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On a result of Cartwright and Field
Journal of Inequalities and Applications volume 2018, Article number: 349 (2018)
Abstract
Let \(M_{n,r}=(\sum_{i=1}^{n}q_{i}x_{i}^{r})^{\frac{1}{r}}\), \(r\neq 0\), and \(M_{n,0}= \lim_{r \rightarrow 0}M_{n,r}\) be the weighted power means of n nonnegative 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 \(r=1\), \(s=0\),
where \(x_{1}=\min \{ x_{i} \}\), \(x_{n}=\max \{ x_{i} \}\), \(\sigma _{n}= \sum_{i=1}^{n}q_{i}(x_{i}M_{n,1})^{2}\). In this paper, we determine all the pairs \((r,s)\) such that the righthand side inequality above holds and all the pairs \((r,s)\), \(1/2 \leq s \leq 1\) such that the lefthand side inequality above holds.
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.
The following elegant refinement of the wellknown arithmeticgeometric mean inequality is given by Cartwright and Field in [1]:
Naturally, one considers the following generalization of (1.1) on bounds for the differences of means:
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 \((rs)/2\) 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 righthand side of (1.2) holds and on all the pairs \((r,s)\), \(1/2 \leq s \leq 1\) such that the lefthand 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 righthand 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 lefthand 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 righthand side of (1.2) holds for \(s=0\) if and only if \(0< r \leq 2\), the lefthand 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 \(0< y \leq 1\),
$$\begin{aligned} g_{r,s}(y)=y^{1/r1}\frac{(rs)(1y)}{2}. \end{aligned}$$(2.1)Then \(g_{r,s}(y)\) is minimized at
$$\begin{aligned} y_{0}(r,s)= \biggl(\frac{2(11/r)}{rs} \biggr)^{1/(21/r)}. \end{aligned}$$(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 \(r+2/rs>r+2/r>2\). This implies that \(\frac{2(11/r)}{rs} <1\), 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. □
We define
It is easy to see that the righthand side of (1.2) is equivalent to \(F \leq 0\) for \(1=x_{1}< x_{2}<\cdots <x_{n1}<x_{n}\) and the lefthand side of (1.2) is equivalent to \(F \geq 0\) for \(0< x_{1}< x_{2}<\cdots <x_{n1}<x_{n} =1 \). 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 righthand 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 lefthand 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
Note first that it is shown in [2, Lemma 3.1] that a necessary condition for either inequality of (1.2) to hold is that \(0 \leq r+s \leq 3\). Now we let \(n=2\), \(x_{1}=x\), \(x_{2}=1\), \(q_{1}=q\), and F be defined as in (2.3) to see that
As the first (second) righthand side expression above is positive when \(r>2\) (\(s<1\)) and \(x \rightarrow +\infty \), we conclude that in order for the righthand side of (1.2) to hold, it is necessary to have \(r \leq 2\) and \(s \geq 1\). Moreover, the first (second) righthand side expression above is negative when \(s>0\), \(rs>2\) (\(r<1\)), and \(x=0\). We then conclude that in order for the lefthand side of (1.2) to hold, it is necessary to have \(r \geq 1\) and \(rs \leq 2\) (note that when \(s < 0\), this condition is also satisfied).
On the other hand, when \(s<0\), we have
For the lefthand side of (1.2) to hold for \(s<0\), the expression above needs to be nonnegative. By setting \(y=1q\), we see that this is equivalent to showing \(g_{r,s}(y)\) is nonnegative for \(0 < y \leq 1\), where \(g_{r,s}(y)\) is given in (2.1). Note that the case \(r=1\) of (2.4) implies that \(s \geq 1\), a condition already given in [2, Theorem 3.2], we may further assume that \(r>1\). It follows from Lemma 2.2 that \(g_{r,s}(y)\) is minimized at \(y_{0}(r,s)\), where \(y_{0}(r,s)\) is given in (2.2). Substituting this value in (2.1), one checks easily that it is necessary to have (2.4) in order for the expression in (2.1) to be nonnegative for \(0< y \leq 1\) and the assertion of the lemma now follows. □
We remark here that inequality (2.4) implies that it is not possible for the lefthand side of (1.2) to hold for \(r>1\) and all \(s<0\). In fact, by setting \(z=11/r\), one sees from part (ii) of Lemma 2.2 that the righthand 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 lefthand side of (1.2) to hold, it is necessary to have \(4 \geq (rs)/2 \geq (ss)/2=s\), 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
We let \(y =x^{s}\) and \(\alpha =\alpha _{0}/s\), so that \(0 \leq \alpha <1\). It suffices to show that \(g(y) \geq 0\) for \(x_{0} \leq y \leq 1\), where
It is easy to see that \(g(y)\) is a concave function of y and \(g(1)=g( x_{0}) =0\), hence the desired result follows. □
Lemma 2.5
Let \(1 \leq s<0\), \(2< r \leq 3s\), \(1s^{2} \leq (r1)(r2)\). 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
As the expressions in (2.5) are linear functions of 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 \(\alpha _{2}/s \leq 1\), it is easy to see that \(h(y)\) is minimized at \(y=1\) with a positive value and this completes the proof. □
For \(r>s\), \(r^{2}+s^{2} \neq 0\), \(0 \leq q<1\), \(x>0\), we define
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 \(1 < s<0\), \(2 < r \leq 3s\), \(0< x \leq 1\), \(0 \leq a, b<1\). 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
As in this case \(\lim_{x \rightarrow 0^{+}}F_{2}(x,q)>0\), \(F_{2}(1,q)=0\), we only need to show that the values of \(F_{2}\) at points satisfying
are nonnegative.
Calculation shows that at these points we have
If \(q(r+1)x^{r}+(r2)(1q) \leq 0\), then no such points exist. Hence we may assume that \(q(r+1)x^{r}+(r2)(1q) > 0\). Applying (2.8) in (2.10), we find that
We write \(\alpha _{3}=\frac{13s}{s}\alpha _{1}+\alpha _{2}\) so that the above inequality implies that
We now apply the arithmeticgeometric inequality and the above estimation to see that
The assertion of the lemma now follows easily. □
Lemma 2.7
Let \(1 < s<0\), \(2 < r \leq 3s \). 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
It is easy to see that \(c_{i}\), \(0 \leq i \leq 3\), are all convex functions of \(r \geq 2\), where \(c_{0}(r,s)\) is defined in (2.9). Also, \(c_{4}\) is a convex function of \(r \geq 3\). Thus, it suffices to show that \(c_{i}\), \(1 \leq i \leq 3\), are nonpositive for \(1/2 \leq s<0\), \(r=2, 3s\) and that \(c_{4}\) is nonpositive for \(1/2 \leq s<0\), \(2 < r \leq 3\), \(r=3s\). One checks directly that \(\max_{i=1,2} \{ c_{i}(2,s), c_{i}(3s,s) \} \leq 0\), \(c_{3}(2,s) \leq c _{0}(2,s) \leq 0\), and \(\max_{2 < r \leq 3} c_{4}(r,s) \leq \max_{2 < r \leq 3} c_{0}(r,s) \leq \max \{ c_{0}(2,s), c_{0}(3,s) \} =0\). We also have
As both expressions on the righthand side above are decreasing functions of \(s<0\), and one checks directly that \(c_{3}(3s,s)<0\), \(c _{4}(3s,s) <0\) for \(s=1/2\), so that \(\max_{i=3,4} \{ c_{i}(3s,s) \} \leq 0\) 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.
As the “only if” part of Theorem 1.1 follows from Lemma 2.3, it remains to prove the “if” part of Theorem 1.1. We consider the righthand side of (1.2) first. Let 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
Now the righthand side of (1.2) follows from \(F \leq 0\), which in turn follows from \(F_{0} \leq 0\) as it implies \(F(\mathbf{x}; \mathbf{q} ) \leq \lim_{x_{n} \rightarrow x_{n1}}F(\mathbf{x}; \mathbf{q})\). By adjusting the value of \(q_{n1}\) in the expression of \(\lim_{x_{n} \rightarrow x_{n1}}F(\mathbf{x};\mathbf{q})\) and repeating the process, it follows easily that \(F \leq 0\).
When \(n \geq 3\), we regard \(x_{1}=1\), \(x_{n}\) as fixed and assume that \(F_{0}\) is maximized at some point \((\mathbf{x}';\mathbf{q}')=(x'_{1}, \ldots , x'_{n}, q'_{1}, \ldots , q'_{n})\) with \(x'_{1}=x_{1}\), \(x'_{n}=x _{n}\). Then at this point we must have
Thus, the \(x'_{i}\), \(2 \leq i \leq n1\), are solutions of the equation
It is easy to see that the above equation can have at most two different positive roots.
On the other hand, by applying the method of Lagrange multipliers, we let
where λ is a constant. Then at \((\mathbf{x}';\mathbf{q}')\) we must have
Thus, the \(x'_{i}\), \(1 \leq i \leq n\), are solutions of the equation
As \(f'_{2}(x)=f_{1}(x)\), it follows from the mean value theorem that there is a solution of \(f_{1}(x)=0\) between any two adjacent \(x'_{i}\), \(x'_{i+1}\), \(1 \leq i \leq n1\), as they are solutions of \(f_{2}(x)=0\). But when \(n \geq 3\), we have at least \(x'_{2}\) as a solution of \(f_{1}(x)=0\). This would imply that \(f_{1}(x)=0\) has at least three different positive solutions (for example, one in between \(x'_{1}\) and \(x'_{2}\), one in between \(x'_{2}\) and \(x'_{3}\), and \(x'_{2}\) itself), a contradiction.
Therefore, it remains to show \(F_{0} \leq 0\) for \(n=2\). In this case, we let \(1=x_{1}< x_{2}=x\), \(0< q_{2}=q <1\), \(q_{1}=1q\) to see that \(F_{0}=F_{1}(x,q)\), where \(F_{1}(x,q)\) is defined in (2.6).
Note that \(F_{2}(x,q) = (1q)^{1}\partial F_{1}/\partial x\), 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 \(x \geq 1\).
We now divide the proof of the righthand side of (1.2) for \(r>s \neq 0\) satisfying \(0 \leq r+s \leq 3\), \(r \leq 2\), \(s \geq 1\) 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 we show in these cases \(F_{2}(x,q) \leq 0\) or equivalently, \(F_{2}(x,q)/(rs)+1 \leq 1\). Note that
One checks that in all the following cases, we have \((r1)(1s) \leq 0\). Therefore, it follows from the arithmeticgeometric mean inequality with nonpositive weights that the righthand side expression in (3.1) is less than or equal to
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 lefthand side expression in (3.2) is ≤1, hence their product is ≤1.
Case 2. \(0< s< r \leq 1/2\).
It is well known that \(r \mapsto M_{n,r}\) is an increasing function of r and \(r<s\). So we have
We also have \((12r)(1r) \leq (12s)(1s)\). Therefore,
which implies that the lefthand side expression of (3.2) is ≤1.
Case 3. \(1/2 \leq s< r <1\).
Note that
so that the lefthand side expression of (3.2) is less than or equal to
This implies that the lefthand side expression of (3.2) is ≤1.
Case 4. \(1< s< r \leq 3s=\min \{ 2, 3s \}\).
Note that
It follows that
which implies that the righthand side expression of (3.2) is ≤1.
Case 5. \(s<0<r<1\), \(r+s \geq 0\).
When \(r \geq 1/2\), each factor of the lefthand side of (3.2) is ≤1, hence their product is ≤1. If \(0< r <1/2\), then again it follows from the fact that \(r \mapsto M_{n,r}\) is an increasing function of r that
As \((12r)(r1) \leq 0\) and \(32(r+s) \geq 0\), it follows that
This now completes the proof for all the cases for the righthand side of (1.2).
Next, we prove the lefthand 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 righthand side of (1.2) above, it suffices to show that \(F_{2}(x,q) \geq 0\) for \(0< x \leq 1\).
Case 1. \(1/2 \leq s<1\), \(2< r \leq 3s\).
As \(r1>0\), it follows from the arithmeticgeometric mean inequality that the righthand side expression of (3.1) is greater than or equal to the expressions in (3.2). As the factors of the righthand side expression of (3.2) are all ≥1, it follows that \(F_{2}(x,q) \geq 0\).
For the remaining cases, one checks easily that we have \(\lim_{x \rightarrow 0^{+}}F_{2}(x,q)>0\), \(F_{2}(1,q)=0\) so that it suffices to show the values of \(F_{2}(x,q)\) at points satisfying (2.10) are nonnegative, assuming that \(q(r+1)x^{r}+(r2)(1q) > 0\). Hence, in what follows, we shall only evaluate \(F_{2}(x,q)\) at these points satisfying the above assumption. We then note that at these points we have
This is seen by noting that the expressions in (3.3) are linear functions of q, hence it suffices to check the validity of inequality (3.3) at \(q=0,1\).
It then follows from (3.3) and (2.10) that at these points we have
an inequality we shall assume in what follows.
Case 2. \(0< s<1/2\), \(2< r \leq 3s\).
Similar to the previous case, the righthand side expression of (3.1) is greater than or equal to the expressions in (3.2). From (3.4) we deduce that
Using this, we see that the righthand side expression of (3.2) is greater than or equal to
When \(1/3 \leq s < 1/2\), we see that the first factor and the last factor above is ≥1, and we write the product of the two factors in the middle as
Note that the first factor above is now ≥1 and it is easy to see that
This implies that the second factor in (3.5) is also ≥1. Hence the righthand side expression of (3.2) is greater than or equal to 1 and it follows that \(F_{2}(x,q) \geq 0\).
When \(0 < s < 1/3\), it follows from (3.4) that
If the righthand side expression above is ≥1, then we have
If the righthand side expression of (3.6) is ≤1, then it implies that
Thus,
where the last inequality above follows from the observation that the function \(r \mapsto (1+2s)(12r)/r+r2\) is an increasing function of \(r \geq 2\) and hence is maximized at \(r=3s\), in which case its value is easily shown to be negative. It follows from (3.7) that \(F_{2}(x,q) \geq 0\) in this case.
Case 3. \(1/2 \leq s<0\), \(2< r \leq 3s\).
We divide this case into a few subcases.
Subcase 1. \(0< q \leq 1/2\).
As \(r \mapsto M_{n,r}\) is an increasing function of r and \(s \leq r\) since \(r+s \geq 0\), we have
As \(0< q \leq 1/2\), we also have
We then deduce from (3.4), (3.8), and (3.9) that
It follows that (note that \(23(r+s) \leq 0\) for \(r \geq 2\), \(s \geq 1\))
With \(c_{1}(r,s)\) being defined in (2.11), we then deduce that
where the first inequality above follows from (3.8) and (3.9), the second inequality above follows from the arithmeticgeometric inequality, and the last inequality above follows from (3.10). It follows from Lemma 2.7 that \(F_{2}(x,q) \geq 0\) in this case.
Subcase 2. \(1/2 \leq q \leq 1\), \((1+s)x^{s}(2s) \geq 0\) or \(1s^{2} \geq (r1)(r2)\).
One checks that if \((1+s)x^{s}(2s) \geq 0\), then the function \(q \mapsto (1s)(q(1+s)x^{s}+(2s)(1q))(qx^{r}+1q)(r1)(q(r+1)x ^{r}+(r2)(1q))\) is a concave function of q and hence is minimized at \(q=0,1\), with values ≥1.
If \(1s^{2} \geq (r1)(r2)\), then
It follows that
Thus, in either case, we deduce from the above and (2.10) that we have
From this we apply the arithmeticgeometric inequality to see that
where \(c_{2}(r,s)\) is defined in (2.11). Now Lemma 2.7 implies that \(F_{2}(x,q) \geq 0\) in this case.
Subcase 3. \((1+s)x^{s}(2s) \leq 0\), \(1s^{2} \leq (r1)(r2)\), and \(1/2 \leq q_{0} \leq q<1\), where \(q_{0}\) is defined by
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. \((1+s)x^{s}(2s) \leq 0\), \(1s^{2} \leq (r1)(r2)\), and \(1/2 \leq q < q_{0}\), where \(q_{0}\) is defined by (3.12).
In this case, we set \(a=1/2\) and \(b=q_{0}\) in Lemma 2.6. Note that as \(r \leq 3s\), it follows from this and (3.12) that when \(s \geq 1/2\),
We then deduce that \(q_{0} \leq 5/6\). Note also that we have \(x^{s} \geq (1+s)/(2s) \geq 1/5\) when \(s \geq 1/2\). Thus, we can take \(q_{1}=5/6\), \(x_{0}=1/5\) in Lemma 2.4 and \(q_{2}=1/2\) in Lemma 2.5 to see that (2.8) is satisfied by \(\alpha _{1}=0.0889s\), \(\alpha _{2}=s(1s^{2})/((r1)(r2))\). It follows from Lemma 2.6 that \(F_{2}(x,q) \geq 0\) as long as \(c_{4}(r,s) \leq 0\), where \(c_{4}(r,s)\) is given in (2.11) and \(F_{2}(x,q) \geq 0\) in this case again follows from Lemma 2.7.
4 Further discussions
We point out that Theorem 1.1 determines all the pairs \((r,s)\), \(r>s\) such that the righthand side of (1.2) holds and all the pairs \((r,s)\), \(1/2 \leq s \leq 1\) such that the lefthand side of (1.2) holds. However, less is known for the lefthand 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 lefthand 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 lefthand side of (1.2) for \(s \geq 1/2\), one can in fact extend the validity of the lefthand 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 lefthand side of (1.2) holds when \((r1)(r2) \leq 1s^{2}\) or when \(1 < s <1/2\), \(2< r<3s\), \(\max_{1 \leq i \leq 4} \{ c_{i}(r,s) \} \leq 0\), 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
Once again we omit the discussions on the conditions of equality. As in the proof of Theorem 1.1, it suffices to prove \(F_{2}(x,q) \geq 0\), where \(F_{2}(x,q)\) is defined in (2.7). When \((r1)(r2) \leq 1s^{2}\), it follows from the expression for \(F_{2}(x,q)/(rs)\) in (3.11) that
When \(1< s <1/2\), our assertion follows by simply combining the arguments in all the subcases of case 3 in the proof of the lefthand side of (1.2) in Sect. 3. This completes the proof. □
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Gao, P. On a result of Cartwright and Field. J Inequal Appl 2018, 349 (2018). https://doi.org/10.1186/s1366001819488
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DOI: https://doi.org/10.1186/s1366001819488