ( x ) = + ( 150 152 x 2 + 142 x 4 + 611 x 6 1 , 215 x 8 ) 1 x 2 arcsin ( x ) ,
(2.37)
ϵ 2 ( x ) = x ( 1 x 2 ) ( 75 26 x 2 6 x 4 40 x 6 ) ϵ 2 ( x ) = + ( 75 76 x 2 94 x 4 + 278 x 6 360 x 8 ) 1 x 2 arcsin ( x ) .
(2.38)

Note that

150 152 x 2 + 142 x 4 + 611 x 6 1 , 215 x 8 > 150 152 × ( 0.7 ) 2 1 , 215 × ( 0.7 ) 8 = 5.477 > 0
(2.39)

for x(0,0.7).

It follows from (2.10), (2.37) and (2.39) that

ϵ 1 ( x ) > x ( 150 202 x 2 15 x 4 68 x 6 + 135 x 8 ) + ( 150 152 x 2 + 142 x 4 + 611 x 6 1 , 215 x 8 ) ( x x 3 3 x 5 3 ) = x 5 3 [ 1 , 183 4 + 709 ( 1 4 x 4 ) + 2 , 047 x 2 ( 1 2 x 2 ) + 604 x 6 + 1 , 215 x 8 ] > 0
(2.40)

for x(0,0.7).

We claim that

ϵ 2 (x)<0
(2.41)

for x(0,1). Indeed, let q(x)=7576 x 2 94 x 4 +278 x 6 360 x 8 , then q(0.8009)=0.000171 , q(0.80091)=0.00356 and

q (x)=4x [ 38 + 10 , 759 x 2 320 + 720 x 2 ( x 2 139 480 ) ] <0

for x(0,1). Therefore, there exists unique x 0 =0.80090(0,1) such that q(x)>0 for x(0, x 0 ) and q(x)0 for [ x 0 ,1). This in conjunction with (2.11) and (2.38) leads to

ϵ 2 ( x ) < x ( 1 x 2 ) ( 75 26 x 2 6 x 4 40 x 6 ) + ( 75 76 x 2 94 x 4 + 278 x 6 360 x 8 ) ( x x 3 3 2 x 5 15 ) = 2 x 5 15 [ 1 , 897 , 305 , 741 27 , 436 , 644 + 2 , 619 ( x 2 2 , 651 5 , 238 ) 2 + 2 x 4 ( 1 x 2 ) ( 491 + 180 x 2 ) ] < 0

for x(0, x 0 ) and ϵ 2 (x)x(1 x 2 )(7526 x 2 6 x 4 40 x 6 )<0 for x[ x 0 ,1).

Therefore, inequality (2.31) follows from (2.33), (2.35), (2.36) and (2.40), and inequality (2.32) follows from (2.33)-(2.36) and (2.41). □

Lemma 2.6 Let

μ(x)= 1 + 3 x 2 ( x + x 3 ) 2 1 ( 1 + x 2 ) [ sinh 1 ( x ) ] 2 x ( 1 + x 2 ) 3 / 2 sinh 1 ( x ) .

Then μ(x)<0.2 for x[0.7,1).

Proof Let

μ 1 (x)= 1 x 2 1 [ sinh 1 ( x ) ] 2 , μ 2 (x)= 2 1 + x 2 x sinh 1 ( x ) .

Then

μ(x)= μ 1 ( x ) 1 + x 2 + μ 2 ( x ) ( 1 + x 2 ) 3 / 2 .
(2.42)

Lemma 2.2 together with x> sinh 1 (x) gives μ 1 (x)<0 and

μ 1 (x)= 2 x 3 [ sinh 1 ( x ) ] 3 [ x 3 1 + x 2 ( sinh 1 ( x ) ) 3 ] >0

for x(0,1). This in turn implies that

[ μ 1 ( x ) 1 + x 2 ] = μ 1 ( x ) ( 1 + x 2 ) 2 x μ 1 ( x ) ( 1 + x 2 ) 2 >0
(2.43)

for x(0,1).

On the other hand, from the expression of μ 2 (x), we get

μ 2 (1)=0.2796>0,
(2.44)
μ 2 (x)= 2 x ( 1 + x 2 ) 3 / 2 + μ 2 ( x ) [ sinh 1 ( x ) ] 2 ,
(2.45)

where

μ 2 (x)= x 1 + x 2 sinh 1 (x),
(2.46)
μ 2 (0)=0,
(2.47)
μ 2 (x)= x 2 ( 1 + x 2 ) 3 / 2 <0
(2.48)

for x(0,1).

From (2.44)-(2.48) we clearly see that μ 2 (x)<0 and μ 2 (x)>0 for x(0,1). This in turn implies that

[ μ 2 ( x ) ( 1 + x 2 ) 3 / 2 ] = μ 2 ( x ) ( 1 + x 2 ) 3 / 2 3 x 1 + x 2 μ 2 ( x ) ( 1 + x 2 ) 3 <0
(2.49)

for x(0,1).

Equation (2.42) and inequalities (2.43) and (2.49) lead to the conclusion that

μ(x) μ 1 ( 1 ) 2 + μ 2 ( 0.7 ) [ 1 + ( 0.7 ) 2 ] 3 / 2 =0.167<0.2

for x[0.7,1). □

Lemma 2.7 Let

ν(x)= 1 + 3 x 2 ( x + x 3 ) 2 + 1 ( 1 x 2 ) arcsin 2 ( x ) x ( 1 x 2 ) 3 / 2 arcsin ( x ) .

Then ν(x)<1.48 for x[0.7,1).

Proof Differentiating ν(x) yields

ν (x)= ( x + x 3 ) 3 arcsin ( x ) ν 1 ( x ) + ( 1 x 2 ) ν 2 ( x ) x 3 ( 1 x 2 ) 5 / 2 ( 1 + x 2 ) 3 arcsin 3 ( x ) ,
(2.50)

where

ν 1 (x)=3x 1 x 2 ( 1 + 2 x 2 ) arcsin(x),
(2.51)
ν 2 (x)=2 ( 1 + 3 x 2 + 6 x 4 ) [ 1 x 2 arcsin ( x ) ] 3 2 ( x + x 3 ) 3 .
(2.52)

Equation (2.51) leads to

ν 1 (0.7)=0.03558,
(2.53)
ν 1 (x)= 2 8 x 2 4 x 1 x 2 arcsin ( x ) 1 x 2 <0
(2.54)

for x[0.7,1).

Therefore,

ν 1 (x)<0
(2.55)

for x[0.7,1) follows from (2.53) and (2.54).

It follows from (2.52) and (2.11) that

ν 2 ( x ) < 2 ( 1 + 3 x 2 + 6 x 4 ) ( x x 3 3 ) 3 2 ( x + x 3 ) 3 = 2 x 5 27 ( 27 9 x 2 + 163 x 4 51 x 6 + 6 x 8 ) < 0
(2.56)

for x[0.7,1).

Equation (2.50) together with inequalities (2.55) and (2.56) leads to the conclusion that ν(x) is strictly decreasing on [0.7,1). This in turn implies that

ν(x)ν(0.7)=1.48798<1.48

for x[0.7,1). □

Lemma 2.8 Let λ 0 =[2log(log(1+ 2 ))+log2]/[2logπlog2]=0.2760 , and Θ(x)=Φ(x)+ λ 0 ϒ(x), where Φ(x) and ϒ(x) are defined as in Lemmas 2.4 and 2.5, respectively. Then the function Θ(x) is strictly decreasing on [0.7,1).

Proof Let μ(x) and ν(x) be defined as in Lemmas 2.6 and 2.7, respectively. Then differentiating Θ(x) yields

Θ (x)= Φ (x)+ λ 0 ϒ (x)=μ(x)+ λ 0 ν(x)<0.21.48 λ 0 =0.208<0

for x[0.7,1). This in turn implies that Θ(x) is strictly decreasing on [0.7,1). □

3 Main result

Theorem 3.1 The double inequality

P α (a,b) Q 1 α (a,b)<M(a,b)< P β (a,b) Q 1 β (a,b)

holds for all a,b>0 with ab if and only if α1/2 and β[2log(log(1+ 2 ))+log2]/[2logπlog2]=0.2760 .

Proof Since P(a,b), M(a,b) and Q(a,b) are symmetric and homogeneous of degree 1, without loss of generality, we assume that a>b. Let p(0,1), λ 0 =[2log(log(1+ 2 ))+log2]/[2logπlog2] and x=(ab)/(a+b). Then x(0,1),

P ( a , b ) A ( a , b ) = x arcsin ( x ) , M ( a , b ) A ( a , b ) = x sinh 1 ( x ) , Q ( a , b ) A ( a , b ) = 1 + x 2 , log [ Q ( a , b ) ] log [ M ( a , b ) ] log [ Q ( a , b ) ] log [ P ( a , b ) ] = log ( 1 + x 2 ) 2 log x + 2 log [ sinh 1 ( x ) ] log ( 1 + x 2 ) 2 log x + 2 log [ arcsin ( x ) ]
(3.1)

and

lim x 0 + log ( 1 + x 2 ) 2 log x + 2 log [ sinh 1 ( x ) ] log ( 1 + x 2 ) 2 log x + 2 log [ arcsin ( x ) ] = 1 2 ,
(3.2)
lim x 1 log ( 1 + x 2 ) 2 log x + 2 log [ sinh 1 ( x ) ] log ( 1 + x 2 ) 2 log x + 2 log [ arcsin ( x ) ] = λ 0 .
(3.3)

The difference between the convex combination of log[P(a,b)], log[Q(a,b)] and log[M(a,b)] is given by

p log [ P ( a , b ) ] + ( 1 p ) log [ Q ( a , b ) ] log [ M ( a , b ) ] = p log [ x arcsin ( x ) ] + 1 p 2 log ( 1 + x 2 ) log [ x sinh 1 ( x ) ] : = D p ( x ) .
(3.4)

Equation (3.4) leads to

D p ( 0 + ) =0, D p ( 1 ) =log [ 2 log ( 1 + 2 ) ] plog ( π 2 ) , D λ 0 ( 1 ) =0,
(3.5)
D p (x)= p 1 x 2 arcsin ( x ) ( 1 p ) x ( 1 + x 2 ) + 1 1 + x 2 sinh 1 ( x ) =Φ(x)+pϒ(x),
(3.6)

where Φ(x) and ϒ(x) are defined as in Lemmas 2.4 and 2.5, respectively.

From Lemmas 2.4 and 2.5, we clearly see that

D 1 / 2 ( x ) = Φ ( x ) + 1 2 ϒ ( x ) < 2 x 3 34 x 3 45 + 4 x 5 5 1 2 ( 4 x 3 34 x 3 45 + 8 x 5 9 ) = 16 x 3 45 ( 17 16 x 2 ) < 0
(3.7)

for x(0,1), and

D λ 0 ( x ) = Φ ( x ) + λ 0 ϒ ( x ) > 2 x 3 34 x 3 45 + 754 x 5 945 x 7 λ 0 ( 4 x 3 34 x 3 45 + 3 x 5 2 ) = x [ 2 ( 1 2 λ 0 ) 3 34 ( 1 λ 0 ) 45 x 2 + ( 754 945 3 λ 0 2 ) x 4 x 6 ] : = x F λ 0 ( x ) > 0
(3.8)

for x(0,0.7).

Note that

F λ 0 (0)=2(12 λ 0 )/3>0, F λ 0 (0.7)=0.00513>0
(3.9)

and

F λ 0 ( x ) = 30 [ ( x 2 1 , 508 2 , 835 λ 0 9 , 450 ) 2 + 2 , 224 , 136 + 4 , 052 , 160 λ 0 8 , 037 , 225 λ 0 2 89 , 302 , 500 ] < 0
(3.10)

for x(0,0.7).

Inequalities (3.8)-(3.10) lead to the conclusion that

D λ 0 (x)>0
(3.11)

for x(0,0.7).

It follows from Lemma 2.8 and (3.6) that D λ 0 (x) is strictly decreasing in [0.7,1). Then from (3.11) and D λ 0 (0.7)=0.0626 together with D λ 0 ( 1 )=, we know that there exists x (0.7,1) such that D λ 0 (x) is strictly increasing on (0, x ] and strictly decreasing on [ x ,1). This in conjunction with (3.5) implies that

D λ 0 (x)>0
(3.12)

for x(0,1).

Equations (3.4), (3.5), (3.7) and (3.12) lead to the conclusion that

M(a,b)< P λ 0 (a,b) Q 1 λ 0 (a,b)
(3.13)

and

M(a,b)> P 1 / 2 (a,b) Q 1 / 2 (a,b).
(3.14)

Therefore, Theorem 3.1 follows from (3.13) and (3.14) together with the following statements:

  • If α<1/2, then (3.1) and (3.2) imply that there exists δ 1 (0,1) such that M(a,b)< P α (a,b) Q 1 α (a,b) for all a,b>0 with (ab)/(a+b)(0, δ 1 ).

  • If β> λ 0 , then (3.1) and (3.3) imply that there exists δ 2 (0,1) such that M(a,b)> P β (a,b) Q 1 β (a,b) for all a,b>0 with (ab)/(a+b)(1 δ 2 ,1).

 □

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Acknowledgements

This research was supported by the Natural Science Foundation of China under Grants 11071069 and 11171307, and the Natural Science Foundation of Zhejiang Province under Grants LY13H070004 and LY13A010004.

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Correspondence to Yu-Ming Chu.

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The authors declare that they have no competing interests.

Authors’ contributions

W-MG provided the main idea and carried out the proof of Theorem 3.1. X-HS carried out the proof of Lemmas 2.1-2.5. Y-MC carried out the proof of Lemmas 2.6-2.8. All authors read and approved the final manuscript.

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Gong, WM., Shen, XH. & Chu, YM. Optimal bounds for the Neuman-Sándor mean in terms of the first Seiffert and quadratic means. J Inequal Appl 2013, 552 (2013). https://doi.org/10.1186/1029-242X-2013-552

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Keywords

  • Neuman-Sándor mean
  • first Seiffert mean
  • quadratic mean