# On an Inequality of Diananda—Part IV

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

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## Abstract

We extend the results in parts I–III on certain inequalities involving the
weighted power means as well as the symmetric means. These inequalities can be largely viewed
as concerning the bounds for ratios of differences of means and can be traced back to the work of
Diananda.


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## Body

## 1. Introduction

LetMn,r(x;q) be the weighted power 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+, x=(x1,…,xn), q=(q1,…,qn), and qi>0 (1≤i≤n) are positive real numbers with ∑i=1nqi=1. In this paper, we let q=minqi and unless otherwise specified, we assume 0≤x1<x2<⋯<xn.Fork∈{0,1,…,n}, the kth symmetric function En,k of x and its mean Pn,k are defined by
(1.1)En,r(x)=∑1≤i1<⋯<ir≤n∏j=1rxij,Pn,rr(x)=En,r(x)(nr),1≤r≤n;En,0=Pn,0=1.We defineAn(x;q)=Mn,1(x;q), Gn(x)=Mn,0(x;q), and Hn(x;q)=Mn,-1(x;q) and we shall write Mn,r for Mn,r(x;q) and similarly for other means when there is no risk of confusion.For a real numberα and mutually distinct numbers r, s, t, we define
(1.2)Δr,s,t,α=|Mn,rα-Mn,tαMn,rα-Mn,sα|,
where we interpret Mn,r0-Mn,s0 as lnMn,r-lnMn,s. We also define Δr,s,t to be Δr,s,t,1.Forr>s>t≥0, α>0, the author studied in [1–3] inequalities of the following two types:
(1.3)Cr,s,t((1-q)α)≥Δr,s,t,α,(1.4)Δr,s,t,α≥Cr,s,t(qα),
where for 0<x<1,
(1.5)Cr,s,t(x)=1-x1/t-1/r1-x1/s-1/r,t>0;Cr,s,0(x)=11-x1/s-1/r.For any set{a,b,c} with a,b,c mutually distinct and nonnegative, we let r=max{a,b,c}, t=min{a,b,c}, s={a,b,c}∖{r,t}. By saying that (1.3) (resp., (1.4)) holds for the set {a,b,c}, α>0, we mean that (1.3) (resp., (1.4)) holds for r>s>t≥0, α>0.A result of Diananda [4, 5] (see also [6, 7]) shows that (1.3) and (1.4) hold for {1,1/2,0}, α=1. When α=1, the sets {a,b,1}'s for which (1.3) or (1.4) holds have been completely determined by the author in [1–3]. Moreover, it is shown in [3] that (1.4) does not hold in general unless 0∈{a,b,c}.For generalα's, the cases r>s>t≥0 for which inequality (1.3) or (1.4) holds are open (with t=0 in (1.4)). It is our main goal in this paper to study these cases. We first restrict our attention to the case {a,b,0}. This is partially because of the following result in [2] (note that there is a typo in the original statement though).Theorem 1.1 (see [2, Theorem  3.2]).
Letr>s>0. If (1.3) holds for {r,s,0}, α>0, then it also holds for {r,s,0}, kα with 0<k<1. If (1.4) holds for {r,s,0}, α>0, then it also holds for {r,s,0}, kα with k>1.Moreover, for the unweighted caseq1=q2=⋯=qn=1/n, the author [3, Theorem  3.5] has shown that (1.3) holds for {1,1/r,0} with α=n/r when r≥2 and (1.4) holds for {1,1/r,0} with α=n/((n-1)r) when 1<r≤2. The values of α's are best possible here; namely, no larger α's can make (1.3) hold for {1,1/r,0} and similarly no smaller α's can make (1.4) hold for {1,1/r,0}.More generally, for arbitrary weights{qi}'s, by using similar arguments as in [3], one sees that the largest α that can make (1.3) hold for {1,1/r,0} is 1/(qr) and the smallest α that can make (1.4) hold for {1,1/r,0} is 1/((1-q)r).In Section2, we will extend Theorem  3.5 of [3] to the case of arbitrary weights. Namely, we will prove the following.Theorem 1.2.
Forr≥2, 0<p≤1/q,
(1.6)Anp≤(1-q)(r-1)p/rMn,rp+(1-(1-q)(r-1)p/r)Gnp.
For 1<r≤2, p≥1/(1-q),
(1.7)Anp≥q(r-1)p/rMn,rp+(1-q(r-1)p/r)Gnp.We note here that by a change of variablesxi→xi1/r, one can easily rewrite (1.6) (resp., (1.7)) in the form of (1.3) (resp., (1.4)).After studying (1.3) and (1.4) for the set {a,b,0}, we move on in Section 3 to the case {a,b,c} with min(a,b,c)>0. Our remark earlier allows us to focus on (1.3) only. In this case, we can recast (1.3), via a change of variables, as the following:
(1.8)Anα≤λr,s((1-q)α)Mn,rα+(1-λr,s((1-q)α))Mn,sα,
where α>0, r>1>s>0 and
(1.9)λr,s(x)=1-1Cr,1,s(x).
We will show that (1.8) holds for all n if and only if it holds for the case n=2. Based on this, we will then be able to prove (1.8) for certain r, s, α's satisfying a natural condition.One certainly expects that analogues of (1.3) and (1.4) hold with weighted power means replaced by the symmetric means; one such example is given by the following result in [8].Theorem 1.3.
Letqi=1/n, then for any integer 2≤k≤n,
(1.10)(∑i=1nxi)k≤(nk-λ̃2,k(n)(nk))Mn,2k+λ̃2,k(n)En,k,
where for 2≤r≤k≤n (with (n-1n)=0 here),
(1.11)λ̃r,k(n)=nk(1-1/n)k/r-(n-1)k(nk)(1-1/n)k/r-(n-1k).As was pointed out in [3], the proof given in [8] for the above theorem is not quite correct. In Section 4, we will study inequalities involving the symmetric means and our results include a proof of Theorem 1.3.


## 2. Proof of Theorem1.2

In view of Theorem1.1, one only needs to prove (1.6) for p=1/q and similarly (1.7) for p=1/(1-q). In this proof we assume that 0<x1≤⋯≤xn. The case x1=0 will follow by taking the limit. We first prove (1.6) and we define
(2.1)f(x)=An1/q-(1-q)(r-1)/qrMn,r1/qGn1/q.
If x1=⋯=xn, then f=0; otherwise we may assume n≥2 and 0<x=x1=⋯=xk<xk+1 for some 1≤k<n, then
(2.2)∂f∂x=∑i=1k∂f∂xi.
We want to show that the right-hand side above is nonnegative. It suffices to show that each single term in the sum is nonnegative. Without loss of generality, we now show that
(2.3)∂f∂x1≥0.
We have
(2.4)qx1Gn1/qq1∂f∂x1=An1/q-1(x1-An)-(1-q)(r-1)/qrMn,r1/q-r(x1r-Mn,rr).
Now we set
(2.5)y(r)=((q1-q)x1r+∑i=2nqixir1-q)1/r,
so that
(2.6)An1/q-1(x1-An)=(1-q)(qx1+(1-q)y(1))1/q-1(x1-y(1))≥(1-q)(qx1+(1-q)y(r))1/q-1(x1-y(r)).
Hence
(2.7)qGn1/qq1(1-q)1/qx11/q-1∂f∂x1≥(q1-q+y(r)x1)1/q-1(1-y(r)x1)-(q1-q+yr(r)x1r)1/qr-1(1-yr(r)x1r).
We want to show that the right-hand side expression above is nonnegative, and by setting z=y(r)/x1, this is equivalent to show that
(2.8)g(z,q)=(q/(1-q)+z)1/q-1(z-1)(q/(1-q)+zr)1/qr-1(zr-1)≤1
for z≥1, 0≤q≤1/2 and calculation yields that
(2.9)((q/(1-q)+zr)1/qr-1(zr-1))2(q/(1-q)+z)1/q-2(q/(1-q)+zr)1/qr-2∂g∂z=1q((z-1+q1-q)(zr-1)(zr+q1-q)-(z+q1-q)(zr-zr-1)(zr+rq1-q-1)).
We now set s=q/(1-q) with 0≤s≤1 and we consider
(2.10)a(z,s)=(z-1+s)(zr-1)(zr+s)-(z+s)(zr-zr-1)(zr+rs-1).
By Cauchy's mean value theorem,
(2.11)∂2a∂s2=2(zr-1-rzr-1(z-1))≤0.
It follows that for 0≤s≤1,
(2.12)a(z,s)≥min{a(z,0),a(z,1)}=min{0,a(z,1)}.
It follows from the discussion in [3] (see the function a(y,1) defined in the proof of Theorem 3.5 there and note that we have r≥2 here) that a(z,1)≥0. This implies that a(z,s)≥0 so that g(z,q) is an increasing function of z and we then deduce that
(2.13)g(z,q)≤limz→+∞g(z,q)=1.
This shows that ∂f/∂x1≥0 and hence ∂f/∂x≥0 and by letting x→xk+1 and repeating the above argument, we conclude that f(x)≤f(xn,xn,…,xn)=1-(1-q)(r-1)/qr which completes the proof for (1.6).Now, to prove (1.7), we consider
(2.14)h(x)=An1/(1-q)-q(r-1)/(1-q)rMn,r1/(1-q)Gn1/(1-q).
Similar to our discussion above, it suffices to show ∂h/∂xn≥0. Now
(2.15)xnGn1/(1-q)(1-q)qn∂h∂xn=An1/(1-q)-1(xn-An)-q(r-1)/(1-q)rMn,r1/(1-q)-r(xnr-Mn,rr).
Now we set
(2.16)w(r)=((qn-q)xnr+∑i=1n-1qixir1-q)1/r,
so that
(2.17)An1/(1-q)-1(xn-An)=(1-q)(qxn+(1-q)w(1))1/(1-q)-1(xn-w(1))≥(1-q)(qxn+(1-q)w(r))1/(1-q)-1(xn-w(r)),
where the inequality follows from the observation that the function
(2.18)z↦(qxn+(1-q)z)1/(1-q)-1(xn-z)
is decreasing for 0<z<xn.We then deduce that(2.19)xnGn1/(1-q)qnq1/(1-q)-1w1/(1-q)(r)∂h∂xn≥(xnw(r)+1-qq)1/(1-q)-1(xnw(r)-1)-(xnrwr(r)+1-qq)1/(1-q)r-1(xnrwr(r)-1).
By proceeding similarly as in the proof of (1.6) above, one is then able to establish (1.7) and we shall omit all the details here.


## 3. A General Discussion on (1.8)

Theorem 3.1.
For fixedα>0, r>1>s>0, (1.8) holds for all n if and only if it holds for n=2.Proof.
We consider the function(3.1)fn(x;q,q):=λr,s((1-q)α)Mn,rα+(1-λr,s((1-q)α))Mn,sα-Anα.
The theorem asserts that in order to show fn≥0, it suffices to check the case n=2. To see this, we may assume by homogeneity that 0≤x1≤x2≤⋯≤xn-1≤xn=1 and we let a=(a1,…,an)∈[0,1]n-1 be the point in which the absolute minimum of fn is reached. We may assume that 0≤a1<a2<⋯<an-1<an=1 for otherwise if ai=ai+1 for some 1≤i≤n-1, by combining ai with ai+1 and qi with qi+1, and noticing that λr,s(x) is an increasing function of x by [2, Lemma  2.1], we have
(3.2)fn(a;q,q)≥fn-1(a′;q′,q′),
where a′=(a1,…,ai-1,ai+ai+1,ai+2,…,an), q′=(q1,…,qi-1,qi+qi+1,qi+2,…,qn), and q′=min(q1,…,qi-1,qi+qi+1,qi+2,…,qn). We can then reduce the determination of the absolute minimum of fn to that of fn-1 with different weights.
Ifa is a boundary point of [0,1]n-1, then a1=0, and in this case we show that fn(a;qn,q)≥0 follows from fn-1(a′′;q′′,q)≥0, where a′′=(a2,…,an) and q′′=(q2,…,qn)/(1-q1). On writing fn-1(a′′;q′′,q)≥0 explicitly, we get
(3.3)(∑i=2nqiai)α≤λr,s((1-q)α)(1-q1)αMn-1,rα(a′′;q′′)+(1-λr,s((1-q)α))(1-q1)αMn-1,sα(a′′;q′′).
Meanwhile, fn(a;qn,q)≥0 is equivalent to
(3.4)(∑i=2nqiai)α≤λr,s((1-q)α)(1-q1)α/rMn-1,rα(a′′;q′′)+(1-λr,s((1-q)α))(1-q1)α/sMn-1,sα(a′′;q′′).
Thus, it amounts to show that
(3.5)λr,s((1-q)α)(1-q1)αMn-1,rα(a′′;q′′)+(1-λr,s((1-q)α))(1-q1)αMn-1,sα(a′′;q′′)≤λr,s((1-q)α)(1-q1)α/rMn-1,rα(a′′;q′′)+(1-λr,s((1-q)α))(1-q1)α/sMn-1,sα(a′′;q′′),
which is equivalent to
(3.6)(1-λr,s((1-q)α))λr,s((1-q1)α)(1-λr,s((1-q1)α))λr,s((1-q)α)Mn-1,sα≤Mn-1,rα.
Now the above inequality follows from Mn-1,s≤Mn-1,r and
(3.7)(1-λr,s((1-q)α))λr,s((1-q1)α)≤(1-λr,s((1-q1)α))λr,s((1-q)α),
since λr,s(x) is an increasing function of x.
Thusfn(a;qn,q)≥0 follows from fn-1(a′′;q′′,q)≥0. Moreover, on writing q′′=min(q2/(1-q1),…,qn/(1-q1)) and noticing that q′′>q, we deduce that fn-1(a′′;q′′,q)≥fn-1(a′′;q′′,q′′). Hence the determination of fn≥0 can be reduced to the determination of fn-1≥0 with different weights.
It remains to show the casea1>0, so that a is an interior point of [0,1]n-1. In this case we have
(3.8)∇fn(a;q,q)=0.
Thus a1,…,an-1 solve the equation
(3.9)g(x)=λr,s((1-q)α)Mn,rα-rxr-1+(1-λr,s((1-q)α))Mn,sα-sxs-1-Anα-1=0.
Note that
(3.10)fn(a;q,q)=∑i=1nqiaig(ai)=qnang(an).
Thus if g(an)≥0, then fn≥0. If g(an)<0, we note that g(x)=0 can have at most two roots in (0,an] since it is easy to see that g′(x)=0 can have at most one positive root. As limx→0+g(x)=+∞, this implies that g(x)=0 has only one root in (0,an]. As a1,…,an-1 are the distinct roots of g(x)=0, this implies that we must have n-1=1 so that we only need to show fn≥0 for the case n=2 and this completes the proof.In what follows, we will apply Theorem3.1 to establish (1.8) for certain r,s,α's. Before we proceed, we note that there is a natural condition to be satisfied by r, s, α in order for (1.8) to hold, namely, if we take n=2 and rewrite it as
(3.11)M2,rα-M2,sαM2,rα-A2α≤11-λr,s((1-q)α).
On taking x1→x2, we conclude that
(3.12)r-sr-1≤11-λr,s((1-q)α).
Before we prove our next result, we need two lemmas.Lemma 3.2.
Fixingu<0, the function
(3.13)f(t)=1-t1-tu,0<t≠1
is concave for u<-1 and is convex for -1<u<0.Proof.
Calculation yields that(3.14)f′′(t)=utu-2(1-tu)3g(t),
where
(3.15)g(t)=-(u-1)tu+1+(1+u)tu-(1+u)t+u-1.
Now
(3.16)g′′(t)=u(u+1)(u-1)tu-2(1-t).
Thus if -1<u<0, then g′′(t)>0 for 0<t<1 and g′′(t)<0 for t>1. Since g′(1)=0, this implies that g′(t)<0 for 0<t≠1. As g(1)=0, we then conclude that g(t)>0 for 0<t<1 and g(t)<0 for t>1. It follows from this that f(t) is convex for -1<u<0 and the other assertion can be shown similarly, which completes the proof.Lemma 3.3.
Forr>1, 0<q1, q2<1, q1+q2=1, the function
(3.17)f(t)=(q1t1/(r-1)+q2)α-1(q1tr/(r-1)+q2)1-α/r
is convex for t>0 when r≥2, 1>α>0, or α=1, r>1.Proof.
Direct calculation shows that(3.18)f′′(t)=q1r-1(q1t1/(r-1)+q2)α-3(q1tr/(r-1)+q2)-α/r-1t1/(r-1)-2g(t),
where
(3.19)g(t)=(1-α)(r-2)r-1q12q2t2+2/(r-1)+(1-α)(r-α)r-1q1q22t2+1/(r-1)+r-αr-1q12q2t1+2/(r-1)+2(1-(1-α)2r-1)q1q22t1+1/(r-1)+r-αr-1q23t+(1-α)(r-α)r-1q1q22t1/(r-1)+(1-α)(r-2)r-1q23.
One then easily deduce the assertion of the lemma from the above expression of g(t) and this completes the proof.Theorem 3.4.
Inequality (1.8) holds for the cases r≥2, 1>α>0 or α=1, r>1, r+s≥2, provided that (3.12) holds with strict inequality.Proof.
By Theorem3.1, it suffices to prove the theorem for the case n=2. We write λ=λr,s((1-q)α) for short in this proof. What we need to prove is the following:
(3.20)(q1x1+q2x2)α≤λ(q1x1r+q2x2r)α/r+(1-λ)(q1x1s+q2x2s)α/s.
Without loss of generality, we may assume that q1=q, q2=1-q and define
(3.21)f(t)=(q1ts+q2)-α/s((q1t+q2)α-λ(q1tr+q2)α/r),
so that we need to show that f(t)≤f(1) for t≥0. We have
(3.22)f′(t)=αq1q2(q1ts+q2)-α/s-1((q1t+q2)α-1(1-ts-1)+λ(q1tr+q2)α/r-1(ts-1-tr-1))=αq1q2(q1ts+q2)-α/s-1(q1tr+q2)α/r-1(1-ts-1)g(tr-1),
where
(3.23)g(t)=(q1t1/(r-1)+q2)α-1(q1tr/(r-1)+q2)1-α/r+λ1-t1-t(s-1)/(r-1)-λ.
By Lemmas 3.2 and 3.3, we see that g(t) is a strictly convex function for r≥2, 1>α>0 or α=1, r>1, r+s≥2. Note that by our assumption (3.12) with strict inequality,
(3.24)limt→1g(t)=1+λr-ss-1<0.
On the other hand, note that λ satisfies
(3.25)(1-q)α=λ(1-q)α/r+(1-λ)(1-q)α/s,
so that
(3.26)limt→0+g(t)=(1-q)α-α/r-λ=(1-λ)(1-q)α/s-α/r>0.
As g(t) is strictly convex, this implies that there are exactly two roots t1, t2 of g(t)=0 satisfying that t1∈(0,1) and t2>1. Note further that f(0)=f(1) and limt→0+f′(t)<0, which implies that f(t)≤f(1) for 0≤t≤1. Similarly, we note that f′(t)<0 for t∈(1,1+ϵ) with ϵ>0 being small enough. This combined with the observation that limt→+∞f(t)≤f(1) as λr,s(x) is an increasing function of x implies that f(t)≤f(1) for t≥1 which completes the proof.We remark here that if condition (3.12) is satisfied for some r, s, α, then it is also satisfied for r, s, α′ with 0<α′<α. Thus it is not surprising to expect a result like Theorem 3.4 to hold.To end this section, we prove a variant of (1.3) which is motivated by the following inequality due to Mesihović [9] (with qi=1/n here):
(3.27)(1-1n)An+1nGn≥(1n∑i=1nxi1-1/n)(1n∑i=1nxi1/n).
We now generalize the above inequality to the arbitrary weight case.Theorem 3.5.
For0≤x1≤x2≤⋯≤xn,
(3.28)(1-q)An+qGn≥(∑i=1nqixi1-q)(∑i=1nqixiq)
with equality holding if and only if x1=⋯=xn or x1=0, q1=q, x2=⋯=xn, or n=2, q=1/2.Proof.
Define(3.29)Dn(x)=(1-q)An+qGn-(∑i=1nqixi1-q)(∑i=1nqixiq).
We need to show Dn≥0 and we have
(3.30)1qn∂Dn∂xn=1-q+qGnxn-(1-q)(xn-q∑i=1nqixiq)-q(xnq-1∑i=1nqixi1-q).
By a change of variables: xi/xn→xi, 1≤i≤n, we may assume 0≤x1≤x2≤⋯≤xn=1 in (3.30) and rewrite it as
(3.31)gn(x1,…,xn-1):=1-q+qGn-(1-q)(∑i=1nqixiq)-q(∑i=1nqixi1-q).
We want to show gn≥0. Let a=(a1,…,an-1)∈[0,1]n-1 be the point in which the absolute minimum of gn is reached. We may assume a1≤a2≤⋯≤an-1. If a1=0, then a is a boundary point of [0,1]n-1, and in this case we have
(3.32)gn(a1=0,…,an-1)=1-q-(1-q)(∑i=2nqiaiq)-q(∑i=2nqiai1-q)≥1-q-(1-q)(1-q1)-q(1-q1)=q1-q≥0
with equality holding if and only if q1=q, a2=⋯=an=1. Now suppose a1>0 and am-1<am=⋯=an=1 for some 1≤m≤n, then a1,…,am-1 solve the equation
(3.33)∇gn(a1,…,an-1)=0.
Equivalently, a1,…,am-1 solve the equation
(3.34)Gn=(1-q)(xq+x1-q).
As the right-hand side expression above is an increasing function of x, the above equation has at most one root (regarding Gn as constant). So we only need to show gn≥0 for the case a1=⋯=am-1=x<am=⋯=an=1 in (3.31) for some 1≤m≤n. In this case we regard gn as a function of x and we recast it as
(3.35)h(ω,x)=1-q+qxω-(1-q)(ωxq+1-ω)-q(ωx1-q+1-ω)=ω-q+qxω-(1-q)ωxq-qωx1-q.
Here 0<x≤1 and q≤ω≤1-q. Note first that when q=1/2, h(ω,x)=0 so that we may now assume 0<q<1/2. We have
(3.36)∂h∂ω|ω=q=1+qxqlnx-(1-q)xq-qx1-q:=d(x).
Now
(3.37)d′(x)=qxq-1e(x),
where
(3.38)e(x)=1+qlnx-(1-q)-(1-q)x1-2q.
Note that e′(x)=0 has one root (1-q)(1-2q)x01-2q=q so that if 0<x0<1, then at this point
(3.39)e(x0)=qlnx0+q-q1-2q<0.
Note also that limx→0+e(x)<0, e(1)=2q-1<0. This implies that e(x)<0 for 0<x≤1. Hence d(x)≥d(1)=0 for 0<x≤1. As it is easy to see that h(ω,x) is a convex function of ω for fixed x, we conclude that h(ω,x) is an increasing function of q≤ω≤1-q for fixed x. Thus for 0<x≤x, q≤ω≤1-q,
(3.40)h(ω,x)≥h(q,x)=q2(xq-x1-q)≥0
with equality holding if and only if x=1.
Thus we have showngn≥0; hence ∂Dn/∂xn≥0 with equality holding if and only if x1=⋯=xn or x1=0, q1=q, x2=⋯=xn or n=2, q1=1/2. By letting xn tend to xn-1, we have Dn≥Dn-1 (with weights q1,…,qn-2, qn-1+qn) with equality holding if and only if xn=xn-1 or n=2 and either q=1/2 or x1=0, q1=q. It follows by induction that Dn≥0 with equality holding if and only if x1=⋯=xn or x1=0, q1=q, x2=⋯=xn or n=2, q=1/2 and this completes the proof.We remark here that if we define(3.41)S(β)=(∑i=1nqixi1-β)(∑i=1nqixiβ),
then for 1≤β≤1/2,
(3.42)dSdβ=∑1≤i≤j≤nqiqjxiβxjβ(xj1-2β-xi1-2β)ln(xixj)≤0.
Hence Theorem 3.5 improves (1.3) for the case {1,1/2,0}, α=1.


## 4. Inequalities Involving the Symmetric Means

In this section, we setqi=1/n, 1≤i≤n. As an analogue of (1.8) (or (1.3)), we first consider
(4.1)Anα≤λα,r,k(n)Mn,rα+(1-λα,r,k(n))Pn,kα,
where α>0, r>1, n≥k≥2, and
(4.2)λα,r,k(n)=(1-1/n)α-((n-k)/n)α/k(1-1/n)α/r-((n-k)/n)α/k.
The case r=2, α=k in (4.1) is just Theorem 1.3. In what follows, we will give a proof of Theorem 1.3 by combining the methods in [8, 10]. Before we prove our result, we would like to first recast (4.1) for the case α=k as
(4.3)(∑i=1nxi)k≤(nk-λ̃r,k(n)(nk))Mn,rk+λ̃r,k(n)En,k,
where λ̃r,k(n) is defined as in the statement of Theorem 1.3. Now we need two lemmas.Lemma 4.1.
For2≤r≤k≤n-1,
(4.4)λ̃r,k(n)≤λ̃r,k(n-1).Proof.
We follow the method in the proof of Lemma2 in [8]. We write λ̃r,k(n) as
(4.5)λ̃r,k(n)=k!(nk(1-1/n)k/r-(n-1)k)(∏i=1k-1(n-i))(n(1-1/n)k/r-n+k).
From the above we see that in order for (4.4) to hold, it suffices to show that
(4.6)f(t)=k!((1-t)k/r-(1-t)k)(∏i=1k-1(1-it))((1-t)k/r-1+kt)
is increasing on (0,1/k). The logarithmic derivative of f(t) is
(4.7)f′(t)f(t)=kr(1-t)(1-r+r-11-(1-t)k(1-1/r)-r-1+(k-r)t(1-t)k/r-1+kt)+∑i=1k-1i1-it.
Note that for 0<t<1, we have
(4.8)(1-t)k/r-1+kt≥1-(1-t)k(1-1/r)>0
by considering the Taylor expansions to the order of t2. It follows from this that
(4.9)f′(t)f(t)≥kr(1-t)(1-r+r-11-(1-t)k(1-1/r)-r-1+(k-r)t1-(1-t)k(1-1/r))+∑i=1k-1i1-t=k1-t(1-rr-(k-r)t/r1-(1-t)k(1-1/r)+k-12).
It is easy to see that the function
(4.10)t↦t1-(1-t)k(1-1/r)
is an increasing function for 0<t≤1. Hence
(4.11)1-rr-(k-r)t/r1-(1-t)k(1-1/r)+k-12≥1-rr-k-rr+k-12=(k-1)(12-1r)≥0.Lemma 4.2.
Inequality (4.1) holds for all x in the case n≥k≥r≥2, α=k if it holds for x=(a,b,…,b) with 0<a≤b.Proof.
In this proof we assume that0≤x1≤x2≤⋯≤xn. We prove the lemma by induction on n. When n=k, the assertion holds as a special case of (1.6). Now assume that the assertion holds for n-1, and we proceed to show it also holds for n. If x1=0, we use the equivalent form (4.3) of (4.1) for the case α=k to see that what we need to prove is
(4.12)(∑i=2n-1xi)k≤(nk-λ̃r,k(n)(nk))(n-1n)k/rMn-1,rk(x2,…,xn)+λ̃r,k(n)En-1,k(x2,…,xn).
By the induction case n-1, we have
(4.13)(∑i=2n-1xi)k≤((n-1)k-λ̃r,k(n-1)(n-1k))Mn-1,rk(x2,…,xn)+λ̃r,k(n-1)En-1,k(x2,…,xn).
Note that
(4.14)(nk-λ̃r,k(n)(nk))(n-1n)k/r=(n-1)k-λ̃r,k(n)(n-1k).
Using this with Lemma 4.1 together with the observation that
(4.15)En-1,k(n-1k)=Pn-1,kk≤Mn-1,rk,
we see that inequality (4.12) follows from (4.13).
Thus from now on we may focus on the casex1>0. Since both sides of (4.1) are homogeneous functions, it suffices to show that
(4.16)minx∈Δ{λk,r,k(n)Mn,rk+(1-λk,r,k(n))Pn,kk}≥1,
where
(4.17)Δ={x=(x1,…,xn):xi≥0,1≤i≤n,∑i=1nxi=n}.
Assume that λk,r,k(n)Mn,rk+(1-λk,r,k(n))Pn,kk attains its minimum at some point (a1,…,an) with ai>0, 1≤i≤n. If a1=a2=⋯=an, then (4.16) holds. Furthermore, if n=2, then (4.16) also holds, being a special case of (1.6). Thus without loss of generality, we may assume n≥3 and an>an-1≥an-2 here. We may also assume that when r=2, k>2 since otherwise inequality (4.1) becomes an identity. Consider the function
(4.18)f(x,y):=λk,r,k(n)Mn,rk(a1,…,an-2,x,y)+(1-λk,r,k(n))Pn,kk(a1,…,an-2,x,y)
on the set
(4.19){(x,y):x≥0,y≥0,x+y=n-∑i=1n-2ai}.
It is minimized at (an-1,an). It is easy to see that f has the form
(4.20)λk,r,k(n)(1n(xr+yr)+B)k/r+C(a1,…,an-2)xy+D,
where B, C, and D are nonnegative constants with C depends on a1,…,an-2. We now set x+y=c and xy=z with 0≤z=xy≤(x+y)2/4=c2/4. Note here that equality holds if and only if xy=0 or x=y. We regard the above function as a function of z=xy and recast it as
(4.21)h(z)=λk,r,k(n)(1n((c+c2-4z2)r+(c-c2-4z2)r)+B)k/r+C(a1,…,an-2)z+D.
For y>x>0, calculation yields
(4.22)h′(xy)=-λk,r,k(n)kn(1n(xr+yr)+B)k/r-1(yr-1-xr-1y-x)+C(a1,…,an-2).
Since an>an-1>0, we must have h′(an-1an)=0 and we can further recast this as
(4.23)C(a1,…,an-2)=λk,r,k(n)knMn,rk-r(a1,…,an)(anr-1-an-1r-1an-an-1).
Now if an-2>0, we can repeat the same argument for the pair (an-1,an-2). By a slightly abuse of notation, we obtain
(4.24)h′(an-2an-1)=C(a1,…,an-3,an)-λk,r,k(n)knMn,rk-r(a1,…,an)(an-1r-1-an-2r-1an-1-an-2).
It is easy to see that (since we assume ai>0 and k≥3 when r=2)
(4.25)C(a1,…,an-3,an)>C(a1,…,an-2).
Moreover, one checks easily that the function
(4.26)(x,y)↦yr-1-xr-1y-x
is increasing with respect to each variable x,y>0 when r≥2. It follows that when r≥2,
(4.27)h′(an-2an-1)>h′(an-1an)=0,
which implies that by decreasing the value of an-2an-1 while keeping an-2+an-1 fixed, one is able to get a smaller value for λk,r,k(n)Mn,rk+(1-λk,r,k(n))Pn,kk, contradicting the assumption that it attains its minimum at (a1,…,an).
Hence we conclude thatλk,r,k(n)Mn,rk+(1-λk,r,k(n))Pn,kk is minimized at (a,b,…,b) with 0<a≤b satisfying a+(n-1)b=n. In this case (4.16) holds by our assumption which completes the proof.Now we are ready to prove a slightly generalization of Theorem1.3.Corollary 4.3.
Inequality (4.1) holds in the cases n≥k≥r=2, α=k and n≥α=r=k≥2.Proof.
The first case is just Theorem1.3 and by Lemma 4.2, it suffices to show that inequality (4.1) holds for the case x=(a,b,…,b) with 0<a≤b and this has been already treated in the proof of Theorem 2 in [8]. For the case n≥α=r=k≥2, by Lemma 4.2 again, we only need to check the case x=(a,b,…,b) with 0<a≤b. In this case we define
(4.28)f(a,b)=λk,k,k(n)(akn+(n-1)bkn)+(1-λk,k,k(n))(n-knbk+knabk-1)-(an+(n-1)bn)k.
As in the proof of Theorem 1.2, it suffices to show that
(4.29)nk(n-1)bk-1∂f∂b=λk,k,k(n)+(1-λk,k,k(n))(n-kn-1+k-1n-1ab)-(anb+n-1n)k-1≥0.
By a change of variables a/b→a, we can recast inequality (4.29) as
(4.30)g(a)=λk,k,k(n)+(1-λk,k,k(n))(n-kn-1+k-1n-1a)-(n-1n+an)k-1.
As g(1)=0 and g(0)=0, we conclude that g(a)≥0 for 0≤a≤1 as g(a) is a concave function for 0≤a≤1 and this completes the proof.We recall a result of Kuczma [10].Theorem 4.4.
Forn≥3, 1≤k≤n-1, Pn,k≤Mn,η(k) with
(4.31)η(k)=k(lnn-ln(n-1))lnn-ln(n-k),
and the result is best possible.The above theorem combined with Corollary4.3 immediately yields the following.Corollary 4.5.
Letqi=1/n with n≥1 being an integer. Then for any integer n-1≥k≥r=2 or n-1≥k=r≥2,
(4.32)Ank≤λk,r,k(n)Mn,rk+(1-λk,r,k(n))Mn,η(k)k,
where λk,r,k(n) is defined as in (4.2) and η(k) is defined as in Theorem 4.4.Next, we consider the following inequality:(4.33)Pn,lα≤μα,k,l(n)Anα+(1-μα,k,l(n))Pn,kα,
where α>0,n≥k>l≥2 and
(4.34)μα,k,l(n)=((n-l)/n)α/l-((n-k)/n)α/k((n-1)/n)α-((n-k)/n)α/k.
We note here that it is easy to check that the function
(4.35)x↦(n-xn)1/x
is a decreasing function for 1≤x<n so that we have 0<μα,k,l(n)<1.As an analogue of Theorem1.1, one can show similarly the following.Proposition 4.6.
Letn=k>l≥2; if (4.33) holds for α0>0, then it also holds for any 0<α<α0.The casen=k, α=1 of (4.33) was established in [11]. In the case of n=k, one possible way of establishing (4.33) is to combine Theorems 4.4 and 1.2 together. However, this is not always applicable as one checks via certain change of variables that one needs to have 1/η(k)≥2 in order to apply Theorem 1.2, a condition which is not always satisfied. We now proceed directly to show the following.Theorem 4.7.
Inequality (4.33) holds for n=k>l≥2 and 0<α≤n.Proof.
In view of Proposition4.6, it suffices to prove the theorem for the case α=n. We write μ(n)=μn,n,l(n) in this proof and note that since both sides of (4.33) are homogeneous functions, it suffices to show that
(4.36)maxx∈Δ{Pn,ln-(1-μ(n))Gnn}≤μ(n),
where
(4.37)Δ={x=(x1,…,xn):xi≥0,1≤i≤n,∑i=1nxi=n}.
Assume that Pn,ln-(1-μ(n))Gnn attains its maximum at some point (a1,…,an) with ai≥0, 1≤i≤n. Consider the function
(4.38)f(x,y):=Pn,ln(x,y,a3,…,an)-xy(1-μ(n))∏i=3nai
on the set
(4.39){(x,y):x≥0,y≥0,x+y=n-∑i=3nai}.
It is maximized at (a1,a2). It is easy to see that f has the form
(4.40)(Axy+B)n/l-Cxy,
where A, B, and C are nonnegative constants. The above function is certainly convex with respect to xy. As 0≤xy≤(x+y)2/4 with equality holding if and only if xy=0 or x=y, f is maximized at xy=0 or x=y. Repeating the same argument for other pairs (ai,aj), we conclude that in order to show (4.33) for α=n=k, it suffices to check that it holds for x being of the following form (0,…,0,a…,a) or (a,…,a) for some positive constant a. It is easy to see that (4.33) holds when x is of the second form and when x is of the first form, let m denote the number of a's in x; if m<l, then it is easy to see that (4.33) holds. So we may now assume that l≤m≤n=k and we need to show that
(4.41)((ml)m1/l)k≤μk,k,l(n)((nl)1/ln)k.
Certainly the left-hand side above increases with m; hence one only needs to verify the above inequality for the case m=n-1, which becomes an identity and this completes the proof.We note here that Alzer [12] has shown that for n≥3,
(4.42)Pn,n-1n-1≤nn+1Ann-1+1n+1Gnn-1.
The case α=l=n-1 of Theorem 4.7 now improves the above result, namely, for n≥3,
(4.43)Pn,n-1n-1≤nn-2(n-1)n-1Ann-1+(1-nn-2(n-1)n-1)Gnn-1,
as one checks easily that for n≥3,
(4.44)nn-2(n-1)n-1≤nn+1.Similar to Theorem4.7, we have the following.Theorem 4.8.
For1≤k≤n, qi=1/n,
(4.45)Pn,kk≤n-kn-1Mn,kk+k-1n-1Gnk.Proof.
We define(4.46)f(x)=n-kn-1Mn,kk+k-1n-1Gnk-Pn,kk.
We now set
(4.47)P̃n-1,k-1=Pn-1,k-1(x1,…,xn-1),G̃n-1=Gn-1(x1,…,xn-1),
so that
(4.48)∂f∂xn=knn-kn-1xnk-1+knk-1n-1Gnkxn-knP̃n-1,k-1k-1,
where we have also used the following relation:
(4.49)Pn,kk=knxnPn-1,k-1k-1(x1,…,xn-1)+n-knPn-1,kk(x1,…,xn-1).
Similar to the proof of Theorem 1.2, it suffices to show that ∂f/∂xn≥0. By a change of variables xi→xi/xn, it suffices to show that for all 0≤xi≤1,
(4.50)P̃n-1,k-1k-1≤n-kn-1+k-1n-1G̃n-1(n-1)k/n.
As a consequence of Lemma 3.2 in [3], one checks easily that
(4.51)P̃n-1,k-1k-1≤n-kn-1+k-1n-1G̃n-1n-1.
The above inequality then implies (4.50) and we then conclude that ∂f/∂xn≥0 which completes the proof.We remark here that once again one may hope to establish the above theorem by combining Theorems4.4 and 1.2 together. However, (1.6) is not applicable in this case since one checks readily that k/η(k)≤n is not satisfied in general.We now want to establish some inequalities involving the symmetric means in the forms similar to (1.4). Before we state our result, let us first recall that for two real finite decreasing sequences x=(x1,x2,…,xn) and y=(y1,y2,…,yn), x is said to be majorized by y (which we denote by x≤majy) if
(4.52)x1+x2+⋯+xj≤y1+y2+⋯+yj(1≤j≤n-1),x1+x2+⋯+xn=y1+y2+⋯+yn.
For a fixed positive sequence x=(x1,x2,…,xn) and a nonnegative sequence α=(α1,α2,…,αn), we define
(4.53)F(α)=1n!∑σxσ(1)α1xσ(2)α2⋯xσ(n)αn,
where the sum is over all the permutations of x. A well-known result of Muirhead states the follwing.Theorem 4.9 (see [13, Theorem  45]).
Letα and α′ be two nonnegative decreasing sequences. Then F(α)≤F(α′) for any positive sequence if and only if α≤majα′.We now use the above result to show the following.Theorem 4.10.
For1≤k≤n, qi=1/n,
(4.54)Ank≥1nk-1Mn,kk+(1-1nk-1)Pn,kk.Proof.
On expandingAnk out, we can write it as
(4.55)Ank≥1nk-1Mn,kk+linearcombinationsofvarioustermsoftheformF(α),
where F(α) is defined as in (4.53) with αi≥1, ∑i=1nαi=n. It is then easy to see via Theorem 4.9 that for any such α appearing in (4.55), we have Pn,kk≤F(α). Hence one deduces that
(4.56)Ank≥1nk-1Mn,kk+cPn,kk
for some constant c, which can be easily identified to be 1-1/nk-1 by taking x=(1,…,1) and noticing that we get identities in all the steps above.Our next result gives a generalization of the above one and we shall need the following two lemmas in our next proof.Lemma 4.11 (Hadamard's inequality).
Letf(x) be a convex function on [a,b], then
(4.57)f(a+b2)≤1b-a∫abf(x)dx≤f(a)+f(b)2.The next lemma is similar to that in [10].Lemma 4.12.
LetA,B,C,D>0 be arbitrary constants and let k≥r=2 or k≥r≥3. The maximum value of f(x,y)=A(xr+yr+B)k/r+Cxy on the set {(x,y):x≥0,y≥0,x+y=2D} is attained either when x=y or when xy=0.Proof.
We setz=xy and note that 0≤z≤D2 with equality holding if and only if x=y or xy=0. Moreover,
(4.58)x,y=D±D2-z.
This allows us to rewrite f(x,y)=Ag(z)+CZ where
(4.59)g(z)=((D+D2-z)r+(D-D2-z)r+B)k/r.
It suffices to show that g(z) is convex for 0≤z≤D2. Note that 2g′(z)=k·h(D2-z) where h(w)=p(w)q(w) with
(4.60)p(w)=((D+w)r+(D-w)r+B)k/r-1,q(w)=((D-w)r-1-(D+w)r-1)w-1.
As the derivative of D2-z is negative for 0≤z<D2, it suffices to show h′(w)≤0 for 0<w<D. Note that p(w)≥0, q(w)≤0 and it is easy to check that p′(w)≥0 for 0<w<D. Hence it suffices to show that q′(w)≤0 for 0<w<D. Calculation shows that
(4.61)w2q′(w)=-(r-1)((D+w)r-2+(D-w)r-2)w-(D-w)r-1+(D+w)r-1=(r-1)(∫D-wD+wur-2du-w((D+w)r-2+(D-w)r-2))≤0
by Lemma 4.11. This completes the proof.Now we are ready to prove the following.Theorem 4.13.
Letqi=1/n and let r be a real number, r=2 or r≥3. Then for integers n, k, n≥k≥r,
(4.62)Ank≥1nk-k/rMn,rk+(1-1nk-k/r)Pn,kk.Proof.
Since both sides of (4.62) are homogeneous functions, it suffices to show that
(4.63)maxx∈Δ{1nk-k/rMn,rk+(1-1nk-k/r)Pn,kk}≤1,
where
(4.64)Δ={x=(x1,…,xn):xi≥0,1≤i≤n,∑i=1nxi=n}.
Assume that (1/nk-k/r)Mn,rk+(1-1/nk-k/r)Pn,kk attains its maximum at some point (a1,…,an) with ai≥0, 1≤i≤n. Consider the function
(4.65)f(x,y):=1nk-k/rMn,rk(x,y,a3,…,an)+(1-1nk-k/r)Pn,kk(x,y,a3,…,an)
on the set
(4.66){(x,y):x≥0,y≥0,x+y=n-∑i=3nai}.
It is maximized at (a1,a2). Clearly, f has the form
(4.67)A(xr+yr+B)k/r+Cxy+(constant),
where A, B, and C are nonnegative constants. By Lemma 4.12, f attains its maximum value at either x=0 or y=0 or x=y. Repeating the same argument for other pairs (ai,aj), we conclude that in order to show (4.62), it suffices to check that it holds for x being of the following form (0,…,0,a…,a) or (a,…,a) for some positive constant a. It is easy to see that (4.62) holds when x is of the second form and when x is of the first form, let m denote the number of a's in x; if m<k, then it is easy to see that (4.62) holds. So we may now assume that k≤m≤n and we need to show that
(4.68)(mn)k≥1nk-k/r(mn)k/r+(1-1nk-k/r)(mk)(nk).
Equivalently, we need to show
(4.69)1nk≥1nkmk/r-k+(1-1nk-k/r)(mk)m-k(nk).
Note that
(4.70)(mk)m-k=1k!∏i=1k-1(1-im).
If we define for 1/n≤u≤1/(k-1),
(4.71)g(u)=1nkuk-k/r+(1-1nk-k/r)1(nk)1k!∏i=1k-1(1-iu),
then it is easy to check that g′′(u)≥0 for r≥k/(k-1). Further note that
(4.72)g(1n)=1nk,g(1k-1)=1nk(1k-1)k-k/r≤1nk.
This implies that g(u)≤1/nk for 1/n≤u≤1/(k-1), and hence it follows that (4.69) holds for k≤m≤n and this completes the proof.



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