#### Abstract

We obtain inequalities of Hölder and Minkowski type with weights generalizing both the case of weights with alternating signs and the classical case of nonnegative weights.

#### 1. Introduction

Recently Chunaev [1] obtained Hölder and Minkowski type inequalities with alternating signs. His results are a supplement to Jensen type inequalities with alternating signs obtained earlier by Szegö [2], Bellman [3, 4], Brunk [5], and others (see [6–11], [12, Section 5.38], and also Remark 7).

In this paper, we intend to give inequalities of Hölder and Minkowski type with more general weights, including both the case of weights with alternating signs and the classical case of nonnegative weights (see, e.g., [12, Section 4.2] and [1, 13]). Namely, weights , satisfying the property are considered. We follow proofs in [1] with several changes in order to obtain our results.

In what follows, we denote nonnegative sequences of real numbers in bold print; for example, or , where is a positive integer or infinity. Expressions like mean that all elements of equal . In proofs we use several well-known inequalities for and :

#### 2. Hölder Type Inequalities

In this section, we show that there is not a direct analog of Hölder’s inequality in the case of our weights, but one of reverse Hölder’s inequalities exists. Note that reverse Hölder’s inequalities for nonnegative weights are well studied (see [13]).

Theorem 1. *Let and be nonincreasing such that
**
where . If, moreover, , and , , then
**
The left hand side of (6) should be read as there exists no positive constant, depending on , or , which bounds the fraction in (6) from below.*

Before the proof of Theorem 1, we establish the following fact.

Lemma 2. *Let be nonincreasing and nondecreasing such that for . If, moreover, for , then
*

*Proof. *Applying the Abel transformation, we have
where the latter expression is nonnegative since the sequences and are nonincreasing, and . The equality holds, for example, if .

*Proof. *We denote the fraction in (6) by . Applying the Abel transformation to the numerator and the denominator of easily yields . But we prove even more, namely, that there exist no positive constants bounding from below. Following [1], let , , where is even, and let be positive and nondecreasing. The sequence is arbitrary except such that for all . It follows that
Thus cannot be bounded from below by a positive absolute constant or a constant depending on ,, maximum or minimum elements of and .

Now we prove the right hand side of (6). Here denotes the numerator of . First we apply the Abel transformation:
By the right hand side of (4) and the Abel transformation
where and are arbitrary positive constants. Therefore, (3) after several simplifications gives
In the latter expression, is nondecreasing and is nonincreasing, because and are nonincreasing. Hence by Lemma 2
It is easily seen that, in order to get the smallest constant in the latter inequality, we must choose . It gives the right hand side of (6). Note that the constant there belongs to .

*Remark 3. *From Theorem 1, it is seen that the constant in the right hand side of (6) tends to infinity as or (note that this constant is better than in [1]). Now we give an example of sequences confirming this [1]. In Theorem 1 we suppose that , , , and in . It gives
From the left hand side of (4) we deduce
where . Therefore, for a fixed positive the sum in the denominator can be made sufficiently small by an appropriate choice of . Consequently, can be arbitrarily large. The same is for .

It is clear that if then the constant in the right hand side of (6) equals . Now we give a more precise constant belonging to for the case when and satisfy several additional conditions.

Proposition 4. *Let and be nonincreasing such that the sequence is monotone and . If, moreover, for , then
**
The left hand side of (16) should be read as there exists no positive constant, depending on and , which bounds the fraction in (16) from below.*

*Proof. *The left hand side inequality follows by the same method as in the proof of Theorem 1. To prove the right hand side we denote the numerator of the fraction in (16) by . First we suppose to be nondecreasing, so . Applying (3) with yields
In the latter expression, the sequence , where , is nondecreasing. Indeed, is nondecreasing and moreover . Since is convex for and has a minimum at , the sequence is nondecreasing. From this by Lemma 2
where . Supposing to be nonincreasing and taking into account that , we obtain the right hand side of (16) by the same technique.

It is easily seen that equality in (16) holds, for example, if . The fact that the constant in the right hand side of (16) belongs to is obvious.

From the well-known weighted inequality of arithmetic and geometric means (see, e.g., [14, Chapter 2]) supposing and , we have This is a multivariable version of Young’s inequality (3). From this we obtain a multivariable version of Theorem 1 (but with less precise constant).

Proposition 5. *Let be nonincreasing sequences such that , where . If, moreover, , , and , , are such that , then
**
The left hand side of (20) should be read as there exists no positive constant, depending on , , and , which bounds the fraction in (20) from below.*

*Proof. *Set is the fraction in (20). Nonexistence of a positive constant bounding from below follows from Theorem 1. To prove the right hand side we denote the numerator of by . By the Abel transformation
The right hand side of (4) and the Abel transformation yields
Supposing in (19), we obtain
where it is obvious that is nondecreasing and that is nonincreasing. Thus by Lemma 2
Several simplifications give the right hand side of (20).

#### 3. Minkowski Type Inequalities

In this section we prove precise Minkowski type inequalities with our weights. As we have already mentioned, these generalize both the case of weight with alternating signs and the case of nonnegative weights (see [1]).

Theorem 6. *Let and be nonnegative nonincreasing sequences, and for . Then for **
The constant is best possible. The left hand side of (25) should be read as there exists no positive constant, depending on only , which bounds the fraction in (25) from below.*

*Proof. *Throughout the proof, denotes the fraction in (25). Applying the Abel transformation for the numerator and the denominator of easily yields . Moreover, there exists no positive constant depending on only that bounds from below. Indeed [1], for each there exists a sequence such that tends to zero. Supposing that , , and with some , from the left hand side of (4) we deduce
In this way as since for all .

Now we prove the right hand side of (25). From (2) we have
Now, before extraction of the th root, it is enough to show that
Inequality (28) by the Abel transformation is equivalent to
where . The latter inequality holds since for all and for . Indeed, for the function , where , , and , we have and . Therefore,
This completes the proof of (28).

The precision of the constant comes out from the following observation from [1]. If for all , (first elements are units) and , then after several simplifications we get
where positive as .

*Remark 7. *The following Jensen-Steffensen type statement was proved in [15] (see also [12, Section 2.2]).

Let be a nonincreasing positive sequence and a function convex on such that . Then the necessary and sufficient condition on weights in order that
is , .

From this point of view, the sufficient condition , , in Theorems 1 and 6 seems to be quite close to the necessary one.

#### 4. Further Generalizations

Now we give integral versions of Lemma 2 and Theorems 1 and 6. In what follows, we use the notation and suppose that all functions of are integrable and differentiable on .

Lemma 8. *For , let be nonnegative and nonincreasing, and let be nondecreasing such that , and . Then
*

*Proof. *Applying integration by parts gives
Here we took into account that ; , , , are nonnegative and is nonpositive for . It is easily seen that equality holds, for example, if .

Using Lemma 8 and integration by parts instead of the Abel transformation, we obtain the following results by essential repeating proofs of Theorems 1 and 6. We emphasize that may be negative here in contrast to the classical case.

Theorem 9. *For , let and be nonincreasing and let
**
If, moreover, , , and , , then
**
The left hand side of (37) should be read as there exists no positive constant, depending on , and , which bounds the fraction in (37) from below.*

Theorem 10. *For , let and be nonnegative and nonincreasing, and . Then
**
The constant is best possible. The left hand side of (38) should be read as there exists no positive constant, depending only on , which bounds the fraction in (38) from below.*

In conclusion we give several examples concerning Theorems 9 and 10. Let and in (33); then , and thus Appropriate discretization yields inequalities with alternating signs obtained earlier in [1] (the case in Theorems 1 and 6).

If is nondecreasing for (i.e., is nonnegative), Theorems 9 and 10 give the classical case of nonnegative weights, for which we can put instead of in the left hand sides of (37) and (38) due to Hölder’s and Minkowski’s inequalities.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgment

The research of P. Chunaev was partially supported by Grant no. MTM 2011-27637.