# On Satnoianu-Wu’s Inequality

Authors: Bo-Yan Xi
Journal: The Scientific World Journal
 (2013)

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## Abstract

By applying techniques in the theory of convex functions and Schur-geometrically convex functions, the author investigates a conjecture of Satnoianu on an algebraic inequality and generalizes some known results in recent years.


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## Body

## 1. Introduction

Forn∈ℕ={1,2,3,…} and xi∈ℝ+=(0,∞) with i=1,2,…,n, let
(1)An(x)=1n∑i=1n‍xi,Gn(x)=(∏i=1n‍xi)1/n,Hn(x)=n∑i=1n(1/xi).
These quantities are, respectively, called the arithmetic, geometric, and harmonic means of a positive sequence x=(x1,x2,…,xn). For more information on the theory of means, see the monograph [1] or the papers [2–5] and plenty of references therein.For convenience, in what follows, we use the notationxq=(x1q,x2q,…,xnq) for
(2)0<m≤x1q,x2q,…,xnq≤M,
and q∈ℝ=(-∞,∞).In [6], Satnoianu posed the following conjecture.Conjecture 1.
Forn≥2, xi>0, and λ≥nn-1-1, it is valid that
(3)∑i=1n‍(xin-1xin-1+λ∏k=1,k≠inxk)1/(n-1)≥n(1+λ)1/(n-1).This conjecture has been solved and researched in [7–10]. Among them, Wu obtained in [10] the following result.Theorem 2 (see [10, Theorem  1]).
Letn be a positive integer and let α, β, and xi for i=1,2,…,n be positive numbers. If p≠0 and
(4)β≥(nmax{p,1}-1)α,
then
(5)∑i=1n‍[xiqαxiq+βG(xq)]1/p≥n(α+β)1/p.It is clear that inequality (5) generalizes (3). Therefore, we would like to call inequality (5) Satnoianu-Wu’s inequality.Now we naturally pose the following questions:(1)
Can one improve the condition (4)?(2)
Does the reversed inequality of (5) exist?(3)
Are there other types of inequalities of Satnoianu-Wu type?The goal of this paper is to answer these questions.


## 2. Definitions and Lemmas

We need the following definitions and lemmas.Definition 3 (see [11, page 8]).
Letx=(x1,x2,…,xn)∈ℝn and y=(y1,y2,…,yn)∈ℝn. We say that x is majorized by y (in symbols x≺y) if
(6)∑i=1k‍x[i]≤∑i=1k‍y[i],∑i=1n‍xi=∑i=1n‍yi,
for k=1,2,…,n-1, where x[1]≥⋯≥x[n] and y[1]≥⋯≥y[n] are rearrangements of x and y in a descending order.Definition 4 (see [12]).
LetΩ⊆ℝ+n.(1)
The setΩ is said to be geometrically convex if (x1λy11-λ,…,xnλyn1-λ)∈Ω for every x,y∈Ω and λ∈[0,1].(2)
A functionφ:Ω→ℝ+ is said to be Schur-geometrically convex on Ω if lnx=(lnx1,…,lnxn)≺lny=(lny1,…,lnyn) implies φ(x)≤φ(y) for every x,y∈Ω.(3)
A functionφ:Ω→ℝ+ is said to be Schur-geometrically concave on Ω if lnx=(lnx1,…,lnxn)≺lny=(lny1,…,lnyn) implies φ(x)≥φ(y) for every x,y∈Ω.Lemma 5 (see [12]).
LetΩ⊆ℝ+n be a symmetric and geometrically convex set with inner points and φ:Ω→ℝ+ a symmetric and differentiable function in Ω∘. Then φ is a Schur-geometrically convex (or Schur-geometrically concave, resp.) function on Ω if and only if
(7)(lnx1-lnx2)[x1∂φ(x)∂x1-x2∂φ(x)∂x2]≥0(or≤0,resp.),x∈Ω∘.Lemma 6 (see [1, page 4, Bernoulli's inequality]).
The inequality(8)(1+t)r≥1+rt
holds for r≥1 and t≥-1 or for r≤0 and t>-1. If  0<r<1 and t≥-1, inequality (8) is reversed.


## 3. Main Results

Now we start off to demonstrate our main results.Theorem 7.
Letm,M be defined as in (2) and n≥2, let α,β,xi∈ℝ+ for i=1,2,…,n, and let p,q∈ℝ with p≠0. If (1-p)mβ≥2pMα, then
(9)∑i=1n‍[xiqαxiq+βAn(xq)]1/p≥n(α+β)1/p.
If (1-p)Mβ≤2pmα, inequality (9) is reversed.Proof.
ForM′>m′>0 and r∈ℝ with r≠0, since
(10)[(t1+t)r]′′=r(r-1-2t)tr-1(1+t)r+2,t∈[m′,M′],
then(1)
forr<0, or for r>1 and r-1-2M′>0, the function (t/(1+t))r is convex on [m′,M′];(2)
for0<r≤1, or for r>1 and r-1-2m′<0, the function (t/(1+t))r is concave on [m′,M′].By Jensen’s inequality, when r<0, or when r>1 and r-1-2M′>0, we have
(11)∑i=1n‍(ui1+ui)r≥n[An(u)1+An(u)]r,
where u=(u1,u2,…,un)∈[m′,M′]n. When 0<r≤1, or when r>1 and r-1-2m′<0, inequality (11) is reversed.
Letui=αxiq/βAn(xq) for i=1,2,…,n, r=1/p, m′=αm/βM, and M′=αM/βm. Then it is easy to show that An(u)=α/β. Making use of (11), we obtain inequality (9). The proof of Theorem 7 is complete.Corollary 8.
Under the conditions of Theorem7 and when M=(n-1)m, if (1-p)β>2p(n-1)α, then
(12)∑i=1n‍[xiqαxiq+βAn(xq)]1/p≥n(α+β)1/p;
if (1-p)(n-1)β<2pα, inequality (12) is reversed.Corollary 9.
Letn≥2, α,β,xi∈ℝ+ for i=1,2,…,n, and p,q∈ℝ. If p<0, then
(13)∑i=1n‍[xiqαxiq+βAn(xq)]1/p≥n(α+β)1/p;
if p≥1, inequality (13) is reversed.Proof.
This follows from Theorem7 by considering that if p<0, we have (1-p)mβ>2pMα and that if p≥1, we have (1-p)Mβ<2pmα.Theorem 10.
Letn≥2, α,β,xi∈ℝ+ for  i=1,2,…,n, p,q∈ℝ with p≠0, and m,M defined by (2). If  mβ≥pMα, then
(14)∑i=1n‍[xiqαxiq+βGn(xq)]1/p≥n(α+β)1/p;
if Mβ≤mpα, inequality (14) is reversed.Proof.
ForM′>m′>0, let
(15)f(u)=∑i=1n‍(1+ui)r,u∈[m′,M′]n.
Then
(16)∂f(u)∂ui=r(1+ui)r-1,∂∂ui[ui∂f(u)∂ui]=r(1+ui)r-2(1+rui),
for 1≤i≤n. When r>0, or when r<0 and 1+rm′<0, we have
(17)(lnu1-lnu2)[u1∂f(u)∂u1-u2∂f(u)∂u2]≥0,u∈[m′,M′]n;
when r<0 and 1+rM′>0, inequality (17) is reversed.
Using Lemma5, we have the following conclusions:(1)
ifr>0, or if r<0 and 1+rm′<0, the function f(u) is Schur-geometrically convex on [m′,M′]n;(2)
ifr<0 and 1+rM′>0, the function f(u) is Schur-geometrically concave on [m′,M′]n.By the fact that
(18)(lnGn(u),lnGn(u),…,lnGn(u))≺(lnu1,lnu2,…,lnun),u∈ℝ+n,
and by Definition 4, if r>0, or if r<0 and 1+rm′<0, we have
(19)∑i=1n‍(1+ui)r≥n(1+Gn(u))r,u∈[m′,M′]n;
if r<0 and 1+rM′>0, inequality (19) is reversed.
Lettingui=βGn(xq)/αxiq for i=1,2,…,n leads to Gn(u)=β/α. Putting r=-1/p, m′=βm/αM and M′=βM/αm in inequality (19) results in inequality (14). The proof of Theorem 10 is complete.Corollary 11.
Under the conditions of Theorem10 and when M=(n-1)m, if β≥p(n-1)α, then
(20)∑i=1n‍[xiqαxiq+βAn(xq)]1/p≥n(α+β)1/p;
if (n-1)β≤pα, inequality (20) is reversed.Remark 12.
Whenn≥2 and p>1, from Lemma 6 and (4), it follows that
(21)β≥(nmax{p,1}-1)α>p(n-1)α.Theorem 13.
Letn≥2, α,β,xi∈ℝ+ for i=1,2,…,n, and p,q∈ℝ with p≠0. If p≥-1, then
(22)∑i=1n‍[xiqαxiq+βHn(xq)]1/p≥n(α+β)1/p;
if p≤-1, inequality (22) is reversed.Proof.
Since(1+t)r is a convex (or concave, resp.) function on ℝ+ for r≥1 or r<0 (or for 0<r≤1, resp.), by Jensen's inequality, if r≥1 or r<0, we have
(23)∑i=1n‍(1+ui)r≥n(1+An(u))r,u∈ℝ+n;
if 0<r≤1, inequality (23) is reversed.
Lettingui=βHn(xq)/αxiq and r=-1/p shows An(u)=β/α. Further from (23), we obtain inequality (22). The proof of Theorem 13 is complete.Remark 14.
It is clear that inequalities (9) and (22) both generalize inequality (3).Corollary 15.
Letn≥2, α,β,xi∈ℝ+ for i=1,2,…,n, p,q∈ℝ with p≠0, and m,M defined as in (2).(1)
When-1≤p<0, one has
(24)∑i=1n[xiqαxiq+βAn(xq)]1/p‍≥∑i=1n‍[xiqαxiq+βGn(xq)]1/p≥∑i=1n‍[xiqαxiq+βHn(xq)]1/p≥n(α+β)1/p.(2)
When0<p<1 and (1-p)mβ>2pMα, one has
(25)∑i=1n‍[xiqαxiq+βHn(xq)]1/p≥∑i=1n‍[xiqαxiq+βGn(xq)]1/p≥∑i=1n‍[xiqαxiq+βAn(xq)]1/p≥n(α+β)1/p.(3)
Whenp>0 and mβ>pMα, one has
(26)∑i=1n‍[xiqαxiq+βHn(xq)]1/p≥∑i=1n‍[xiqαxiq+βGn(xq)]1/p≥n(α+β)1/p.(4)
Whenp>0 and Mβ<mpα, one has
(27)∑i=1n‍[xiqαxiq+βAn(xq)]1/p≤∑i=1n‍[xiqαxiq+βGn(xq)]1/p≤n(α+β)1/p.Proof.
This follows from utilizing the well-known harmonic-geometric-arithmetic mean inequality(28)Hn(x)≤Gn(x)≤An(x)
and Theorems 7 to 13.Corollary 16.
Under the conditions of Corollary15 and when M=(n-1)m,(1)
if0<p<1 and (1-p)β>2p(n-1)α, one has
(29)∑i=1n‍[xiqαxiq+βHn(xq)]1/p≥∑i=1n‍[xiqαxiq+βGn(xq)]1/p≥∑i=1n‍[xiqαxiq+βAn(xq)]1/p≥n(α+β)1/p;(2)
ifp>0 and β>p(n-1)α, one has
(30)∑i=1n[xiqαxiq+βHn(xq)]1/p≥∑i=1n‍[xiqαxiq+βGn(xq)]1/p≥n(α+β)1/p;(3)
ifp>0 and (n-1)β<pα, one has
(31)∑i=1n‍[xiqαxiq+βAn(xq)]1/p≤∑i=1n‍[xiqαxiq+βGn(xq)]1/p≤n(α+β)1/p.



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*Source: 645201-2013-07-16.xml*