PENG GAO
Received 12 May 2005 and in revised form 20 October 2005
The study of the behavior of means under equal increments of their variables provides a new approach to Ky Fan-type inequalities. Via this approach we are able to prove some new results on Ky Fan-type inequalities. We also prove some inequalities involving the symmetric means.
1. Introduction
Let Mn,r(x) be the generalized weighted power means:Mn,r(x)=(ni=1ωixri)1/r, where ωi>0, 1≤i≤n, withni=1ωi=1 andx=(x1,x2,. . .,xn). HereMn,0(x) denotes the limit ofMn,r(x) asr→0+. Unless specified otherwise, we always assume 0< x1≤x2··· ≤xn. We denoteσn=n
i=1ωi(xi−An)2.
To any givenx,t≥0 we associatex=(1−x1, 1−x2,. . ., 1−xn),xt=(x1+t,. . .,xn+t).
When there is no risk of confusion, we will writeMn,rforMn,r(x),Mn,r,tforMn,r(xt), and Mn,r forMn,r(x) ifxn<1. We also defineAn=Mn,1,Gn=Mn,0,Hn=Mn,−1and similarly forAn,Gn,Hn,An,t,Gn,t,Hn,t.
To simplify expressions, we define
∆r,s,t,α=Mn,r,tα −Mn,s,tα
Mn,rα −Mn,sα , ∆r,s=Mn,r −Mn,s
Mn,r−Mn,s (1.1)
with∆r,s,t,0=(ln(Mn,r,t/Mn,s,t))/(ln(Mn,r/Mn,s)). We also write∆r,s,t for∆r,s,t,1. In order to include the case of equality for various inequalities in our discussions, for any given inequality, we define 0/0 to be the number which makes the inequality an equality.
Recently, the author [14, Theorem 2.1] has proved the following result.
Theorem1.1. Forr > s, the following inequalities are equivalent:
r−s
2x1 σn≥Mn,r−Mn,s≥r−s
2xnσn, (1.2)
xn
1−xn≥∆r,s≥ x1
1−x1, (1.3)
where in (1.3) we requirexn<1.
Copyright©2005 Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences 2005:22 (2005) 3551–3574 DOI:10.1155/IJMMS.2005.3551
Cartwright and Field [9] first proved the validity of (1.2) forr=1,s=0. For other extensions and refinements of (1.2), see [3,13, 18,19]. Inequality (1.3) is commonly referred to as the additive Ky Fan’s inequality. We refer the reader to the survey article [2]
and the references therein for an account of Ky Fan’s inequality.
The study of the behavior of means under equal increments of their variables was initiated by Hoehn and Niven [16]. Acz´el and P´ales [1] proved∆1,s,t≤1 for anys=1. We can interpret their result as an assertion of the monotonicity ofAn,t−Mn,s,tas a function oft. The asymptotic behavior oft(Mn,r,t−An,t) was studied by Brenner and Carlson [7].
The same idea of [14] can be used to show that both (1.2) and (1.3) are equivalent to xn
t+x1 ≥∆r,s,t≥ x1
t+xn, (1.4)
which holds for allt≥0,
In Section 3, we will study the monotonicities of (t+xn)(Mn,r,t−Mn,s,t) and (t+ x1)(Mn,r,t−Mn,s,t) as functions oftforr=1 ors=1 and then apply the result to in- equalities of the type (1.2).
The study of the behavior of means under equal increments of their variables provides us with a new approach of studying Ky Fan-type inequalities. InSection 4, we use this approach to show that some of the inequalities we have studied are actually equivalent.
The following inequality connecting three classical means (withωi=1/nhere) is due to W. L. Wang and P. F. Wang [24] (left-hand side inequality) and Alzer et al. [5] (right- hand side inequality):
An An
n−1
Hn Hn ≥
Gn Gn
n
≥ Hn
Hn n−1
An
An. (1.5)
The above inequality was refined in [14] and inSection 5we will give another refine- ment of the above inequality.
Alzer [4] has given a counterexample to show thatAαn−Gαn and Anα−Gnα are not comparable in general forα >1. However, Peˇcari´c and Alzer [22] (see also [2, Theorem 7.2] proved the following result.
Theorem1.2. Forωi=1/n,xn≤1/2,
Ann−Gnn≥Ann−Gnn. (1.6) Theorem 1.2suggests thatAnα−Gnα≥Aαn−Gαnforα=1/qwithq=min{ωi}, a result we will establish inSection 6. A similar result is also proved there.
Letr∈ {0, 1,. . .,n}; therth symmetric functionEn,rofxand its meanPn,rare defined by
En,r(x)=
1≤i1<···<ir≤n
r j=1
xij, 1≤r≤n, En,0=1, Prn,r(x)=En,r(x)
n r
. (1.7)
The usage ofEn,r,En,r,En,r,t,Pn,r,Pn,r,Pn,r,tis similar to the case for the power means.
Many of the results we will obtain for power means also have their analogues for sym- metric means, which we will spend the last section to explore. For example, the result of W. L. Wang and P. F. Wang mentioned above is more general; they have shown the following.
Theorem1.3. For1≤r≤n−1,xi∈(0, 1/2],1≤i≤n,
lnPn,r−lnPn,r+1≥lnPn,r −lnPn,r+1 . (1.8) We also note the following result of Bullen and Marcus [8].
Theorem1.4. For1≤k≤r≤n,
(r+ 1)lnPn+1,k−lnPn+1,r+1 ≥rlnPn,k−lnPn,r (1.9)
with equality holding if and only ifx1= ··· =xn+1.
InSection 7, we will provide a refinement of Theorems1.3and1.4fork=1.
2. Lemmas
Lemma2.1. LetJ(x)be the smallest closed interval that contains all ofxiand letf(x),g(x)∈ C2(J(x))be two twice differentiable functions, then
n
i=1ωif(xi)−fni=1ωixi
n
i=1ωig(xi)−gni=1ωixi = f(ξ)
g(ξ) (2.1)
for someξ∈J(x), provided that the denominator of the left-hand side is nonzero.
Lemma 2.1and the following consequence of it are due to Mercer [17].
Lemma2.2. Forw > u,w=v,u=v,
u(u−v) w(w−v)
1
x1w−u≥
Mn,uu −Mun,v Mn,ww −Mn,vw
≥
u(u−v) w(w−v)
1
xwn−u (2.2) with equality holding if and only ifx1= ··· =xn.
Lemma 2.3. Lett≥0, q=(q1,. . .,qn)with qi≥1,1≤i≤n, then for n≥2, f(x;q)≤ f(xt;q), where
f(x;q)=x2n n
i=1qi −1 n
i=1qixi + 1
x21ni=1qi/xi − 1 x1
, (2.3)
and equality holds if and only if one of the following conditions holds:(i)t=0;(ii)n=2, q1=q2=1;(iii)x1=x2= ··· =xn. Moreover, when (ii) or (iii) happens, it also holds that
f(x;q)= f(xt;q)=0.
Proof. It is routine to check f(x;q)= f(xt;q) whent=0 holds. So now we may assume thatt >0. Then one checks easily that
n
i=1qi −1 n
i=1qixi − 1
x1+ 1
x12ni=1qi/xi
= n
i=1qi
x1−xi −x1
x1
n
i=1qixi + 1
x21
n
i=1qi/xi
= n
i=1qix1−xi −x1 x1
n
i=1qi/xi +ni=1qixi x21
n
i=1qixi n i=1qi/xi
.
(2.4)
Note that
n
i=1
qix1−xi −x1
x1
n
i=1
qi
xi
+
n i=1
qixi
= n
i=1
qi x1−xi
x1
n
i=1
qi
xi
− n i=1
qi x21−x2i
xi
= n i=1
qi x1−xi
q1−1 +
j=1,i
qjx1
xj +
qi−1 x1
xi
.
(2.5)
Hence, f(x;q)=
n
i=1qix1−xi q1−1 xn/x1+j=1,iqjxn/xj +qi−1 xn/xi n
i=1qixi/xn ni=1qix1/xi ≤0. (2.6) Sincex/ y >(x+t)/(y+t) whenx > y >0,t >0, it follows from the above expression that f(x;q)≤f(xt;q). The conditions for equality can be checked easily and this completes
the proof.
Lemma 2.4. Let t≥0,q=(q1,. . .,qn)with qi≥1,1≤i≤n, then forn≥2,g(x;q)≤ g(xt;q), where
g(x;q)=x2n 1
x1− 1
n i=1qixi−
n
i=1qi −1 x21ni=1qi/xi
, (2.7)
and equality holds if and only if one of the following conditions holds:(i)t=0;(ii)n=2, q1=q2=1;(iii)x1=x2= ··· =xn. Moreover, when (ii) or (iii) happens, it also holds that g(x;q)=g(xt;q)=0.
Proof. It is routine to checkg(x;q)=g(xt;q) whent=0 holds. So now we may assume thatt >0. Then one checks easily that
1 x1−
n
i=1qi −1 x21
n
i=1qi/xi − 1
n i=1qixi
= n
i=1qi
x1−xi /xi+ 1 x12ni=1qi/xi −
n1
i=1qixi
= n
i=1qi
x1−xi /xi+ 1 ni=1qixi −x12ni=1qi/xi
x21ni=1qixi ni=1qi/xi .
(2.8)
Note that
n
i=1
qi
x1−xi
xi + 1 n
i=1
qixi
−x21 n
i=1
qi
xi
= n
i=1
qi
x1−xi
xi
n
i=1
qixi
− n i=1
qi
x21−x2i xi
= n i=1
qi
x1−xi
xi
q1−1 x1+
j=1,i
qjxj+qi−1 xi
.
(2.9)
Hence, g(x;q)=
n
i=1qi
x1−xi xn/xi q1−1 +j=1,iqjxj/x1 +qi−1 xi/x1
n
i=1qixi/xn n
i=1qix1/xi ≤0.
(2.10) Sincex/ y >(x+t)/(y+t) whenx > y >0,t >0, it follows from the above expression that g(x;q)≤g(xt;q). The conditions for equality can be checked easily and this completes
the proof.
Lemma2.5. For0< q <1, when0< y < x≤1,
f(q)=2qx1/q−y1/q (2.11)
is an increasing function ofq. When1≤y < x,f(q)is a decreasing function ofq.
Proof. We have
f(q)=2x1/q−y1/q −2lnx1/q x1/q−lny1/q y1/q . (2.12) So it is enough to show thatu−ulnuincreases for 0< u≤1 and decreases foru≥1 and
this is easy to check.
Lemma2.6. For0< q <1,(1−q)1/q−1is an increasing function ofq, in particular, (1−q)1/q−1≤1
2 (2.13)
when0< q≤1/2and the above inequality reverses when1/2≤q <1. In either case, equality holds if and only ifq=1/2.
Proof. It suffices to show thatf(q)>0 for 0< q <1 withf(q)=(1/q−1) ln(1−q). Now f(q)= −h(q)/q2withh(q)=q+ ln(1−q)<0 for 0< q <1 and this completes the proof.
The following lemma is due to Wu et al. [25] (see also [2, pages 317-318]).
Lemma2.7. Let2≤r≤n,x=(x1,. . .,xn),x1≤x2≤ ··· ≤xn. There existsy=(y1,. . .,yr) withx1≤y1≤ ··· ≤yr≤xnsuch thatPn,i(x)=Pr,i(y),0≤i≤r. Moreover, ifx1,. . .,xn
are not all equal, theny1,. . .,yrare also not all equal.
3. Some monotonicity properties Theorem3.1. Letr > s,t≥0.
(i)If∆r,s,t,α≤1, then∆r,s,t,β≤1forβ≤α. If∆r,s,t,α≥1, then∆r,s,t,β≥1forβ≥α.
(ii)Letα≤1. If∆r,s,t≤xn/(t+xn), then∆r,s,t,α≤(xn/(t+xn))2−α. If∆r,s,t≥x1/(t+x1), then∆r,s,t,α≥(x1/(t+x1))2−α.
(iii)∆r,s,t,α≤1forα≤0and for anys=1,α≤1,∆1,s,t,α≤1.
(iv)For any r = 1, min(((t+ xn)/xn)r−2, ((t+ x1)/x1)r−2) ≤ ∆r,1,t,r ≤ max(((t + xn)/xn)r−2, ((t+x1)/x1)r−2).
(v)For−1≤s=1≤2,
xn
t+xn≥∆1,s,t≥ x1
t+x1 (3.1)
with equality holding if and only ift=0orx1= ··· =xn.
Proof. (i) Let f(t)= |Mn,r,tα −Mn,s,tα |. Sincexis arbitrary,∆r,s,t,α≤1 is then equivalent to f(0)≤0 or the second inequality below:
Mn,rβ−r
Mβn,s−s
≤Mαn,r−r Mn,sα−s ≤
Mn,r1−−r1
M1n,s−−s1. (3.2)
Now∆r,s,t,β≤1 follows from the first inequality above. This proves the first assertion and the second assertion follows similarly.
(ii) We will prove the first assertion for 0< α <1 and the other proofs are similar. Let f(t)=(t+xn)2−α(Mn,r,tα −Mαn,s,t), then it suffices to show f(0)≤0, or equivalently,
(2−α)Mn,rα −Mn,sα ≤αxn
Mαn,s−1 Mn,s
Mn,s−1
1−s
−Mn,rα−1 Mn,r
Mn,r−1
1−r
. (3.3)
We also have M1n,s−α
α
Mαn,r−Mn,sα ≤Mn,r−Mn,s≤xn
Mn,s
Mn,s−1
1−s
− Mn,r
Mn,r−1
1−r
, (3.4)
where the first inequality above follows from the mean value theorem and the second inequality follows from∆r,s,t≤xn/(t+xn). Similarly, by using the mean value theorem, we get
Mn,rα −Mn,sα Mαn,s−1−Mαn,r−1≤
α
1−αMn,r≤ α 1−αxn
Mn,r Mn,r−1
1−r
, (3.5)
where the last inequality follows fromMrn,r=n
i=1ωixri ≤n
i=1ωixnxri−1=xnMn,rr−1−1. Now (ii) follows by rewriting (3.4) and (3.5) as
Mαn,r−Mn,sα ≤αMn,sα−1xn
Mn,s Mn,s−1
1−s
− Mn,r
Mn,r−1
1−r ,
(1−α)Mn,rα −Mαn,s ≤αxn
Mn,sα−1−Mn,rα−1
Mn,r Mn,r−1
1−r (3.6)
and adding the above two inequalities.
(iii)∆1,s,t,α≤1, fors=1,α≤1, follows from the result of Acz´el and P´ales [1] and (i). Again by (i), in order to show∆r,s,t,α≤1 forα≤0, it suffices to show∆r,s,t,0≤1. Let f(t)=lnMn,r,t−lnMn,s,t, it then suffices to show f(0)≤0 orRn,s≤Rn,r, whereRn,r:= Mn,rr /Mn,rr−1−1. The last inequality holds by a result of Beckenbach [6, Theorem 1].
(iv) This follows by applyingLemma 2.1tof(x)=(t+x)r,g(x)=xr,r=0 and f(x)= ln(t+x),g(x)=lnxwhenr=0.
(v) We will prove the left-hand side inequality of (3.1) and the other proofs are similar.
For 0≤s <1, let
Dn(x,t)=xnAn−Mn,s −
t+xn An,t−Mn,s,t . (3.7)
We want to showDn≥0 here. We can assumex1< x2<···< xnand prove by induction that the casen=1 is clear, so we will start withn >1 variables assuming the inequality holds forn−1 variables. Then
∂Dn
∂xn =
An−Mn,s −
An,t−Mn,s,t +ωnAn−Mn,s1−sxns −
An,t−Mn,s,t1−st+xn s
≥ωn
An−Mn,s −
An,t−Mn,s,t +An−Mn,s1−sxns −
An,t−Mn,s,t1−st+xn s
=ωn
Mn,s,t1−st+xn s
+Mn,s,t−2t−Mn,s−Mn,s1−sxsn ,
(3.8) where the inequality follows from∆1,s,t≤1. Now we consider
g(t)=Mn,s,t1−st+xn s+Mn,s,t−2t (3.9)
and we have g(t)=(1−s)
t+xn
Mn,s,t
s Mn,s,t
Mn,s−1,t
1−s
+s Mn,s,t
t+xn
1−s
+
Mn,s,t
Mn,s−1,t
1−s
−2
≥(1−s)ys+sys−1−1≥y(1−s)s+s(s−1)−1=0,
(3.10)
wherey=(t+xn)/Mn,s,t≥1 and the first inequality above follows from (Mn,s,t/Mn,s−1,t)1−s
≥1. The last inequality above follows from the arithmetic-geometric mean inequality.
Thusg(t)≥0, henceg(t)≥g(0)=Mn,s+Mn,s1−sxsnand it follows∂Dn/∂xn≥0 and by let- tingxntend toxn−1, we haveDn≥Dn−1(with weightsω1,. . .,ωn−2,ωn−1+ωn) and thus the right-hand side inequality of (3.1) holds by induction. It is easy to see that the equality holds if and only ift=0 orx1= ··· =xn.
For−1≤s <0, we have 1
ω1
∂Dn
∂x1 = −t−xn
Mn,s
x1
1−s
+t+xn
Mn,s,t
t+x1
1−s
:= −t−fx1 . (3.11) Consider
fx1 = −(1−s) n j=2
ωj
Mn,s x1
1−2s
·xnxsj x1s+1 −
Mn,s,t t+x1
1−2s
t+xn t+xj s
t+x1 s+1
≤0.
(3.12) The last inequality holds, since when−1≤s <0, 2≤j≤n, we have
xn
t+xn· xj
t+xj
s
≥ xj
t+xj
1+s
≥ x1
t+x1
1+s
, (3.13)
andxj/x1≥(t+xj)/(t+x1) so that Mn,s
x1
1−2s
≥ Mn,s,t
t+x1
1−2s
. (3.14)
Thus by a similar argument as above, we deduce that f(x1)≥ −tand∂Dn/∂x1≤0, which further impliesDn≥0 with equality holding if and only ift=0 orx1= ··· =xn.
For 1< s≤2, it suffices to show∂Dn/∂t≤0 or, equivalently, Mn,ss−1
xn ≤
Mn,ss−1−Msn,s−1−1
Mn,s−An . (3.15)
The above inequality certainly holds fors=2, otherwise it follows fromMsn,s−1/xn≤xns−2
andLemma 2.2withu=s−1,v=s,w=1.
We remark here that one cannot compareMn,r,t−Mn,s,tandMn,r−Mn,sin general. For example, let f(t)=Gn,t−Hn,t. Then f(0)=Gn/Hn−Hn2/Mn,2−2. By a change of variables xi→1/xn−i+1, we can rewrite this as f(0)=(A3n−GnMn,22 )/(A2nGn) and by considering the casen=2 withq1=q2=1/2,x2=1, it is easy to see that f(0)<0 ifx1=1/2 and
f(0)>0 ifx1=0.
Corollary3.2. Inequality (1.2) holds forr=1,−1≤s <1and1< r≤2,s=1.
Proof. This follows fromTheorem 3.1(v) and that (1.2) and (1.4) are equivalent.
The above result was first proved by the author in [14, Theorem 3.2]; in fact it was shown there that those are the only cases (1.2) can hold forr=1 ors=1. Thus by notic- ing the equivalence of (1.2) and (1.4), we obtain the following.
Corollary3.3. Inequality (3.1) holds for allt≥0if and only if−1≤s=1≤2.
Corollary3.4. For−1≤s <1, xn
M1n,s−−s1
≥
An−Mn,s Mn,s1−s−M1n,s−−s1
≥ x1
M1n,s−−s1
. (3.16)
Proof. Theorem 3.1(v) implies f(t)=(t+xn)(An,t−Mn,s,t) is a decreasing function oft and f(0)≤0 implies the left-hand side inequality of (3.16). The proof of the right-hand
side inequality of (3.16) is similar.
By a change of variables xi→1/xn−i+1, the left-hand side inequality of (3.16) when s= −1 gives
An−Hn≤ Hn
x1Anσn, (3.17)
a refinement of the left-hand side inequality of (1.2) forr=1,s= −1.
4. Some equivalent inequalities
Theorem4.1. The following inequalities are equivalent:
(i)An−Gn≥σn/2xn; (ii)An−Gn≤σn/2x1;
(iii)An−Gn≤(xn/Hn)(Gn−Hn);
(iv)An−Gn≥(x1/Hn)(Gn−Hn);
(v)Gn−Hn≥Hnσn/2x2n; (vi)Gn−Hn≤Hnσn/2x21.
In particular, inequalities(i)–(vi)are all valid since(i)holds byCorollary 3.2.
Proof. We first show (ii)⇒(iii)⇒(i) and similarly one can show (i)⇒(iv)⇒(ii).
(iii)⇒(i). Let f(t)=(xn+t)(An,t−Gn,t),t≥0. It is easy to see that limt→∞f(t)=σn/2.
Thus it suffices to show f(t) is a decreasing function oftin order to prove (i). Sincexis arbitrary, it suffices to have f(0)=An−Gn+xn(1−Gn/Hn)≤0, which is just (iii).
(ii)⇒(iii). Let f(t)=An,t−Gn,t+ (xn+t)(1−Gn,t/Hn,t), t≥0. It is easy to see that limt→∞f(t)=0, so it suffices to showf(t)≥0 in order to prove (iii). Sincexis arbitrary,
