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Optimal power mean bounds for Yang mean

Abstract

In this paper, we prove that the double inequality M p (a,b)<U(a,b)< M q (a,b) holds for all a,b>0 with ab if and only if p2log2/(2logπlog2)=0.8684 and q4/3, where U(a,b) and M r (a,b) are the Yang and r th power means of a and b, respectively.

MSC:26E60.

1 Introduction

Let pR and a,b>0 with ab. Then the p th power mean M p (a,b) of a and b is given by

M p (a,b)= ( a p + b p 2 ) 1 / p (p0), M 0 (a,b)= a b .

The main properties for the power mean are given in [1]. It is well known that M p (a,b) is strictly increasing with respect to pR for fixed a,b>0 with ab. Many classical means are the special cases of the power mean, for example, M 1 (a,b)=2ab/(a+b)=H(a,b) is the harmonic mean, M 0 (a,b)= a b =G(a,b) is the geometric mean, M 1 (a,b)=(a+b)/2=A(a,b) is the arithmetic mean, and M 2 (a,b)= ( a 2 + b 2 ) / 2 =Q(a,b) is the quadratic mean.

Let L(a,b)=(ba)/(logbloga), P(a,b)=(ab)/[2arcsin((ab)/(a+b))], M(a,b)=(ab)/[2 sinh 1 ((ab)/(a+b))], I(a,b)= ( a a / b b ) 1 / ( a b ) /e and T(a,b)=(ab)/[2arctan((ab)/(a+b))] be the logarithmic, first Seiffert, Neuman-Sándor, identric, and second Seiffert means of two distinct positive real numbers a and b, respectively. Then it is well known that the inequalities

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

hold for all a,b>0 with ab.

Recently, the bounds for certain bivariate means in terms of the power mean have been the subject of intensive research. Seiffert [2] proved that the inequalities

2 π M 1 (a,b)<P(a,b)< M 1 (a,b)<T(a,b)< M 2 (a,b)

hold for all a,b>0 with ab.

Jagers [3] proved that the double inequality

M 1 / 2 (a,b)<P(a,b)< M 2 / 3 (a,b)

holds for all a,b>0 with ab.

In [4, 5], Hästö established that

P(a,b)> M log 2 / log π (a,b),P(a,b)> 2 2 π M 2 / 3 (a,b)

for all a,b>0 with ab.

Witkowski [6] proved that the double inequality

2 2 π M 2 (a,b)<T(a,b)< 4 π M 1 (a,b)

holds for all a,b>0 with ab.

In [7], Costin and Toader presented the result that

M log 2 / ( log π log 2 ) (a,b)<T(a,b)< M 5 / 3 (a,b)

for all a,b>0 with ab.

Chu and Long [8] proved that the double inequality

M p (a,b)<M(a,b)< M q (a,b)

holds for all a,b>0 with ab if and only if plog2/log[2log(1+ 2 )]=1.224 and q4/3.

The following sharp bounds for the logarithmic and identric means in terms of the power means can be found in the literature [916]:

M 0 ( a , b ) < L ( a , b ) < M 1 / 3 ( a , b ) , M 2 / 3 ( a , b ) < I ( a , b ) < M log 2 ( a , b ) , M 0 ( a , b ) < L 1 / 2 ( a , b ) I 1 / 2 ( a , b ) < M 1 / 2 ( a , b ) , M log 2 / ( 1 + log 2 ) ( a , b ) < L ( a , b ) + I ( a , b ) 2 < M 1 / 2 ( a , b )

for all a,b>0 with ab.

Recently, Yang [17] introduced the Yang mean U(a,b) of two distinct positive real numbers a and b as follows:

U(a,b)= a b 2 arctan a b 2 a b ,

and he proved that the inequalities

P ( a , b ) < U ( a , b ) < T ( a , b ) , G ( a , b ) T ( a , b ) A ( a , b ) < U ( a , b ) < P ( a , b ) Q ( a , b ) A ( a , b ) , Q 1 / 2 ( a , b ) [ 2 G ( a , b ) + Q ( a , b ) 3 ] 1 / 2 < U ( a , b ) < Q 2 / 3 ( a , b ) [ G ( a , b ) + Q ( a , b ) 2 ] 1 / 3 , G ( a , b ) + Q ( a , b ) 2 < U ( a , b ) < [ 2 3 ( G ( a , b ) + Q ( a , b ) 2 ) 1 / 2 + 1 3 Q 1 / 2 ( a , b ) ] 2

hold for all a,b>0 with ab.

In [18], Yang et al. presented several sharp bounds for the Yang mean U(a,b) in terms of the geometric mean G(a,b) and quadratic mean Q(a,b).

The main purpose of this article is to find the greatest value p and the least value q such that the double inequality

M p (a,b)<U(a,b)< M q (a,b)

holds for all a,b>0 with ab.

2 Lemmas

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

Lemma 2.1 Let f 1 :(0,1)×RR be defined by

f 1 (x,p)= ( 1 x 2 ) ( 1 + x p ) x ( 1 + x 2 ) ( 1 + x p 1 ) 2 arctan 1 x 2 x .
(2.1)

Then

  1. (1)

    f 1 (x,p) is strictly decreasing with respect to x on (0,1) if and only if p4/3;

  2. (2)

    f 1 (x,p) is strictly increasing with respect to x on (0,1) if and only if p1/2.

Proof It follows from (2.1) that

f 1 ( x , p ) x = ( 1 x ) x p 1 / 2 2 ( 1 + x 2 ) 2 ( x + x p ) 2 f 2 (x,p),
(2.2)

where

f 2 ( x , p ) = x 1 p ( 1 + x 5 x 2 3 x 3 ) + x p ( 3 + 5 x x 2 + x 3 ) ( 2 p 1 ) + 4 x 4 x 3 + ( 2 p 1 ) x 4 .
(2.3)
  1. (1)

    If f 1 (x,p) is strictly decreasing with respect to x on (0,1), then (2.2) leads to the conclusion that f 2 (x,p)<0 for all x(0,1). In particular, from (2.3) we have

    lim x 1 f 2 ( x , p ) 1 x =24 ( p 4 3 ) 0.
    (2.4)

Therefore, p4/3 follows from (2.4).

If p4/3, then it follows from (2.3) that

f 2 ( x , p ) p = [ x p ( x 3 x 2 + 5 x + 3 ) + x 1 p ( 3 x 3 + 5 x 2 x + 1 ) ] log x 2 ( 1 x 4 ) < 0
(2.5)

for all x(0,1).

Equation (2.3) and inequality (2.5) lead to the conclusion that

f 2 ( x , p ) f 2 ( x , 4 3 ) = x 1 / 3 3 ( 1 x 2 / 3 ) 3 × ( 3 x 8 / 3 + 5 x 7 / 3 + 9 x 2 + 12 x 5 / 3 + 6 x 4 / 3 + 12 x + 9 x 2 / 3 + 5 x 1 / 3 + 3 ) < 0
(2.6)

for all x(0,1).

Therefore, f 1 (x,p) is strictly decreasing with respect to x on (0,1) follows from (2.2) and (2.6).

  1. (2)

    If f 1 (x,p) is strictly increasing with respect to x on (0,1), then (2.2) leads to the conclusion that f 2 (x,p)>0 for all x(0,1). In particular, we have f 2 ( 0 + ,p)0 and we assert that p1/2. Indeed, from (2.3) we clearly see that f 2 ( 0 + ,p)= for p>1, f 2 ( 0 + ,1)=2, f 2 ( 0 + ,0)=4, f 2 ( 0 + ,p)= for p<0, and f 2 ( 0 + ,p)=12p for 0<p<1.

If p1/2, then inequality (2.5) holds again. It follows from (2.3) and (2.5) that

f 2 (x,p) f 2 ( x , 1 2 ) =2 x 1 / 2 (1x) ( x 2 + 2 x 3 / 2 + 4 x + 2 x 1 / 2 + 1 ) >0
(2.7)

for all x(0,1).

Therefore, f 1 (x,p) is strictly increasing with respect to x on (0,1) follows from (2.2) and (2.7). □

Lemma 2.2 Let f 1 :(0,1)×RR be defined by (2.1). Then

  1. (1)

    f 1 (x,p)>0 for all x(0,1) if and only if p4/3;

  2. (2)

    f 1 (x,p)<0 for all x(0,1) if and only if p1/2.

Proof (1) If f 1 (x,p)>0 for all x(0,1), then from (2.1) and the L’Hôpital rules we have

lim x 1 f 1 ( x , p ) ( 1 x ) 3 = 1 12 (3p4)0

and p4/3.

If p4/3, then (2.1) and Lemma 2.1(1) lead to the conclusion that f 1 (x,p)> f 1 (1,p)=0 for all x(0,1).

  1. (2)

    If f 1 (x,p)<0 for all x(0,1), then f 1 ( 0 + ,p)0. We claim that p1/2. Indeed, it follows from (2.1) that f 1 ( 0 + ,p)=+ if p>1/2.

If p1/2, then (2.1) and Lemma 2.1(2) lead to the conclusion that f 1 (x,p)< f 1 (1,p)=0 for all x(0,1). □

Lemma 2.3 Let f 3 :(0,1)×RR be defined by

f 3 ( x , p ) = x 1 2 p + x 2 2 p 5 x 3 2 p 3 x 4 2 p + 3 + 5 x x 2 + x 3 ( 2 p 1 ) x p + 4 x 1 p 4 x 3 p + ( 2 p 1 ) x 4 p .
(2.8)

Then 4 f 3 (x,p)/ x 4 <0 for all x(0,1) if p(1,4/3).

Proof It follows (2.8) that

x p + 4 4 f 3 ( x , p ) x 4 = x 1 p ( a 3 x 3 + a 2 x 2 + a 1 x + a 0 ) + b 4 x 4 + b 3 x 3 + b 1 x+ b 0 ,
(2.9)

where

a 3 =3(2p1)(2p2)(2p3)(2p4)<0,
(2.10)
a 2 =10p(2p1)(2p2)(2p3)>0,
(2.11)
a 1 =2p(2p1)(2p+1)(2p2)>0,
(2.12)
a 0 =2p(2p1)(2p+1)(2p+2)<0,
(2.13)
b 4 =(2p1)(p1)(p2)(p3)(p4)<0,
(2.14)
b 3 =4p(p1)(p2)(p3)<0,
(2.15)
b 1 =4p(p1)(p+1)(p+2)>0,
(2.16)
b 0 =p(2p1)(p+1)(p+2)(p+3)<0.
(2.17)

From (2.11)-(2.13) and (2.16) together with (2.17) we get

a 2 x 2 + a 1 x+ a 0 < a 2 + a 1 + a 0 =4p(2p1) ( 10 p 2 21 p + 17 ) <0,
(2.18)
b 1 x+ b 0 < b 1 + b 0 =p(p+2)(p+1) ( 2 p 2 + p + 1 ) <0
(2.19)

for all x(0,1).

Therefore, Lemma 2.3 follows easily from (2.9), (2.10), (2.14), (2.15), (2.18), and (2.19). □

Lemma 2.4 Let f 3 :(0,1)×RR be defined by (2.8). Then 2 f 3 (x,p)/ x 2 <0 for all x(0,1) if p(1/2,4/3).

Proof It follows from (2.8) that

x p + 2 2 f 3 ( x , p ) x 2 = 6 x p + 3 + ( 2 p 1 ) ( p 3 ) ( p 4 ) x 4 2 x p + 2 3 ( 2 p 3 ) ( 2 p 4 ) x 4 p 5 ( 2 p 2 ) ( 2 p 3 ) x 3 p + ( 2 p 1 ) ( 2 p 2 ) x 2 p 2 p ( 2 p 1 ) x 1 p 4 ( p 2 ) ( p 3 ) x 3 + 4 p ( p 1 ) x p ( 2 p 1 ) ( p + 1 ) ,
(2.20)
2 f 3 ( x , p ) x 2 | x = 1 =48 ( 4 3 p ) ( 3 2 p ) <0,
(2.21)
3 f 3 ( x , p ) x 3 | x = 1 =88 p 3 300 p 2 +380p144.
(2.22)

We divide the proof into two cases.

Case 1. p(1/2,1]. Then from

( 2 p 1 ) ( p 3 ) ( p 4 ) > 0 , 3 ( 2 p 3 ) ( 2 p 4 ) < 0 , 5 ( 2 p 2 ) ( 2 p 3 ) 0 , ( 2 p 1 ) ( 2 p 2 ) 0 , 2 p ( 2 p 1 ) < 0 , 4 ( p 2 ) ( p 3 ) < 0 , 4 p ( p 1 ) < 0 , p ( 2 p 1 ) ( p + 1 ) < 0 , 0 < x 4 x p + 3 < x 4 p x 3 x p + 2 < x 3 p x 2 < x 2 p x < x 1 p 1

and (2.20) we clearly see that

x p + 2 2 f 3 ( x , p ) x 2 < [ 6 + ( 2 p 1 ) ( p 3 ) ( p 4 ) ] x 4 p + [ 2 3 ( 2 p 3 ) ( 2 p 4 ) 5 ( 2 p 2 ) ( 2 p 3 ) + ( 2 p 1 ) ( 2 p 2 ) 2 p ( 2 p 1 ) 4 ( p 2 ) ( p 3 ) + 4 p ( p 1 ) p ( 2 p 1 ) ( p + 1 ) ] x 4 p = 8 ( 3 p 4 ) ( 2 p 3 ) x 4 p < 0

for all x(0,1).

Case 2. p(1,4/3]. Then (2.22) leads to

3 f 3 ( x , p ) x 3 | x = 1 =88(p1) ( p 53 44 ) 2 + 887 22 (p1)+24>0.
(2.23)

It follows from Lemma 2.3 and (2.23) that 2 f 3 (x,p)/ x 2 is strictly increasing with respect to x on (0,1).

Therefore, 2 f 3 (x,p)/ x 2 <0 for all x(0,1) follows from (2.21) and the monotonicity of the 2 f 3 (x,p)/ x 2 with respect to x on the interval (0,1). □

Lemma 2.5 Let f 1 :(0,1)×RR be defined by (2.1). Then there exists λ(0,1) such that f 1 (x,p) is strictly decreasing with respect to x on the interval (0,λ] and strictly increasing with respect to x on the interval [λ,1) if p(1/2,4/3).

Proof Let f 2 (x,p) and f 3 (x,p) be defined by (2.3) and (2.8), respectively. Then from (2.8) we clearly see that

f 3 (1,p)=0, f 3 ( 0 + , p ) =,
(2.24)
f 3 ( x , p ) x | x = 1 =8(3p4)<0, lim x 0 + f 3 ( x , p ) x =+.
(2.25)

It follows from Lemma 2.4 and (2.25) that there exists λ 0 (0,1) such that f 3 (x,p) is strictly increasing with respect to x on (0, λ 0 ] and strictly decreasing with respect to x on [ λ 0 ,1). This in conjunction with (2.24) leads to the conclusion that there exists λ(0,1) such that f 3 (x,p)<0 for x(0,λ) and f 3 (x,p)>0 for x(λ,1).

Note that

f 2 (x,p)= x p f 3 (x,p).
(2.26)

Therefore, Lemma 2.5 follows from (2.2) and (2.26) together with the piecewise positive and negative of f 3 (x,p) on (0,1). □

Lemma 2.6 Let f:(0,1)×RR be defined by

f(x,p)=log U ( 1 , x ) M p ( 1 , x ) =log 1 x 2 arctan 1 x 2 x 1 p log 1 + x p 2 (p0),
(2.27)
f(x,0)= lim p 0 log U ( 1 , x ) M p ( 1 , x ) =log 1 x 2 arctan 1 x 2 x 1 2 logx.
(2.28)

Then the following statements are true:

  1. (1)

    f(x,p) is strictly increasing with respect to x on (0,1) if and only if p4/3;

  2. (2)

    f(x,p) is strictly decreasing with respect to x on (0,1) if and only if p1/2;

  3. (3)

    If 1/2<p<4/3, then there exists μ(0,1) such that f(x,p) is strictly increasing with respect to x on (0,μ] and strictly decreasing with respect to x on [μ,1).

Proof It follows from (2.27) and (2.28) that

f ( x , p ) x = 1 + x p 1 2 ( 1 x ) ( 1 + x p ) arctan 1 x 2 x f 1 (x,p),
(2.29)

where f 1 (x,p) is defined by (2.1).

Therefore, parts (1) and (2) follow from Lemma 2.2 and (2.29).

Next, we prove part (3). If 1/2<p<4/3, then (2.1) leads to

f 1 ( 0 + , p ) =+, f 1 (1,p)=0.
(2.30)

From Lemma 2.5 and (2.30) we clearly see that there exists μ(0,1) such that f 1 (x,p)>0 for x(0,μ) and f 1 (x,p)<0 for x(μ,1).

Therefore, part (3) follows from (2.29) and the fact that f 1 (x,p)>0 for x(0,μ) and f 1 (x,p)<0 for x(μ,1). □

3 Main results

Theorem 3.1 The double inequality

M p (a,b)<U(a,b)< M q (a,b)

holds for all a,b>0 with ab if and only if p p 0 =2log2/(2logπlog2)=0.8684 and q4/3.

Proof Since both the Yang mean U(a,b) and the r th power mean M r (a,b) are symmetric and homogeneous of degree 1, without loss of generality, we assume that a=1 and b=x(0,1).

We first prove that the inequality U(1,x)< M q (1,x) holds for all x(0,1) if and only if q4/3.

If q=4/3, then from (2.27) and Lemma 2.6(1) we get

log U ( 1 , x ) M 4 / 3 ( 1 , x ) =f ( x , 4 3 ) <f ( 1 , 4 3 ) =0
(3.1)

for all x(0,1).

Therefore, U(1,x)< M q (1,x) for all x(0,1) and q4/3 follows from (3.1) and the monotonicity of the function q M q (1,x).

If U(1,x)< M q (1,x), then (2.27) and (2.28) lead to f(x,q)<0 for all x(0,1). In particular, we have

lim x 1 f ( x , q ) ( 1 x ) 2 = 1 8 ( 4 3 q ) 0

and q4/3.

Next, we prove that the inequality U(1,x)> M p (1,x) holds for all x(0,1) if and only if p p 0 .

If U(1,x)> M p (1,x) holds for all x(0,1), then (2.27) leads to f(x,p)>0 for all x(0,1). In particular, we have

f ( 0 + , p ) = ( 1 p + 1 2 ) log2logπ0.
(3.2)

We claim that p p 0 . Indeed, p p 0 follows from (3.2) if p>0, and p< p 0 is obvious if p<0.

If p= p 0 , then (2.27) leads to

f ( 0 + , p 0 ) =f(1, p 0 )=0.
(3.3)

It follows from (2.27) and (3.3) together with Lemma 2.6(3) that

log U ( 1 , x ) M p 0 ( 1 , x ) =f(x, p 0 )>0
(3.4)

for all x(0,1).

Therefore, U(1,x)> M p (1,x) for all x(0,1) and p p 0 follows from (3.4) and the monotonicity of the function p M p (1,x). □

Theorem 3.2 Let a,b>0 with ab. Then the double inequality

2 5 / 4 π M 4 / 3 (a,b)<U(a,b)< 2 5 / 2 π M 1 / 2 (a,b)

holds with the best possible constants 2 5 / 4 /π and 2 5 / 2 /π.

Proof It follows from Lemma 2.6(1) and (2) together with (2.27) that

log U ( 1 , x ) M 1 / 2 ( 1 , x ) =f ( x , 1 2 ) <f ( 0 + , 1 2 ) =log 2 5 / 2 π
(3.5)

and

log U ( 1 , x ) M 4 / 3 ( 1 , x ) =f ( x , 4 3 ) >f ( 0 + , 4 3 ) =log 2 5 / 4 π
(3.6)

for all x(0,1).

Therefore, 2 5 / 4 /π M 4 / 3 (1,x)<U(1,x)< 2 5 / 2 /π M 1 / 2 (1,x) for all x(0,1) follows from (3.5) and (3.6), and the optimality of the parameters 2 5 / 4 /π and 2 5 / 2 /π follows from the monotonicity of the functions f(x,1/2) and f(x,4/3). □

Remark 3.1 For all a 1 , a 2 , b 1 , b 2 >0 with a 1 / b 1 < a 2 / b 2 <1. Then from Lemma 2.6(1) and (2) together with (2.27) we clearly see that the Ky Fan type inequalities

M p ( a 2 , b 2 ) M p ( a 1 , b 1 ) < U ( a 2 , b 2 ) U ( a 1 , b 1 ) < M q ( a 2 , b 2 ) M q ( a 1 , b 1 )

hold if and only if p4/3 and q1/2.

Let pR and L p (a,b)=( a p + 1 + b p + 1 )/( a p + b p ) be the p th Lehmer mean of two positive real numbers and a and b. Then the function f 1 (x,p) defined by (2.1) can be rewritten as

f 1 (x,p)=(1x) [ A ( 1 , x ) L p 1 ( 1 , x ) G ( 1 , x ) Q 2 ( 1 , x ) 1 U ( 1 , x ) ] .
(3.7)

From Lemma 2.2 and (3.7) we get Remark 3.2.

Remark 3.2 The double inequality

G ( a , b ) Q 2 ( a , b ) A ( a , b ) L p 1 ( a , b ) <U(a,b)< G ( a , b ) Q 2 ( a , b ) A ( a , b ) L q 1 ( a , b )

holds for all a,b>0 with ab if and only if p4/3 and q1/2.

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Acknowledgements

This research was supported by the Natural Science Foundation of China under Grants 11171307 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

Z-HY provided the main idea and carried out the proof of Lemmas 2.1 and 2.2. L-MW carried out the proof of Lemmas 2.3-2.6. Y-MC carried out the proof of Theorems 3.1 and 3.2. All authors read and approved the final manuscript.

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Yang, Z., Wu, L. & Chu, Y. Optimal power mean bounds for Yang mean. J Inequal Appl 2014, 401 (2014). https://doi.org/10.1186/1029-242X-2014-401

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Keywords

  • Yang mean
  • power mean
  • Neuman-Sándor mean