# Schur quadratic concavity of the elliptic Neuman mean and its application

## Abstract

For $x,y>0$ and $k∈[0,1]$, we prove that the elliptic Neuman mean $N k (x,y)$ is strictly Schur quadratically concave on $(0,∞)×(0,∞)$ if and only if $k∈[ 2 /2,1]$. As an application, the bounds for elliptic Neuman mean $N k (x,y)$ in terms of the quadratic mean $Q(x,y)= ( x 2 + y 2 ) / 2$ are presented.

MSC:26B25, 26E60.

## 1 Introduction

Let $(x,y)∈(0,∞)×(0,∞)$ and $k∈[0,1]$. Then the elliptic Neuman mean $N k (x,y)$, see [1], is defined by

$N k (x,y)= { y 2 − x 2 c n − 1 ( x / y , k ) , x < y , x , x = y , x 2 − y 2 n c − 1 ( x / y , k ) , y < x ,$
(1.1)

where $c n − 1 (x,k)= ∫ x 1 d u ( 1 − u 2 ) ( k ′ 2 + k 2 u 2 )$ and $n c − 1 (x,k)= ∫ 1 x d u ( u 2 − 1 ) ( k 2 + k ′ 2 u 2 )$ are the inverse functions of Jacobian elliptic functions cn and nc, see [2, 3], respectively, and $k ′ = 1 − k 2$. In particular, $c n − 1 (0,k)=K(k)= ∫ 0 π / 2 d t 1 − k 2 sin 2 t$ is the well-known complete elliptic integral of the first kind.

In [1] Neuman proved that $N k (x,y)$ is symmetric and homogeneous on $(0,∞)×(0,∞)$, and strictly decreasing with respect to $k∈[0,1]$ for fixed $(x,y)∈(0,∞)×(0,∞)$ with $x≠y$. In this context let us note that if a mean is homogeneous, then the order of its homogeneity must be 1; see [4].

Let us recall the notion of Schur quadratic convexity (concavity) [57] for a real-valued function on $(0,∞)×(0,∞)$.

A real-valued function $f:(0,∞)×(0,∞)→R$ is said to be strictly Schur quadratically convex on $(0,∞)×(0,∞)$ if $f( x 1 , x 2 ) for each pair of 2-tuples $( x 1 , x 2 ),( y 1 , y 2 )∈(0,∞)×(0,∞)$ with $max{ x 1 , x 2 } and $x 1 2 + x 2 2 = y 1 2 + y 2 2$. f is said to be strictly Schur quadratically concave if −f is strictly Schur quadratically convex.

The main purpose of this paper is to present the range of k such that the elliptic Neuman mean $N k (x,y)$ is strictly Schur quadratically concave on $(0,∞)×(0,∞)$. As an application, an inequality between the elliptic Neuman mean $N k (x,y)$ and the quadratic mean $Q(x,y)= ( x 2 + y 2 ) / 2$ is also given.

## 2 Two lemmas

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

Lemma 2.1 (See [[5], Corollary 2.1], [[6], Corollary 1], [[7], Corollary 1])

Suppose that $f:(0,∞)×(0,∞)→(0,∞)$ is a continuous symmetric function. If f is differentiable in $(0,∞)×(0,∞)$, then f is strictly Schur quadratically convex on $(0,∞)×(0,∞)$ if and only if

$(x−y) ( y ∂ f ( x , y ) ∂ x − x ∂ f ( x , y ) ∂ y ) >0$
(2.1)

for all $x,y∈(0,∞)$ with $x≠y$, and f is strictly Schur quadratically concave on $(0,∞)×(0,∞)$ if and only if inequality (2.1) is reversed.

Lemma 2.2 Let $t∈(0,1)$, $k∈[0,1]$, and

$f k (t)=c n − 1 (t,k)− ( 1 + t 2 ) 1 − t 2 2 t 1 − k 2 + k 2 t 2 .$
(2.2)

Then $f k (t)<0$ for all $t∈(0,1)$ if and only if $2 /2≤k≤1$, and there exists $λ=λ(k)∈(0,1)$ such that $f k (t)<0$ for $t∈(0,λ)$ and $f k (t)>0$ for $t∈(λ,1)$ if $k∈[0, 2 /2)$.

Proof We distinguish for the proof two cases.

Case 1. $k=1$. Then from (2.2) one has

$f 1 ( t ) = c n − 1 ( t , 1 ) − ( 1 + t 2 ) 1 − t 2 2 t 2 = cosh − 1 ( 1 t ) − ( 1 + t 2 ) 1 − t 2 2 t 2 = log ( 1 + 1 − t 2 ) − log t − ( 1 + t 2 ) 1 − t 2 2 t 2 , f 1 ( 1 − ) = 0 ,$
(2.3)
$f ′ (t)= ( 1 − t 2 ) ( 2 − t 2 ) 2 t 3 1 − t 2 >0$
(2.4)

for all $t∈(0,1)$. (Here and in the sequel, $f( t − )$ and $f( t + )$ denote, respectively, the left and right limit of f at t.)

From (2.3) and (2.4) we clearly see that $f 1 (t)<0$ for all $t∈(0,1)$.

Case 2. $0≤k<1$. Then (2.2) leads to

$f k ( 0 + ) =−∞,$
(2.5)
$f k ( 1 − ) =0,$
(2.6)
$f k ′ ( t ) = − 1 ( 1 − t 2 ) ( 1 − k 2 + k 2 t 2 ) − − k 2 t 6 + ( 3 k 2 − 2 ) t 4 + ( 1 − 3 k 2 ) t 2 + k 2 − 1 2 t 2 1 − t 2 ( 1 − k 2 + k 2 t 2 ) 3 / 2 = − − k 2 t 6 + ( 5 k 2 − 2 ) t 4 + ( 3 − 5 k 2 ) t 2 + k 2 − 1 2 t 2 1 − t 2 ( 1 − k 2 + k 2 t 2 ) 3 / 2 = − 1 − t 2 [ k 2 t 4 + ( 2 − 4 k 2 ) t 2 + k 2 − 1 ] 2 t 2 ( 1 − k 2 + k 2 t 2 ) 3 / 2 .$
(2.7)

Let

$g k (t)= k 2 t 4 + ( 2 − 4 k 2 ) t 2 + k 2 −1.$
(2.8)

$g k (0)= k 2 −1<0,$
(2.9)
$g k (1)=2 ( 2 2 − k ) ( 2 2 + k ) ,$
(2.10)
$g k ′ (t)=4 k 2 t 3 +4 ( 1 − 2 k 2 ) t,$
(2.11)
$g k ′ (0)=0,$
(2.12)
$g k ′ (1)=4(1−k)(1+k)>0,$
(2.13)
$g k ″ (t)=12 k 2 t 2 +4 ( 1 − 2 k 2 ) ,$
(2.14)
$g k ″ (0)=8 ( 2 2 − k ) ( 2 2 + k ) ,$
(2.15)
$g k ″ (1)=4 ( 1 + k 2 ) >0.$
(2.16)

We distinguish for the proof three subcases.

Subcase 2.1. $k= 2 /2$. Then (2.7) leads to the conclusion that

$f 2 / 2 ′ (t)= 2 1 − t 2 ( 1 − t 4 ) 2 t 2 ( 1 + t 2 ) 3 / 2 >0$
(2.17)

for all $t∈(0,1)$.

Therefore, $f 2 / 2 (t)<0$ for all $t∈(0,1)$ follows from (2.6) and (2.17).

Subcase 2.2. $2 /2. Then (2.10) and (2.15) lead to

$g k (1)<0,$
(2.18)
$g k ″ (0)<0.$
(2.19)

It follows from (2.14) that $g k ″$ is strictly increasing on $(0,1)$, then (2.16) and (2.19) lead to the conclusion that there exists $λ 0 ∈(0,1)$ such that $g k ′$ is strictly decreasing on $(0, λ 0 ]$ and strictly increasing on $[ λ 0 ,1)$.

From (2.12) and (2.13) together with the piecewise monotonicity of $g k ′$ we clearly see that there exists $λ 1 ∈( λ 0 ,1)$ such that $g k$ is strictly decreasing on $(0, λ 1 ]$ and strictly increasing on $[ λ 1 ,1)$. Then (2.9) and (2.18) lead to the conclusion that

$g k (t)<0$
(2.20)

for all $t∈(0,1)$.

It follows from (2.7) and (2.8) together with (2.20) that $f k$ is strictly increasing on $(0,1)$. Therefore, $f k (t)<0$ for all $t∈(0,1)$ follows easily from (2.6) and the monotonicity of $f k$.

Subcase 2.3. $0≤k< 2 /2$. Then from (2.10) and (2.11) we know that $g k$ is strictly increasing on $(0,1)$ and

$g k (1)>0.$
(2.21)

It follows from (2.9) and (2.21) together with the monotonicity of $g k$ that there exists $μ 0 ∈(0,1)$ such that $g k (t)>0$ for $t∈(0, μ 0 )$ and $g k (t)<0$ for $t∈( μ 0 ,1)$. Then (2.7) and (2.8) lead to the conclusion that $f k$ is strictly increasing on $(0, μ 0 ]$ and strictly decreasing on $[ μ 0 ,1)$. Therefore, there exists $λ=λ(k)∈(0, μ 0 )⊂(0,1)$ such that $f k (t)<0$ for $t∈(0,λ)$ and $f k (t)>0$ for $t∈(λ,1)$ follows from (2.5) and (2.6) together with the piecewise monotonicity of $f k$. □

## 3 Main results

Theorem 3.1 The elliptic Neuman mean $N k (x,y)$ is strictly Schur quadratically concave on $(0,∞)×(0,∞)$ if and only if $k∈[ 2 /2,1]$, and $N k (x,y)$ is not Schur quadratically convex on $(0,∞)×(0,∞)$ if $k∈[0, 2 /2)$.

Proof Since $N k (x,y)$ is symmetric and homogeneous of degree 1, without loss of generality, we assume that $x. Let $t=x/y∈(0,1)$, then

$N k (x,y)=y N k (t,1), ∂ t ∂ y =− x y 2 , ∂ t ∂ x = 1 y ,$
(3.1)
$∂ N k ( x , y ) ∂ y = N k (t,1)−t d N k ( t , 1 ) d t , ∂ N k ( x , y ) ∂ x = d N k ( t , 1 ) d t .$
(3.2)

Note that

$d N k ( t , 1 ) d t =− t 1 − t 2 c n − 1 ( t , k ) + 1 ( c n − 1 ( t , k ) ) 2 1 − k 2 + k 2 t 2 .$
(3.3)

It follows from (1.1) and (3.2) together with (3.3) that

$( y − x ) ( x ∂ N k ( x , y ) ∂ y − y ∂ N k ( x , y ) ∂ x ) = x ( y − x ) [ N k ( t , 1 ) − ( t + 1 t ) d N k ( t , 1 ) d t ] = 2 x ( y − x ) 1 − t 2 ( c n − 1 ( t , k ) ) 2 ( c n − 1 ( t , k ) − ( 1 + t 2 ) 1 − t 2 2 t 1 − k 2 + k 2 t 2 ) = 2 x ( y − x ) 1 − t 2 ( c n − 1 ( t , k ) ) 2 f k ( t ) ,$
(3.4)

where $f k (t)$ is defined as in Lemma 2.2.

Therefore, Theorem 3.1 follows easily from Lemmas 2.1 and 2.2 together with (3.4). □

Theorem 3.2 The elliptic Neuman mean $N k (x,y)$ is strictly Schur quadratically concave (or convex, respectively) on $(0,∞)×(0,∞)$ if and only if the function $N k (t,1)/Q(t,1)$ is strictly increasing (or decreasing, respectively) in $(0,1)$, where $Q(x,y)= ( x 2 + y 2 ) / 2$ is the quadratic mean of x and y.

Proof Without loss of generality, we assume that $x. Let $t=x/y∈(0,1)$, then from (3.1) and (3.2) together with (3.4) we get

$d ( N k ( t , 1 ) / Q ( t , 1 ) ) d t = − 2 t ( t 2 + 1 ) 3 / 2 ( N k ( t , 1 ) − ( t + 1 t ) d N k ( t , 1 ) d t ) = − 2 y ( y − x ) ( t 2 + 1 ) 3 / 2 ( y − x ) ( x ∂ N k ( x , y ) ∂ y − y ∂ N k ( x , y ) ∂ x ) .$
(3.5)

Therefore, Theorem 3.2 follows easily from Lemma 2.1 and (3.5). □

Theorem 3.3 The inequalities

$Q(x,y)> N 2 / 2 (x,y)$
(3.6)

and

$Q(x,y)< 2 K ( k ) 2 N k (x,y)$
(3.7)

hold for all $x,y>0$ with $x≠y$, and $k∈[0,1]$, and $N 2 / 2 (x,y)$ is the best possible lower elliptic Neuman mean bound for the quadratic mean $Q(x,y)$.

Proof Without loss of generality, we assume that $y>x>0$. Let $t=x/y∈(0,1)$ and $L k (t)= N k (t,1)/Q(t,1)$. Then

$L k (t)= N k ( x , y ) Q ( x , y ) ,$
(3.8)
$L k (0)= 2 K ( k ) ,$
(3.9)
$L k (1)=1.$
(3.10)

We distinguish for the proof two cases.

Case 1. $k∈[ 2 /2,1]$. Then from Theorems 3.1 and 3.2 we clearly see that $L k$ is strictly increasing on $(0,1)$. Then (3.8)-(3.10) lead to the conclusion that

$2 K ( k ) < N k ( x , y ) Q ( x , y ) <1.$
(3.11)

In particular, for $k= 2 /2$ we have

$N 2 / 2 ( x , y ) Q ( x , y ) <1.$
(3.12)

Therefore, inequalities (3.6) and (3.7) follow from (3.11) and (3.12).

Case 2. $k∈[0, 2 /2)$. Then (3.4) and (3.5) together with the Subcase 2.3 in Lemma 2.2 lead to the conclusion that there exists $λ=λ(k)∈(0,1)$ such that $L k ′ (t)>0$ for $t∈(0,λ)$ and $L k ′ (t)<0$ for $t∈(λ,1)$, hence $L k$ is strictly increasing on $(0,λ]$ and strictly decreasing on $[λ,1)$. Therefore, $N k (x,y)>Q(x,y)$ for all $x,y>0$ with $x/y∈(λ,1)$ follows from (3.8) and (3.10) together with the monotonicity of $L k$ on $[λ,1)$, and the optimality of inequality (3.6) follows.

Note that

$L k (0)= 2 K ( k ) < 2 K ( 0 ) = 2 2 π =0.90031⋯<1.$
(3.13)

From (3.8), (3.9), (3.13), and the piecewise monotonicity of $L k$ we clearly see that

$N k ( x , y ) Q ( x , y ) = L k (t)> L k (0)= 2 K ( k ) .$
(3.14)

Therefore, inequality (3.7) follows from (3.14). □

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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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All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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Chu, Y., Zhang, Y. & Qiu, S. Schur quadratic concavity of the elliptic Neuman mean and its application. J Inequal Appl 2014, 397 (2014). https://doi.org/10.1186/1029-242X-2014-397