Comparison of orders generated by Ky Fan type inequalities for bivariate means

In this paper, we deal with seven types of Ky Fan type relations between bivariate, symmetric and homogeneous means. For each relation we determine necessary and sufficient conditions for means to be in this relation. Additionally, we investigate the dependencies between these relations.


A little bit of history
Let A n , G n and H n denote the arithmetic, geometric and harmonic means of positive arguments x = (x 1 , . . ., x n ) .
It is well known that for arbitrary x the inequalities hold In the 1960s it was discovered that for x ∈ (0, where 1 − x = (1 − x 1 , . . ., 1 − x n ).Beckenbach and Bellman [5] attributed the right-hand side of this elegant result to Ky Fan, while the left inequality was proved by Wang and Wang [15].Later on, similar results were obtained for other multivariate, homogeneous means, also in weighted cases, see e.g.[8,14].
As the three classical means are members of the monotone family of power means A r,n (x) = A 1/r n (x r ), x r = (x r 1 , . . ., x r n ), where A 0,n = G n , it is natural to ask whether similar inequalities hold for them.Chan et al. [6] discovered, that for r < s the inequality A r,n (x) A r,n (1 − x) ≤ A s,n (x) A s,n (1 − x) , hold only if the number of variables equals 2. For more details see the review paper by Alzer [4].
Neuman and Sándor [11] proved the following sequence of inequalities for bivariate means.For x, y ∈ (0, where L, P, NS, T stand for the logarithmic, first Seiffert, Neuman-Sándor and second Seiffert means. The following inequalities look like intriguing companion of ( 1) The outermost inequality is due to Sándor [13], while the refinement was found by Alzer [2].
In [12] Neuman and Sándor developed a method that allowed them to deduce the following chain of inequalities from (2) .
The additive counterparts of ( 1) and (3) were found by Alzer [1], [3] G and What is even more important from our point of view, Alzer also found out that the analogous inequality between the geometric and harmonic means is not valid.Three more types of inequalities between classical means have been investigated by mathematicians.For x = (x 1 , . . ., In [7] we find the counterpart of the classical Ky Fan inequalities There is also a version for differences of reciprocal of means proven in [9] 1 Finally, the additive analogue follows from Jensen's inequality for log(1 + exp y) by substitution y i = log x i [10, 3.2.34],while can be obtained applying the Jensen inequality and substitution y i = 1/x i to the function y 1+y .Like in the Ky Fan case, similar inequality between the geometric and harmonic means cannot be established.

What is this paper about?
In the previous section we presented seven types of inequalities between classical means: arithmetic, geometric and harmonic.We also know that similar inequalities can be established for other bivariate means.
In this paper we focus on the family of bivariate, symmetric and homogeneous means defined on R + .We shall denote this family by S. Each type of inequalities defines certain relation S, e.g. the standard inequality M ≤ N leads to the relation {(M, N ): ∀x, y > 0 M(x, y) ≤ N (x, y)}.
The aim of this paper is twofold: (a) for each relation we determine necessary and sufficient conditions for means to be in this relation, and (b) investigate the dependencies between these relations.
The conditions in (a) will be expressed in a uniform language of Seiffert's functions (see Definition 3.1), which will allow them to be easily compared and to prove dependencies.
The first results in this direction were established by one of the authors in [17].
Another essential class of functions we use in this paper is the class of Seiffert functions.
is called Seiffert function.
If for x = y we denote z = |x−y| x+y , then the important relation between means and Seiffert functions is given by the following identity from [16]: This shows the one-to-one correspondence between means and Seiffert functions given by the formula The name of Seiffert function comes from Heinz-Jürgen Seiffert, who was the first to show that The means will be denoted by uppercase letters, while lowercase ones will denote the corresponding Seiffert functions.We shall use sansserif font to denote the well-known means and their Seiffert functions.In particular formula (4) can be written as which reflects the condition fulfilled by every mean.
Let us note two obvious properties of Seiffert function the follow immediately from (4): If there is no risk of ambiguity, we shall skip the argument of means.Next we recall the definition of Ky Fan functions from [17].
Clearly every nondecreasing function is a Ky Fan functions, but this class is much broader.For example, if f is nondecreasing in (0, 1/3) and f (x) ≥ f (1/3 − ) for x ≥ 1/3, than f is also a Ky Fan function.The following property will be useful

Property 3.2 If f is a Ky Fan function and lim x→0
The Ky Fan functions will play an important role in our considerations due to the following fact proven in [17].Theorem 3.1 For a function f : (0, 1) → R the following conditions are equivalent We shall give here another useful equivalence.
Theorem 3.2 For a function f : (0, 1) → R the following conditions are equivalent , which shows that |(1+x)−(1+y)| (1+x)+(1+y) assumes all values between 0 and a as x varies from zero to infinity and a remains fixed.Therefore (ii) can be written as follows: for each 0 < a < 1 sup t<a f (t) ≤ f (a), which yields (i).
∧-shaped functions will appear quite often in our examples, so it is good to have an easy criterion to check if such a function is a Ky Fan.

Lemma 3.1 For a ∧-shaped function f the following conditions are equivalent
Proof Choose arbitrary 0 < a < 1 and 0 < t < a 2a+1 and suppose (b) holds.This means . This means f is Ky Fan.If (b) does not hold, then by continuity of f one can easily find an a close to 1 such that f ( a 2a+1 ) is strictly greater than f (a), so the Ky Fan condition is not satisfied.
Let us note one more fact, quite surprising, but extremely important in our considerations.

Lemma 3.2 Every Seiffert function is a Ky Fan function.
Proof Choose arbitrary 0 < a < 1 and 0 < t < a 2a+1 .Then for every Seiffert function f one has Let us now define the relations between means.

Definition 3.4
For means M, N we define the following relations

Necessary and sufficient conditions
In this section, we will show how the different types of inequality between means can be expressed by the properties of their Seiffert functions.These results will be the basis for comparing the relationships between different types of inequality.
Recall that by M, N we denote the means, and by m, n their respective Seiffert functions.Let's start with the most basic.Proof Follows immediately from (5).
Following three theorems concern Ky Fan type inequalities.
and application of Theorem 3.1 completes the proof.
And now three theorems where M(1 + x, 1 + y) are involved.

Examples
Before formulation of our results regarding the dependencies between relations ( 6)-( 12) let us consider some examples of means and investigate their properties.
Example 5.1 Consider two Seiffert functions and their corresponding means .
Consider now the family of weighted harmonic means of K 0 and K 1 : , where 0 < t < 1.
The reader will easily verify that their Seiffert functions are We will investigate relations between K t and the arithmetic mean A (recall that a(z) = z).
Let us begin with the function (k t − a)(z).We have . We have k t a (0 , so it strictly increases for t ≤ 1/3 and is ∧-shaped otherwise.Using Lemma 3.1 we see that k t a is Ky Fan if, and only if t ≤ 3/7.Finally we came to the functions 1 We easily calculate that a is Ky Fan for all t ∈ (0, 1).Note that it is negative for 0 < t < 1/2 and changes sign for 1/2 < t < 1.
Using Theorems 4.1-4.7 we can summarize this example as follows Our second example will be quite similar, but the differences are essential.

Example 5.2 Consider two Seiffert functions and their corresponding means
and as in Example 5.1, we create a family of weighted harmonic means , where 0 < t < 1, with Seiffert functions We have (a ), so the function a − j t is increasing for t ≤ 3/8 and ∧-shaped otherwise.Thus we conclude that a − j t is nonnegative if, and only if t ≤ 1/2 and is Ky Fan function if, and only if t ≤ 117/238 ≈ 0.4916.

Let us establish the properties of
) 2 we see that the function a j t is strictly increasing for t ≤ 1/3 and ∧-shaped otherwise.Using Lemma 3.1 we discover that it is Ky Fan for t ≤ 9/19 ≈ 0.4737.
Finally we consider 1 We see that 1 j t − 1 a (0 + ) = 0. We shall show that it is or increasing or ∧-shaped.Let us calculate the derivative , it has a real negative root.Our goal is to show, that is does not have more than one root in the interval (0, 1).Consider two cases: If for some t, the polynomial L t had two additional roots in the interval (0, 1), then its derivative would be positive to the right of the largest of them.But it is impossible because and changes sign once otherwise.Therefore 1 ≈ 0.2753 and is ∧-shaped otherwise.
Using Lemma 3.1 we find that Example 5. 3 We rewrite inequalities (4) in an equivalent form and use it to define the Seiffert function.
The mean generated by m c is given by the formula Let us recall that a(z) = z, so The expression d dz z 2 (2c − z) has two zeroes at z 0 = 0 and z 1 = 4 3 c.Therefore a m c is strictly increasing if c ≥ 3  4 and is ∧-shaped otherwise.Applying Lemma 3.1 we see that it is a Ky Fan function if and only if c > 13  24 .Consider now the difference .
We are interested in the sign of which is the same as the sign of u c (z) = z 4 − 4cz 3 + 4c 2 z 2 − 4c 2 z + 6c 3 .We have and lim x→0 + m n (x) = 1.This fact, Property 3.2 and properties of monotonic functions in conjunction with Theorems 4.2, 4.5, 4.3 and 4.6 show that the relations ≺ C , ≺ R , ≺ + C , ≺ + R are also antisymmetric.Therefore ≤, ≺ C , ≺ R , ≺ + C , ≺ + R define partial orders on the set of means.Surprisingly, the relations ≺ A and ≺ + A are not reflexive.Let us spend a while investigating this interesting phenomenon.
Let us rewrite the definition of Seiffert functions (4) in the following form: The relations ≺ A and ≺ + A are determined by the difference of reciprocals of Seiffert functions.For c ∈ By Theorems 4.4 and 4.7, for any c . Now we shall prove some lemmas about properties of our relations.
Proof Follows immediately from Theorems 3.1, 3.2 and the fact, that every nondecreasing function is a Ky Fan function.Proof Suppose M ≺ R N does not hold.Then for some 0 < a < 1 and 0 < t < a 2a+1 the difference of corresponding Seiffert functions satisfies This yields n(t) > n(a) which is impossible, since n is a Seiffert function and thus Ky Fan.
Example 5.5 shows that the additive relations do not imply nor classical nor reciprocal relations.One of the main reasons for that is two means can be in an additive relations even if they are not comparable in the ordinary sense.But looking at Example 5.1 (0.48 < t < 0.5) we see that even if the difference of reciprocals of Seiffert functions preserves sign, the classical and reciprocal relations may not hold.As we shall see from the Lemma below, the reason is that the means in Example 5.1 are comparable in a wrong direction.
x+y and T(x, y) = |x − y| 2 arctan |x−y| x+y are means (called today the first and the second Seiffert means).

Theorem 4 . 2 [ 17 ,Theorem 4 . 4
Theorem 3]  Let M, N be means and m, n their Seiffert functions.The following conditions are equivalent:1.M ≺ C N , 2.m n is a Ky Fan function.Theorem 4.3 [17, Theorem 4] Let M, N be means and m, n their Seiffert functions.The following conditions are equivalent: (a) M ≺ R N , (b) m − n is a Ky Fan function.Let M, N be means and m, n their Seiffert functions.The following conditions are equivalent: (a) M ≺ A N , (b) 1 m − 1 n is a Ky Fan function.Proof By (5) the inequality in (a) can be written as |x − y| 2m |x−y| 2

Theorem 4 . 5 Theorem 4 . 6
Let M, N be means and m, n their Seiffert functions.The following conditions are equivalent:(a) M ≺ + C N , (b) mn is a nondecreasing function.Proof Using (5) the inequalityM(x, y) M(x + 1, y + 1) ≤ N (x, y) N (x + 1, y + 1)can be written as m n|(1 + x) − (1 + y)| (1 + x) + (1 + y) ≤ m n |x − y| x + yand application of Theorem 3.2 completes the proof.Proofs of the following two theorems are similar.Let M, N be means and m, n their Seiffert functions.The following conditions are equivalent: (a) M ≺ + R N , (b) m − n is a nondecreasing function.Theorem 4.7 Let M, N be means and m, n their Seiffert functions.The following conditions are equivalent: is strictly increasing for c ≥ 1 and-by Lemma 3.1-is a Ky Fan function for c ≥ 2 3 .The definition of m c implies that for all c we have m c ≤ a and m c (z) = c 2 z c 2 +z 2 (2c−z) .Let us investigate the function

Lemma 6 . 2 2 . 6 . 3
If M, N are means and M ≺ R N or M ≺ + R N , then M ≤ N .Proof By Theorem 4.3 the difference of corresponding Seiffert functions m − n is Ky Fan, and since lim z→0 + (m − n)(z) = 0 we have m − n ≥ 0 by Property 3.Lemma If M, N are means and M ≺ C N or M ≺ + C N , then M ≺ R N .

Lemma 6 . 4
If M, N are means and M ≺ A N and N ≤ M, then N ≺ C M and N ≺ R M. Proof If 0 < a < 1 and 0 < t < a 2a+1 , then for corresponding Seiffert functions we have 0 ≤ n(t) − m(t) m(t)n(t) ≤ n(a) − m(a) m(a)n(a) .Multiplying this by n(t) ≤ n(a) and adding 1 we get n m (t) ≤ n m (a), which is equivalent to M ≺ C N .By Lemma 6.3 M ≺ R N is also valid.

Table 1
shows the correspondences between the relations.Symbol in row R 1 and column R 2 means that a R 1 b does not imply a R 2 b nor bR 2 a for all a, b.Symbol ⇒ in row R 1 and column R 2 means that a R 1 b implies a R 2 b for all a, b.Symbol ⇐ in row R 1 and column R 2 means that a R 1 bimplies bR 2 a for all a, b.Letters refer to justifications listed below.The last line has been added to illustrate the special case described in Lemma 6.4.Take t = 1/2 in Example 5.2, b Lemma 6.2, c Take t = 0.48 in Example 5.2, d Lemma 6.1 (every nondecreasing functions is Ky Fan), e Lemma 6.3, f Take t = 0.42 in Example 5.2, g Example 5.5, h Take t = 0.3 in Example 5.2, i Example 5.7, j Take c=0.65 in Example 5.3, k Take c=0.9 in Example 5.3, l Example 5.8, m Take t = 0.35 in Example 5.2, n Take t = 0.4 in Example 5.2, o Lemma 6.4.

Table 1
Correspondences between relations