FENG QI
Abstract. Letn, m∈Nand{ai}ni=1+mbe an increasing, logarithmically con-cave, positive, and nonconstant sequence such that the sequence
iai+1 ai −
1 ni=1+m−1 is increasing. Then the following inequality between ratios of the power means and of the geometic means holds:
1
n
n
X
i=1
ari
1
n+m
n+m
X
i=1
ari
!1/r <
n
√
an!
n+mp an+m!
,
where r is a positive number,ai! denotes the sequence factorial defined by
Qn
i=1ai. The upper bound is the best possible.
1. Introduction
It is well-known that the following inequality
n n+ 1 <
1
n
n
X
i=1 ir
1
n+ 1
n+1 X
i=1 ir
!1/r <
n
√
n!
n+1p
(n+ 1)! (1)
holds for r > 0 andn ∈ N. We call the left-hand side of this inequality Alzer’s inequality [1], and the right-hand side Martins’s inequality [7].
Alzer’s inequality has invoked the interest of several mathematicians, we refer the reader to [5, 9,16,18] and the references therein.
Recently, F. Qi and L. Debnath in [15] proved that: Letn, m∈N and{ai}∞i=1
be an increasing sequence of positive real numbers satisfying
(k+ 2)ar
k+2−(k+ 1)a
r k+1
(k+ 1)ar
k+1−ka
r k
≥
ak+2 ak+1
r
(2)
for a given positive real numberrandk∈N. Then
an
an+m ≤
(1/n)Pn
i=1a
r i
(1/(n+m))Pn+m
i=1 a
r i
1/r
. (3)
The lower bound of (3) is the best possible.
2000Mathematics Subject Classification. Primary 26D15.
Key words and phrases. Martins’s inequality, Alzer’s inequality, K¨onig’s inequality, increasing sequence, logarithmically concave, ratio, power mean, geometric mean.
The author was supported in part by NSF (#10001016) of China, SF for the Prominent Youth of Henan Province (#0112000200), SF of Henan Innovation Talents at Universities, NSF of Henan Province (#004051800), SF for Pure Research of Natural Science of the Education Department of Henan Province (#1999110004), Doctor Fund of Jiaozuo Institute of Technology, China.
This paper was typeset usingAMS-LATEX.
In [12,13,14,19,20,21], F. Qi and others proved the following inequalities and other more general results:
n+k+ 1
n+m+k+ 1 <
n+k
Y
i=k+1 i
!1/n, n+m+k
Y
i=k+1 i
!1/(n+m)
≤
r
n+k
n+m+k, (4)
a(n+k+ 1) +b a(n+m+k+ 1) +b <
h Qn+k
i=k+1(ai+b) i1n
h
Qn+m+k
i=k+1 (ai+b) in+1m
≤
s
a(n+k) +b
a(n+m+k) +b, (5)
wheren, m∈N,kis a nonnegative integer,aa positive constant, andba nonegative constant. The equalities in (4) and (5) is valid forn= 1 andm= 1.
In [17], the following monotonicity results for the gamma function were obtained: The function
[Γ(x+y+ 1)/Γ(y+ 1)]1/x
x+y+ 1 (6)
is decreasing inx≥1 for fixed y≥0. Then, for positive real numbersxandy, we have
x+y+ 1
x+y+ 2 ≤
[Γ(x+y+ 1)/Γ(y+ 1)]1/x
[Γ(x+y+ 2)/Γ(y+ 1)]1/(x+1). (7)
In [11,15], F. Qi proved that: Letnandmbe natural numbers,ka nonnegative integer. Then
n+k n+m+k <
1
n
n+k
X
i=k+1 ir
1
n+m
n+m+k
X
i=k+1 ir
!1/r
, (8)
whereris a given positive real number. The lower bound is the best possible. In [4,18], some more general results for the lower bound of ratio of power means
1
n
Pn
i=1a
r i
1
n+m
Pn+m
i=1 a
r i
1/r
for positive sequence{ai}i∈Nwere obtained.
An open problem in [10,11] asked for the validity of the following inequality:
1
n
n+k
X
i=k+1 ir
1
n+m
n+m+k
X
i=k+1 ir
!1/r <
n p
(n+k)!/k!
n+mp
(n+m+k)!/k!, (9)
wherer >0,n, m∈N,k∈Z+.
Let{ai}i∈N be a positive sequence. If ai+1ai−1 ≥a2i fori≥2, we call {ai}i∈N
a logarithmically convex sequence; if ai+1ai−1 ≤ a2i for i ≥ 2, we call {ai}i∈N a
logarithmically concave sequence. See [8, p. 284].
In [3], the open problem mentioned above was solved and generalized affirma-tively: Let {ai}ni=1+m be an increasing, logarithmically concave, positive, and
non-constant sequence satisfying (a`+1/a`)` ≥ (a`/a`−1)`−1 for any positive integer
` >1, then n1Pn
i=1a
r i
1
n+m
Pn+m
i=1 a
r i
1/r
< √na
n!
n+mp
an+m!, where ris a
posi-tive number,n, m∈N, and ai! denotes the sequence factorialQ n
i=1ai. The upper
bound is best possible.
The purpose of this paper is to give a new generalization of inequality (9) as follows.
Theorem 1. Letn, m∈Nand{ai}ni=1+mbe an increasing, logarithmically concave,
positive, and nonconstant sequence such that the sequence iai+1 ai −1
n+m−1
increasing. Then the following inequality between ratios of the power means and of the geometic means holds:
1
n
n
X
i=1 ari
1
n+m
n+m
X
i=1 ari
!1/r <
n
√
an!
n+mp an+m!
, (10)
where r is a positive number and ai! denotes the sequence factorial Q n
i=1ai. The
upper bound is the best possible.
As an easy consequence of Theorem1 by taking {ai}in=1+m={(i+k+b)
α}n+m i=1
for a positive constantα, we have
Corollary 1. Letαbe a positive real number,ka nonnegative integer andba real number such that k+b >0, andm, n∈N. If the sequence
i
1 + 1
i+k+b α
−1
n+m−1
i=1
(11)
is increasing, then for any real number r >0, we have
1
n
Pn+k
i=k+1[(i+b)
α]r
1
n+m
Pn+m+k
i=k+1 [(i+b)α]r !1/r
< n q
Qn+k
i=k+1(i+b)α
n+mqQn+m+k
i=k+1 (i+b)α
. (12)
The upper bound is the best possible.
Remark 1. By lettingα= 1 andb= 0 in (12), we recover inequality (9). Takingα= 2 in Corollary1 leads to the following
Corollary 2. Letkbe a nonnegative integer, andba real number such thatk+b≥
1
2, andm, n∈N. Then, for any real number r >0, we have
1
n
Pn+k
i=k+1[(i+b) 2]r
1
n+m
Pn+m+k
i=k+1 [(i+b)2]r !1/r
< n q
Qn+k
i=k+1(i+b)2
n+mqQn+m+k
i=k+1 (i+b)2
. (13)
The upper bound is the best possible.
Considering {ai}i∈N=
eiα
i∈N in Theorem1 and standard argument gives us
the following
Corollary 3. Letm, n∈N. If the constant 0< α <1such that inequality
e(1+x)α
−exα
xα−1−(1 +x)α−1 ≥αxe
(1+x)α (14)
holds with xon[1,∞), then, for any real numberr >0, we have
1
n
Pn
i=1e
iαr
1
n+m
Pn+m
i=1 ei αr
!1/r
<exp
1
n
n
X
i=1
iα− 1
n+m
n+m
X
i=1 iα
. (15)
2. Lemmas
To prove our main results, the following lemmas are neccessary.
Lemma 1. Let n, m ∈ N, and {ai}ni=1+m+1 a nonconstant positive sequence such
that the sequence iai+1
ai −1
n+m
i=1 is increasing, then the sequence
( i
√
ai!
ai+1 )n+m
i=1
(16)
is decreasing. As a simple consequence, we have the following
n
√
an!
n+mp an+m!
> an+1 an+m+1
, (17)
wherean! denotes the sequence factorial defined byQ n i=1ai.
Proof. For 1≤i≤n+m−1, the monotonicity of the sequence (16) is equivalent to the following
i
√
ai!
ai+1
≥
i+1p ai+1! ai+2
, (18)
⇐⇒
i
Y
k=1 ak
ai+1 !1/i
≥ i+1 Y
k=1 ak
ai+2
!1/(i+1)
,
⇐⇒ 1
i
i
X
k=1
ln ak
ai+1
≥ 1
i+ 1
i+1 X
k=1
ln ak
ai+2 ,
⇐⇒ i
i+ 1
i+1 X
k=1
ln ak
ai+2
≤ i
X
k=1
ln ak
ai+1
. (19)
Since lnxis concave on (0,∞), by definition of concaveness, it follows that, for 1≤k≤i,
k i+ 1ln
ak+1 ai+2
+i−k+ 1
i+ 1 ln
ak
ai+2
≤ln k
i+ 1 ·
ak+1 ai+2
+i−k+ 1
i+ 1 ·
ak
ai+2 !
(20)
= ln
kak+1+ (i−k+ 1)ak
(i+ 1)ai+2
.
Since the sequence iai+1
ai −1
n+m
i=1 is increasing, we have, for 1≤i≤n+m−1
and 1≤k≤i, the following
(i+ 1)ai+2 ai+1
−(i+ 1)≥ iai+1
ai −i,
⇐⇒ (i+ 1)ai+2
ai+1
−(i+ 1)≥ kak+1
ak −k,
⇐⇒ kak+1+ (i−k+ 1)ak
ak
≤ (i+ 1)ai+2
ai+1 ,
⇐⇒ kak+1+ (i−k+ 1)ak
(i+ 1)ai+2
≤ ak
Combining the last line above with (20) yields
k i+ 1ln
ak+1 ai+2
+i−k+ 1
i+ 1 ln
ak
ai+2
≤ln ak
ai+1
. (21)
Summing up on both sides of (21) with k from 1 to i and simplifying reveals inequality (19). The monotonicity follows.
Since {ai}ni=1+m+1 is a nonconstant positive sequence, there exists at least one
number 1 ≤i0 ≤ n+m−1 such that ai0 6= ai0+1. The function lnx is strictly
concave on (0,∞). Then, for anyisuch thati0≤i≤n+m−1, we have
i0 i+ 1ln
ai0+1
ai+2
+i−i0+ 1
i+ 1 ln
ai0
ai+2
<ln i0
i+ 1 ·
ai0+1
ai+2
+i−i0+ 1
i+ 1 ·
ai0
ai+2 !
= ln
i
0ai0+1+ (i−i0+ 1)ai0
(i+ 1)ai+2
≤ln ai0
ai+1 ,
(22)
notice that the last line follows from the sequence iai+1
ai −1
n+m
i=1 being increasing.
Therefore, for anyisuch thati0≤i≤n+m−1, inequality (18) is strict. Inequality (17) is proved.
The proof is complete.
Lemma 2. Let n >1 be a positive integer and{ai}ni=1 an increasing nonconstant
positive sequence such that iai+1
ai −1
n−1
i=1 is increasing. Then the sequence
( ai
(ai!)
1/i
)n
i=1
(23)
is increasing, and, for any positive integer `satisfying 1≤` < n,
a`
an
< (a`!) 1/`
(an!)
1/n, (24)
wherean! denotes the sequence factorial Q n i=1ai.
Proof. For 1≤`≤n−1, the monotonicity of the sequence (23) is equivalent to
a`
(a`!)1/`
≤ a`+1
(a`+1!)1/(`+1) ,
⇐⇒
`
Y
j=1 aj
a`
!1`
≥ `+1 Y
j=1 aj
a`+1 !`+11
,
⇐⇒ 1
`
`−1 X
j=1
lnaj
a` ≥ 1
`+ 1
`
X
j=1
ln aj
a`+1 ,
⇐⇒
`−1 X
j=1
lnaj
a` ≥ `
`+ 1
`
X
j=1
ln aj
a`+1
Since lnxis concave on (0,∞), by definition of concaveness, it follows that, for 1≤j ≤`,
j `+ 1ln
aj+1 a`+1
+`−j+ 1
`+ 1 ln
aj
a`+1
≤ln
j
`+ 1·
aj+1 a`+1
+`−j+ 1
`+ 1 ·
aj
a`+1
(26)
= ln
ja
j+1+ (`−j+ 1)aj
(`+ 1)a`+1
.
Straightforward computation gives us
`
X
j=1 j
`+ 1ln
aj+1 a`+1
+`−j+ 1
`+ 1 ln
aj
a`+1
= `
`+ 1
`
X
j=1
ln aj
a`+1
+
`
X
j=1 j
`+ 1ln
aj+1 a`+1
− `
X
j=1 j−1
`+ 1ln
aj
a`+1
= `
`+ 1
`
X
j=1
ln aj
a`+1
+
`+1 X
j=2 j−1
`+ 1ln
aj
a`+1
− `
X
j=1 j−1
`+ 1ln
aj
a`+1
= `
`+ 1
`
X
j=1
ln aj
a`+1 .
(27)
From combining of (25), (26) and (27), it suffices to prove for 1≤j≤`
jaj+1+ (`−j+ 1)aj
(`+ 1)a`+1
≤ aj
a`
,
⇐⇒ jaj+1+ (`−j+ 1)aj
aj
≤ (`+ 1)a`+1
a`
,
⇐⇒ jaj+1
aj
+`−j+ 1≤ (`+ 1)a`+1
a`
,
⇐⇒ (`+ 1)ha`+1
a`
−1i≥jhaj+1 aj
−1i. (28)
Since the sequences{ai}ni=1and
iai+1 ai −1
n−1
i=1 are increasing, the inequality
(28) holds.
Moreover, the sequence{ai}ni=1is nonconstant positive, then there exists at least
one number 1 ≤ i1 ≤ n−1 such that ai1 6= ai1+1. The function lnx is strictly
concave on (0,∞). Then, for any`such thati1< `≤n−1, we have
i1 `+ 1ln
ai1+1
a`+1
+`−i1+ 1
`+ 1 ln
ai1
a`+1
<ln
i 1 `+ 1 ·
ai1+1 a`+1
+`−i1+ 1
`+ 1 ·
ai1
a`+1
= ln
i1a
i1+1+ (`−i1+ 1)ai1
(`+ 1)a`+1
≤lnai1
a`
.
Therefore, for any`such thati1+ 1≤` < n, inequality (25) is strict, and
a`
a`+1
< (a`!) 1/`
(a`+1!)
1/(`+1), (30)
and then inequality (24) is strict. The proof is complete.
Remark 2. Some problems similar to Lemma1and Lemma2were discussed in [19] by the author and B.-N. Guo.
The methods proving Lemma1 and Lemma2had been used in [18] and others.
Lemma 3 (K¨onig’s inequality [2, p. 149] and [7,22]). Let {ai}ni=1 and{bi}ni=1 be
decreasing nonnegativen-tuples such that
k
Y
i=1 bi≤
k
Y
i=1
ai, 1≤k≤n, (31)
then, for r >0, we have
k
X
i=1 bri ≤
k
X
i=1
ari, 1≤k≤n. (32)
Remark 3. Lemma 3 is a well-known result due to K¨onig used to give a proof of Weyl’s inequality (cf. Corollary 1.b.8 of [6, p. 24]).
3. Proofs of Theorem1
Inequality (10) holds for n = 1 by the power mean inequality and its case of equality.
Forn≥2, inequality (10) is equivalent to
1
n
n
X
i=1 ari
1
n+ 1
n+1 X
i=1 ari
!1/r <
n
√
an!
n+1p an+1!
, (33)
which is equivalent to
1
n
n
X
i=1
a
i
n
√
an!
r < 1
n+ 1
n+1 X
i=1
ai
n+1p an+1!
!r
. (34)
Set
bjn+1=bjn+2=· · ·=bjn+n=
an+1−j
n+1p an+1!
, 0≤j≤n; (35)
cj(n+1)+1=cj(n+1)+2=· · ·=cj(n+1)+(n+1)= an−j
n
√
an!
, 0≤j ≤n−1. (36)
Direct calculation yields
n(n+1) X
i=1 bri =
n
X
j=0
n
X
k=1 brjn+k
=n
n
X
j=0
an+1−j
n+1p an+1!
!r
=n
n+1 X
i=1
ai
n+1p an+1!
!r
and
n(n+1) X
i=1
cri = (n+ 1)
n
X
i=1
a
i
n
√
an!
r
. (38)
Since{ai}ni=1+1 is increasing, the sequence{bi} n(n+1)
i=1 and{ci} n(n+1)
i=1 are
decreas-ing. Therefore, by Lemma 3, to obtain inequality (34), it is sufficient to prove inequality
bm!≥cm! (39)
for 1≤m≤n(n+ 1).
It is easy to see thatbn(n+1)! =cn(n+1)! = 1. Thus, inequality (39) is equivalent
to
n(n+1) Y
i=m
bi≤ n(n+1)
Y
i=m
ci (40)
for 2≤m≤n(n+ 1).
For 0≤`≤nand 0≤j ≤n−2, we have 2≤(n−`)n+ (n−j) = (n−`)(n+ 1) + (`−j)≤n(n+ 1). Then
n(n+1) Y
i=(n−`)n+(n−j) bi=
(a`+1)j+1(a`!)n
(an+1!) `n+j+1
n+1
; (41)
n(n+1) Y
i=(n−`)(n+1)+(`−j) ci=
(a`)n−`+j+2(a`−1!)n+1
(an!)
`n+j+1
n
, ` > j; (42)
n(n+1) Y
i=(n−`)(n+1)+(`−j) ci=
n(n+1) Y
i=(n−`−1)(n+1)+(n+1+`−j) ci
= (a`+1)
j−`+1(a
`!)n+1
(an!)
`n+j+1
n
, `≤j;
(43)
wherea0= 1.
The last term in (43) is bigger than the right term in (42), so, without loss of generality, we can assumej < `. Therefore, from formulae (41) and (42), inequality (40) is reduced to
(a`+1)j+1(a`!)n(an+1!) `−j−1
n+1
(an+1!)`
≤ (a`)
n−`+j+2(a
`−1!)n+1
(an!)`(an!)
j+1
n
, (44)
that is
(a`+1)j+1(an+1!) `−j−1
n+1
(a`!)(a`)j−`+1
≤ (an+1) `(a
n!) −`
n
(an!)
j−`+1
n
, (45)
this is further equivalent to
(a`+1)j+1(an+1!) `−j−1
n+1
a`!(a`)j−`+1(an!)
`−j−1
n
≤(an+1) `
(an!)
` n
, (46)
which can be rearranged as
a`+1 a`
·
n
√
an!
n+1pa n+1!
!j+1`
≤
`
√
a`!
a`
· an+1
n+1pa n+1!
Utilizing Lemma 2 and the logarithmical concaveness of the sequence {ai}ni=1+1
yields
n
√
an!
n+1p an+1!
> an an+1
≥ a`
a`+1
. (48)
Since j+1` ≤1 and a`+1 a` ·
n
√
an! n+1√a
n+1!
>1 by (48), thus, to obtain (47), it suffices
to prove
a`+1n p
an!< an+1` p
a`!, (49)
this follows from Lemma1.
Since the sequence{ai}ni=1+mis nonconstant, the inequality (10) is strict.
By the L’Hospital rule, easy calculation produces
lim
r→∞
1
n
n
X
i=1 ari
1
n+m
n+m
X
i=1 ari
1/r
=
n
√
an!
n+mpa
n+m!
, (50)
thus, the upper bound is the best possible. The proof is complete.
Remark 4. Recently, some new inequalities for the ratios of the mean values of functions were established in [23].
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(F. Qi)Department of Applied Mathematics and Informatics, Jiaozuo Institute of Technology, #142, Mid-Jiefang Road, Jiaozuo City, Henan 454000, China
E-mail address:[email protected]
