# Optimal power mean bounds for Yang mean

## Abstract

In this paper, we prove that the double inequality $M p (a,b) holds for all $a,b>0$ with $a≠b$ if and only if $p≤2log2/(2logπ−log2)=0.8684⋯$ and $q≥4/3$, 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 $p∈R$ and $a,b>0$ with $a≠b$. Then the p th power mean $M p (a,b)$ of a and b is given by

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

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 $p∈R$ for fixed $a,b>0$ with $a≠b$. 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 $L(a,b)=(b−a)/(logb−loga)$, $P(a,b)=(a−b)/[2arcsin((a−b)/(a+b))]$, $M(a,b)=(a−b)/[2 sinh − 1 ((a−b)/(a+b))]$, $I(a,b)= ( a a / b b ) 1 / ( a − b ) /e$ and $T(a,b)=(a−b)/[2arctan((a−b)/(a+b))]$ 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 $a≠b$.

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)

hold for all $a,b>0$ with $a≠b$.

Jagers [3] proved that the double inequality

$M 1 / 2 (a,b)

holds for all $a,b>0$ with $a≠b$.

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 $a≠b$.

Witkowski [6] proved that the double inequality

$2 2 π M 2 (a,b)

holds for all $a,b>0$ with $a≠b$.

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

$M log 2 / ( log π − log 2 ) (a,b)

for all $a,b>0$ with $a≠b$.

Chu and Long [8] proved that the double inequality

$M p (a,b)

holds for all $a,b>0$ with $a≠b$ if and only if $p≤log2/log[2log(1+ 2 )]=1.224⋯$ and $q≥4/3$.

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 $a≠b$.

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 $a≠b$.

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 $Q(a,b)$.

$M p (a,b)

holds for all $a,b>0$ with $a≠b$.

## 2 Lemmas

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

Lemma 2.1 Let $f 1 :(0,1)×R→R$ 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 $p≥4/3$;

2. (2)

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

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, $p≥4/3$ follows from (2.4).

If $p≥4/3$, 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 $p≤1/2$. 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 $f 2 ( 0 + ,p)=1−2p$ for $0.

If $p≤1/2$, then inequality (2.5) holds again. It follows from (2.3) and (2.5) that

$f 2 (x,p)≥ f 2 ( x , 1 2 ) =2 x 1 / 2 (1−x) ( x 2 + 2 x 3 / 2 + 4 x + 2 x 1 / 2 + 1 ) >0$
(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 $f 1 :(0,1)×R→R$ be defined by (2.1). Then

1. (1)

$f 1 (x,p)>0$ for all $x∈(0,1)$ if and only if $p≥4/3$;

2. (2)

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

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

$lim x → 1 − f 1 ( x , p ) ( 1 − x ) 3 = 1 12 (3p−4)≥0$

and $p≥4/3$.

If $p≥4/3$, 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 $p≤1/2$. Indeed, it follows from (2.1) that $f 1 ( 0 + ,p)=+∞$ if $p>1/2$.

If $p≤1/2$, 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 $f 3 :(0,1)×R→R$ 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

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

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

$a 2 x 2 + a 1 x+ a 0 < a 2 + a 1 + a 0 =−4p(2p−1) ( 10 p 2 − 21 p + 17 ) <0,$
(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 $f 3 :(0,1)×R→R$ 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)
$∂ 3 f 3 ( x , p ) ∂ x 3 | x = 1 =88 p 3 −300 p 2 +380p−144.$
(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

$∂ 3 f 3 ( x , p ) ∂ x 3 | x = 1 =88(p−1) ( p − 53 44 ) 2 + 887 22 (p−1)+24>0.$
(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 $f 1 :(0,1)×R→R$ 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)
$∂ f 3 ( x , p ) ∂ x | x = 1 =8(3p−4)<0, lim x → 0 + ∂ f 3 ( x , p ) ∂ x =+∞.$
(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 $f:(0,1)×R→R$ be defined by

$f(x,p)=log U ( 1 , x ) M p ( 1 , x ) =log 1 − x 2 arctan 1 − x 2 x − 1 p log 1 + x p 2 (p≠0),$
(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 $p≥4/3$;

2. (2)

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

3. (3)

If $1/2, 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, 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)

holds for all $a,b>0$ with $a≠b$ if and only if $p≤ p 0 =2log2/(2logπ−log2)=0.8684⋯$ and $q≥4/3$.

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 $q≥4/3$.

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 )
(3.1)

for all $x∈(0,1)$.

Therefore, $U(1,x)< M q (1,x)$ for all $x∈(0,1)$ and $q≥4/3$ 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 $q≥4/3$.

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

$f ( 0 + , p ) = ( 1 p + 1 2 ) log2−logπ≥0.$
(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 $a≠b$. Then the double inequality

$2 5 / 4 π M 4 / 3 (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 )
(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) 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 $p≥4/3$ and $q≤1/2$.

Let $p∈R$ 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

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

holds for all $a,b>0$ with $a≠b$ if and only if $p≥4/3$ and $q≤1/2$.

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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.

## Author information

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### Corresponding author

Correspondence to Yu-Ming Chu.

### Competing interests

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