1 Introduction

For a, b > 0 with ab, the root-square mean S(a, b) and Seiffert mean T(a, b) are defined by

S ( a , b ) = a 2 + b 2 2
(1.1)

and

T ( a , b ) = a - b 2 arctan a - b a + b ,
(1.2)

respectively. In the recent past, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for S and T can be found in the literature [111].

Let A ( a , b ) = ( a + b ) 2,G ( a , b ) = a b , and H(a, b) = 2ab/(a + b) be the classical arithmetic, geometric, and harmonic means of two positive numbers a and b, respectively. In [1], Seiffert proved that

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

for all a, b > 0 with ab.

Taneja [5] presented that

G ( a , b ) < 2 3 H ( a , b ) + 1 3 S ( a , b ) < 1 2 A ( a , b ) + 1 2 H ( a , b ) < 1 2 S ( a , b ) + 1 2 G ( a , b ) (1)  < 1 3 H ( a , b ) + 2 3 S ( a , b ) < A ( a , b ) < S ( a , b ) - G ( a , b ) + H ( a , b ) (2)  (3) 

for all a, b > 0 with a ≠ b.

In [2], the authors find the greatest value p and the least value q such that the double inequality H p (a, b) < T(a, b) < H q (a, b) for all a, b > 0 with ab. Here, H p ( a , b ) = [ ( a p + ( a b ) p 2 + b p ) 3 ] 1 p is the power-type Heron mean of a and b.

Wang, Qiu, and Chu [3] established that

T ( a , b ) < L 1 3 ( a , b )

for all a, b > 0 with ab, where L p (a, b) = (ap+1+ bp+1) = (ap + bp ) is the Lehmer mean of a and b.

The purpose of the paper is to find the greatest values α1 and α2, and the least values β1 and β2, such that the double inequalities α1S(a, b) + (1 - α1)A(a, b) < T (a, b) < β1S(a, b) + (1 - β1)A(a, b) and S α 2 ( a , b ) A 1 - α 2 ( a , b ) <T ( a , b ) < S β 2 ( a , b ) A 1 - β 2 ( a , b ) hold for all a, b > 0 with ab.

2 Main results

Theorem 2.1. The double inequality α1S(a, b)+(1 - α1)A(a, b) < T (a, b) < β1S(a, b) + (1 - β1)A(a, b) holds for all a, b > 0 with ab if and only if α 1 ( 4 - π ) [ ( 2 - 1 ) π ] = 0 . 6 5 9 and β1 ≥ 2/3.

Proof. Firstly, we prove that

T ( a , b ) < 2 3 S ( a , b ) + 1 3 A ( a , b ) ,
(2.1)
T ( a , b ) > 4 - π π ( 2 - 1 ) S ( a , b ) + 2 π - 4 π ( 2 - 1 ) A ( a , b )
(2.2)

for all a, b > 0 with a ≠ b.

Without loss of generality, we assume that a > b. Let t= a b >1 and p 2 3 , ( 4 - π ) [ ( 2 - 1 ) π ] , then from (1.1) and (1.2) we have

T ( a , b ) - [ p S ( a , b ) + ( 1 - p ) A ( a , b ) ] = b [ 2 p 1 + t 2 + ( 1 - p ) ( 1 + t ) ] 2 arctan t - 1 t + 1 × t - 1 2 p 1 + t 2 + ( 1 - p ) ( 1 + t ) - a r c t a n t - 1 t + 1 .
(2.3)

Let

f ( t ) = t - 1 2 p 1 + t 2 + ( 1 - p ) ( 1 + t ) - a r c t a n t - 1 t + 1 ,
(2.4)

then simple computations lead to

f ( 1 ) = 0 ,
(2.5)
lim t + f ( t ) = 1 ( 2 - 1 ) p + 1 - π 4 ,
(2.6)
f ( t ) = f 1 ( t ) ( t 2 + 1 ) [ 2 p 1 + t 2 + ( 1 - p ) ( 1 + t ) ] 2 ,
(2.7)

where

f 1 ( t ) = 2 p ( 2 p - 1 ) ( t + 1 ) t 2 + 1 - [ ( 3 p 2 - 1 ) t 2 + 2 ( p - 1 ) 2 t + 3 p 2 - 1 ]
(2.8)

We divide the proof into two cases.

Case 1. p = 2/3. Then, we clearly see that

2 p - 1 = 3 p 2 - 1 = 1 3 > 0 ,
(2.9)

and

[ 2 p ( 2 p - 1 ) ( t + 1 ) 1 + t 2 ] 2 - [ ( 3 p 2 - 1 ) t 2 + 2 ( p - 1 ) 2 t + 3 p 2 - 1 ] 2 = - ( t - 1 ) 4 8 1 < 0
(2.10)

for t > 1.

Therefore, inequality (2.1) follows from (2.3)-(2.5) and (2.7)-(2.10).

Case 2. p= ( 4 - π ) [ ( 2 - 1 ) π ] =0.659. Then, simple computations yield that

2 p - 1 > 0 ,
(2.11)
2 - 3 p > 0 ,
(2.12)
3 p 2 - 1 > 0 ,
(2.13)
- p 4 - 8 p 3 + 8 p 2 - 1 = - 0 . 0 0 4 5 6 < 0
(2.14)

and

[ 2 p ( 2 p 1 ) ( t + 1 ) 1 + t 2 ] 2 [ ( 3 p 2 1 ) t 2 + 2 ( p 1 ) 2 t + 3 p 2 1 ] 2 = ( t 1 ) 2 [ ( p 4 8 p 3 + 8 p 2 1 ) t 2 + 2 ( p 4 4 p 3 + 6 p 2 4 p + 1 ) t p 4 8 p 3 + 8 p 2 1 ] .
(2.15)

Let

g ( t ) = ( - p 4 - 8 p 3 + 8 p 2 - 1 ) t 2 + 2 ( p 4 - 4 p 3 + 6 p 2 - 4 p + 1 ) t - p 4 - 8 p 3 + 8 p 2 - 1 ,
(2.16)

then from (2.11) and (2.12) together with (2.14), we get

g ( 1 ) = 4 p ( 2 p - 1 ) ( 2 - 3 p ) > 0 ,
(2.17)
lim t + g ( t ) = - ,
(2.18)
g ( t ) = 2 ( - p 4 - 8 p 3 + 8 p 2 - 1 ) t + 2 ( p 4 - 4 p 3 + 6 p 2 - 4 p + 1 ) , g ( 1 ) = 4 p ( 2 p - 1 ) ( 2 - 3 p ) > 0 ,
(2.19)
lim t + g ( t ) = -
(2.20)

and

g ( t ) = 2 ( - p 4 - 8 p 3 + 8 p 2 - 1 ) = - 0 . 0 0 9 1 2 < 0 .
(2.21)

It follows from (2.19)-(2.21) that there exists t0 > 1 such that g'(t) > 0 for t ∈ [1, t0) and g'(t) < 0 for t ∈ (t0, ∞). Hence, g(t) is strictly increasing in [1, t0] and strictly decreasing in [t0, ∞).

From (2.17) and (2.18) together with the piecewise monotonicity of g(t), we clearly see that there exists t1> t0> 1 such that g(t) > 0 for t ∈ [1, t1) and g(t) < 0 for t ∈ (t1, ∞). Then, from (2.8), (2.11), (2.13), (2.15), and (2.16), we know that f1(t) > 0 for t ∈ [1, t1) and f1(t) < 0 for t ∈ (t1, ∞). It follows from (2.7) that f(t) is strictly increasing in [1, t1] and strictly decreasing in [t1, ∞).

Note that (2.6) becomes

lim t + f ( t ) = 0 .
(2.22)

Therefore, inequality (2.2) follows from (2.3)-(2.5) and (2.22) together with the piecewise monotonicity of f(t).

Secondly, we prove that 2S(a, b)/3 + A(a, b)/3 is the best possible upper convex combination bound of root-square and arithmetic means for the Seiffert mean T(a, b).

Letting x > 0 (x → 0) and making use of the Taylor expansion, one has

β 1 S ( 1 , 1 + x ) + ( 1 β 1 ) A ( 1 , 1 + x ) T ( 1 , 1 + x ) = β 1 [ [ 1 + 1 2 x + 1 8 x 2 + o ( x 2 ) ] + ( 1 β 1 ) ( 1 + x 2 ) [ 1 + 1 2 x + 1 12 x 2 + o ( x 2 ) ] = 1 24 ( 3 β 1 2 ) x 2 + o ( x 2 ) .
(2.23)

Equation (2.23) implies that for any β1 < 2/3, there exists δ1 = δ1(β1) > 0, such that β1S(1, 1 + x) + (1 - β1)A(1, 1 + x) < T (1, 1 + x) for x ∈ (0, δ1).

Finally, we prove that ( 4 - π ) S ( a , b ) [ ( 2 - 1 ) π ] + ( 2 π - 4 ) A ( a , b ) [ ( 2 - 1 ) π ] is the best possible lower convex combination bound of root-square and arithmetic means for the Seiffert mean T (a, b).

For any α 1 > ( 4 - π ) [ ( 2 - 1 ) π ] , it follows from (1.1) and (1.2) that

lim x + α 1 S ( 1 , x ) + ( 1 - α 1 ) A ( 1 , x ) T ( 1 , x ) = ( 2 - 1 ) α 1 + 1 4 π > 1 .
(2.24)

Inequality (2.24) implies that for any α 1 > ( 4 - π ) [ ( 2 - 1 ) π ] , there exists X1 = X1(α1) > 1 such that α1S(1, x) + (1 α1)A(1, x) > T (1, x) for x ∈ (X1, ∞).

Theorem 2.2. The double inequality S α 2 ( a , b ) A 1 - α 2 ( a , b ) <T ( a , b ) < S β 2 ( a , b ) A 1 - β 2 ( a , b ) holds for all a, b > 0 with ab if and only if α2 ≤ 2/3 and β2 ≥ 4 - 2 log = log 2 = 0.697⋯.

Proof. Firstly, we prove that

T ( a , b ) > S 2 3 ( a , b ) A 1 3 ( a , b ) ,
(2.25)
T ( a , b ) < [ S ( a , b ) ] 4 - 2 log π log 2 [ A ( a , b ) ] 2 log π log 2 - 3
(2.26)

for all a, b > 0 with ab.

Without loss of generality, we assume that a > b. Let t= a b >1 and q ∈ {2/3, 4 - 2 log π /log 2}, then from (1.1) and (1.2), we have

log T ( a , b ) - [ q log S ( a , b ) + ( 1 - q ) log A ( a , b ) ] = log t - 1 2 arctan t - 1 t + 1 - q 2 log 1 + t 2 2 - ( 1 - q ) log 1 + t 2
(2.27)

Let

F ( t ) = log t - 1 2 arctan t - 1 t + 1 - q 2 log 1 + t 2 2 - ( 1 - q ) log 1 + t 2 ,
(2.28)

then simple computations lead to

lim t 1 F ( t ) = 0 ,
(2.29)
lim t + F ( t ) = log 4 π - q 2 log 2 ,
(2.30)
F ( t ) = ( 2 - q ) t 2 + 2 q t + 2 - q ( t 4 - 1 ) arctan t - 1 t + 1 F 1 ( t )
(2.31)

where

F 1 ( t ) = a r c t a n t - 1 t + 1 - t 2 - 1 ( 2 - q ) t 2 + 2 q t + 2 - q ,
(2.32)
F 1 ( 1 ) = 0 ,
(2.33)
lim t + F 1 ( t ) = π 4 - 1 2 - q ,
(2.34)
F 1 ( t ) = ( t - 1 ) 2 ( 1 + t 2 ) [ ( 2 - q ) t 2 + 2 q t + 2 - q ] 2 F 2 ( t ) ,
(2.35)

where

F 2 ( t ) = ( q 2 - 6 q + 4 ) t 2 - 2 q 2 t + q 2 - 6 q + 4 ,
(2.36)
F 2 ( 1 ) = 4 ( 2 - 3 q ) ,
(2.37)
F 2 ( t ) = 2 ( q 2 - 6 q + 4 ) t - 2 q 2 ,
(2.38)
F 2 ( 1 ) = 4 ( 2 - 3 q ) .
(2.39)

We divide the proof into two cases.

Case 1. q = 2/3. Then, we clearly see that

2 - 3 q = 0 ,
(2.40)
q 2 - 6 q + 4 = 4 9 > 0 .
(2.41)

From (2.38)-(2.41), we know that F 2 ( t ) >0 for t ∈ (1, ∞). Hence, F2(t) is strictly increasing in [1, ∞). It follows from (2.35), (2.37), (2.40), and the monotonicity of F2(t) that F1(t) is strictly increasing in [1, ∞).

Therefore, inequality (2.25) follows from (2.27)-(2.29), (2.31), (2.33), and the monotonicity of F1(t).

Case 2. q = 4 - 2 log π /log 2 = 0:697⋯. Then, simple computations lead to

log 4 π - q 2 log 2 = 0 ,
(2.42)
π 4 - 1 2 - q = 0 . 0 1 7 9 > 0 ,
(2.43)
q 2 - 6 q + 4 = 0 . 3 0 3 > 0 ,
(2.44)
2 - 3 q = - 0 . 0 9 1 0 2 < 0 .
(2.45)

It follows from (2.36) and (2.38) together with (2.44) that

lim t + F 2 ( t ) = + ,
(2.46)
lim t + F 2 ( t ) = + .
(2.47)

From (2.38) and (2.44), we clearly see that F 2 ( t ) is strictly increasing in [1, ∞). Then, (2.39) and (2.45) together with (2.47) imply that there exists λ0 > 1 such that F 2 ( t ) <0 for t ∈ [1, λ0) and F 2 ( t ) >0 for t ∈ (λ0, ∞). Hence, F2(t) is strictly decreasing in [1, λ0] and strictly increasing in [λ0, ∞).

From (2.37), (2.45), (2.46), and the piecewise monotonicity of F2(t), we know that there exists λ1 > λ0 > 1 such that F2(t) < 0 for t ∈ [1, λ1) and F2(t) > 0 for t ∈ (λ1, ∞). Then (2.35) implies that F1(t) is strictly decreasing in [1, λ1] and strictly increasing in [λ1, ∞).

From (2.33), (2.34), (2.43), and the piecewise monotonicity of F1(t), we conclude that there exists λ2 > λ1 > 1 such that F1(t) < 0 for t ∈ (1, λ2) and F1(t) > 0 for t ∈ (λ2, ∞). Then, (2.31) implies that F(t) is strictly decreasing in (1, λ2] and strictly increasing in [λ2, ∞).

Therefore, inequality (2.26) follows from (2.27)-(2.30) and (2.42) together with the piecewise monotonicity of F(t).

Secondly, we prove that S2/3(a, b)A1/3(a, b) is the best possible lower geo-metric combination bound of root-square and arithmetic means for the Seiffert mean T(a, b).

Letting x > 0 (x → 0) and making use of the Taylor expansion, one has

S α 2 ( 1 , 1 + x ) A 1 - α 2 ( 1 , 1 + x ) - T ( 1 , 1 + x ) = 1 + α 2 2 x + α 2 2 8 x 2 + o ( x 2 ) 1 + 1 - α 2 2 x + α 2 ( α 2 - 1 ) 8 x 2 + o ( x 2 ) - 1 + 1 2 x + 1 1 2 x 2 + o ( x 2 ) = 1 2 4 ( 3 α 2 - 2 ) x 2 + o ( x 2 ) .
(2.48)

Equation (2.48) implies that for any α2 > 2/3, there exists δ2 = δ2(α2) > 0, such that S α 2 ( 1 , 1 + x ) A 1 - α 2 ( 1 , 1 + x ) >T ( 1 , 1 + x ) for x ∈ (0, δ2).

Finally, we prove that [S(a, b)]4-2 logπ /log2[A(a, b)]2 log π/log 2-3is the best possible upper geometric combination bound of root-square and arithmetic means for the Seiffert mean T (a, b).

For any β2 < 4 - 2 log π /log 2 and x > 0, from (1.1) and (1.2), one has

lim x + S β 2 ( 1 , x ) A 1 - β 2 ( 1 , x ) T ( 1 , x ) = 2 β 2 2 × π 4 < 1 .
(2.49)

Inequality (2.49) implies that for any β2 < 4 - 2 log π /log 2, there exists X2 = X2(β2) > 1 such that T ( 1 , x ) > S β 2 ( 1 , x ) A ( 1 - β 2 ) ( 1 , x ) for x ∈ (X2, +∞).