Research Article | Open Access
Peng Gao, "On an Inequality of Diananda—Part IV", International Journal of Mathematics and Mathematical Sciences, vol. 2009, Article ID 468290, 24 pages, 2009. https://doi.org/10.1155/2009/468290
On an Inequality of Diananda—Part IV
We extend the results in parts I–III on certain inequalities involving the weighted power means as well as the symmetric means. These inequalities can be largely viewed as concerning the bounds for ratios of differences of means and can be traced back to the work of Diananda.
Let be the weighted power means: , where denotes the limit of as , , , and () are positive real numbers with . In this paper, we let and unless otherwise specified, we assume .
For , the th symmetric function of and its mean are defined by
We define , , and and we shall write for and similarly for other means when there is no risk of confusion.
For a real number and mutually distinct numbers , , , we define where we interpret as . We also define to be .
A result of Diananda [4, 5] (see also [6, 7]) shows that (1.3) and (1.4) hold for , . When , the sets 's for which (1.3) or (1.4) holds have been completely determined by the author in [1–3]. Moreover, it is shown in  that (1.4) does not hold in general unless .
For general 's, the cases for which inequality (1.3) or (1.4) holds are open (with in (1.4)). It is our main goal in this paper to study these cases. We first restrict our attention to the case . This is partially because of the following result in  (note that there is a typo in the original statement though).
Moreover, for the unweighted case , the author [3, Theorem 3.5] has shown that (1.3) holds for with when and (1.4) holds for with when . The values of 's are best possible here; namely, no larger 's can make (1.3) hold for and similarly no smaller 's can make (1.4) hold for .
Theorem 1.2. For , , For , ,
After studying (1.3) and (1.4) for the set , we move on in Section 3 to the case with . Our remark earlier allows us to focus on (1.3) only. In this case, we can recast (1.3), via a change of variables, as the following: where , and We will show that (1.8) holds for all if and only if it holds for the case . Based on this, we will then be able to prove (1.8) for certain , , 's satisfying a natural condition.
Theorem 1.3. Let , then for any integer , where for (with here),
As was pointed out in , the proof given in  for the above theorem is not quite correct. In Section 4, we will study inequalities involving the symmetric means and our results include a proof of Theorem 1.3.
2. Proof of Theorem 1.2
In view of Theorem 1.1, one only needs to prove (1.6) for and similarly (1.7) for . In this proof we assume that . The case will follow by taking the limit. We first prove (1.6) and we define If then ; otherwise we may assume and for some , then We want to show that the right-hand side above is nonnegative. It suffices to show that each single term in the sum is nonnegative. Without loss of generality, we now show that We have Now we set so that Hence We want to show that the right-hand side expression above is nonnegative, and by setting , this is equivalent to show that for , and calculation yields that We now set with and we consider By Cauchy's mean value theorem, It follows that for , It follows from the discussion in  (see the function defined in the proof of Theorem there and note that we have here) that . This implies that so that is an increasing function of and we then deduce that This shows that and hence and by letting and repeating the above argument, we conclude that which completes the proof for (1.6).
Now, to prove (1.7), we consider Similar to our discussion above, it suffices to show . Now Now we set so that where the inequality follows from the observation that the function is decreasing for .
3. A General Discussion on (1.8)
Theorem 3.1. For fixed , , (1.8) holds for all if and only if it holds for .
Proof. We consider the function
The theorem asserts that in order to show , it suffices to check the case . To see this, we may assume by homogeneity that and we let be the point in which the absolute minimum of is reached. We may assume that for otherwise if for some , by combining with and with , and noticing that is an increasing function of by [2, Lemma 2.1], we have
where , , and . We can then reduce the determination of the absolute minimum of to that of with different weights.
If is a boundary point of , then , and in this case we show that follows from , where and . On writing explicitly, we get Meanwhile, is equivalent to Thus, it amounts to show that which is equivalent to Now the above inequality follows from and since is an increasing function of .
Thus follows from . Moreover, on writing and noticing that , we deduce that . Hence the determination of can be reduced to the determination of with different weights.
It remains to show the case , so that is an interior point of . In this case we have Thus solve the equation Note that Thus if then . If , we note that can have at most two roots in since it is easy to see that can have at most one positive root. As , this implies that has only one root in . As are the distinct roots of , this implies that we must have so that we only need to show for the case and this completes the proof.
In what follows, we will apply Theorem 3.1 to establish (1.8) for certain 's. Before we proceed, we note that there is a natural condition to be satisfied by , , in order for (1.8) to hold, namely, if we take and rewrite it as On taking , we conclude that Before we prove our next result, we need two lemmas.
Lemma 3.2. Fixing , the function is concave for and is convex for .
Proof. Calculation yields that where Now Thus if , then for and for . Since , this implies that for . As , we then conclude that for and for . It follows from this that is convex for and the other assertion can be shown similarly, which completes the proof.
Lemma 3.3. For , , , , the function is convex for when , , or , .
Proof. Direct calculation shows that where One then easily deduce the assertion of the lemma from the above expression of and this completes the proof.
Proof. By Theorem 3.1, it suffices to prove the theorem for the case . We write for short in this proof. What we need to prove is the following: Without loss of generality, we may assume that , and define so that we need to show that for . We have where By Lemmas 3.2 and 3.3, we see that is a strictly convex function for , or , , . Note that by our assumption (3.12) with strict inequality, On the other hand, note that satisfies so that As is strictly convex, this implies that there are exactly two roots , of satisfying that and . Note further that and , which implies that for . Similarly, we note that for with being small enough. This combined with the observation that as is an increasing function of implies that for which completes the proof.
Theorem 3.5. For , with equality holding if and only if or , , , or , .
We need to show and we have
By a change of variables: , , we may assume in (3.30) and rewrite it as
We want to show . Let be the point in which the absolute minimum of is reached. We may assume . If , then is a boundary point of , and in this case we have
with equality holding if and only if , . Now suppose and for some , then solve the equation
Equivalently, solve the equation
As the right-hand side expression above is an increasing function of , the above equation has at most one root (regarding as constant). So we only need to show for the case in (3.31) for some . In this case we regard as a function of and we recast it as
Here and . Note first that when , so that we may now assume . We have
Note that has one root so that if , then at this point
Note also that , . This implies that for . Hence for . As it is easy to see that is a convex function of for fixed , we conclude that is an increasing function of for fixed . Thus for , ,
with equality holding if and only if .
Thus we have shown ; hence with equality holding if and only if or , , or , . By letting tend to , we have (with weights , ) with equality holding if and only if or and either or , . It follows by induction that with equality holding if and only if or , , or , and this completes the proof.
4. Inequalities Involving the Symmetric Means
In this section, we set , . As an analogue of (1.8) (or (1.3)), we first consider where , , , and The case , in (4.1) is just Theorem 1.3. In what follows, we will give a proof of Theorem 1.3 by combining the methods in [8, 10]. Before we prove our result, we would like to first recast (4.1) for the case as where is defined as in the statement of Theorem 1.3. Now we need two lemmas.
Lemma 4.1. For ,
Proof. We follow the method in the proof of Lemma in . We write as From the above we see that in order for (4.4) to hold, it suffices to show that is increasing on . The logarithmic derivative of is Note that for , we have by considering the Taylor expansions to the order of . It follows from this that It is easy to see that the function is an increasing function for . Hence
Lemma 4.2. Inequality (4.1) holds for all in the case , if it holds for with .
Proof. In this proof we assume that . We prove the lemma by induction on . When , the assertion holds as a special case of (1.6). Now assume that the assertion holds for , and we proceed to show it also holds for . If , we use the equivalent form (4.3) of (4.1) for the case to see that what we need to prove is
By the induction case , we have
Using this with Lemma 4.1 together with the observation that
we see that inequality (4.12) follows from (4.13).
Thus from now on we may focus on the case . Since both sides of (4.1) are homogeneous functions, it suffices to show that where Assume that attains its minimum at some point with , . If , then (4.16) holds. Furthermore, if , then (4.16) also holds, being a special case of (1.6). Thus without loss of generality, we may assume and here. We may also assume that when , since otherwise inequality (4.1) becomes an identity. Consider the function on the set It is minimized at . It is easy to see that has the form where , , and are nonnegative constants with depends on . We now set and with . Note here that equality holds if and only if or . We regard the above function as a function of and recast it as For , calculation yields Since , we must have and we can further recast this as Now if , we can repeat the same argument for the pair . By a slightly abuse of notation, we obtain It is easy to see that (since we assume and when ) Moreover, one checks easily that the function is increasing with respect to each variable when . It follows that when , which implies that by decreasing the value of while keeping fixed, one is able to get a smaller value for , contradicting the assumption that it attains its minimum at .
Hence we conclude that is minimized at with satisfying . In this case (4.16) holds by our assumption which completes the proof.
Now we are ready to prove a slightly generalization of Theorem 1.3.
Corollary 4.3. Inequality (4.1) holds in the cases , and .
Proof. The first case is just Theorem 1.3 and by Lemma 4.2, it suffices to show that inequality (4.1) holds for the case with and this has been already treated in the proof of Theorem in . For the case , by Lemma 4.2 again, we only need to check the case with . In this case we define As in the proof of Theorem 1.2, it suffices to show that By a change of variables , we can recast inequality (4.29) as As and , we conclude that for as is a concave function for and this completes the proof.
We recall a result of Kuczma .
Theorem 4.4. For , , with and the result is best possible.
The above theorem combined with Corollary 4.3 immediately yields the following.
Next, we consider the following inequality: where and We note here that it is easy to check that the function is a decreasing function for so that we have .
As an analogue of Theorem 1.1, one can show similarly the following.
Proposition 4.6. Let ; if (4.33) holds for , then it also holds for any .
The case , of (4.33) was established in . In the case of , one possible way of establishing (4.33) is to combine Theorems 4.4 and 1.2 together. However, this is not always applicable as one checks via certain change of variables that one needs to have