# Extensions and Applications of Power-Type Aczél-Vasić-Pečarić’s Inequalities

Authors: Kun Chen; Fei Yan; Hong-Ying Yue
Journal: Mathematical Problems in Engineering
 (2019)

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## Abstract

In this paper, we enrich and develop power-type Aczél-Vasić-Pečarić’s inequalities. First of all, we give some new versions of theorems and corollaries about Aczél-Vasić-Pečarić’s inequalities by quoting some lemmas. Moreover, in combination with Hölder’s inequality, we give some applications of the new version of Aczél-Vasić-Pečarić’s inequality and give its proof process.


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## Body

## 1. Introduction

In 1956, Aczél [1] discovered the Aczél’s inequality.Theorem 1.
Let n be a positive integer and letλs, νs(s=1,2,…,n), n≥2, be positive numbers such that λ12-∑s=2nλs2≥0; ν12-∑s=2nνs2≥0. Then(1)λ12-∑s=2nλs2ν12-∑s=2nνs2≤λ1ν1-∑s=2nλsνs2.It is matter of common observation that Aczél’s inequality (1) is of great significance in the theory of functional equations in non-Euclidean geometry; meanwhile, many authors including Bellman [2], Hu et al.[3], Tian [4, 5], Tian and Ha [6], Tian and Sun [7], Tian and Wu [8], Tian and Wang [9], Tian and Zhou [10], Wu [11], and Wu and Debnath [12, 13] pay more attention to this inequality and its refinements.In 1959, Popoviciu [14] gave a generalization of Aczél’s inequality, as follows.Theorem 2.
Letp>1,q>1, and 1/p+1/q=1, and let λs,νs(s=1,2,…,n) be positive numbers such that λ1p-∑s=2nλsp>0 and ν1p-∑s=2nνsp>0. Then(2)λ1p-∑s=2nλsp1/pν1q-∑s=2nνsq1/q≤λ1ν1-∑s=2nλsνs.In 1979, Vasić and Pečarić [15] proved the following extension of inequality (2).Theorem 3.
Letpt>0(t=1,2,…,m) and ∑t=1m1/pt≥1, and let λst(s=1,2,…,n,t=1,2,…,m) be positive integers such that λ1tpt-∑s=2nλstpt>0. Then(3)∏t=1mλ1tpt-∑s=2nλstpt1/pt≤∏t=1mλ1t-∑s=2n∏t=1mλst.Inequality (3) is known as Aczél- Vasić-Pečarić’s inequality.In 2005, Wu and Debnath [13] generalized inequality (3) in the following form.Theorem 4.
Letpt and λst(s=1,2,…,n,t=1,2,…,m) be positive numbers such that λ1tpt-∑s=2nλstpt>0 for t=1,2,…,m, and let τ=min⁡{∑t=1m1/pt,1}. Then(4)∏t=1mλ1tpt-∑s=2nλstpt1/pt≤n1-τ∏t=1mλt-∑s=2n∏t=1mλt.Later, Wu in [11] established the Aczél-Vasić-Pečarić inequality (3).Theorem 5.
Letp1,p2,…,pm>0 and ρ=1/p1+1/p2+⋯+1/pm, and let λst>0, λ1tpt-∑s=2nλstpt>0(s=1,2,…,n,t=1,2,…,m), and m≥2, n≥2. Then we have (5)∏t=1mλ1tpt-∑s=2nλstpt1/pt≤n1-minρ,1∏t=1mλt-∑s=2n∏t=1mλt-a∑1≤t≤k≤m∑s=2nλstptλ1tpt-λskpkλ1kpk2, where (6)a=∏t=1mλ1tη,η=m-1max⁡p1,p2,…,pm,m2,ρ≥1,nρ-1m2m-1-1max⁡p1,p2,…,pm,1-ρ-1,ρ<1.Moreover, in 2014 Tian [7] also presented an improvement of the Aczél-Vasić-Pečarić inequality (3).Theorem 6.
Letλrt>0, pt>0, and λ1tpt-∑r=2nλrtpt>0(r=1,2,…,n,t=1,2,…,m), let m≥2, and let ρ=min{∑t=1m1/pt,1}. Then (7)∏t=1mλ1tpt-∑r=2nλrtpt1/pt≤n1-ρ1-2mm-1∑1≤i≤t≤m∑r=2nλripiλ1ipi-λrtptλ1tpt2γ∏t=1mλ1t-∑r=2n∏t=1mλrt≤n1-ρ∏t=1mλ1t-∑r=2n∏t=1mλrt, where γ=m/2max⁡{p1,p2,…,pm}.Recently, the power mean has attracted of many researcher ([16–24]), and many remarkable inequalities for the power mean including Hölder-type inequalities can be found in the literature [25–30]. The purpose of the article is to establish some distinctive versions of Aczél-Vasić-Pečarić’s inequality (3) for power mean type. As consequences, several integral inequalities of the obtained results are given.


## 2. Some Power Mean Types of Aczél-Vasić-Pečarić’s Inequality

Lemma 7 (see [13]).
Letλst>0 and 1/pt>0(s=1,2,…,n,t=1,2,…,m), and let τ=min⁡{∑t=1m1/pt,1}. Then (8)∑s=1n∏t=1mλst≤n1-τ∏t=1m∑s=1nλstpt1/pt.Like in [31], the power mean of order r for n positive numbers λ1,λ2,…,λn is defined by (9)
M
n
λ
,
r
=
1
n
∑
k
=
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n
λ
k
r
1
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r
,
r
≠
0
,
λ
1
λ
2
⋯
λ
n
n
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=
0
, where λ=(λ1,λ2,…,λn).Lemma 8 (see [32]).
Letλ1,λ2,…,λn be a positive sequence, n∈N. If r∈R, then the function r↦Mn(λ,r) is increasing for fixed λ1,λ2,…,λn.Lemma 9 (see [31]).
Letλs>0(s=1,2,…,n) be positive numbers, and let p>0. Then(10)∑s=1nλsp≤n1-min⁡p,1∑s=1nλsp.Puttingp≥1, we get the following inequality:(11)∑s=1nλsp≤∑s=1nλsp.Let (12)λt=λ1t,λ2t,…,λnt,t=1,2,…,m,Mmλtp,r=1m∑s=1mλstpr1/r,Mm,nλtp,r=1n-m+1∑s=mnλstpr1/r,Mm,nχλj,r=1n-m+1∑i=mnλstrχ/r.The main conclusions of this paper are as follows.Theorem 10.
Letr>0, pt>0, M1pt(λt,r)-M2,npt(λt,r)>0, and ∏t=1mλt=(∏t=1mλ1t,∏t=1mλ2t,…,∏t=1mλnt), and let τ=min⁡{∑t=1m1/pt,1} and α=1+∑t=1mmin{1/r,1/pt}-∑t=1m1/pt-m/r. Then(13)∏t=1mM1λtpt,r-M2,nptλt,r1/pt≤n1-τM1∏t=1mλt,1-n-1αM2,n∏t=1mλt,1.Proof.
Simple computations lead to(14)∏t=1mM1λtpt,r-M2,nptλt,r1/pt=∏t=1mλ1tpt-1n-1∑s=2nλstrpt/r1/pt. Since pt>0, λ1tpt-(1/n-1∑s=2nλstr)pt/r>0, let(15)λ1tpt-1n-1∑s=2nλstrpt/r=ξtpt,ξt>0,t=1,2,…,m-1,(16)∏t=1mλ1t-nτ-1n-1∑t=1mmin⁡1/r,1/pt-∑t=1m1/pt-m/r∑s=2n∏t=1mλst=nτ-1∏t=1mξt.Combining (15) and inequalities (10) and (8), we have(17)ξmpm+1n-1∑s=2nλsmrpm/r1/pm∏t=1m-1λ1t=∏t=1mξtpt+1n-1∑s=2nλstrpt/r1/pt≥∏t=1mξtpt+n-1min⁡pt/r,1-pt/r-1∑s=2nλstpt1/pt=∏t=1mξtpt+∑s=2nn-1min⁡1/r,1/pt-1/r-1/ptλstpt1/pt≥nτ-1∏t=1mξt+∑s=2n∏t=1mn-1min⁡1/r,1/pt-1/r-1/ptλst=nτ-1∏t=1mξt+n-1∑t=1mmin⁡1/r,1/pt-∑t=1m1/pt-m/r∑s=2n∏t=1mλst=∏t=1mλ1t.Therefore, from (15), (16), and (17), we obtain(18)ξmpm+1n-1∑s=2nλsmrpm/r1/pm∏t=1m-1λ1t≥∏t=1mλ1t. Hence, we obtain(19)ξm≥λ1mpm-1n-1∑s=2nλsmrpm/r1/pm. Therefore, we have(20)∏t=1mξt≥λ1mpm-1n-1∑s=2nλsmrpm/r1/pm∏t=1m-1ξt=∏t=1mλ1tpt-1n-1∑s=2nλstrpt/r1/pt.By using (16), we immediately obtain(21)∏t=1mλ1tpt-1n-1∑s=2nλstrpt/r1/pt≤n1-τ∏t=1mλ1t-n-1α1n-1∑s=2n∏t=1mλst, which leads to inequality (13). The proof of Theorem 10 is accomplished.According to Theorem10 we can get the following corollaries.Corollary 11.
Letr≥pt>0 and λ1tpt-(1/n-1∑s=2nλstr)pt/r>0(s=2,3,…,n,t=1,2,…,m), and let τ=min{∑t=1m1/pt,1}. Then(22)∏t=1mλ1tpt-1n-1∑s=2nλstrpt/r1/pt≤n1-τ∏t=1mλ1t-n-1-∑t=1m1/pt∑s=2n∏t=1mλst.Proof.
From inequality (21), we can get(23)∏t=1mλ1tpt-1n-1∑s=2nλstrpt/r1/pt≤n1-τ∏t=1mλ1t-n-1α1n-1∑s=2n∏t=1mλst≤n1-τ∏t=1mλ1t-n-1α-1∑s=2n∏t=1mλst≤n1-τ∏t=1mλ1t-n-1-∑t=1m1/pt∑s=2n∏t=1mλst. The proof of Corollary 11 is accomplished.Corollary 12.
Let1≥pt>0 and λ1tpt-(1/n-1∑s=2nλst)pt>0(s=2,3,…,n,t=1,2,…,m). Then(24)∏t=1mλ1tpt-1n-1∑s=2nλstpt1/pt≤∏t=1mλ1t-n-1-∑t=1m1/pt∑s=2n∏t=1mλst.If we put∑t=1m1/pt=1 in Corollary 11, the conclusions that we can draw from this are as follows.Corollary 13.
Letr≥pt>0 and λ1tpt-(1/n-1∑s=2nλstr)pt/r>0(s=2,3,…,n,t=1,2,…,m), and let ∑t=1m1/pt=1. Then(25)∏t=1mλ1tpt-1n-1∑s=2nλstrpt/r1/pt≤∏t=1mλ1t-1n-1∑s=2n∏t=1mλst.Letpt=r(t=1,2,…,m); then from Theorem 10, we getCorollary 14.
Letr>0, λ1tr-1/n-1∑s=2nλstr>0(s=2,3,…,n,t=1,2,…,m), and let τ=min⁡{m/r,1}. Then(26)∏t=1mλ1tr-1n-1∑s=2nλstr1/r≤n1-τ∏t=1mλ1t-n-1-m/r∑s=2n∏t=1mλst.Using the obtained inequality of inequality (26), we can get the following result of Wu [13].Corollary 15.
Letr>0 and ψ1tr-∑s=2nψstr>0(s=2,3,…,n,t=1,2,…,m), and let τ=min⁡{m/r,1}. Then(27)∏t=1mψ1tr-∑s=2nψstr1/r≤n1-τ∏t=1mψ1t-∑s=2n∏t=1mψst.Theorem 16.
Letr>0, pt>0, and M1(λtpt,r)-M2,n(λtpt,r)>0, let λt=(λ1t,λ2t,…,λnt) and ∏t=1mλt=(∏t=1mλ1t,∏t=1mλ2t,…,∏t=1mλnt), and denote τ=min⁡{∑t=1m1/pt,1}. Then(28)∏t=1mM1λtpt,r-M2,nλtpt,r1/pt≤21-τM1∏t=1mλt,r-n-1τ-∑t=1m1/pt/rM2,n∏t=1mλt,r.Proof.
For the hypothesespt>0 and M1(λtpt,r)-M2,n(λtpt,r)>0(t=1,2,…,m), we have (29)λ1tpt-1n-1∑s=2nλstrpt1/r>0. Denote(30)λ1tpt-1n-1∑s=2nλstrpt1/r=ξtpt,ξt>0,t=1,2,…,m-1,(31)∏t=1mλ1t-2τ-1n-11/rτ-∑t=1m1/pt1n-1∑s=2n∏t=1mλstr1/r=2τ-1∏t=1mξt.
By using inequality (8), simple computations lead to(32)∏t=1mξtpt+1n-1∑s=2nλstrpt1/r1/pt≥2τ-1∏t=1mξt+∏t=1m1n-1∑s=2nλstrpt1/rpt=2τ-1∏t=1mξt+n-1-1/r∑t=1m1/pt∏t=1m∑s=2nλstrpt1/rpt≥2τ-1∏t=1mξt+n-11/rτ-∑t=1m1/pt1n-1∑s=2n∏t=1mλstr1/r.Therefore, from (30), (31), and (32), we obtain (33)ξmpm+1n-1∑s=2nλsmrpm1/r1/pm∏t=1m-1ξtpt+1n-1∑s=2nλstrpt1/r1/pt=ξmpm+1n-1∑s=2nλsmrpm1/r1/pm∏t=1m-1λ1t≥2τ-1∏t=1mξt+n-11/rτ-∑t=1m1/pt1n-1∑s=2n∏t=1mλstr1/r=∏t=1mλ1t. Hence (34)ξm≥λ1mpm-1n-1∑s=2nλsmrpm1/r1/pm. Thus(35)∏t=1mξt≥λ1mpm-1n-1∑s=2nλsmrpm1/r1/pm∏t=1m-1ξt=∏t=1mλ1tpt-1n-1∑s=2nλstrpm1/r1/pt.
Combining (31) and inequality (35), we get (36)∏t=1mλ1tpt-1n-1∑s=2nλstrpm1/r1/pt≤21-τ∏t=1mλ1t-n-11/rτ-∑t=1m1/pt1n-1∑s=2n∏t=1mλstr1/r, which is equivalent to inequality (28). Successful proof of Theorem 16 is as follows.Corollary 17.
Letλst,pt>0 and λ1tpt-(1/n-1∑s=2nλstrpt)1/r>0(s=2,3,…,n,t=1,2,…,m), and let ∑t=1m1/pt>1. Then(37)∏t=1mλ1tpt-1n-1∑s=2nλstrpt1/r1/pt≤∏t=1mλ1t-n-11-∑t=1m1/pt/r1n-1∑s=2n∏t=1mλstr1/r.In particular, puttingr=1 in Corollary 17, we obtain a new version of the Aczél-Vasić-Pečarić inequality (3).(38)∏t=1mλ1tpt-1n-1∑s=2nλstpt1/pt≤∏t=1mλ1t-n-1-∑t=1m1/pt∑s=2n∏t=1mλst.Corollary 18.
Letλst,pt>0 and λ1tpt-(1/n-1∑s=2nλstrpt)1/r>0(s=2,3,…,n,t=1,2,…,m), and let ∑t=1m1/pt≤1. Then(39)∏t=1mλ1tpt-1n-1∑s=2nλstrpt1/r1/pt≤21-∑t=1m1/pt∏t=1mλ1t-1n-1∑s=2n∏t=1mλstr1/r.Corollary 19.
Letλst,pt>0 and λ1tpt-1/n-1∑s=2nλstpt>0(s=2,3,…,n,t=1,2,…,m), and let ∑t=1m1/pt≤1. Then(40)∏t=1mλ1tpt-1n-1∑s=2nλstpt1/pt≤21-∑t=1m1/pt∏t=1mλ1t-1n-1∑s=2n∏t=1mλst.If we putr=1 in Theorem 16, then we have the following corollary.Corollary 20.
Letλst,pt>0 and λ1tpt-1/n-1∑s=2nλstpt>0(s=2,3,…,n,t=1,2,…,m), and let τ=min⁡{∑t=1m1/pt,1}. Then(41)∏t=1mλ1tpt-1n-1∑s=2nλstpt1/pt≤21-τ∏t=1mλ1t-n-1τ-∑t=1m1/pt-1∑s=2n∏t=1mλst.Using the substitutionsψ1t→λ1t and (n-1)1/ptψst→λst(s=2,3,…,n,t=1,2,…,m) in Corollary 20, we obtain the following refinement of the Aczél-Vasić-Pečarić inequality (3).Corollary 21.
Letψst,pt>0 and ψ1tpt-∑s=2nψstpt>0(s=1,2,…,n,t=1,2,…,m), and let τ=min{∑t=1m1/pt,1}. Then(42)∏t=1mψ1tpt-∑s=2nψstpt1/pt≤21-τ∏t=1mψ1t-n-1τ-1∑s=2n∏t=1mψst.If we put∑t=1m1/pt≥1 in Corollary 21, then we can derive inequality (3).In particular, puttingm=2, p1=p, p2=q, λs1=λs, and λs2=νs(s=1,2,…,n) in Corollary 21, we obtain a fresh improvement and promotion of nonequality (2).Corollary 22.
Letλs,νs>0, p,q>0, and τ=min⁡{1/p+1/q,1}, and let λsp-∑s=2nλsp>0 and νsq-∑s=2nνsq>0(s=1,2,…,n,t=1,2,…,m). Then(43)λsp-∑s=2nλsp1/pνsq-∑s=2nνsq1/q≤21-τλ1ν1-n-1τ-1∑s=2nλsνs.Theorem 23.
Letλst,pt>0(s=1,2,…,n,t=1,2,…,m) and λ1tpt-∏s=2nλstpt/n-1>0, and let ∑t=1m1/pt≤1. Then(44)∏t=1mλ1tpt-∏s=2nλstpt/n-11/pt≤21-∑t=1m1/pt∏t=1mλ1t-∏s=2n∏t=1mλst1/n-1.Proof.
For∑t=1m1/pt≤1, by inequality (28), we obtain(45)∏t=1mλ1tpt-1n-1∑s=2nλstrpt1/r1/pt≤21-∑t=1m1/pt∏t=1mλ1t-1n-1∑s=2n∏t=1mλstr1/r.We get this inequality (44) by adding r→0 on both sides of inequality (45).For the hypotheses∑t=1m1/pt≤1 and λ1tpt-∏s=2nλstpt/n-1>0 in Theorem 23, denote ζt=(∏s=2nλst/λ1t)1/n-1, then we have 0<ζtpt<ζt<1. Thus, we obtain the following corollaries.Corollary 24.
Letpt>0, 0<ζt<1(t=1,2,…,m), and let ∑t=1m1/pt<1. Then(46)∏t=1m1-ζtpt1/pt+∏t=1mζt≤21-∑t=1m1/pt.DenoteΘ(r)=∏t=1m[M1(λtp,r)-M2,n(λtp,r)]1/pt, by Lemma 8, we realize that Θ(r) is decreasing on (0,+∞). Combining Theorems 16 and 23, we obtain the following corollary.Corollary 25.
Letλst>0 and pt>0(s=1,2,…,n,t=1,2,…,m), and let λ1tpt-1/n-1∑s=2nλstpt>0. If ∑t=1m1/pt<1, then(47)∏t=1mλ1tpt-1n-1∑s=2nλstpt1/pt≤∏t=1mλ1tpt-∏s=2nλstpt/n-11/pt≤21-∑t=1m1/pt∏t=1mλ1t-∏s=2n∏t=1mλst1/n-1.


## 3. Applications

As is well known, analytic inequalities have various important applications in many branches of mathematics [33–37]. In this section, we will get some applications for this new inequality in Section 2.Theorem 26.
Letψt>0, r≥pt>0, ∑t=1m1/pt=1, (t=1,2,…,m), and ψtpt-(∫01ftr(x)dx)pt/r>0. Then(48)∏t=1mψtpt-∫01ftrxdxpt/r1/pt≤∏t=1mψt-∫01∏t=1mftxdx,where ft(x) can be integrated on [0,1] and ft(x)>0.Proof.
For all positive integersn, the equidistance of the partition of [0,1] is selected as(49)0<1n<⋯<n-1n<1,xs=sn,△xs=1n,s=0,1,…,n-1. Noting that ψtpt-(∫01ftr(x)dx)pt/r>0(t=1,2,…,m), we have (50)ψtpt-limn→∞∑s=0n-1ftr0+sn1npt/r>0t=1,2,…,m. Therefore, there is a positive integer N, like that (51)ψtpt-∑s=0n-1ftr0+sn1npt/r>0, for all n>N and t=1,2,…,m.
Moreover, for anyn>N, it follows from Corollary 13 that(52)∏t=1mψtpt-1n∑s=0n-1ftrsnpt/r1/pt≤∏t=1mψt-1n∑s=0n-1∏t=1mftsn.
We may find thatft(x)(t=1,2,…,m) are positive Riemann integrable functions on [0,1]; we know that ∏t=1mft(x) and ftr(x) are also integrable on [0,1]. Letting n→∞ on both sides of inequality (52), we get the desired inequality (48). The proof of Theorem 26 is achieved.Theorem 27.
Letψt>0 and r≥m, and let ft(x)(t=1,2,…,m) be positive Riemann integrable functions on [0,1] such that ψtr-∫01ftr(x)dx>0. Then(53)∏t=1mψtr-∫01ftrxdx1/r≤∏t=1mψt-∫01∏t=1mftxdx.Proof.
Applying Corollary14 and along the lines of the proof Theorem 26, Theorem 27 is simply given.Theorem 28.
Letr>0, ψt>0, pt>0, ∑t=1m1/pt≤1, (t=1,2,…,m), and ψtpt-(∫01ftrpt(x)dx)1/r>0. Then(54)∏t=1mψtpt-∫01ftrptxdx1/r1/pt≤21-∑t=1m1/pt∏t=1mψt-∫01∏t=1mftrxdx1/r,where ft(x) can be integrated on [0,1] and ft(x)>0.Proof.
Combining the proof Theorem26 and Corollary 18, it is easy to get Theorem 28.Corollary 29.
Letpt>0, ∑t=1m1/pt≤1, (t=1,2,…,m), and ψtpt-∫01ftpt(x)dx>0. Then(55)∏t=1mψtpt-∫01ftptxdx1/pt≤21-∑t=1m1/pt∏t=1mψt-∫01∏t=1mftxdx,where ft(x) can be integrated on [0,1] and ft(x)>0.A direct result from Theorem28 is given by us. Putting m=2, p1=p>0, p2=q>0, ψ1=λ1, ψ2=ν1, f1=f, and f2=g in (54), we can get the following corollaries.Corollary 30.
Letr>0, λ1,ν1>0, p,q>0, 1/p+1/q≤1, λ1p-(∫01frp(x)dx)1/r>0, and ν1q-(∫01grq(x)dx)1/r>0. Then(56)λ1p-∫01frpxdx1/r1/pν1q-∫01grqxdx1/r1/q≤21-1/p-1/qλ1ν1-∫01fxgxrdx1/r,where f(x),g(x) can be integrated on [0,1] and f(x)>0,g(x)>0.If we setr=1 in Corollary 30, then the following inequality holds.Corollary 31.
Letλ1,ν1>0, p,q>0, 1/p+1/q≤1, λ1p-∫01fp(x)dx>0, and ν1q-∫01gq(x)dx>0. Then(57)λ1p-∫01fpxdx1/pν1q-∫01gqxdx1/q≤21-1/p-1/qλ1ν1-∫01fxgxdx,where f(x),g(x) can be integrated on [0,1] and f(x)>0,g(x)>0.



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