# On a Class of Ky Fan-Type Inequalities

Authors: Peng Gao
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## Abstract

We study one class of Ky Fan-type inequalities, which has ties with the
original Ky Fan inequality. Our result extends the known ones.


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## Body

## 1. Introduction

LetMn,r(x;q) be the generalized weighted means: Mn,r(x;q)=(∑i=1nqixir)1/r, where Mn,0(x;q) denotes the limit of Mn,r(x;q) as r→0+. Here x=(x1,…,xn), q=(q1,…,qn) with qi>0(1≤i≤n) satisfying ∑i=1nqi=1. In this paper, we always assume 0<x1≤x2≤⋯≤xn. To any given x and t≥0, we set x′=(1-x1,…,1-xn),xt=(x1+t,…,xn+t).We defineAn(x;q)=Mn,1(x;q),Gn(x;q)=Mn,0(x;q),Hn(x;q)=Mn,-1(x;q), and we shall write Mn,r for Mn,r(x;q), Mn,r,t for Mn,r(xt;q), and Mn,r′ for Mn,r(x′;q) if xn<1 and similarly for other means when there is no risk of confusion. We further denote σn=∑i=1nqi(xi-An)2.Whenxn<1, we define(1.1)Δr,s,α,t=(Mn,r,t'α-Mn,s,t'α)/α(Mn,r,tα-Mn,s,tα)/α,
where we set Mn,r0/0=lnMn,r and we shall write Δr,s,α for Δr,s,α,0 and Δr,s for Δr,s,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. The author [1, Theorem 2.1] has shown the following (in fact, only the case α=1 is shown there but one can easily extend the result to all α≤2 following the method there).Theorem 1.1.
Forr>s and α≤2, the following inequalities are equivalent:
(1.2)r-s2x12-ασn≥(Mn,rα-Mn,sα)α≥r-s2xn2-ασn,(1.3)(xn1-xn)2-α≥Δr,s,α≥(x11-x1)2-α,
where in (1.3) one requires xn<1.In fact, one can further show that (see [2]) the two inequalities in Theorem 1.1 are equivalent to(1.4)(xnt+x1)2-α≥Δr,s,α,t≥(x1t+xn)2-α
being valid for all t≥0. We point out here that when inequality (1.2) holds for some r,s, one can often expect for a better result than (1.4), namely,(1.5)(xnt+xn)2-α≥Δr,s,α,t≥(x1t+x1)2-α.We note that inequality (1.2) does not hold for all pairs r,s (see [1]). Cartwright and Field [3] first proved the validity of (1.2) for r=1,s=0,α=1. For other extensions and refinements of (1.2), see [2, 4–8]. When α=1, inequality (1.3) is commonly referred as the additive Ky Fan’s inequality. We refer the reader to the survey article [9] and the references therein for an account of Ky Fan's inequality.In this paper, we will focus on the special caseα=0 of (1.2), which has ties with the following result of Ky Fan that initiated the study of the whole subject.Theorem 1.2 (see [10,page 5]).
Forxi∈(0,1/2], Δ1,0,0≤1, with equality holding if and only if x1=···=xn.A nice result of Wang and Chen [11] determines all the pairs r,s with r>s such that Δr,s,0≤1 is satisfied when xi∈(0,1/2]. Their result is contained in the following.Theorem 1.3.
Forr>s,xi∈(0,1/2], Δr,s,0≤1 holds if and only if |r+s|≤3,2s/s≥2r/r when s>0, s2s≤r2r when r<0.We note here that Theorem1.2 follows from the left-hand side inequality of (1.3) for the case r=1,s=0,andα=0, which in turn is a consequence of the above mentioned result of Cartwright and Field. In fact, we have the following result which is contained implicitly in [12].Theorem 1.4.
If either side of inequality (1.2) holds for r,s,α≤2, then the same side inequality of (1.2) also holds for r,s and any β≤α. Moreover, the above assertion also holds when applied to (1.3) or (1.4).On combining the above result with the result of Cartwright and Field we see that (1.2) holds for r=1,s=α=0 and consequently (1.3) holds for r=1,s=α=0 in virtue of Theorem 1.1.Now, it is natural to be motivated by the result of Wang and Chen, in view of the discussions above, to ask whether one can determine all the pairsr,s with r>s such that either one of the inequalities (1.2)–(1.4) holds for α=0. It is our goal in this paper to investigate such a problem. Before we proceed, we would like to summarize the known results in this area. On taking l=2,t=1 in [5, Proposition 2.3], we deduce with the help of Theorem 1.4 that (1.2) holds for -1≤s≤1,s≤r≤1+s,α=0. On the other hand, [5, Corollary 3.2] combined with Theorem 1.4 implies that (1.3) holds for α=0,0≤s≤r≤1 and r-1≤s≤1,0≤r≤2. We also observe that if (1.2) holds for r>s and s>s′, then it also holds for r>s′. As (1.2) and (1.3) are equivalent, we conclude that when α=0, (1.2) holds for any r>s,0≤r≤2,-1≤s≤1.


## 2. The Main Theorem

Lemma 2.1.
Letr>s, I1=(0,1],I2=[1,+∞) and let E denote the region E={(q1,q2)∣q1≥0,q2≥0,q1+q2=1}. Define
(2.1)Dr,s(t;q1,q2)=tr-1-ts-1+(r-s)(1-t)(q1+q2tr)(q1+q2ts).
Then for s≥0, Dr,s(t;q1,q2)≤0 holds for all (t,q1,q2)∈I1×E if and only if s≤1 and r+s≤3 and Dr,s(t;q1,q2)≤0 holds for all (t,q1,q2)∈I2×E if and only if r≤2 and r+s≤3.
Fors<0, if r≤0, then Dr,s(t;q1,q2)≤0 holds for all (t,q1,q2)∈I1×E if and only if -1≤r≤0 and -3≤r+s≤0 and Dr,s(t;q1,q2)≤0 holds for all (t,q1,q2)∈I2×E if and only if s≥-2 and -3≤r+s≤0.
Fors<0<r, Dr,s(t;q1,q2)≤0 holds for all (t,q1,q2)∈I2×E if and only if r≤2 and r+s≥0 or s≥-2 and r+s≤0.Proof.
Whens≥0, in order for Dr,s(t;q1,q2)≤0 to hold for all (t,q1,q2)∈I1×E, one just needs to check the case q1=1,q2=0. In this case we can rewrite Dr,s(t;1,0) as
(2.2)f(t)=tr-1-ts-1+(r-s)(1-t).
Note that f(1)=f′(1)=0; hence in order for f(t)≤0 to hold for all 0<t≤1, it is necessary that f′′(1)≤0. Note that f′′(t)=(r-1)(r-2)tr-3-(s-1)(s-2)ts-3 and from this one checks easily that f′′(1)≤0 is equivalent to r+s≤3. On the other hand, on taking t→0+, we see that one needs to have s≤1 in order for f(t)≤0 to hold for all 0<t≤1. Now, it also follows from s≤1 that t3-sf′′(t)=(r-1)(r-2)tr-s-(s-1)(s-2)≤max{(r-1)(r-2)-(s-1)(s-2),-(s-1)(s-2)}≤0. Hence one deduces via Taylor expansion of f(t) at 1 that f(t)≤0 for all 0<t≤1.
Similarly, whens≥0, in order for Dr,s(t;q1,q2)≤0 to hold for all (t,q1,q2)∈I2×E, one just needs to check f(t)≤0 for t≥1. As f(1)=f′(1)=0, certainly it is necessary to have f′′(1)≤0 and limt→+∞f(t)≤0. These imply that r≤2 and r+s≤3 and one checks easily that these conditions are also sufficient.
As a consequence of the above discussion, one can deduce the assertion of the lemma for the cases<0 and r≤0 by noting that Dr,s(t;q1,q2)=tr+sD-s,-r(t;q2,q1).
It remains to treat the caser>0>s. We let g(q)=(1-q+qtr)(1-q+qts), and note that g′′(q)=2(tr-1)(ts-1)≤0. It follows from this that in order for Dr,s(t;q1,q2)≤0 to hold for (t,q1,q2)∈I2×E, it suffices to check the cases q2=0,1. When r+s≥0, we only need to check the case q2=0 and in this case one can discuss similarly to the case s≥0 above to conclude the assertion of the lemma. We just point out here that as s<0<r≤2, we have r+s<2. When r+s≤0, it suffices to check the case q2=1 and in this case one uses the relation Dr,s(t;0,1)=tr+sD-s,-r(t;1,0) to convert this to the previous case that has been discussed.Theorem 2.2.
Letr>s. The right-hand side inequality of (1.2) holds for α=0 when 0≤s≤1,r+s≤3 or s<0,-1≤r≤0 and -3≤r+s≤0. The left-hand side inequality of (1.2) holds for α=0 when -2≤s≤0,-3≤r+s≤0.Proof.
To prove the first assertion of the theorem, we may assumer>2 or -1≤r≤0 in view of our discussion in the last paragraph of Section 1 and for the case r>2, we define
(2.3)gn(q,x)=lnMn,r-lnMn,s-r-s2xn2σn.
Similar to the proof of Theorem 5.1 [2], it suffices to show that ∂gn/∂x1≤0. Calculation shows that
(2.4)1q1∂gn∂x1=x1r-1Mn,rr-x1s-1Mn,ss-r-sxn2(x1-An):=fn(q,x).
We now show by induction on n that fn(q,x)≤0. When n=1, there is nothing to prove. When n=2, this becomes
(2.5)1q2f2(q,x)=x2r+s-1Dr,s(x1/x2;q2,q1)M2,rrM2,ss≤0,
by Lemma 2.1.
Suppose nown≥3; in order to show fn(q,x)≤0, we may assume that 0<x1<xn are being fixed and it suffices to show that the maximum value of fn(q,x) is non-positive on the region Rn×Sn-2, where Rn={(q1,q2,…,qn):0≤qi≤1,1≤i≤n,∑i=1nqi=1} and Sn-2={(x2,…,xn-1):xi∈[x1,xn],2≤i≤n-1}.
Let(q′,x′) be a point of Rn×Sn-2 in which the absolute maximum of fn is reached. If xi′=xi+1′ for some 1≤i≤n-1, by combining xi′ with xi+1′ and qi′ with qi+1′, we are back to the case of n-1 variables with different weights. If qi′=1 for some i, then we have
(2.6)x1r-1Mn,rr-x1s-1Mn,ss-r-sxn2(x1-An)=x1r-1xir-x1s-1xis-r-sxn2(x1-xi)≤x1r-1xir-x1s-1xis-r-sxi2(x1-xi)=1xiDr,s(x1xi;1,0)≤0,
by Lemma 2.1. If qi′=0 for some 1<i<n, we are back to the case of n-1 variables. If qn′=0, then we may assume that qn-1′≠0 and note that we have Mn,r=Mn-1,r,Mn,s=Mn-1,s,An=An-1 and that
(2.7)x1r-1Mn,rr-x1s-1Mn,ss-r-sxn2(x1-An)≤x1r-1Mn-1,rr-x1s-1Mn-1,ss-r-sxn-12(x1-An-1),
and we are again back to the case n-1. If q1′=0, then similarly we may assume that q2′≠0 and if we can show that (again with Mn,r=Mn-1,r,Mn,s=Mn-1,s and An=An-1 here)
(2.8)x1r-1Mn,rr-x1s-1Mn,ss-r-sxn2(x1-An)≤x2r-1Mn-1,rr-x2s-1Mn-1,ss-r-sxn2(x2-An-1),
then we are back to the case of n-1 variables. Note that the above inequality will follow if the function
(2.9)x↦xr-1Mn,rr-xs-1Mn,ss-r-sxn2x
is an increasing function for 0<x≤Mn,r (in fact, one only needs this for 0<x≤x2) and its derivative is
(2.10)(r-1)xr-2Mn,rr+(1-s)xs-2Mn,ss-r-sxn2≥(r-1)xr-2xnr+(1-s)xs-2xns-r-sxn2:=h(x),
with the inequality holding for the case r>2 (note that together with r+s≤3, this implies that s<1). It also follows from r+s≤3 that h′(x)=0 has no root in (0,xn). One then deduces from h(xn)=0 and limx→0+h(x)=+∞ that h(x)≥0 for 0<x≤xn.
So from now on it remains to consider the caseqi′≠0,1,xi′≠xj′ for 1≤i,j≤n,i≠j and this implies that (q′,x′) is an interior point of Rn×Sn-2. We will now show that this cannot happen.
We define(2.11)p(x)=-x1r-1xrMn,r2r+x1s-1xsMn,s2s+(r-s)xxn2-λ.
Note here in the definition of p(x) that Mn,randMn,s are not functions of x, they take values at some point (q,x) to be specified, and λ is also a constant to be specified.
As(q′,x′) is an interior point of Rn×Sn-2, we may use the Lagrange multiplier method to obtain a real number λ so that at (q′,x′),
(2.12)∂fn∂qi=λ∂∂qi(∑i=1nqi-1),1qj∂fn∂xj=0
for all 1≤i≤n and 2≤j≤n-1.
By (2.12), a computation shows that each x'i (1≤i≤n) is a root of p(x)=0 (where Mn,r,Mn,s take their values at (q′,x′)) and each x'i (2≤i≤n-1) is a root of p′(x)=0. Now n≥3 implies p(x2)=0. As p(x1)=p(x2)=p(xn)=0, it follows from Rolle’s Theorem that there must be two numbers x1<a<x2<b<xn such that p′(a)=p′(x2)=p′(b)=0. However, it is easy to see that p′(x)=0 has at most two positive roots and this contradiction implies the first assertion of the theorem for the case 0≤s≤1.
Now to show the right-hand side inequality hold of (1.2) for the case s<0,-1≤r≤0 and -3≤r+s≤0, once again it suffices to show that the function gn(q,x) defined above is nonnegative for any integer n≥1. We note that when n=1, this is obvious and when n=2, this follows again from ∂g2/∂x1≤0 by Lemma 2.1.
Suppose nown≥3; in order to show gn(q,x)≥0, we may assume that 0<x1<xn are being fixed and it suffices to show the minimum value of gn(q,x) is nonnegative on the region Rn×Sn-2, where Rn and Sn-2 are defined as above.
Let(q′,x′) be a point of Rn×Sn-2 in which the absolute minimum of gn is reached. Note that σn=Mn,22-An2; thus if xi′=xi+1′ for some 1≤i≤n-1, by combining xi′ with xi+1′ and qi′ with qi+1′, we are back to the case of n-1 variables with different weights. Similarly, if qi′=1 for some i, then we are back to the case n=1. If qi′=0 for some 1≤i<n, we are back to the case of  n-1 variables. If qn′=0, then we may assume that qn-1′≠0 and note that we have Mn,r=Mn-1,r,Mn,s=Mn-1,s,Mn,2=Mn-1,2,An=An-1 and that
(2.13)lnMn,r-lnMn,s-r-s2xn2σn=lnMn,r-lnMn,s-r-s2xn2(Mn,22-An2)≥lnMn-1,r-lnMn-1,s-r-s2xn-12(Mn-1,22-An-12)=lnMn-1,r-lnMn-1,s-r-s2xn-12σn-1,
and we are again back to the case of n-1 variables.
So from now on it remains to consider the caseqi′≠0,1,xi′≠xj′ for 1≤i,j≤n,i≠j, and this implies that (q′,x′) is an interior point of Rn×Sn-2. We will now show that this cannot happen.
We define(2.14)a(x)=xrrMn,rr-xssMn,ss-(r-s)(x2-2Anx)2xn2-λ.
Here we define x0/0=lnx. Also note here in the definition of a(x), that Mn,r,Mn,s,andAn are not functions of x, they take values at some point (q,x) to be specified, and λ is also a constant to be specified.
As(q′,x′) is an interior point of Rn×Sn-2, we may use the Lagrange multiplier method to obtain a real number λ so that at (q′,x′),
(2.15)∂gn∂qi=λ∂∂qi(∑i=1nqi-1),1qj∂gn∂xj=0
for all 1≤i≤n and 2≤j≤n-1.
By (2.15), a computation shows that each xi′ (1≤i≤n) is a root of a(x)=0 (where Mn,r,Mn,s,andAn take their values at (q′,x′)) and each xi′ (2≤i≤n-1) is a root of a′(x)=0. Now n≥3 implies a(x2)=0. As a(x1)=a(x2)=a(xn)=0, it follows from Rolle’s Theorem that there must be two numbers x1<c<x2<d<xn such that a′(c)=a′(x2)=a′(d)=0. However, we have
(2.16)a′(x)=xr-1Mn,rr-xs-1Mn,ss-(r-s)(x-An)xn2.
It is easy to see that a′′′(x)=0 has at most one positive root, which implies that a′(x)=0 has at most three positive roots. As r≤0, it follows from limx→0+a′(x)=-∞ and limx→+∞a′(x)=-∞ that a′(x)=0 has even numbers of roots so that a′(x)=0 can have at most two positive roots. This contradiction now establishes the right-hand side inequality of (1.2) for the case s<0,-1≤r≤0, and -3≤r+s≤0.
One can show the second assertion of the theorem using an argument similar to the above and we shall leave this to the reader.


## 3. Further Discussions

As we have pointed out in Section1 that if either one of the inequalities (1.2)–(1.4) holds for some r,s,α≤2, then one often expects inequality (1.5) to hold as well for the same r,s,α. In view of this, one may ask whether it is feasible to prove so for those pairs r,s,α=0 satisfying Theorem 2.2. We now prove a special case here.Theorem 3.1.
Let-3≤r≤3,r≠0,t≥0, then the following inequality holds:
(3.1)xn2|lnGn-lnMn,r|≥(xn+t)2|lnGn,t-lnMn,r,t|.Proof.
We first prove the theorem for the case-3≤r<0. For this, we may assume that t>0 is fixed and replace r with -r so that 0<r≤3 in what follows. We define
(3.2)fn(q,x)=xn2(lnGn-lnMn,-r)-(xn+t)2(lnGn,t-lnMn,-r,t).
As in the proof of Theorem 2.2, it suffices to show that ∂fn/∂x1≤0 and calculation shows
(3.3)-1q1∂fn∂x1=gn(q,x)-gn(q,xt),
where
(3.4)gn(q,x)=xn2(∑i=1nqi(xir-x1r)/xirx1∑i=1nqi(x1/xi)r).
It is easy to check that
(3.5)xnxi≥xn+txi+t,x1xi≤x1+txi+t.
In view of (3.5), the inequality ∂fn/∂x1≤0 will follow from
(3.6)d1(xi)=xn2(xir-x1r)x1xir-(xn+t)2((xi+t)r-(x1+t)r)(x1+t)(xi+t)r≥0
for x1≤xi≤xn. We may assume that xn>x1 here and it is easy to see that d1′(x)=0 can have at most one root x0 in between x1 and xn. This combined with the observation that d1(x1)=0,d1′(x1)>0 implies that d1(x) reaches its local maximum at x0 if it exists. Hence we are left to check that d1(xn)≥0. In this case we note that xn-x1=(xn+t)-(x1+t) and we rewrite d1(xn) as
(3.7)d1(xn)=xn2(xnr-x1r)x1xnr(xn-x1)-(xn+t)2((xn+t)r-(xn+t)r)(x1+t)(xn+t)r((xn+t)-(x1+t))=e(xnx1)-e(xn+tx1+t),
where
(3.8)e(x)=xr-1xr-2(x-1).
In view of (3.5) again, we just need to show that e(x) is an increasing function for x>1. Note that
(3.9)e′(x)=xr-3(xr+1-2xr+(r-1)x-(r-2))(xr-2(x-1))2,
and it is easy to see that the function xr+1-2xr+(r-1)x-(r-2) is non-negative for x≥1 when 0<r≤3 by considering its Taylor expansion at x=1 and this completes the proof for the assertion of the theorem for the case -3≤r<0.
To prove the theorem for the case0<r≤3, we may again assume that t>0 is fixed and define
(3.10)un(q,x)=xn2(lnMn,r-lnGn)-(xn+t)2(lnMn,r,t-lnGn,t).
Again it suffices to show that ∂un/∂x1≤0 and calculation shows
(3.11)-1q1∂un∂x1=vn(q,x)-vn(q,xt),
where
(3.12)vn(q,x)=xn2(∑i=1nqi(xir-x1r)/xnrx1∑i=1nqi(xi/xn)r).
In view of (3.5), the inequality ∂un/∂x1≤0 will follow from
(3.13)d2(xi)=xn2(xir-x1r)x1xnr-(xn+t)2((xi+t)r-(x1+t)r)(x1+t)(xn+t)r≥0
for x1≤xi≤xn. We may assume that xn>x1 here and it is easy to see that d2′(x)=0 can have at most one root x0 in between x1 and xn. This combined with the observation that d2(x1)=0,d2′(x1)>0 implies that d1(x) reaches its local maximum at x0 if it exists. Hence we are left to check that d2(xn)≥0. As d2(xn)=d1(xn), this completes the proof for the remaining case 0<r≤3 of the theorem.Now we show that, in general, it is not true that for-3≤r≤3,r≠0,t≥0,(3.14)x12|lnGn-lnMn,r|≤(x1+t)2|lnGn,t-lnMn,r,t|.
To proceed, we first look at the following related inequalities (with r>s here):(3.15)lnMn,r-lnMn,s-(r-s)σn2x12≤lnMn,r,t-lnMn,s,t-(r-s)σn2(x1+t)2,(3.16)lnMn,r-lnMn,s-(r-s)σn2xn2≥lnMn,r,t-lnMn,s,t-(r-s)σn2(xn+t)2.
Let fn(q,x,t) denote the right-hand side expression of (3.15); then (3.15) holds if and only if ∂fn(q,x,0)/∂t≥0. As x is arbitrary, we can recast this condition as(3.17)Mn,r-1r-1Mn,rr-Mn,s-1s-1Mn,ss+(r-s)σnx13≥0.
Similarly, (3.16) holds if and only if the following inequality holds:(3.18)Mn,r-1r-1Mn,rr-Mn,s-1s-1Mn,ss+(r-s)σnxn3≤0.
As a first step towards establishing (3.17), we consider the case n=2 here; in this case we let x1=1≤t=x2 and rewrite the left-hand side of (3.17) as(3.19)q1+q2tr-1q1+q2tr-q1+q2ts-1q1+q2ts+(r-s)q1q2(t-1)2=q1q2(1-t)(q1+q2tr)(q1+q2ts)Dr,s(t;q1,q2)
with Dr,s(t;q1,q2) being defined as in Lemma 2.1. Using the same notations as in Lemma 2.1, we see that in order for (3.17) to hold for n=2, we need to have Dr,s(t;q1,q2)≤0 for (t,q1,q2)∈I2×E. Similar treatment of (3.18) shows that in order for it to hold in the case n=2, one needs to have Dr,s(t;q1,q2)≤0 for (t,q1,q2)∈I1×E.It follows from the proof of Lemma2.1 that Dr,0(t;1,0)≤0 fails to hold for all t≥1 when r>2. In another words, there exists x,q such that when r>2,(3.20)Mn,r-1r-1Mn,rr-1Hn+(r-s)σnx13<0
holds. Now we return to the inequality (3.14) and we take r>0 there. Just as in the discussion above, one sees that (3.14) is equivalent to(3.21)2(lnMn,r-lnGn)x1+Mn,r-1r-1Mn,rr-1Hn≥0.
This combined with (3.20) now implies that for r>2,(3.22)lnMn,r-lnGn>(r-s)σn2x12.However, on takingt→+∞ on (3.14), we get the above inequality reversed (with > replaced by ≤) and this leads to a contradiction; hence (3.14) does not hold for r>2 in general.To end this paper, we note that it is an open problem to determine all the triples(r,s,α) so that inequality (1.2) holds. However, when α=0 with r=0 or s=0, the result given in Theorem 2.2 is best possible. In this case Theorem 2.2 implies that for |r|≤3,r≠0,(3.23)|lnMn,r-lnGn|≥|r|2xn2σn.We point out here that inequality (3.23) does not hold in general when |r|>3. To see this, it suffices to consider the case n=2 and in this case we can set 0<x1=t≤x2=1 and consider more generally for r>s, the function g2(q,x) defined in the proof of Theorem 2.2, regarding it as a function f(t) of t. It is easy to check that f(1)=f′(1)=f′′(1)=0; hence by the Taylor expansion of f(t) around t=1, we need f(3)(1)≤0 in order for f(t)≥0 to hold for any 0<t≤1. Calculation shows that(3.24)f(3)(1)=q1(r-s)(r+s-3-3(r+s-1)q1+2(r+s)q12).
On taking q1→0+, one sees immediately that we must have r+s≤3 here in order for f(t)≥0 for all 0<t≤1. On taking s=0, we see that one needs r≤3 in order for (3.23) to hold for positive r. Similarly, one checks easily that in the case n=2, if inequality (3.23) holds for some r, then it also holds for -r by a change of variables xi→1/x2-i+1. Hence one needs r≥-3 in order for (3.23) to hold for any negative r.



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*Source: 293539-2010-01-13.xml*