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Sharp bounds for the Neuman mean in terms of the quadratic and second Seiffert means

Abstract

In this paper, we prove that α=0 and β= 3 π 4 log ( 2 + 3 ) ( 2 π 4 ) log ( 2 + 3 ) =0.29758 are the best possible constants such that the double inequality

αQ(a,b)+(1α)T(a,b)< S C A (a,b)<βQ(a,b)+(1β)T(a,b)

holds for all a,b>0 with ab, where Q(a,b)= ( a 2 + b 2 ) / 2 ,

S C A (a,b)= ( a b ) 3 ( a 2 + b 2 ) + 2 a b 2 ( a + b ) sinh 1 ( ( a b ) 3 ( a 2 + b 2 ) + 2 a b ( a + b ) 2 )

and T(a,b)=(ab)/[2arctan((ab)/(a+b))] are the quadratic, Neuman and second Seiffert means of a and b, respectively.

MSC:26E60.

1 Introduction

For a,b>0 with ab, the Neuman mean S C A (a,b) [1, 2] derived from the Schwab-Borchardt mean [3, 4], the quadratic mean Q(a,b) and the second Seiffert mean T(a,b) [5] are given by

S C A (a,b)= ( a b ) 3 ( a 2 + b 2 ) + 2 a b 2 ( a + b ) sinh 1 ( ( a b ) 3 ( a 2 + b 2 ) + 2 a b ( a + b ) 2 ) ,
(1.1)
Q(a,b)= a 2 + b 2 2
(1.2)

and

T(a,b)= a b 2 arctan ( a b a + b ) ,
(1.3)

respectively, where sinh 1 (x)=log(x+ 1 + x 2 ) is the inverse hyperbolic sine function. Recently, the Neuman, quadratic and second Seiffert means have been the subject of intensive research. In particular, many remarkable inequalities for these means can be found in the literature [14, 615].

Let A(a,b)=(a+b)/2 and C(a,b)=( a 2 + b 2 )/(a+b) be the arithmetic and contraharmonic means of a and b, respectively. Then Neuman [1] proved that the inequalities

A(a,b)<T(a,b)< S C A (a,b)<Q(a,b)<C(a,b)
(1.4)

hold for any a,b>0 with ab.

In [1, 2], Neuman found that α 1 =[ 3 log(2+ 3 )]/log(2+ 3 )=0.315, β 1 =1/3, α 2 =1/3, β 2 =[log32log(log(2+ 3 ))]/(2log2)=0.395, α 3 =2log(2+ 3 )/31=0.520 and β 3 =2/3 are the best possible constants such that the double inequalities

α 1 C ( a , b ) + ( 1 α 1 ) A ( a , b ) < S C A ( a , b ) < β 1 C ( a , b ) + ( 1 β 1 ) A ( a , b ) , C α 2 ( a , b ) A 1 α 2 ( a , b ) < S C A ( a , b ) < C β 2 ( a , b ) A 1 β 2 ( a , b )

and

α 3 A ( a , b ) + 1 α 3 C ( a , b ) < 1 S C A ( a , b ) < β 3 A ( a , b ) + 1 β 3 C ( a , b )

hold for any a,b>0 with ab.

He et al. [16] proved that α=1/2+ 3 / log ( 2 + 3 ) 1 /2 and β=1/2+ 3 /6 are the best possible constants in [1/2,1] such that the double inequality

C [ α a + ( 1 α ) b , α b + ( 1 α ) a ] < S C A (a,b)<C [ β a + ( 1 β ) b , β b + ( 1 β ) a ]

holds for any a,b>0 with ab.

In [17, 18], the authors proved that the double inequalities

α [ 1 3 C ( a , b ) + 2 3 A ( a , b ) ] + ( 1 α ) C 1 / 3 ( a , b ) A 2 / 3 ( a , b ) < S C A ( a , b ) < β [ 1 3 C ( a , b ) + 2 3 A ( a , b ) ] + ( 1 β ) C 1 / 3 ( a , b ) A 2 / 3 ( a , b )

and

λA(a,b)+(1λ)Q(a,b)< S C A (a,b)<μA(a,b)+(1μ)Q(a,b)

hold for any a,b>0 with ab if and only if α 3 [ 2 3 log ( 2 + 3 ) 3 ] ( 3 2 3 4 ) log ( 2 + 3 ) =0.7528, β4/5, λ1/3 and μ 2 log ( 2 + 3 ) 3 ( 2 1 ) log ( 2 + 3 ) =0.2390.

The main purpose of this paper is to present the best possible constants α and β such that the double inequality

αQ(a,b)+(1α)T(a,b)< S C A (a,b)<βQ(a,b)+(1β)T(a,b)

holds for any a,b>0 with ab. All numerical computations are carried out using MATHEMATICA software.

2 Lemmas

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

Lemma 2.1 The double inequality

2 x 3 + 16 x 3 45 2 x 5 7 < x ( 1 + x 2 ) arctan 2 x 1 arctan x < 2 x 3 + 16 x 3 45
(2.1)

holds for x(0,0.6).

Proof Let

ϕ 1 (x)=x ( 1 + x 2 ) arctanx+ ( 2 x 3 16 x 3 45 + 2 x 5 7 ) ( 1 + x 2 ) arctan 2 x,
(2.2)
ϕ 2 (x)=x ( 1 + x 2 ) arctanx+ ( 2 x 3 16 x 3 45 ) ( 1 + x 2 ) arctan 2 x.
(2.3)

Then we only need to show that ϕ 1 (x)>0 and ϕ 2 (x)<0 for x(0,0.6).

Taking the differentiation of ϕ 1 (x) yields

ϕ 1 (0)=0,
(2.4)
ϕ 1 (x)= 2 arctan x 315 ϕ 1 (x),
(2.5)

where

ϕ 1 (x)= ( 105 + 147 x 2 55 x 4 + 315 x 6 ) arctanxx ( 105 + 112 x 2 90 x 4 ) ,
(2.6)
ϕ 1 (0)=0,
(2.7)
ϕ 1 (x)= x 1 + x 2 ϕ 1 (x),
(2.8)

where

ϕ 1 (x)=2 ( 147 + 37 x 2 + 835 x 4 + 945 x 6 ) arctanxx ( 294 59 x 2 765 x 4 ) .
(2.9)

It is well known that the inequality

arctanx>x x 3 3
(2.10)

holds for all x(0,1).

Equation (2.9) and inequality (2.10) lead to the conclusion that

ϕ 1 ( x ) > 2 ( 147 + 37 x 2 + 835 x 4 + 945 x 6 ) ( x x 3 3 ) x ( 294 59 x 2 765 x 4 ) = x 3 3 [ 105 + 7 , 231 x 2 + 2 , 110 x 4 + 1 , 890 x 4 ( 1 x 2 ) ] > 0
(2.11)

for x(0,0.6).

Therefore, ϕ 1 (x)>0 for x(0,0.6) follows easily from (2.4)-(2.8) and (2.11).

Differentiating ϕ 2 (x) leads to

ϕ 2 (0)=0,
(2.12)
ϕ 2 (x)= 2 arctan x 45 ϕ 2 (x),
(2.13)

where

ϕ 2 (x)= ( 15 x + 16 x 3 ) ( 15 + 21 x 2 40 x 4 ) arctanx.
(2.14)

It is well known that the inequality

arctanx<x x 3 3 + x 5 5
(2.15)

holds for all x(0,1).

Equation (2.14) and inequality (2.15) lead to the conclusion that

ϕ 2 ( x ) > ( 15 x + 16 x 3 ) ( 15 + 21 x 2 40 x 4 ) ( x x 3 3 + x 5 5 ) = x 5 15 ( 660 263 x 2 + 120 x 4 ) > 0
(2.16)

for x(0,0.6).

Therefore, ϕ 2 (x)<0 for x(0,0.6) follows from (2.12) and (2.13) together with (2.16). □

Lemma 2.2 The double inequality

x 1 + x 2 + x ( 1 + x 2 ) arctan 2 x 1 arctan x > x 3 x 3 6
(2.17)

holds for x(0,0.6).

Proof A simple computation leads to

( 1 x 2 2 + x 4 4 ) 2 ( 1 + x 2 ) = 1 x 4 16 [ 8 ( 2 2 + x ) ( 2 2 x ) + 2 x 4 + x 4 ( 1 x 2 ) ] < 1

for x(0,0.6). This implies

x 1 + x 2 >x x 3 2 + x 5 4
(2.18)

for x(0,0.6).

From Lemma 2.1 and (2.18) we clearly see that

x 1 + x 2 + x ( 1 + x 2 ) arctan 2 x 1 arctan x > ( x x 3 2 + x 5 4 ) + ( 2 x 3 + 16 x 3 45 2 x 5 7 ) = x 3 13 x 3 90 x 5 28 = x 3 x 3 6 + x 3 28 ( 28 45 + x ) ( 28 45 x ) > x 3 x 3 6

for x(0,0.6). □

Lemma 2.3 The inequality

x [ sinh 1 ( x 2 + x 2 ) ] 2 1 + x 2 2 + x 2 sinh 1 ( x 2 + x 2 ) > x 3 + 2 x 3 45 x 5 63
(2.19)

holds for x(0,1).

Proof Let

φ ( x ) = x 2 + x 2 ( 1 + x 2 ) sinh 1 ( x 2 + x 2 ) + ( x 3 2 x 3 45 + x 5 63 ) [ sinh 1 ( x 2 + x 2 ) ] 2 2 + x 2 .
(2.20)

Then we only need to show that φ(x)>0 for x(0,1).

Differentiating (2.20) leads to

φ(0)=0,
(2.21)
φ (x)= 2 x sinh 1 ( x 2 + x 2 ) 315 ( 1 + x 2 ) φ 1 (x),
(2.22)

where

φ 1 ( x ) = 105 133 x 2 18 x 4 + 10 x 6 + 3 ( 35 + 56 x 2 + 20 x 4 + 4 x 6 + 5 x 8 ) sinh 1 ( x 2 + x 2 ) x 2 + x 2 .
(2.23)

We claim that

sinh 1 ( x 2 + x 2 ) x 2 + x 2 >1 x 2 3 + 2 x 4 15 2 x 6 35
(2.24)

for x(0,1). Indeed, let

ω(x)= sinh 1 ( x 2 + x 2 ) x 2 + x 2 ( 1 x 2 3 + 2 x 4 15 2 x 6 35 ) ,

then ω(x)>0 for x(0,1) follows from the fact that

ω(0)=0, ω (x)= 16 x 8 35 2 + x 2 >0.

It follows from (2.23) and (2.24) that

φ 1 ( x ) > 105 133 x 2 18 x 4 + 10 x 6 + 3 ( 35 + 56 x 2 + 20 x 4 + 4 x 6 + 5 x 8 ) ( 1 x 2 3 + 2 x 4 15 2 x 6 35 ) = x 6 35 [ 644 + 90 x 2 + 16 x 6 + ( 1 x 2 ) ( 239 x 2 + 30 x 6 ) ] > 0
(2.25)

for x(0,1).

Therefore, φ(x)>0 for x(0,1) follows from (2.21) and (2.22) together with (2.25). □

Lemma 2.4 The inequality

arctanx> π 4 + x 1 2 2 ( x 1 ) 2 7 > π 4 + 3 ( x 1 ) 4
(2.26)

holds for x[0.55,1).

Proof Let

ν(x)=arctanx [ π 4 + x 1 2 2 ( x 1 ) 2 7 ] .
(2.27)

Then simple computations lead to

ν(0.55)=0.00030219,ν(1)=0,
(2.28)
ν (x)= ν 1 ( x ) 14 ( 1 + x 2 ) ,
(2.29)
ν 1 (x)=1+8x15 x 2 +8 x 3 ,
(2.30)
ν 1 (0.55)=0.1935, ν 1 (1)=0,
(2.31)
ν 1 (x)=24 ( x 15 33 24 ) ( x 15 + 33 24 ) .
(2.32)

From (2.32) and (15 33 )/24=0.385643<0.55 together with 0.55<(15+ 33 )/24=0.864357<1, we clearly see that ν 1 (x) is strictly decreasing on [0.55,(15+ 33 )/24] and strictly increasing on [(15+ 33 )/24,1). This in conjunction with (2.31) implies that there exists x 1 (0.55,1) such that ν 1 (x)>0 for x[0.55, x 1 ) and ν 1 (x)<0 for x( x 1 ,1). Then equation (2.29) leads to the conclusion that ν(x) is strictly increasing on [0.55, x 1 ] and strictly decreasing on [ x 1 ,1].

Therefore, ν(x)>0 for x[0.55,1) follows from (2.28) and the piecewise monotonicity of ν(x). Moreover, the second inequality in (2.26) follows from

x 1 2 2 ( x 1 ) 2 7 > 3 ( x 1 ) 4 + ( 1 x ) ( 8 x 1 ) 28 > 3 ( x 1 ) 4 .

 □

Lemma 2.5 The inequality

xarctanx< 7 20 x arctan 2 x
(2.33)

holds for x[0.55,1).

Proof Let

μ(x)=xarctanx 7 20 x arctan 2 x.
(2.34)

Then it suffices to show μ(x)<0 for x[0.55,1).

Differentiating μ(x) yields

μ (x)= μ 1 ( x ) 20 ( 1 + x 2 ) ,
(2.35)

where

μ 1 (x)=20 x 2 14xarctanx7 arctan 2 x7 x 2 arctan 2 x.
(2.36)

It is well known that

arctanx>x x 3 3 + x 5 5 x 7 7
(2.37)

for x(0,1).

For x[0.55,0.7], it follows from (2.36) and (2.37) that

μ 1 ( x ) < 20 x 2 14 x ( x x 3 3 + x 5 5 x 7 7 ) 7 ( x x 3 3 + x 5 5 x 7 7 ) 2 7 x 2 ( x x 3 3 + x 5 5 x 7 7 ) 2 = x 2 1 , 575 μ ( x 2 ) ,
(2.38)

where

μ ( x ) = 1 , 575 + 3 , 675 x 2 , 695 x 2 + 2 , 135 x 3 + 3 , 129 x 4 861 x 5 + 405 x 6 225 x 7 ,
(2.39)
μ (0.49)=9.99966.
(2.40)

Differentiating μ (x) yields

μ ( x ) = ( 3 , 675 5 , 390 x + 6 , 405 x 2 ) + ( 12 , 516 x 3 4 , 305 x 4 ) + ( 2 , 430 x 5 1 , 575 x 6 ) > 0
(2.41)

for x[0.3025,0.49].

Therefore, μ (x)<0 for x[0.3025,0.49] follows from (2.40) and (2.41). This in conjunction with (2.35) and (2.38) implies that μ(x) is strictly decreasing on [0.55,0.7]. Therefore, we get μ(x)μ(0.55)=0.00151709<0 for x[0.55,0.7].

It follows from Lemma 2.4 that

μ(x)<x [ π 4 + x 1 2 2 ( x 1 ) 2 7 ] 7 20 [ π 4 + x 1 2 2 ( x 1 ) 2 7 ] 2 = μ 2 ( x ) 2 , 240
(2.42)

for x(0.7,1), where

μ 2 ( x ) = ( 1 , 760 560 π ) + ( 308 π 49 π 2 644 ) x + ( 1 , 960 420 π ) x 2 + ( 112 π 1 , 252 ) x 3 + 480 x 4 64 x 5 .
(2.43)

Differentiating μ 2 (x) yields

μ 2 (0.7)=1.68877, μ 2 (1)=2.9025,
(2.44)
μ 2 ( x ) = ( 644 + 308 π 49 π 2 ) + ( 3 , 920 840 π ) x + ( 336 π 3 , 756 ) x 2 + 1 , 920 x 3 320 x 4 ,
(2.45)
μ 2 (0.7)=4.73674, μ 2 (1)=20.6372,
(2.46)
μ 2 (x)=8 ( 490 105 π 939 x + 84 π x + 720 x 2 160 x 3 ) ,
(2.47)
μ 2 (0.7)=116.173, μ 2 (1)=360.212,
(2.48)
μ 2 ( x ) = 24 ( 28 π 313 + 480 x 160 x 2 ) > 24 ( 28 π 313 + 480 × 0.7 160 × ( 0.7 ) 2 ) = 781.55 > 0 .
(2.49)

It follows from (2.48) and (2.49) that there exists x 2 (0.7,1) such that μ 2 (x) is strictly decreasing on (0.7, x 2 ] and strictly increasing on [ x 2 ,1). This in conjunction with (2.46) implies that there exists x 3 (0.7,1) such that μ 2 (x) is strictly decreasing on (0.7, x 3 ] and strictly increasing on [ x 3 ,1). From (2.44) and the piecewise monotonicity of μ 2 (x), we know that μ 2 (x)<0 for x(0.7,1); this in conjunction with (2.42) implies μ(x)<0 for x(0.7,1). □

Lemma 2.6 The function

σ(x)= 1 + x 2 arctan 3 x 2 ( x arctan x ) ( 1 + x 2 ) 2 arctan 3 x

is strictly decreasing on [0.55,1). Moreover, σ(x)<0.236 for x[0.55,1).

Proof Differentiating σ(x) yields

σ (x)= σ 1 ( x ) ( 1 + x 2 ) 3 arctan 4 x ,
(2.50)

where

σ 1 (x)=6(xarctanx)+6 x 2 arctanx8x arctan 2 x3x 1 + x 2 arctan 4 x.
(2.51)

From Lemma 2.5 and (2.51) we clearly see that

σ 1 (x)<6 x 2 arctanx 59 10 x arctan 2 x3x arctan 4 x=xarctanx σ 2 (x)
(2.52)

for x[0.55,1), where

σ 2 (x)=6x 59 10 arctanx3 arctan 3 x.
(2.53)

Differentiating σ 2 (x) leads to

σ 2 (0.55)=0.0482086, σ 2 (1)=0.0872684,
(2.54)
σ 2 (x)= σ 3 ( x ) 10 ( 1 + x 2 ) ,
(2.55)
σ 3 (x)=1+60 x 2 90 arctan 2 x,
(2.56)
σ 3 (0.55)=3.60662, σ 3 (1)=5.48348,
(2.57)
σ 3 (x)= 60 σ 4 ( x ) 1 + x 2 ,
(2.58)
σ 4 (x)=2x+2 x 3 3arctanx,
(2.59)
σ 4 (0.55)=0.0757796, σ 4 (1)=1.64381,
(2.60)
σ 4 (x)= 1 + 8 x 2 + 6 x 4 1 + x 2 >0.
(2.61)

It follows from (2.58)-(2.61) that there exists x 4 (0.55,1) such that σ 3 (x) is strictly decreasing on (0.55, x 4 ] and strictly increasing on [ x 4 ,1). This in conjunction with (2.55)-(2.57) implies that there exists x 5 (0.55,1) such that σ 2 (x) is strictly decreasing on (0.55, x 5 ] and strictly increasing on [ x 5 ,1). Then from (2.54) we clearly see that σ 2 (x)<0 for x(0.55,1).

Therefore, it follows from (2.50) and (2.52) that σ(x) is strictly decreasing on [0.55,1). Moreover, σ(x)σ(0.55)=0.235477<0.236 for x[0.55,1). □

Lemma 2.7 The function

κ(x)= 2 ( 4 + 3 x 2 ) sinh 1 ( x 2 + x 2 ) 8 x 2 + x 2 ( 2 + x 2 ) [ sinh 1 ( x 2 + x 2 ) ] 3

is strictly decreasing on [0.55,1). Moreover, κ(x)<0.771 for x[0.55,1).

Proof Simple computations lead to

κ(0.55)=0.770758,
(2.62)
κ (x)= 8 κ 1 ( x ) ( 2 + x 2 ) 2 [ sinh 1 ( x 2 + x 2 ) ] 4 ,
(2.63)

where

κ 1 (x)=6x ( 2 + x 2 ) 3 ( 2 + x 2 ) 3 / 2 sinh 1 ( x 2 + x 2 ) +x [ sinh 1 ( x 2 + x 2 ) ] 2 .
(2.64)

We claim that

2 x x 3 6 2 < sinh 1 ( x 2 + x 2 ) < 2 x
(2.65)

for x(0,1). Indeed, let

η 1 (x)= sinh 1 ( x 2 + x 2 ) 2 x+ x 3 6 2 ,
(2.66)
η 2 (x)= sinh 1 ( x 2 + x 2 ) 2 x.
(2.67)

Then we clearly see that

η 1 (0)= η 2 (0)=0,
(2.68)
η 1 (x)= 2 2 + x 2 + 2 4 x 2 2 ,
(2.69)
η 2 (x)= 2 2 + x 2 2 <0,
(2.70)
η 1 (0)=0,
(2.71)
η 1 (x)=x ( 1 2 2 ( 2 + x 2 ) 3 / 2 ) >0.
(2.72)

Therefore, the double inequality (2.65) follows easily from (2.68)-(2.72).

Equation (2.64) and inequality (2.65) imply that

κ 1 (x)<6x ( 2 + x 2 ) 3 ( 2 + x 2 ) 3 / 2 ( 2 x x 3 6 2 ) +x ( 2 x ) 2 = x 4 κ 2 (x),
(2.73)

where

κ 2 (x)=16 ( 3 + 2 x 2 ) 2 ( 12 x 2 ) ( 2 + x 2 ) 3 / 2 .
(2.74)

Let u= 2 + x 2 , then x 2 = u 2 2, 2 <u< 3 and κ 2 (x) becomes

κ ˜ (u)=16+32 u 2 14 2 u 3 + 2 u 5 .
(2.75)

Equation (2.75) leads to

κ ˜ ( 2 )=0,
(2.76)
κ ˜ (u)=u ( 64 42 2 u + 5 2 u 3 ) =u κ ˜ 1 (u),
(2.77)
κ ˜ 1 (u)=6442 2 u+5 2 u 3 , κ ˜ 1 ( 2 )=0, κ ˜ 1 ( 3 )=2.1362,
(2.78)
κ ˜ 1 (u)=15 2 ( u 14 5 ) ( u + 14 5 ) .
(2.79)

From (2.79) we clearly see that κ ˜ 1 (u)<0 for u( 2 , 14 / 5 ) and κ ˜ 1 (u)>0 for u( 14 / 5 , 3 ). This in conjunction with (2.77) implies that κ ˜ (u) is strictly decreasing on ( 2 , 14 / 5 ] and strictly increasing on [ 14 / 5 , 3 ). Thus κ ˜ (u)<0 for u( 2 , 3 ) follows from (2.78) and the piecewise monotonicity of κ ˜ (u).

Therefore, κ 2 (x)= κ ˜ (u)<0 follows from (2.76). This in conjunction with (2.63) and (2.73) implies that κ(x) is strictly decreasing on [0.55,1). Moreover, it follows from (2.62) that κ(x)κ(0.55)=0.770758<0.771 for x[0.55,1). □

Lemma 2.8 The function

τ(x)= 2 ( x arctan x ) ( 1 + x 2 ) 2 arctan 3 x 2 x ( 3 + x 2 ) ( 2 + x 2 ) 3 / 2 sinh 1 ( x 2 + x 2 ) <0.88

for x[0.55,1).

Proof We first prove

2 + x 2 sinh 1 ( x 2 + x 2 ) <2x+ x 3 3
(2.80)

for x(0,1). Let

ε(x)= 2 + x 2 sinh 1 ( x 2 + x 2 ) ( 2 x + x 3 3 ) .

Then ε(x)<0 follows from ε(0)=0 and the fact that

ε (x)= x 2 + x 2 ( sinh 1 ( x 2 + x 2 ) x 2 + x 2 ) < x 2 + x 2 ( 2 x x 2 + x 2 ) <0,

where the second term follows from (2.65).

From Lemma 2.5 and (2.10) we clearly see that

x arctan x arctan 3 x < 7 x 20 arctan x < 21 20 ( 3 x 2 )
(2.81)

for x[0.55,1).

It follows from (2.80) and (2.81) that

τ(x)< 21 10 ( 1 + x 2 ) 2 ( 3 x 2 ) 6 ( 3 + x 2 ) ( 2 + x 2 ) ( 6 + x 2 ) =: τ 1 (x)
(2.82)

for x[0.55,1).

Simple computation yields

τ 1 (0.55)=0.906585, τ 1 (1)=0.880357,
(2.83)
τ 1 (x)= 3 x 5 ( x 2 3 ) 2 ( 1 + x 2 ) 3 ( 2 + x 2 ) 2 ( 6 + x 2 ) 2 τ ˜ (x),
(2.84)

where

τ ˜ ( x ) = 2 , 880 + 2 , 424 x 2 + 6 , 052 x 4 + 1 , 468 x 6 939 x 8 219 x 10 + 60 x 12 + 20 x 14 ,
(2.85)
τ ˜ (0.55)=1,560.68, τ ˜ (1)=5,986,
(2.86)
τ ˜ ( x ) = 2 x ( 2 , 424 + 12 , 104 x 2 + 4 , 404 x 4 3 , 756 x 6 1 , 095 x 8 + 360 x 10 + 140 x 12 ) > 0 .
(2.87)

From (2.85)-(2.87) we know that there exists x 6 (0.55,1) such that τ ˜ (x)<0 for x(0.55, x 6 ) and τ ˜ (x)>0 for x( x 6 ,1). This in conjunction with (2.84) implies that τ 1 (x) is strictly decreasing on [0.55, x 6 ) and strictly increasing on [ x 6 ,1).

Therefore, τ(x)< τ 1 (x)max{ τ 1 (0.55), τ 1 (1)}=0.880357<0.88 follows from (2.83) and the piecewise monotonicity of τ 1 (x). □

3 Main result

Theorem 3.1 The double inequality

αQ(a,b)+(1α)T(a,b)< S C A (a,b)<βQ(a,b)+(1β)T(a,b)
(3.1)

holds for all a,b>0 with ab if and only if α0 and β β 0 = 3 π 4 log ( 2 + 3 ) ( 2 π 4 ) log ( 2 + 3 ) =0.29758.

Proof Since the Neuman mean S C A (a,b), the quadratic mean Q(a,b) and the second Seiffert mean T(a,b) are symmetric and homogeneous of degree 1, without loss of generality, we assume that a>b. Let v=(ab)/(a+b)(0,1), then from (1.1)-(1.3) one has

S C A (a,b)=A(a,b) v 2 + v 2 sinh 1 ( v 2 + v 2 ) ,
(3.2)
T(a,b)=A(a,b) v arctan ( v ) ,Q(a,b)=A(a,b) 1 + v 2 .
(3.3)

Equations (3.2) and (3.3) lead to

S C A ( a , b ) T ( a , b ) Q ( a , b ) T ( a , b ) = v 2 + v 2 sinh 1 ( v 2 + v 2 ) v arctan ( v ) 1 + v 2 v arctan ( v ) .
(3.4)

It is easy to find that

lim v 0 + v 2 + v 2 sinh 1 ( v 2 + v 2 ) v arctan ( v ) 1 + v 2 v arctan ( v ) =0,
(3.5)
lim v 1 v 2 + v 2 sinh 1 ( v 2 + v 2 ) v arctan ( v ) 1 + v 2 v arctan ( v ) = β 0 .
(3.6)

We investigate the difference between the convex combination of Q(a,b), T(a,b) and S C A (a,b) as follows:

p Q ( a , b ) + ( 1 p ) T ( a , b ) S C A ( a , b ) = A ( a , b ) [ p 1 + v 2 + ( 1 p ) v arctan ( v ) v 2 + v 2 sinh 1 ( v 2 + v 2 ) ] .
(3.7)

Let

D p (v)=p 1 + v 2 +(1p) v arctan ( v ) v 2 + v 2 sinh 1 ( v 2 + v 2 ) .
(3.8)

Then simple computations lead to

D p ( 0 + ) =0, D p ( 1 ) =p ( 2 4 π ) + 4 π 3 log ( 2 + 3 ) , D β 0 ( 1 ) =0,
(3.9)
D p ( v ) = p [ v 1 + v 2 + v ( 1 + v 2 ) arctan 2 v 1 arctan v ] + v ( sinh 1 ( v 2 + v 2 ) ) 2 1 + v 2 2 + v 2 sinh 1 ( v 2 + v 2 ) v ( 1 + v 2 ) arctan 2 v + 1 arctan v ,
(3.10)
D p ( v ) = p 1 + v 2 arctan 3 v 2 ( v arctan v ) ( 1 + v 2 ) 2 arctan 3 v + 2 ( 4 + 3 v 2 ) sinh 1 ( v 2 + v 2 ) 8 v 2 + v 2 ( 2 + v 2 ) ( sinh 1 ( v 2 + v 2 ) ) 3 + 2 ( v arctan v ) ( 1 + v 2 ) 2 arctan 3 v 2 v ( 3 + v 2 ) ( 2 + v 2 ) 3 / 2 sinh 1 ( v 2 + v 2 ) = p σ ( v ) + κ ( v ) + τ ( v ) ,
(3.11)

where σ(x), κ(x) and τ(x) are defined as in Lemmas 2.6, 2.7 and 2.8, respectively.

From Lemmas 2.1-2.3 and (3.10) we clearly see that

D β 0 ( v ) > β 0 ( v 3 v 3 6 ) v 3 + 2 v 3 45 v 5 63 + 2 v 3 16 v 3 45 = v 630 [ 210 ( 1 + β 0 ) 7 ( 28 + 15 β 0 ) v 2 10 v 4 ] > v 630 [ 210 ( 1 + 0.29758 ) 7 ( 28 + 15 × 0.29759 ) × ( 0.55 ) 2 10 × ( 0.55 ) 4 ] = v 630 × 202.83 > 0
(3.12)

for v(0,0.55].

It follows from Lemmas 2.6-2.8 and (3.11) that

D β 0 (v)= β 0 σ(v)+κ(v)+τ(v)<0.236 β 0 +0.7710.88=0.0387709
(3.13)

for v[0.55,1). Then from D β 0 (0.55)=0.0139552 and D β 0 (1)=0.0650268 we know that there exists v 0 (0.55,1) such that D β 0 (v)>0 for v[0.55, v 0 ) and D β 0 (x)<0 for v( v 0 ,1). This in conjunction with (3.13) leads to the conclusion that D β 0 (v) is strictly increasing on [0.55, v 0 ] and strictly decreasing on [ v 0 ,1).

Therefore, D β 0 (v)>0 for v(0,1) follows from (3.9) and the monotonicity of D β 0 (v). In other words, we obtain

β 0 Q(a,b)+(1 β 0 )T(a,b)> S C A (a,b)
(3.14)

for a,b>0 with ab.

Obviously, if α=0, then (1.4) gives

T(a,b)< S C A (a,b)
(3.15)

for a,b>0 with ab.

Therefore, Theorem 3.1 follows from (3.14) and (3.15) together with the following statements:

  • If α>0, then (3.4) and (3.5) imply that there exists δ 1 (0,1) such that S C A (a,b)<αQ(a,b)+(1α)T(a,b) for all a,b>0 with (ab)/(a+b)(0, δ 1 ).

  • If β< β 0 , then (3.4) and (3.6) imply that there exists δ 2 (0,1) such that S C A (a,b)>βQ(a,b)+(1β)T(a,b) for all a,b>0 with (ab)/(a+b)(1 δ 2 ,1).

 □

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Acknowledgements

The research was supported by the Natural Science Foundation of China under Grants 11301127 and 61374086, and the Natural Science Foundation of Zhejiang Province under Grant 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

Y-MC provided the main idea and carried out the proof of Theorem 3.1. HW carried out the proof of Lemmas 2.1-2.4. T-HZ carried out the proof of Lemmas 2.5-2.8 and drafted the manuscript. All authors read and approved the final manuscript.

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Chu, YM., Wang, H. & Zhao, TH. Sharp bounds for the Neuman mean in terms of the quadratic and second Seiffert means. J Inequal Appl 2014, 299 (2014). https://doi.org/10.1186/1029-242X-2014-299

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Keywords

  • Neuman mean
  • quadratic mean
  • second Seiffert mean