# On a result of Cartwright and Field

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## Abstract

*n*non-negative numbers \(x_{i}\), \(1 \leq i \leq n\), with \(q_{i} > 0\) satisfying \(\sum^{n}_{i=1}q_{i}=1\). For \(r>s\), a result of Cartwright and Field shows that when \(r=1\), \(s=0\),

## Keywords

Arithmetic-geometric mean Inequality Power means## MSC

26D15## 1 Introduction

Let \(M_{n,r}(\mathbf{x}; \mathbf{q})\) be the weighted power means: \(M_{n,r}(\mathbf{x}; \mathbf{q})=(\sum_{i=1}^{n}q_{i}x_{i}^{r})^{ \frac{1}{r}}\), where \(M_{n,0}(\mathbf{x}; \mathbf{q})\) denotes the limit of \(M_{n,r}(\mathbf{x}; \mathbf{q})\) as \(r\rightarrow 0\), \(\mathbf{x}=(x _{1},\ldots, x_{n})\), \(\mathbf{q}=(q_{1},\ldots, q_{n})\) with \(x_{i} \geq 0\), \(q_{i}>0\) for all \(1 \leq i \leq n\) and \(\sum_{i=1}^{n}q _{i}=1\). We further define \(A_{n}(\mathbf{x};\mathbf{q})=M_{n,1}( \mathbf{x};\mathbf{q})\), \(G_{n}(\mathbf{x};\mathbf{q})=M_{n,0}( \mathbf{x};\mathbf{q})\), \(\sigma _{n}=\sum_{i=1}^{n}q_{i}(x_{i}-A_{n})^{2}\). We shall write \(M_{n,r}\) for \(M_{n,r}(\mathbf{x};\mathbf{q})\) and similarly for other means when there is no risk of confusion.

It is shown in [2, Theorem 3.2] that when \(r = 1\) (resp. \(s=1\)), inequalities (1.2) hold if and only if \(-1 \leq s <1\) (resp. \(1< r \leq 2\)). Moreover, it is shown in [2] that the constant \((r-s)/2\) is best possible when either inequality in (1.2) is valid. However, neither inequality in (1.2) is valid for all *r*, *s* and a necessary condition on *r*, *s* such that either inequality of (1.2) is valid is given in Lemma 2.3 in Sect. 2.

In this paper, we determine all the pairs \((r,s)\) such that the right-hand side of (1.2) holds and on all the pairs \((r,s)\), \(-1/2 \leq s \leq 1\) such that the left-hand side of (1.2) holds. In Sect. 3 we will prove the following theorem.

### Theorem 1.1

*Let*\(r>s\)*and*\(x_{1}=\min \{ x_{i} \}\), \(x_{n}=\max \{ x_{i} \}\). *The right*-*hand side of* (1.2) *holds if and only if*\(0 \leq r+s \leq 3\), \(r \leq 2\), \(s \geq -1\). *When*\(-1/2 \leq s \leq 1\), *the left*-*hand side of* (1.2) *holds if and only if*\(0 \leq r+s \leq 3\), \(r \geq 1\). *Moreover*, *in all these cases we have equality if and only if*\(x_{1}=x_{2}=\cdots =x_{n}\).

## 2 Lemmas

Our first lemma gathers known results on inequalities (1.2).

### Lemma 2.1

*Let*\(r>s\)*and*\(x_{1}=\min \{ x_{i} \}\), \(x_{n}=\max \{ x_{i} \}\). *Both inequalities in* (1.2) *hold when*\(1 \leq r \leq 2\), \(-1 \leq s \leq 1\). *The right*-*hand side of* (1.2) *holds for*\(s=0\)*if and only if*\(0< r \leq 2\), *the left*-*hand side of* (1.2) *holds for*\(s=0\)*if and only if*\(1 \leq r \leq 3\). *Moreover*, *in all these cases we have equality if and only if*\(x_{1}=x_{2}=\cdots =x_{n}\).

### Proof

As shown in [2, Theorem 3.2] both inequalities in (1.2) are valid when \(-1 \leq s <1=r\) and \(s=1 < r \leq 2\). The first assertion of the lemma follows from the observation that when either inequality in (1.2) is valid for \(r>r'\) and \(r'>s\), then it is valid for \(r>s\). The second assertion of the lemma is [3, Theorem 2]. The cases for equalities also follow from [2, Theorem 3.2] and [3, Theorem 2]. □

Our next lemma establishes some auxiliary results needed in the proof of (and remarks on) Lemma 2.3.

### Lemma 2.2

- (i).
*Let*\(r>1\), \(s<0\).*We define*,*for*\(0< y \leq 1\),$$\begin{aligned} g_{r,s}(y)=y^{1/r-1}-\frac{(r-s)(1-y)}{2}. \end{aligned}$$(2.1)*Then*\(g_{r,s}(y)\)*is minimized at*$$\begin{aligned} y_{0}(r,s)= \biggl(\frac{2(1-1/r)}{r-s} \biggr)^{1/(2-1/r)}. \end{aligned}$$(2.2) - (ii).
*The function*\(h(z)= (1+z)^{1+z}z^{-z}\)*is an increasing function of*\(z>0\).

### Proof

Note first that as \(s<0\), we have \(r+2/r-s>r+2/r>2\). This implies that \(\frac{2(1-1/r)}{r-s} <1\), which in turn implies that \(0< y_{0}(r,s) < 1\). Now one checks that \(y_{0}(r,s)\) is the only root of \(g'_{r,s}(y)=0\). As it is easy to see that \(g''_{r,s}(y_{0}(r,s))>0\), the assertion of the lemma on (i) follows from this. To prove (ii), one calculates directly the logarithmic derivative of the function \(h(z)\) is positive for \(z>0\). This completes the proof of the lemma. □

It is easy to see that the right-hand side of (1.2) is equivalent to \(F \leq 0\) for \(1=x_{1}< x_{2}<\cdots <x_{n-1}<x_{n}\) and the left-hand side of (1.2) is equivalent to \(F \geq 0\) for \(0< x_{1}< x_{2}<\cdots <x_{n-1}<x_{n} =1 \). We expect the extreme values of *F* occur at \(n=2\) with one of the \(x_{i}\) or \(q_{i}\) taking a boundary value. Based on this consideration, to establish inequalities (1.2), we prove the following necessary condition.

### Lemma 2.3

*Let*\(r>s \neq 0\).

*A necessary condition for the right*-

*hand side of*(1.2)

*to hold is that*\(0 \leq r+s \leq 3\), \(r \leq 2\), \(s \geq -1\).

*A necessary condition for the left*-

*hand side of*(1.2)

*to hold is that*\(0 \leq r+s \leq 3\), \(r \geq 1\), \(rs \leq 2\),

*and*

*when*\(s<0\),

*where we define*\(0^{0}=1\).

### Proof

*F*be defined as in (2.3) to see that

We remark here that inequality (2.4) implies that it is not possible for the left-hand side of (1.2) to hold for \(r>1\) and all \(s<0\). In fact, by setting \(z=1-1/r\), one sees from part (ii) of Lemma 2.2 that the right-hand side of (2.4) is an increasing function of *z*, hence is maximized at \(z=1\), with value 4. It follows then from (2.4) and the condition \(r+s \geq 0\) that in order for the left-hand side of (1.2) to hold, it is necessary to have \(4 \geq (r-s)/2 \geq (-s-s)/2=-s\), which implies that \(s \geq -4\).

### Lemma 2.4

*Let*\(-1 \leq s<0\), \(0 \leq q \leq q_{1} \leq 1\), \(0< x_{0} <1\), \(x_{0} \leq x^{-s} \leq 1\).

*Then*

*for any*\(\alpha _{1} \geq 0\)

*satisfying*\(\alpha _{1} \leq \alpha _{0}\),

*where*

### Proof

*y*and \(g(1)=g( x_{0}) =0\), hence the desired result follows. □

### Lemma 2.5

*Let*\(-1 \leq s<0\), \(2< r \leq 3-s\), \(1-s^{2} \leq (r-1)(r-2)\).

*Suppose that there exists a number*\(q_{2}\), \(1/2 \leq q_{2} \leq 1\)

*such that*

*Then*,

*for*\(q_{2} \leq q \leq 1\), \(0< x \leq 1\),

### Proof

*q*, it suffices to prove inequality (2.5) for \(q=q_{2}, 1\). The case \(q=1\) is trivial and when \(q=q_{2}\), we set \(y=x^{s}\) to see that inequality (2.5) follows from \(h(y) \geq 0\) for \(y \geq 1\), where

As we can see in some part of Theorem 1.1, we need \(F_{2}(x,q) \geq 0\) for \(0 < x \leq 1\) and various *q*. The following lemma gives a sufficient condition for this.

### Lemma 2.6

*Let*\(-1 < s<0\), \(2 < r \leq 3-s\), \(0< x \leq 1\), \(0 \leq a, b<1\).

*If for*\(a \leq q < b\),

*Then*\(F_{2}(x,q) \geq 0\)

*for*\(a \leq q< b\)

*when*\(c(r,s, \alpha _{1}, \alpha _{2}) \leq 0\),

*where*\(F_{2}(x,q)\)

*is defined in*(2.7)

*and*

### Proof

### Lemma 2.7

*Let*\(-1 < s<0\), \(2 < r \leq 3-s \).

*Let*\(c(r,s,\alpha _{1}, \alpha _{2})\)

*be defined as in Lemma*2.6.

*Define*

*Then*\(\max_{1 \leq i \leq 4} \{ c_{i}(r,s)\} \leq 0\)

*when*\(-1/2 \leq s<0\).

### Proof

As both expressions on the right-hand side above are decreasing functions of \(s<0\), and one checks directly that \(c_{3}(3-s,s)<0\), \(c _{4}(3-s,s) <0\) for \(s=-1/2\), so that \(\max_{i=3,4} \{ c_{i}(3-s,s) \} \leq 0\) and this completes the proof. □

## 3 Proof of Theorem 1.1

Throughout this section, we assume that \(r>s\). We omit the discussions on the conditions for equality in each inequality, we shall prove as one checks easily that the desired conditions hold by going through our arguments in what follows. As the case \(s=0,1\) or \(r=1\) has been proven in [2, Theorem 3.2] and [3, Theorem 2], we further assume \(r \neq 1\), \(s \neq 0,1\) in what follows.

*F*be defined as in (2.3) and \(x_{1}=1< x_{2}<\cdots <x_{n}\), \(q_{i}>0\), \(1 \leq i \leq n\). We have

*λ*is a constant. Then at \((\mathbf{x}';\mathbf{q}')\) we must have

Therefore, it remains to show \(F_{0} \leq 0\) for \(n=2\). In this case, we let \(1=x_{1}< x_{2}=x\), \(0< q_{2}=q <1\), \(q_{1}=1-q\) to see that \(F_{0}=F_{1}(x,q)\), where \(F_{1}(x,q)\) is defined in (2.6).

Note that \(F_{2}(x,q) = (1-q)^{-1}\partial F_{1}/\partial x\), where \(F_{2}(x,q)\) is defined in (2.7). As \(F_{1}(1,q)=0\), we see that it suffices to show that \(F_{2}(x,q) \leq 0\) for \(x \geq 1\).

Thus, it suffices to show that either side expression in (3.2) is ≤1.

Case 1. \(0< s \leq 1/2 \leq r < 1\).

Each factor of the left-hand side expression in (3.2) is ≤1, hence their product is ≤1.

Case 2. \(0< s< r \leq 1/2\).

*r*and \(-r<-s\). So we have

Case 3. \(1/2 \leq s< r <1\).

Case 4. \(1< s< r \leq 3-s=\min \{ 2, 3-s \}\).

Case 5. \(s<0<r<1\), \(r+s \geq 0\).

*r*that

Next, we prove the left-hand side of (1.2) for \(0 \leq r+s \leq 3\), \(-1/2 \leq s \leq 1\), \(r \geq 1\). In this case, it suffices to show \(F \geq 0\) provided that we assume \(0< x_{1}< x_{2}<\cdots <x_{n}=1\). Similar to our discussions above, one shows easily that this follows from \(\partial F/\partial x_{1} \leq 0\) for \(n=2\), which is equivalent to \(F_{1}(x,q) \leq 0\) for \(0< x \leq 1\). Again we divide the proof into several cases. As the case \(-1 \leq s \leq 1 \leq r \leq 2\) follows directly from Lemma 2.1, we only consider the remaining cases in what follows and similar to our proof of the right-hand side of (1.2) above, it suffices to show that \(F_{2}(x,q) \geq 0\) for \(0< x \leq 1\).

Case 1. \(1/2 \leq s<1\), \(2< r \leq 3-s\).

As \(r-1>0\), it follows from the arithmetic-geometric mean inequality that the right-hand side expression of (3.1) is greater than or equal to the expressions in (3.2). As the factors of the right-hand side expression of (3.2) are all ≥1, it follows that \(F_{2}(x,q) \geq 0\).

*q*, hence it suffices to check the validity of inequality (3.3) at \(q=0,1\).

Case 2. \(0< s<1/2\), \(2< r \leq 3-s\).

Case 3. \(-1/2 \leq s<0\), \(2< r \leq 3-s\).

We divide this case into a few subcases.

Subcase 1. \(0< q \leq 1/2\).

*r*and \(-s \leq r\) since \(r+s \geq 0\), we have

Subcase 2. \(1/2 \leq q \leq 1\), \((1+s)x^{s}-(2-s) \geq 0\) or \(1-s^{2} \geq (r-1)(r-2)\).

One checks that if \((1+s)x^{s}-(2-s) \geq 0\), then the function \(q \mapsto (1-s)(q(1+s)x^{s}+(2-s)(1-q))(qx^{r}+1-q)-(r-1)(-q(r+1)x ^{r}+(r-2)(1-q))\) is a concave function of *q* and hence is minimized at \(q=0,1\), with values ≥1.

In this case, Lemma 2.5 with \(q_{2}=q_{0}\) implies that (2.8) is satisfied by \(\alpha _{1}=0\) and \(\alpha _{2}=s\), where we set \(a=q_{0}\) and \(b=1\) in Lemma 2.6. It follows from Lemma 2.6 that \(F_{2}(x,q) \geq 0\) as long as \(c_{3}(r,s) \leq 0\), where \(c_{3}(r,s)\) is given in (2.11). As Lemma 2.7 implies that \(c_{3}(r,s) \leq 0\), we see that \(F_{2}(x,q) \geq 0\) in this case.

Subcase 4. \((1+s)x^{s}-(2-s) \leq 0\), \(1-s^{2} \leq (r-1)(r-2)\), and \(1/2 \leq q < q_{0}\), where \(q_{0}\) is defined by (3.12).

## 4 Further discussions

We point out that Theorem 1.1 determines all the pairs \((r,s)\), \(r>s\) such that the right-hand side of (1.2) holds and all the pairs \((r,s)\), \(-1/2 \leq s \leq 1\) such that the left-hand side of (1.2) holds. However, less is known for the left-hand side of (1.2) when \(r>s>1\) or \(s<-1/2\). This is partially due to the fact that our approach in the proof of Theorem 1.1 relies on showing \(F_{1}(x,q) \leq 0\) (via \(F_{2}(x,q) \geq 0\)) for \(0< x\leq 1\), \(0< q<1\), where \(F_{1}\), \(F_{2}\) are defined in (2.6) and (2.7). However, it is easy to see that \(F_{1}(0, q) >0\) when \(r>s>1\) and \(\lim_{x \rightarrow 0^{+}}F_{2}(x,q) < 0\) when \(r>2\), \(s<-1\). It also follows from this that in order to show \(F_{2}(x,q) \geq 0\) when \(s<-1\), we must have \(r \leq 2\). As Lemma 2.3 implies a necessary condition for the left-hand side of (1.2) to hold is \(r \geq 1\), \(0 \leq r+s \leq 3\), we then deduce that when \(s \leq -1\), one can only expect to show \(F_{2}(x,q) \geq 0\) for \(1 \leq r \leq 2\), \(s \geq -r \geq -2\).

On the other hand, though Theorem 1.1 only establishes the validity of the left-hand side of (1.2) for \(s \geq -1/2\), one can in fact extend the validity of the left-hand side of (1.2) for certain \(r>s\), \(s <-1/2\) by going through the proof of Theorem 1.1. This is given in the following theorem.

### Theorem 4.1

*Let*\(r>s\)*and*\(x_{1}=\min \{ x_{i} \}\), \(x_{n}=\max \{ x_{i} \}\). *The left*-*hand side of* (1.2) *holds when*\((r-1)(r-2) \leq 1-s^{2}\)*or when*\(-1 < s <-1/2\), \(2< r<3-s\), \(\max_{1 \leq i \leq 4} \{ c_{i}(r,s) \} \leq 0\), *where*\(c_{i}(r,s)\), \(1 \leq i \leq 4\)*is defined in* (2.11). *Moreover*, *in all these cases we have equality if and only if*\(x_{1}=x_{2}=\cdots =x_{n}\).

### Proof

## Notes

### Acknowledgements

The author is very grateful to the referees for their many valuable comments and suggestions.

### Authors’ contributions

The author completed the paper and approved the final manuscript.

### Funding

Not applicable.

### Competing interests

The author declares that there are no competing interests.

## References

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**2003**, 995–1002 (2003) MathSciNetCrossRefGoogle Scholar - 3.Gao, P.: A complement to Diananda’s inequality. Math. Inequal. Appl.
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