Abstract
Optimal bounds for the weighted geometric mean of the first Seiffert and logarithmic means by weighted generalized Heronian mean are proved. We answer the question: for what the greatest value and the least value such that the double inequality, , holds for all with are. Here, and denote the first Seiffert, logarithmic, and weighted generalized Heronian means of two positive numbers and respectively.
1. Introduction
Recently, means has been the subject of intensive research. In particular, many remarkable inequalities for the Seiffert, logarithmic, and Heronian mean can be found in the literature [1–11]. In the paper [1], authors proved the following optimal inequalities:
Let , , then is the first Seiffert mean, which was introduced by Seiffert in [9] In [9], Seiffert proved that , where is the identric mean is the logarithmic mean is the weighted geometric mean is the weighted generalized Heronian mean introduced by Janous [7] It is well known, that is a strictly decreasing continuous function of the argument . From this and from results of [1], it is natural to assume that there exist optimal functions , , such that The purpose of this paper is to find the optimal functions. For some other details about means, see [1–11] and the related references cited there in.
2. Main Results
The main result of this paper is the following theorem.
Theorem 1. Let , , . Then where , are the best possible functions.
Proof. First, we prove the left inequality of (8). The inequalities (1) imply that
From for (see (14)) we obtain that is the optimal function.
Without loss of generality, we assume that . Let ; then . The right inequality of (8) can be rewritten as
Simple computations lead to
Then the inequality (11) is equivalent to
Denote
From and we have for . It implies . From , , we obtain and so . It implies that for , . This leads to
If we show for , then will be the best function in (8). Simple computations lead to which is equivalent to
Using the inequality for it suffices to show that
It will be done, if we show and . It follows from being a linear continuous function in the argument
is equivalent to
From it suffices to show that which is equivalent to
It follows from and
where .
Next, we show that
The inequality (21) is equivalent to
So, it suffices to show that
It is easy to see that
Because of
it suffices to prove for . From
, for , we have done it, if we show
on and on .
Simple computation gives
The inequality is equivalent to
From we get is a decreasing function. implies on . So, we obtain is a decreasing function. From we have on . It implies that is a decreasing function. From we get on . So on .
Next, we show on .
Simple computation gives
The inequality is equivalent to
on . From , , , , it suffices to show that on ; on and has only one root in .
First, we show on . From , , we have
where
It is easy to see that for . From on and we have . It implies on .
Next, we show on . Simple computation gives
where
, , imply so on .
Finally, we show that has only one root on . From we obtain is a decreasing function. Because of we have on so is a concave function. From and we have that has only one root on . It implies on . So, the proof of decreasing of is complete.
In what follows, we find the representation of the function .
It is easy to see that
Equation (36) can be rewritten as
where
Simple computations give
where is a suitable function. Similarly we have
where is a suitable function. Denote . Then
where is a suitable function. Using the L’Hospital's rule we obtain
The proof is complete.
Acknowledgments
The work was supported by VEGA Grant no. 1/0530/11 and KEGA Grant no. 0007 TnUAD-4/2013. The author thanks to the faculty FPT TnUAD in Púchov, Slovakia for its kind support and the anonymous referees for their careful reading of the paper and fruitful comments and suggestions. The author would especially like to thank Prof. Walther Janous for his kind reading the manuscript and for his correction of the calculation .