# Refinements of Generalized Aczél's Inequality and Bellman's Inequality and Their Applications

Authors: Jing-Feng Tian; Shu-Yan Wang
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We give some refinements of generalized Aczél's inequality and Bellman's inequality proposed by Tian. As applications, some refinements of integral type of generalized Aczél's inequality and Bellman's inequality are given.


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## Body

## 1. Introduction

The famous Aczél’s inequality [1] states that if ai, bi(i=1,2,…,n) are real numbers such that a12-∑i=2n‍ai2>0 and b12-∑i=2n‍bi2>0, then
(1)(a12-∑i=2n‍ai2)(b12-∑i=2n‍bi2)≤(a1b1-∑i=2n‍aibi)2.It is well known that Aczél’s inequality plays an important role in the theory of functional equations in non-Euclidean geometry. In recent years, various attempts have been made by many authors to improve and generalize the Aczél’s inequality (see [2–19] and references therein). We state here some improvements of Aczél’s inequality.One of the most important results in the references mentioned above is an exponential extension of (1), which is stated in the following theorem.Theorem A.
Letp and q be real numbers such that p, q≠0, and 1/p+1/q=1 and let ai, bi(i=1,2,…,n) be positive numbers such that a1p-∑i=2n‍aip>0 and b1q-∑i=2n‍biq>0. If p>1, then
(2)(a1p-∑i=2n‍aip)1/p(b1q-∑i=2n‍biq)1/q≤a1b1-∑i=2n‍aibi.
If p<1(p≠0), then the reverse inequality in (2) holds.Remark 1.
The casep>1 of Theorem A was proved by Popoviciu [8]. The case p<1 was given in [15] by Vasić and Pečarić.Vasić and Pečarić [16] presented a further extension of inequality (1).Theorem B.
Letarj>0, λj>0, a1jλj-∑r=2n‍arjλj>0, r=1,2,…,n, j=1,2,…,m, and let ∑j=1m‍(1/λj)≥1. Then
(3)∏j=1m‍(a1jλj-∑r=2n‍arjλj)1/λj≤∏j=1m‍a1j-∑r=2n‍∏j=1m‍arj.In a recent paper [18], Wu and Debnath established an interesting generalization of Aczél’s inequality [1] as follows.Theorem C.
Letarj>0, λj>0, a1jλj-∑r=2n‍arjλj>0, r=1,2,…,n, j=1,2,…,m, and let ρ=min{∑j=1m‍(1/λj),1}. Then
(4)∏j=1m‍(a1jλj-∑r=2n‍arjλj)1/λj≤n1-ρ∏j=1m‍a1j-∑r=2n‍∏j=1m‍arj.In 2012, Tian [10] presented the following reversed version of inequality (4).Theorem D.
Letarj>0, λ1≠0, λj<0(j=2,3,…,m), a1jλj-∑r=2n‍arjλj>0, r=1,2,…,n, j=1,2,…,m, and let τ=max{∑j=1m‍(1/λj),1}. Then
(5)∏j=1m‍(a1jλj-∑r=2n‍arjλj)1/λj≥n1-τ∏j=1m‍a1j-∑r=2n‍∏j=1m‍arj.Therefore, applying the above inequality, Tian gave the reversed version of inequality (3) as follows.Theorem E.
Letarj>0, λ1≠0, λj<0(j=2,3,…,m), ∑j=1m‍(1/λj)≤1, a1jλj-∑r=2n‍arjλj>0, and r=1,2,…,n, j=1,2,…,m. Then
(6)∏j=1m‍(a1jλj-∑r=2n‍arjλj)1/λj≥∏j=1m‍a1j-∑r=2n‍∏j=1m‍arj.Moreover, in [10], Tian obtained the following integral form of inequality (5).Theorem F.
Letλ1>0, λj<0(j=2,3,…,m), ∑j=1m‍λj=1, Pj>0(j=1,2,…,m), and let fj(x)(j=1,2,…,m) be positive Riemann integrable functions on [a,b] such that Pjλj-∫ab‍fjλj(x)dx>0. Then
(7)∏j=1m‍(Pjλj-∫ab‍fjλj(x)dx)1/λj≥∏j=1m‍Pj-∫ab‍∏j=1m‍fj(x)dx.Bellman inequality [20] related with Aczél’s inequality is stated as follows.Theorem G.
Letai, bi(i=1,2,…,n) be positive numbers such that a1p-∑i=2n‍aip>0 and b1q-∑i=2n‍biq>0. If p≥1, then
(8)[(a1p-∑i=2n‍aip)1/p+(b1p-∑i=2n‍bip)1/p]p≤(a1+b1)p-∑i=2n‍(ai+bi)p.
If  0<p<1, then the reverse inequality in (8) holds.Remark 2.
The casep>1 of Theorem G was proposed by Bellman [20]. The case 0<p<1 was proved in [15] by Vasić and Pečarić.The main purpose of this work is to give refinements of inequalities (5) and (8). As applications, some refinements of integral type of inequality (5) and (8) are given.


## 2. Refinements of Generalized Aczél’s Inequality and Bellman’s Inequality

Theorem 3.
Letarj>0, λ1≠0, λj<0(j=2,3,…,m), a1jλj-∑r=2n‍arjλj>0, r=1,2,…,n, j=1,2,…,m, and let τ=max{∑j=1m‍(1/λj),1}. Then
(9)∏j=1m‍(a1jλj-∑r=2n‍arjλj)1/λj≥T(s,aij)≥n1-τ∏j=1m‍a1j-∑r=2n‍∏j=1m‍arj,
where
(10)T(s,aij)=(n+1-s)1-τ∏j=1m‍(a1jλj-∑r=2s‍arjλj)1/λj-∑r=s+1n‍∏j=1m‍arj.Proof
Case (I). When ∑j=1m‍(1/λj)<1, then τ=1. On the one hand, we split the left-hand side of inequality (9) as follows:
(11)∏j=1m‍(a1jλj-∑r=2n‍arjλj)1/λj=∏j=1m‍(a1jλj-∑r=2s‍arjλj-∑r=s+1n‍arjλj)1/λj=∏j=1m‍(Ajλj-∑r=s+1n‍arjλj)1/λj,
where
(12)Aj=(a1jλj-∑r=2s‍arjλj)1/λj,j=1,2,…,m.
From this hypothesis, it is immediate to obtain the inequality
(13)Ajλj≥∑r=s+1n‍arjλj.
Thus, by using inequality (6), we have
(14)∏j=1m‍(Ajλj-∑r=s+1n‍arjλj)1/λj≥∏j=1m‍Aj-∑r=s+1n‍∏j=1m‍arj.
On the other hand, by using inequality (6) again, we obtain the inequality
(15)∏j=1m‍Aj=∏j=1m‍(a1jλj-∑r=2s‍arjλj)1/λj≥∏j=1m‍a1j-∑r=2s‍∏j=1m‍arj.
Combining inequalities (11), (14), and (15) we can get inequality (9).
Case (II). When ∑j=1m‍(1/λj)≥1, using Case (I) with ∑j=1m+1‍(1/λj)=1, we have
(16)∏j=1m+1‍(a1jλj-∑r=2n‍arjλj)1/λj≥∏j=1m+1‍(a1jλj-∑r=2s‍arjλj)1/λj-∑r=s+1n‍∏j=1m+1‍arj,
where a1jλj-∑r=2n‍arjλj>0, arj>0, λ1≠0, λj<0(j=2,3,…,m+1).
Hence, takinga1(m+1)λm+1=1, a2(m+1)λm+1=a3(m+1)λm+1=⋯=an(m+1)λm+1=1/n into (16), we obtain
(17)(1n)1-1/λ1-1/λ2-⋯-1/λm∏j=1m‍(a1jλj-∑r=2n‍arjλj)1/λj≥(n+1-sn)1-1/λ1-1/λ2-⋯-1/λm×∏j=1m‍(a1jλj-∑r=2s‍arjλj)1/λj-(1n)1-1/λ1-1/λ2-⋯-1/λm∑r=s+1n‍∏j=1m‍arj,
that is,
(18)∏j=1m‍(a1jλj-∑r=2n‍arjλj)1/λj≥(n+1-s)1-1/λ1-1/λ2-⋯-1/λm×∏j=1m‍(a1jλj-∑r=2s‍arjλj)1/λj-∑r=s+1n‍∏j=1m‍arj.
Therefore, repeating the foregoing arguments, we get
(19)(n+1-s)1-1/λ1-1/λ2-⋯-1/λm∏j=1m‍(a1jλj-∑r=2s‍arjλj)1/λj-∑r=s+1n‍∏j=1m‍arj≥n1-τ∏j=1m‍a1j-∑r=2n‍∏j=1m‍arj.
Combining inequalities (18) and (19) leads to inequality (9) immediately. The proof of Theorem 3 is completed.If we set∑j=1m‍(1/λj)≤1, then from Theorem 3, we obtain the following refinement of inequality (6).Corollary 4.
Letarj>0, λ1≠0, λj<0(j=2,3,…,m), ∑j=1m‍(1/λj)≤1, a1jλj-∑r=2n‍arjλj>0, r=1,2,…,n, and j=1,2,…,m. Then
(20)∏j=1m‍(a1jλj-∑r=2n‍arjλj)1/λj≥∏j=1m‍(a1jλj-∑r=2s‍arjλj)1/λj-∑r=s+1n‍∏j=1m‍arj≥∏j=1m‍a1j-∑r=2n‍∏j=1m‍arj.Puttingm=2, λ1=p≠0, λ2=q<0, ar1=ar, and ar2=br(r=1,2,…,n) in Theorem 3, we obtain the refinement and generalization of Theorem A for p<1.Corollary 5.
Letar>0, br>0(r=1,2,…,n), a1p-∑r=2n‍arp>0, b1q-∑r=2n‍brq>0, p≠0, q<0, and ρ=max{1/p+1/q,1}. Then, the following inequality holds:
(21)(a1p-∑r=2n‍arp)1/p(b1q-∑r=2n‍brq)1/q≥(n+1-s)1-ρ(a1p-∑r=2s‍arp)1/p×(b1q-∑r=2s‍brq)1/q-∑r=s+1n‍arbr≥n1-ρa1b1-∑r=2n‍arbr.Based on the mathematical induction, it is easy to see that the following generalized Bellman’s inequality is true.Theorem 6.
Letarj>0, a1jλj-∑r=2n‍arjλj>0, r=1,2,…,n, j=1,2,…,m, and let 0<p<1. Then
(22)[∏j=1m‍(a1jp-∑r=2n‍arjp)1/p]p≥(∑j=1m‍a1j)p-∑r=2n‍(∑j=1m‍arj)p.Next, we give a refinement of generalized Bellman’s inequality (22) as follows.Theorem 7.
Letarj>0, a1jλj-∑r=2n‍arjλj>0, r=1,2,…,n, j=1,2,…,m, and let 0<p<1. Then
(23)[∏j=1m‍(a1jp-∑r=2n‍arjp)1/p]p≥[∏j=1m‍(a1jp-∑r=2s‍arjp)1/p]p-∑r=s+1n‍(∑j=1m‍arj)p≥(∑j=1m‍a1j)p-∑r=2n‍(∑j=1m‍arj)p.Proof.
The proof of Theorem7 is similar to the one of Theorem 3. Applying generalized Bellman’s inequality (22) twice, we can deduce the inequality (23).


## 3. Application

In this section, we show two applications of the inequalities newly obtained in Section2.Firstly, we give an improvement of inequality (7) by using Theorem 3.Theorem 8.
LetBj>0(j=1,2,…,m), let λ1>0, λj<0(j=2,3,…,m), ∑j=1m‍λj=1, and let fj(x)(j=1,2,…,m) be positive integrable functions defined on [a,b] with Bjλj-∫ab‍fjλj(x)dx>0. Then, for any t∈[a,b), one has
(24)∏j=1m‍(Bjλj-∫ab‍fjλj(x)dx)1/λj≥∏j=1m‍(Bjλj-∫at‍fjλj(x)dx)1/λj-∫tb‍∏j=1m‍fj(x)dx≥∏j=1m‍Bj-∫ab‍∏j=1m‍fj(x)dx.Proof.
We need to prove only the left side of inequality (24). The proof of the right side of inequality (24) is similar. For any positive integers n and l, we choose an equidistant partition of [a,t] and [t,b], respectively, as
(25)a<a+c-an<⋯<a+c-ank<⋯<a+c-an(n-1)<c,c<c+b-cl<⋯<c+b-cli<⋯<c+b-cl(l-1)<b,xk=a+c-ank,Δxk=c-an,k=1,2,…,n,xi=c+b-cli,Δxi=b-cl,i=1,2,…,n.
Noting that Bjλj-∫ab‍fjλj(x)dx=Bjλj-(∫ac‍fjλj(x)dx+∫cb‍fjλj(x)dx)>0(j=1,2,…,m), we have
(26)Bjλj-{limn→∞∑k=1n‍fjλj[a+k(c-a)n]c-an+liml→∞∑i=1l‍fjλj[c+i(b-c)l]b-cl}>0,(j=1,2,…,m).
Consequently, there exists a positive integer N, such that
(27)Bjλj-[∑k=1n‍fjλj(a+k(c-a)n)c-an+∑i=1l‍fjλj(c+i(b-c)l)b-cl]>0
for all n,l>N and j=1,2,…,m.
By using Theorem3, for any n,l>N, the following inequality holds:
(28)∏j=1m‍{Bjλj-[∑k=1n‍fjλj(a+k(c-a)n)c-an+∑i=1l‍fjλj(c+i(b-c)l)b-cl]}1/λj≥∏j=1m‍[Bjλj-∑k=1n‍fjλj(a+k(c-a)n)c-an]1/λj-∑i=1l‍[∏j=1m‍fj(c+i(b-c)l)](b-cl)∑j=1m(1/λj).
Since(29)∑j=1m‍1λj=1,
we have(30)∏j=1m‍{Bjλj-[∑k=1n‍fjλj(a+k(c-a)n)c-an+∑i=1l‍fjλj(c+i(b-c)l)b-cl]}1/λj≥∏j=1m‍[Bjλj-∑k=1n‍fjλj(a+k(c-a)n)c-an]1/λj-∑i=1l‍[∏j=1m‍fj(c+i(b-c)l)](b-cl).
Noting thatfj(x)(j=1,2,…,m) are positive Riemann integrable functions on [a,b], we know that ∏j=1m‍fj(x) and fjλj(x) are also integrable on [a,b]. Letting n→∞ on both sides of inequality (30), we get the left side of inequality (24). The proof of Theorem 8 is completed.We give here a direct consequence from Theorem8. Putting m=2, λ1=p, λ2=q, B1=a1, B2=b1, f1=f, and f2=g in (24), we obtain a special important case as follows.Corollary 9.
Letp and q be real numbers such that p>0, q<0, and (1/p)+(1/q)=1, let a1,b1>0, and let f,g be positive integrable functions defined on [a,b] with a1p-∫ab‍fp(x)dx>0 and b1q-∫ab‍gq(x)dx>0. Then, for any t∈[a,b), one has
(31)(a1p-∫ab‍fp(x)dx)1/p(b1q-∫ab‍gq(x)dx)1/q≥(a1p-∫ac‍fp(x)dx)1/p(b1q-∫ac‍gq(x)dx)1/q≥a1b1-∫ab‍f(x)g(x)dx.Finally, we present a refinement of integral type of generalized Bellman’s inequality.Theorem 10.
LetBj>0(j=1,2,…,m), let 0<p<1, and let fj(x)(j=1,2,…,m) be positive integrable functions defined on [a,b] with Bjλj-∫ab‍fjλj(x)dx>0. Then, for any t∈[a,b), one has
(32)[∑j=1m‍(Bjp-∫ab‍fjp(x)dx)1/p]p≥[∑j=1m‍(Bjp-∫at‍fjp(x)dx)1/p]p-∫tb‍(∑j=1m‍fj(x))pdx≥(∑j=1m‍Bj)p-∫ab‍(∑j=1m‍fj(x))pdx.Proof.
The proof of Theorem10 is similar to the proof of Theorem 8.A special case to the last theorem is as follows.Corollary 11.
Let0<p<1, let a1,b1>0, and let f,g be positive integrable functions defined on [a,b] with a1p-∫ab‍fp(x)dx>0 and b1q-∫ab‍gq(x)dx>0. Then, for any t∈[a,b), one has
(33)[(a1p-∫ab‍fp(x)dx)1/p+(b1p-∫ab‍gp(x)dx)1/p]p≥[(a1p-∫at‍fp(x)dx)1/p+(b1p-∫at‍gp(x)dx)1/p]p-∫tb‍(f(x)+g(x))pdx≥(a1+b1)p-∫ab‍(f(x)+g(x))pdx.



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