Complementary Inequalities Involving the Stolarsky Mean

Let us begin with some definitions. Given the positive real numbers and and the real numbers and , the difference mean or Stolarsky mean of and is defined by (see, e.g., [1] or [2])

(1.1)

The power mean of power corresponding to the real numbers is defined by

(1.2)

The relation between the Stolarsky mean and the power mean can be written as

(1.3)

It is well known that for fixed and , we have the inequality

(1.4)

with equality for (independent of ), or for (see [3–5] or [6]).

Shisha and Mond [7] obtained a complementary result which examines the upper bounds of (1.4) for weighted versions of the power means. Also, we have a considerable amount of work regarding the complementary means done by many authors, including Diaz and Metcalf [8], Beck [9], and Páles [10].

Returning to our problem, by defining the function

(1.5)

we obtain

(1.6)

Using the inequalities between power means (1.4), if and only if therefore if and only if This condition is more general than but there are details in the subsequent proofs which would not be satisfied in the other cases.

As the minimum of over is (possible only for ), it is natural to question what the maximum of is, and, eventually, to find the configuration where this is attained. Since the problem of finding the maximum of only makes sense when all the variables of are restricted to the compact interval .

The first theorem in the next section, deals with finding the maximum and the corresponding optimal configuration. The result enables one to obtain elegant proofs for some related inequalities. In the end of the present work we obtain some asymptotic limits relative to the configuration where the maximum of is attained.

Theorem 2.1.

Given the positive integer , the real numbers and . Consider the function , defined by (1.4). Then the following assertions are true.

1. (1)
   The function attains its maximum at a point if and only if of the variables are equal to while the other are equal to b, where can be

   

   (2.1)
    

provided the limit exists.

As an application of Theorem 2.1, the following problem (see [3, pages 70–72]) is solved.

Corollary 2.2.

Given the positive integer , determine the smallest value of such that the inequality

(2.3)

holds true for all positive real numbers

Theorem 2.3.

Given the positive integer , the smallest value of such that (2.3) holds true for all positive real numbers is

(2.4)

In the following theorem we examine the behavior of when the numbers , in Theorem 2.1, are terms of a sequence with certain properties.

Theorem 2.4.

Consider the sequences and satisfying with and For each define as in (2.1), for the powers and Then the verifies

(2.5)

Lemma 3.1.

The function attains its maximum at the point if and only if for all

Proof of Lemma 3.1.

Since is continuous on the compact interval , there is a point where attains its maximum. If is an interior point of , then for all therefore

(3.1)

which implies

(3.2)

for all However, if , then which clearly is not the maximum of Consequently, lies on the boundary of . Due to symmetry and since there exist and such that

(3.3)

If then For this case, consider the function defined by

(3.4)

If the point where the maximum of is attained is interior to , in virtue of Fermat's theorem, we deduce that

(3.5)

for all This is equivalent to

(3.6)

hence

(3.7)

A simple computation shows that

(3.8)

and for this configuration we have

(3.9)

Let us define the function as

(3.10)

and prove it is increasing. Indeed, one finds

(3.11)

where , and Since it follows that so is increasing and the upper bound is

(3.12)

This finally proves that of the numbers are equal to while the other are equal to as anticipated. This ends the proof of Lemma 3.1.

The only thing to be done is to find the value of for which the expression

(3.13)

attains its maximum.

To do this, consider the function defined by

(3.14)

and find the points where the maximum of is attained in the interval .

The critical points of are found from the equation

(3.15)

so they satisfy

(3.16)

As seen in the definition of the Stolarsky mean for this case,

(3.17)

It is finally found that has a single critical point

(3.18)

which (fortunately) is contained in the interior of

Taking into account that the second derivative of is

(3.19)

the extremal point is a point of maximum for , and also the function is decreasing on the interval . Because , we obtain for , and for Finally, this means that is increasing on and decreasing on .

We conclude that

(3.20)

The maximum of (3.13) is then attained when takes one of the values and , where

(3.21)

The value of this is to be called from now on.

Remark 3.2.

Because in our case

(3.22)

the Stolarsky mean satisfies the strict inequality , so

1. (2)
   Using the properties of the integer part , we obtain

   

   (3.23)
    

so

(3.24)

It is then enough to work out the limit

(3.25)

On the other hand we have

(3.26)

Due to symmetry the partial derivatives are equal, so the desired limit is

(3.27)

Taking the limit in (3.23), we obtain that the limit of as is confined to the interval

Proof of Theorem 2.3.

Considering and in Theorem 2.1, we obtain

(3.28)

Out of here, we can immediately obtain the best constant for which

(3.29)

Following the steps mentioned before, the function gets the maximum only when

(3.30)

where , or .

This proves that the following inequality holds:

(3.31)

so the best constant will be

(3.32)

Remark 3.3.

Although appealing, a result involving arbitrary powers would depend on which the exact value of is (out of the two possibilities). At the same time, the power on the righthand-side can only be obtained for

Proof of Theorem 2.4.

To ease the notations we write and The following relation holds:

(3.33)

Using the notation the limit can be written as

(3.34)

Since the denominator converges to it only remains to examine the limit

(3.35)

which can be written as

(3.36)

It can be proven that the two terms of (3.36) converge to finite limits, and analyze each. From the hypothesis so the limit of the first term is

(3.37)

while second term can be written as

(3.38)

Since

(3.39)

the same argument as above can be used to obtain

(3.40)

where

(3.41)

In the end we obtain

(3.42)