We present some new versions of generalized Hölder’s inequalities. The results are used to improve Minkowski’s inequality and a Beckenbach-type inequality.
1. Introduction
If ak≥0, bk≥0(k=1,2,…,n), p>1, 1/p+1/q=1, then
(1)∑k=1nakbk≤∑k=1nakp1/p∑k=1nbkq1/q.
The sign of the inequality is reversed if p<1, p≠0 (for p<0, we assume that ak,bk>0). Inequality (1) and its reversed version are called Hölder’s inequality.
In 1979, Vasić and Pečarić [1] presented the following result.
Theorem A.
Let Aij>0(i=1,2,…,n, j=1,2,…,m).
If βj>0 and if ∑j=1m(1/βj)≥1, then
(2)∑i=1n∏j=1mAij≤∏j=1m∑i=1nAijβj1/βj.
If βj<0(j=1,2,…,m), then
(3)∑i=1n∏j=1mAij≥∏j=1m∑i=1nAijβj1/βj.
If β1>0, βj<0(j=2,3,…,m), and if ∑j=1m(1/βj)≤1, then
(4)∑i=1n∏j=1mAij≥∏j=1m∑i=1nAijβj1/βj.
Inequalities (2), (3), and (4) are called generalized Hölder’s inequalities. It is well known that Hölder’s inequality and generalized Hölder’s inequalities are important in mathematical analysis and in the field of applied mathematics. For example, Agahi et al. [2] presented generalizations of the Hölder and the Minkowski inequality for pseudointegrals and Liu [3] established a Hölder type inequality. For a discussion on inequalities we refer the reader to [1, 4–9] and the references therein. Although generalized Hölder’s inequalities play an important and basic role in many branches of mathematics, some problems can not be precisely estimated by generalized Hölder’s inequalities. For example, if we set n=3, m=2, A11=1, A21=1, A31=1, A12=1, A22=1, A32=19, β2=1, β1=1/2, then from generalized Hölder’s inequality (2) we obtain 21≤189. It is of interest to develop a refinement of Hölder’s inequality.
In this paper we present new refinements of inequalities (2), (3), and (4) in Section 2. In Section 3, we use our results to improve the Minkowski inequality and a Beckenbach-type inequality.
2. Refinements of Generalized Hölder’s Inequalities
We begin with a known result.
Lemma 1 (see [10]).
If x>-1, α>1, or α<0, then
(5)1+xα≥1+αx.
The inequality is reversed for 0<α<1.
Lemma 2.
Let Xij>0 and 1-∑i=1nXijβj>0(i=1,2,…,n,j=1,2,…,m).
If 0<β1<β2<⋯<βm and if ∑j=1m(1/βj)≥1, then
(6)∏j=1m1-∑i=1nXijβj1/βj+∑i=1n∏j=1mXij≤∏j=1[m/2]1-∑i=1nXi2jβ2j-∑i=1nXi2j-1β2j-121/β2j.
If 0>β1>β2>⋯>βm, then
(7)∏j=1m1-∑i=1nXijβj1/βj+∑i=1n∏j=1mXij≥∏j=1[m/2]1-∑i=1nXi2jβ2j-∑i=1nXi2j-1β2j-121/β2j.
If β1>0, 0>β2>β3>⋯>βm, and if ∑j=1m(1/βj)≤1, then
(8)∏j=1m1-∑i=1nXijβj1/βj+∑i=1n∏j=1mXij≥∏j=1[m/2]1-∑i=1nXi2jβ2j-∑i=1nXi2j-1β2j-121/β2j.
Proof.
(a) Note first that 1/β1>1/β2>⋯>1/βm-1>1/βm>0 and 1/βj-1/βj+1>0 (j=1,2,…,m-1).
Case (I). Let m be even.
Note that 1/β1-1/β2+1/β2+1/β2+1/β3-1/β4+1/β4+1/β4+⋯ + (1/βm-1-1/βm) + 1/βm+1/βm≥1, and, using inequality (2), we have
(9)∏j=1[m/2]1-∑i=1nXi2jβ2j-∑i=1nXi2j-1β2j-121/β2j=∏j=1m/21-∑i=1nXi2j-1β2j-1+∑i=1nXi2jβ2j1/β2jkkkkkkkk×1-∑i=1nXi2jβ2j+∑i=1nXi2j-1β2j-11/β2jkkkkkkkk×1-∑i=1nXi2j-1β2j-1+∑i=1nXi2j-1β2j-11/β2j-1-1/β2j=1-∑i=1nXi1β1+∑i=1nXi2β21/β2×1-∑i=1nXi2β2+∑i=1nXi1β11/β2×1-∑i=1nXi1β1+∑i=1nXi1β11/β1-1/β2×1-∑i=1nXi3β3+∑i=1nXi4β41/β4×1-∑i=1nXi4β4+∑i=1nXi3β31/β4×1-∑i=1nXi3β3+∑i=1nXi3β31/β3-1/β4⋮×1-∑i=1nXim-1βm-1+∑i=1nXimβm1/βm×1-∑i=1nXimβm+∑i=1nXim-1βm-11/βm×1-∑i=1nXim-1βm-1+∑i=1nXim-1βm-11/βm-1-1/βm≥∏j=1m/21-∑i=1nXi2j-1β2j-11/β2j1-∑i=1nXi2jβ2j1/β2jkkkkkkkk×1-∑i=1nXi2j-1β2j-11/β2j-1-1/β2j+∏j=1m/2X12jβ2j1/β2jX12j-1β2j-11/β2jkkkkkkkkk×X12j-1β2j-11/β2j-1-1/β2j+∏j=1m/2X22jβ2j1/β2jX22j-1β2j-11/β2jkkkkkkkkk×X22j-1β2j-11/β2j-1-1/β2j⋮+∏j=1m/2Xn2jβ2j1/β2jXn2j-1β2j-11/β2jkkkkkkkkk×Xn2j-1β2j-11/β2j-1-1/β2j=∏j=1m1-∑i=1nXijβj1/βj+∑i=1n∏j=1mXij,
so (6) holds when m is even.
Case (II). Let m be odd.
Note that (1/β1-1/β2)+1/β2+1/β2+(1/β3-1/β4)+1/β4+1/β4+⋯ + (1/βm-2-1/βm-1) + 1/βm-1+1/βm-1+1/βm≥1, and, using inequality (2), we have
(10)∏j=1[m/2]1-∑i=1nXi2jβ2j-∑i=1nXi2j-1β2j-121/β2j=∏j=1(m-1)/21-∑i=1nXi2jβ2j-∑i=1nXi2j-1β2j-121/β2j=∏j=1m-1/21-∑i=1nXi2jβ2j-∑i=1nXi2j-1β2j-121/β2j×1-∑i=1nXimβm+∑i=1nXimβm1/βm=∏j=1(m-1)/21-∑i=1nXi2j-1β2j-1+∑i=1nXi2jβ2j1/β2jkkkkkkkkkkkkkkk×1-∑i=1nXi2jβ2j+∑i=1nXi2j-1β2j-11/β2jkkkkkkkkkkkkkkk×1-∑i=1nXi2j-1β2j-1kkkkkkkkkkkkkkkkl+∑i=1nXi2j-1β2j-11/β2j-1-1/β2j∏j=1(m-1)/2×1-∑i=1nXimβm+∑i=1nXimβm1/βm≥∏j=1(m-1)/21-∑i=1nXi2j-1β2j-11/β2j1-∑i=1nXi2jβ2j1/β2jkkkkkkkkkkkk×1-∑i=1nXi2j-1β2j-11/β2j-1-1/β2j∏j=1(m-1)/2×1-∑i=1nXimβm1/βm+∏j=1m-1/2X12jβ2j1/β2jX12j-1β2j-11/β2jkkkkkkkkkkkkk×X12j-1β2j-11/β2j-1-1/β2j∏j=1m-1/2×X1mβm1/βm+∏j=1m-1/2X22jβ2j1/β2jX22j-1β2j-11/β2jkkkkkkkkkkkkk×X22j-1β2j-11/β2j-1-1/β2j∏j=1m-1/2×X2mβm1/βm⋮+∏j=1m-1/2Xn2jβ2j1/β2jXn2j-1β2j-11/β2jkkkkkkkkkkkkk×Xn2j-1β2j-11/β2j-1-1/β2j∏j=1m-1/2×Xnmβm1/βm=∏j=1m1-∑i=1nXijβj1/βj+∑i=1n∏j=1mXij,
so (6) holds for m is odd.
(b) Using similar reasoning as in Case (a) and using inequality (3), we obtain inequality (7).
(c) The proof of inequality (8) is similar to the reasoning used to prove inequality (6) so we omit it.
The proof of Lemma 2 is complete.
Next, we present new refinements of inequalities (2), (3), and (4).
Theorem 3.
Let Aij>0(i=1,2,…,n, j=1,2,…,m), and let s be any given natural number (1≤s≤n).
If 0<β1<β2<⋯<βm and if ∑j=1m(1/βj)≥1, then
(11)∑i=1n∏j=1mAij≤∏j=1m∑i=1nAijβj1/βj×∏j=1[m/2]1-As2jβ2j∑k=1nAk2jβ2j-As2j-1β2j-1∑k=1nAk2j-1β2j-121/β2j≤∏j=1m∑i=1nAijβj1/βj.
If β1>0, 0>β2>β3>⋯>βm, and if ∑j=1m(1/βj)≤1, then
(12)∑i=1n∏j=1mAij≥∏j=1m∑i=1nAijβj1/βj×∏j=1[m/2]1-As2jβ2j∑k=1nAk2jβ2j-As2j-1β2j-1∑k=1nAk2j-1β2j-121/β2j≥∏j=1m∑i=1nAijβj1/βj.
If 0>β1>β2>⋯>βm, then
(13)∑i=1n∏j=1mAij≥∏j=1m∑i=1nAijβj1/βj×∏j=1[m/2]1-As2jβ2j∑k=1nAk2jβ2j-As2j-1β2j-1∑k=1nAk2j-1β2j-121/β2j≥∏j=1m∑i=1nAijβj1/βj.
Proof.
(a) Consider the substitution
(14)Xij=Aij∑k=1nAkjβj1/βj(i=1,2,…,n,j=1,2,…,m).
It is easy to see that, for any given natural number s(1≤s≤n), the following inequalities hold:
(15)Xij>0,1-∑1≤i≤n,i≠sXijβj>0.
Consequently, by using substitution (14) and inequality (6), we have
(16)∏j=1m1-∑1≤i≤n,i≠sAijβj∑k=1nAkjβj1/βj+∑1≤i≤n,i≠s∏j=1mAij∑k=1nAkjβj1/βj≤∏j=1m/2∑1≤i≤n,i≠sAi2j-1β2j-1∑k=1nAk2j-1β2j-121-∑1≤i≤n,i≠sAi2jβ2j∑k=1nAk2jβ2jkkkkkkkkkkkkkk-∑1≤i≤n,i≠sAi2j-1β2j-1∑k=1nAk2j-1β2j-12∑1≤i≤n,i≠sAi2j-1β2j-1∑k=1nAk2j-1β2j-121-∑1≤i≤n,i≠sAi2jβ2j∑k=1nAk2jβ2j1/β2j,
and thus we have
(17)∏j=1mAsj∏j=1m∑k=1nAkjβj1/βj+∑1≤i≤n,i≠s∏j=1mAij∏j=1m∑k=1nAkjβj1/βj≤∏j=1[m/2]1-As2jβ2j∑k=1nAk2jβ2j-As2j-1β2j-1∑k=1nAk2j-1β2j-121/β2j;
that is,
(18)∑i=1n(∏j=1mAij)∏j=1m∑k=1nAkjβj1/βj≤∏j=1[m/2]1-As2jβ2j∑k=1nAk2jβ2j-As2j-1β2j-1∑k=1nAk2j-1β2j-121/β2j.
We have the desired inequality (11). The proof of inequalities (12) and (13) is similar to the reasoning used to prove inequality (11) so we omit the proof.
From Theorem 3, we obtain the following new refinements of generalized Hölder inequalities (2), (3), and (4).
Theorem 4.
Let Aij>0(i=1,2,…,n, j=1,2,…,m), and let s be any given natural number (1≤s≤n).
If 0<β1<β2<⋯<βm and if ∑j=1m(1/βj)≥1, then
(19)∑i=1n∏j=1mAij≤∏j=1m∑i=1nAijβj1/βj×min1≤s≤n∏j=1[m/2]As2j-1β2j-1∑k=1nAk2j-1β2j-121-As2jβ2j∑k=1nAk2jβ2jkkkkkkkkkkkkkkkkkkkk-As2j-1β2j-1∑k=1nAk2j-1β2j-121/β2j≤∏j=1m∑i=1nAijβj1/βj.
If β1>0, 0>β2>β3>⋯>βm, and if ∑j=1m(1/βj)≤1, then
(20)∑i=1n∏j=1mAij≥∏j=1m∑i=1nAijβj1/βj×max1≤s≤n∏j=1[m/2]As2j-1β2j-1∑k=1nAk2j-1β2j-121-As2jβ2j∑k=1nAk2jβ2jkkkkkkkkkkkkkkkkkkkk-As2j-1β2j-1∑k=1nAk2j-1β2j-121/β2j≥∏j=1m∑i=1nAijβj1/βj.
If 0>β1>β2>⋯>βm, then
(21)∑i=1n∏j=1mAij≥∏j=1m∑i=1nAijβj1/βj×max1≤s≤n∏j=1[m/2]-As2j-1β2j-1∑k=1nAk2j-1β2j-121-As2jβ2j∑k=1nAk2jβ2jkkkkkkkkkkkkkkkkkkkk-As2j-1β2j-1∑k=1nAk2j-1β2j-121/β2j≥∏j=1m∑i=1nAijβj1/βj.
From Lemma 1 and Theorem 4, we obtain the following refinements of Hölder’s inequality.
Theorem 5.
Let Aij>0(i=1,2,…,n, j=1,2,…,m), and let s be any given natural number (1≤s≤n).
If 0<β1<β2<⋯<βm and if ∑j=1m(1/βj)≥1, then
(22)∑i=1n∏j=1mAij≤∏j=1m∑i=1nAijβj1/βj×min1≤s≤n∏j=1m/2-As2j-1β2j-1∑k=1nAk2j-1β2j-121-1β2jAs2jβ2j∑k=1nAk2jβ2jkkkkkkkkkkkkkkkkkkkkkkkkk-As2j-1β2j-1∑k=1nAk2j-1β2j-12≤∏j=1m∑i=1nAijβj1/βj.
If β1>0, 0>β2>β3>⋯>βm, and if ∑j=1m(1/βj)≤1, then
(23)∑i=1n∏j=1mAij≥∏j=1m∑i=1nAijβj1/βj×max1≤s≤n∏j=1m/2As2j-1β2j-1∑k=1nAk2j-1β2j-121-1β2jAs2jβ2j∑k=1nAk2jβ2jkkkkkkkkkkkkkkkkkkkkkkkkkl-As2j-1β2j-1∑k=1nAk2j-1β2j-12≥∏j=1m∑i=1nAijβj1/βj.
If 0>β1>β2>⋯>βm, then
(24)∑i=1n∏j=1mAij≥∏j=1m∑i=1nAijβj1/βj×max1≤s≤n∏j=1m/2-As2j-1β2j-1∑k=1nAk2j-1β2j-121-1β2jAs2jβ2j∑k=1nAk2jβ2jkkkkkkkkkkkkkkkkkkkkkkklkl-As2j-1β2j-1∑k=1nAk2j-1β2j-12≥∏j=1m∑i=1nAijβj1/βj.
In particular, putting m=2, β2=p, β1=q, Ar1=br, Ar2=ar(r=1,2,…,n) in inequality (19) and putting m=2, β1=p, β2=q, Ar1=ar, Ar2=br(r=1,2,…,n) in inequalities (20) and (21), respectively, we obtain the following corollary.
Corollary 6.
Let ar,br>0(r=1,2,…,n), and let s be any given natural number (1≤s≤n).
If p>q>0, 1/p+1/q≥1, then
(25)∑k=1nakbk≤∑k=1nakp1/p∑k=1nbkq1/q×min1≤s≤n1-asp∑k=1nakp-bsq∑k=1nbkq21/p.
If p>0, q<0, 1/p+1/q≤1, then
(26)∑k=1nakbk≥∑k=1nakp1/p∑k=1nbkq1/q×max1≤s≤n1-asp∑k=1nakp-bsq∑k=1nbkq21/q.
If 0>p>q, then
(27)∑k=1nakbk≥∑k=1nakp1/p∑k=1nbkq1/q×max1≤s≤n1-asp∑k=1nakp-bsq∑k=1nbkq21/q.
Remark 7.
Let n=3, b1=1, b2=1, b3=1, a1=19, a2=1, and a3=1, and let p=1, q=1/2. Then from inequality (25) we obtain 21≤891/7≈127.28571.
Similarly, putting m=2, β2=p, β1=q, Ar1=br, Ar2=ar(r=1,2,…,n) in inequality (22) and putting m=2, β1=p, β2=q, Ar1=ar, Ar2=br(r=1,2,…,n) in inequalities (23) and (24), respectively, we obtain the following corollary.
Corollary 8.
Let ar,br>0(r=1,2,…,n), and let s be any given natural number (1≤s≤n).
If p>q>0, 1/p+1/q≥1, then
(28)∑k=1nakbk≤∑k=1nakp1/p∑k=1nbkq1/q×min1≤s≤n1-1pasp∑k=1nakp-bsq∑k=1nbkq2.
If p>0, q<0, 1/p+1/q≤1, then
(29)∑k=1nakbk≥∑k=1nakp1/p∑k=1nbkq1/q×max1≤s≤n1-1qasp∑k=1nakp-bsq∑k=1nbkq2.
If 0>q>p, then
(30)∑k=1nakbk≥∑k=1nakp1/p∑k=1nbkq1/q×max1≤s≤n[1-1qasp∑k=1nakp-bsq∑k=1nbkq2].
3. Applications
In this section, we give two applications of our new inequalities. Firstly, we present a refinement of Minkowski’s inequality.
Theorem 9.
Let ak>0, bk>0(k=1,2,…,n), and let s be any given natural number (1≤s≤n). If p>1, then
(31)∑k=1nak+bkp1/p≤∑k=1nakp1/p+∑k=1nbkp1/p-max1≤s≤n1p∑k=1nakp1/pasp∑k=1nakp-as+bsp∑k=1n(ak+bk)p2-max1≤s≤n1p∑k=1nbkp1/pbsp∑k=1nbkp-as+bsp∑k=1nak+bkp2≤∑k=1nakp1/p+∑k=1nbkp1/p.
If 0<p<1, then
(32)∑k=1nak+bkp1/p≥∑k=1nakp1/p+∑k=1nbkp1/p+max1≤s≤n1-pp∑k=1nakp1/pasp∑k=1nakp-as+bsp∑k=1nak+bkp2+max1≤s≤n1-pp∑k=1nbkp1/pbsp∑k=1nbkp-as+bsp∑k=1nak+bkp2≥∑k=1nakp1/p+∑k=1nbkp1/p.
Proof.
Consider the following.
Case (i). Let p>1.
Now
(33)∑k=1nak+bkp=∑k=1nakak+bkp-1+∑k=1nbkak+bkp-1,
and apply Corollary 6 with indices p and p/p-1 to each sum on the right so
(34)∑k=1nak+bkp≤∑k=1nakp1/p∑k=1nak+bkp(p-1)/p×min1≤s≤n1-asp∑k=1nakp-(as+bs)p∑k=1n(ak+bk)p21/p+∑k=1nbkp1/p∑k=1nak+bkp(p-1)/p×min1≤s≤n1-bsp∑k=1nbkp-(as+bs)p∑k=1n(ak+bk)p21/p.
From Lemma 1 we have
(35)∑k=1nak+bkp≤∑k=1nakp1/p∑k=1nak+bkp(p-1)/p×min1≤s≤n1-1pasp∑k=1nakp-as+bsp∑k=1nak+bkp2+∑k=1nbkp1/p∑k=1nak+bkp(p-1)/p×min1≤s≤n1-1pbsp∑k=1nbkp-as+bsp∑k=1nak+bkp2.
Dividing both sides by (∑k=1n(ak+bk)p)(p-1)/p, we obtain the desired inequality.
Case (ii). Let 0<p<1.
Similar reasoning as in Case (i) yields inequality (32).
Now, we give a sharpened version of a Beckenbach-type inequality. The Beckenbach inequality [11] was generalized and extended in several directions (see, e.g., [12–15]). In 1983, Wang [16] presented the following Beckenbach-type inequality.
Theorem B.
Let p>q>0, 1/p+1/q=1, let a,b,c be positive numbers, and let f(x), g(x) be positive integrable functions defined on [0,T]. Then
(36)a+c∫0Tφpxdx1/pb+c∫0Tφ(x)g(x)dx≤a+c∫0Tfpxdx1/pb+c∫0Tf(x)g(x)dx,
where φ(x)=(ag(x)/b)q/p. The sign of the inequality in (36) is reversed if 0<p<1.
From Corollary 8, we obtain a new refinement of Beckenbach-type inequality (36).
Theorem 10.
Let a,b,c be positive numbers, and let f(x), g(x) be positive integrable functions defined on [0,T]. If p>q>0, 1/p+1/q=1, then
(37)a+c∫0Tφpxdx1/pb+c∫0Tφ(x)g(x)dx≤a+c∫0Tfpxdx1/pb+c∫0Tf(x)g(x)dx×1-1paa+c∫0Tfpxdxkkkkkkkkkkkkkkl-a-q/pbqa-q/pbq+c∫0Tgqxdx2,
where φ(x)=(ag(x)/b)q/p.
If 0<p<1, q<0, 1/p+1/q=1, then
(38)a+c∫0Tφpxdx1/pb+c∫0Tφ(x)g(x)dx≥a+c∫0Tfpxdx1/pb+c∫0Tf(x)g(x)dx×1-1qaa+c∫0Tfpxdxkkkkkkkkkkklklkk-a-q/pbqa-q/pbq+c∫0Tgqxdx2,
where φ(x)=(ag(x)/b)q/p.
Proof.
We first consider the case p>q>0, 1/p+1/q=1. Using simple computations, we have
(39)a+c∫0Tφpxdx1/pb+c∫0Tφ(x)g(x)dx=a-q/pbq+c∫0Tgqxdx-1/q.
Moreover, from Corollary 8 we obtain
(40)b+c∫0Tf(x)g(x)dx≤b+c∫0Tfpxdx1/p∫0Tgqxdx1/q=a1/p(ba-1/p)+c∫0Tfpxdx1/p∫0Tgqxdx1/q≤a+c∫0Tfpxdx1/pa-q/pbq+c∫0Tgqxdx1/q×1-1paa+c∫0Tfpxdxkkkkkkkkkkkkkl-a-q/pbqa-q/pbq+c∫0Tgqxdx2;
that is,
(41)a-q/pbq+c∫0Tgqxdx-1/q≤a+c∫0Tfpxdx1/pb+c∫0Tf(x)g(x)dx×1-1paa+c∫0Tfpxdxkkkkkkkkkkkklkl-a-q/pbqa-q/pbq+c∫0Tgqxdx2.
Then combining inequalities (39) and (41) yields inequality (37).
Using the above reasoning and applying inequality (26), it is easy to obtain inequality (38).
4. Conclusions
In this paper we presented some new refinements of Hölder’s inequality and we obtained a refinement of Cauchy’s inequality. We improved Minkowski inequality and a Beckenbach-type inequality. In future research we hope to obtain new results using inequalities (11), (12), (13), and (19)–(30).
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
This work was supported by the Fundamental Research Funds for the Central Universities (no. 13ZD19) and the Higher School Science Research of Hebei Province of China (no. Z2013038).
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