# Schurm-Power Convexity of a Class of Multiplicatively Convex Functions and Applications

Authors: Wen Wang; Shiguo Yang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We investigate the conditions under which the symmetric functionsF
n
,
k
(
x
,
r
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
f
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
,
k
=
1,2
,
…
,
n, are Schur m-power convex for x
∈
R
+
+
n and r
>
0. As a consequence, we prove that these functions are Schur
geometrically convex and Schur harmonically convex, which
generalizes some known results. By applying the theory of
majorization, several inequalities involving the pth power mean and
the arithmetic, the geometric, or the harmonic means are presented.


---

## Body

## 1. Introduction

LetR
n
=
{
(
x
1
,
x
2
,
…
,
x
n
)
:
x
i
∈
R
,
i
=
1,2
,
…
,
n
} and R
+
+
n
=
{
(
x
1
,
x
2
,
…
,
x
n
)
:
x
i
>
0
,
i
=
1,2
,
…
,
n
}. In particular, R
+
+
=
(
0
,
∞
). For x
=
(
x
1
,
x
2
,
…
,
x
n
), we denote by
(1)
x
=
(
x
1
,
x
2
,
…
,
x
n
)
,
ln
⁡
(
x
)
=
(
ln
⁡
x
1
,
ln
⁡
x
2
,
…
,
ln
⁡
x
n
)
,
1
x
=
(
1
x
1
,
1
x
2
,
…
,
1
x
n
)
.The Hamy symmetric function [1, 2] is defined as
(2)
F
n
(
x
,
r
)
=
F
n
(
x
1
,
x
2
,
…
,
x
n
;
r
)
=
∑
1
≤
i
1
<
i
2
<
⋯
<
i
r
≤
n
‍
(
∏
j
=
1
r
‍
x
i
j
)
1
/
r
,
mmmmmmmmmmmmmmmmmmmmhh
r
=
1,2
,
…
,
n
.The properties and applications of Hamy symmetric function can be found in the book of Bullen et al. [1].Throughout the paper, letI
⊂
R
+
+ and I
n
⊂
R
+
+
n.In 2007, Guan [3] defined a more general symmetric function: let f
:
I
n
→
R
+
+, and
(3)
∑
n
r
‍
(
f
(
x
)
)
=
∑
1
≤
i
1
<
i
2
<
⋯
<
i
r
≤
n
f
(
∏
j
=
1
r
‍
x
i
j
1
/
r
)
,
‍
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
r
=
1,2
,
…
,
n
,
where i
1
,
i
2
,
…
,
i
n are nonnegative integers, r
∈
N
=
{
1,2
,
3
,
…
}. Guan proved that the above symmetric function is Schur geometrically convex on I
n.In 2010, Rovenţa [4] defined the following symmetric function. Let f
:
I
→
R
+
+ be a log-convex function
(4)
F
k
(
x
)
=
∑
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
∏
j
=
1
k
‍
f
(
x
i
j
)
,
h
h
h
h
h
h
h
h
h
h
h
h
h
k
=
1,2
,
…
,
n
.Rovenţa proved that (4) is a Schur convex function on I
n.In 2010, Meng et al.  [5] proved the dual form of the Hamy symmetric function
(5)
H
n
(
x
,
r
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
r
‍
x
i
j
1
/
r
)
,
mmmmmmmmii
k
=
1,2
,
…
,
n
,
was Schur harmonically convex in R
+
+
n.In 2013, Shi and Zhang [6] investigated the following dual form of F
k
(
x
):
(6)
F
k
*
(
x
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
∑
j
=
1
k
‍
f
(
x
i
j
)
,
k
=
1,2
,
…
,
n
.
They proved that F
k
*
(
x
) is Schur convex, Schur geometrically and harmonically convex on I
n.Recently, Yang [7–9] generalized the notion of Schur convexity to Schur m-power convexity, which contains the Schur convexity, Schur geometrical convexity, and Schur harmonic convexity. Moreover, he discussed Schur m-power convexity of Stolarsky means [7], Gini means [8], and Daróczy means [9]. Wang and Yang showed that generalized Hamy symmetric function [10] and a class of symmetric functions [11] are Schur m-power convex.Now we define the more general dual form of symmetric function.Definition 1.
LetΩ
⊂
R
+
+
n be a symmetric convex set with nonempty interior and f
:
Ω
→
R
+
+ is continuous on Ω and differentiable in the interior of Ω. For r
>
0, define the symmetric functions F
n
,
k
(
x
,
r
) by
(7)
F
n
,
k
(
x
,
r
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
f
(
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
)
,
m
m
m
m
m
m
m
m
m
m
m
m
m
m
k
=
1,2
,
…
,
n
.In this paper, we investigate the Schurm-power convexity of the above more general dual form of symmetric functions. In particular, we obtain that the above more general dual form of symmetric functions is Schur geometrically convex and Schur harmonically convex, which generalizes some known results. As a consequence, we are able to prove a number of new inequalities concerning the pth power mean, the arithmetic mean, and the geometric and the harmonic mean.


## 2. Definitions and Lemmas

We first recall several definitions as follows.Definition 2 (see [12, 13]).
Letx
=
(
x
1
,
x
2
,
…
,
x
n
) and y
=
(
y
1
,
y
2
,
…
,
y
n
)
∈
R
n.(1)
x
≥
y means x
i
≥
y
i for all i
=
1,2
,
…
,
n.
(2)
LetΩ
⊂
R
n, f
:
Ω
→
R is said to be increasing if x
≥
y implies f
(
x
)
≥
f
(
y
). f is said to be decreasing if and only if -
f is increasing.Definition 3 (see [12, 13]).
Suppose thatx
=
(
x
1
,
x
2
,
…
,
x
n
) and y
=
(
y
1
,
y
2
,
…
,
y
n
)
∈
R
n are two n-tuples real numbers.(1)
y majorizes x (in symbols x
≺
y), if ∑
i
=
1
k
‍
x
[
i
]
≤
∑
i
=
1
k
‍
y
[
i
]
,
(
k
=
1,2
,
…
,
n
-
1
) and ∑
i
=
1
n
‍
x
[
i
]
=
∑
i
=
1
n
‍
y
[
i
], where x
[
1
]
≥
x
[
2
]
≥
⋯
≥
x
[
n
]
,
y
[
1
]
≥
y
[
2
]
≥
⋯
≥
y
[
n
] are rearrangements of x and y in a descending order.
(2)
A real-valued functionf
:
Ω
⊂
R
n
→
R is said to be Schur convex on Ω if
(8)
x
≺
y
on
⁡
Ω
⟹
f
(
x
)
≤
f
(
y
)
.
fis a Schur concave function on Ω if and only if -
f is a Schur convex function.Definition 4 (see [14]).
Supposex
=
(
x
1
,
x
2
,
…
,
x
n
) and y
=
(
y
1
,
y
2
,
…
,
y
n
)
∈
R
n are two n-tuples real numbers. Let Ω
⊂
R
+
+
n. A function f
:
Ω
→
R
+
+ is called Schur geometrically convex if
(9)
ln
⁡
x
≺
ln
⁡
y
on
⁡
Ω
⟹
φ
(
x
)
≤
φ
(
y
)
.
f is Schur geometrically concave if -
f is Schur geometrically convex.The following Theorem is basic and plays an important role in the theory of the Schur geometrically convex function.Lemma 5 (see [14]).
Letφ
(
x
)
=
φ
(
x
1
,
x
2
,
…
,
x
n
) be symmetric and continuous on Ω
⊂
R
+
+
n and differentiable in Ω
0. Then φ
:
Ω
→
R
+
+ is Schur geometrically convex (Schur geometrically concave) if and only if
(10)
(
ln
⁡
x
1
-
ln
⁡
x
2
)
(
x
1
∂
φ
∂
x
1
-
x
2
∂
φ
∂
x
2
)
≥
0
(
≤
0
)
.Definition 6 (see [15, 16]).
LetΩ
∈
R
n.(1)
A set Ω is called harmonically convex if x
y
/
(
λ
x
+
(
1
-
λ
y
)
)
∈
Ω for every x
,
y
∈
Ω and λ
∈
[
0,1
], where x
y
=
∑
i
=
1
n
‍
x
i
y
i and 1
/
x
=
(
1
/
x
1
,
…
,
1
/
x
n
).
(2)
A function φ
:
Ω
→
R
+
+ is called Schur harmonically convex on Ω if 1
/
x
≺
1
/
y implies φ
(
x
)
≤
φ
(
y
). f is Schur harmonically concave if -
f is Schur harmonically convex.Lemma 7 (see [15, 16]).
LetΩ
∈
R
+
+
n be a symmetric and harmonically convex set with inner points and let φ
:
Ω
→
R
+
+ be a continuously symmetric function which is differentiable in Ω
0. Then φ is Schur harmonically convex (Schur harmonically concave) on Ω if and only if
(11)
(
x
1
-
x
2
)
(
x
1
2
∂
φ
∂
x
1
-
x
2
2
∂
φ
∂
x
2
)
≥
0
(
≤
0
)
.
Schur convex, Schur geometrically convex, and Schur harmonically convex were introduced by Marshall et al. [13], Zhang [14], and Chu and Sun [15], respectively, and played a key role in analytic inequalities [1–36]. Moreover, the theory of convex functions and Schur convex functions is one of the most important research fields in modern analysis and geometry.Recently, Yang presents the Schurf-convexity in [7] as follows.Definition 8 (see [7–9]).
LetΩ
⊆
R
n be a set with nonempty interior and f a strictly monotone function defined on Ω. Let
(12)
f
(
x
)
=
(
f
(
x
1
)
,
f
(
x
2
)
,
…
,
f
(
x
n
)
)
,
f
(
y
)
=
(
f
(
y
1
)
,
f
(
y
2
)
,
…
,
f
(
y
n
)
)
.
Then function φ
:
Ω
→
R is said to be Schurf-convex on Ω if f
(
x
)
≺
f
(
y
) on Ω implies ψ
(
x
)
≤
ψ
(
y
).
ψ is said to be Schurf-concave if -
ψ is Schurf-convex.Takef
(
x
)
=
x, ln
⁡
x, x
-
1 in Definition 8, it yields the Schur convexity, Schur geometrical convexity, and Schur harmonic convexity. It is clear that the Schurf-convexity is a generalization of the Schur convexity mentioned above. In general, we have the following.Definition 9 (see [7–9]).
Letf
:
R
+
+
→
R be defined by f
(
x
)
=
(
x
m
-
1
)
/
m if m
≠
0 and f
(
x
)
=
ln
⁡
x if m
=
0. Then function ψ
:
Ω
⊆
R
+
+
n
→
R is said to be Schur m-power convex on Ω if f
(
x
)
≺
f
(
y
) on Ω implies ψ
(
x
)
≤
ψ
(
y
).
ψ is said to be Schur m-power concave if -
ψ is Schur m-power convex.Lemma 10 (see [7–9]).
Letψ
:
Ω
⊆
R
+
+
n
→
R be continuous on Ω and differentiable in Ω
0. Then ψ is schur m-power convex (Schur m-power concave) on Ω if and only if ψ is symmetric on Ω and
(13)
x
1
m
-
x
2
m
m
(
x
1
1
-
m
∂
φ
∂
x
1
-
x
2
1
-
m
∂
φ
∂
x
2
)
vvvv
≥
0
(
≤
0
)
,
i
f
m
≠
0
,
(
ln
⁡
x
1
-
ln
⁡
x
2
)
(
x
1
∂
φ
∂
x
1
-
x
2
∂
φ
∂
x
2
)
vvvv
≥
0
(
≤
0
)
,
i
f
m
≠
0
,
hold for any x
=
(
x
1
,
x
2
,
…
,
x
n
)
∈
Ω
0 with x
1
≠
x
2, where Ω
⊆
R
+
+
n is a symmetric set with nonempty interior Ω
0.The following lemma is clearly due to the monotonicity property of the functionx
p on R
+
+.Lemma 11 (see [10]).
Forx
1
,
x
2
>
0 with x
1
≠
x
2, let U be defined by
(14)
U
(
p
;
x
1
,
x
2
)
≔
{
x
1
p
-
x
2
p
p
(
x
1
-
x
2
)
,
i
f
p
≠
0
,
ln
⁡
x
1
-
ln
⁡
x
2
x
1
-
x
2
,
i
f
p
=
0
.
Thensgn
⁡
(
(
x
1
p
-
x
2
p
)
/
p
(
x
1
-
x
2
)
)
=
1, (i.e., U
(
p
;
x
1
,
x
2
)
>
0.)Remark 12 (see [10]).
By Lemma11, we see that
(15)
sgn
⁡
(
x
1
p
-
x
2
p
p
)
=
sgn
⁡
(
x
1
-
x
2
)
if
p
≠
0
,
sgn
⁡
(
ln
⁡
x
1
-
ln
⁡
x
2
)
=
sgn
⁡
(
x
1
-
x
2
)
.
Then the two discrimination inequalities in Lemma 10 are equivalent to
(16)
(
x
1
-
x
2
)
(
x
1
1
-
m
∂
φ
∂
x
1
-
x
2
1
-
m
∂
φ
∂
x
2
)
≥
(
≤
)
0
.Definition 13 (see [3, 17]).
Functionf
:
I
⊂
R
+
+
→
R
+
+ is said to be multipicatively convex if
(17)
x
,
y
∈
I
,
μ
∈
[
0,1
]
⟹
f
(
x
1
-
μ
y
μ
)
≤
f
(
x
)
1
-
μ
f
(
y
)
μ
.The following results have been proven, respectively.Lemma 14 (see [17]).
A continuous functionf
:
I
⊂
R
+
+
→
R
+
+ is multiplicatively convex if and only if
(18)
x
,
y
∈
I
⟹
f
(
x
y
)
≤
f
(
x
)
f
(
y
)
,
or
(19)
x
1
,
…
,
x
n
∈
I
⟹
f
(
x
1
,
…
,
x
n
n
)
≤
f
(
x
1
)
⋯
f
(
x
n
)
n
.Lemma 15 (see [17]).
Assume thatf
:
I
⊂
R
+
+
→
R
+
+ is a differential function. Then the following assertions are equivalent. (i)
f is multiplicatively convex.
(ii)
The functionx
f
′
(
x
)
/
f
(
x
) is nondecreasing.Moreover, iff is twice differentiable, then f is multiplicatively convex if and only if
(20)
x
[
f
(
x
)
f
′′
(
x
)
-
f
′
2
(
x
)
]
+
f
(
x
)
f
′
(
x
)
≥
0
∀
x
>
0
.


## 3. Main Results and Proof

Our main results are stated as follows.Theorem 16.
Iff is increasing and multiplicatively convex, then for m
≤
0 and r
>
0, F
n
,
k
(
x
,
r
) defined in (7) are Schur m-power convex on Ω, where k
=
1,2
,
…
,
n.Takem
=
0
,
-
1 in Theorem 16, we get the following Corollaries.Corollary 17.
Iff is increasing and multiplicatively convex, then for any k
=
1,2
,
…
,
n and r
>
0, F
n
,
k
(
x
,
r
) defined in (7) are Schur geometrically convex on Ω.Corollary 18.
Iff is increasing and multiplicatively convex, then for any k
=
1,2
,
…
,
n and r
>
0, F
n
,
k
(
x
,
r
) defined in (7) are Schur harmonically convex function on Ω.Theorem 19.
Letx
=
(
x
1
,
x
2
,
…
,
x
n
)
∈
R
+
+
n, and ∑
i
=
1
n
‍
x
i
=
s. If f is increasing and multiplicatively convex, then for any k
=
1,2
,
…
,
n and r
>
0, one has
(21)
F
n
,
k
(
n
-
μ
s
-
μ
x
,
r
)
≤
F
n
,
k
(
1
x
,
r
)
,
0
≤
μ
≤
1
;
F
n
,
k
(
n
+
μ
s
+
μ
x
,
r
)
≤
F
n
,
k
(
1
x
,
r
)
,
0
≤
μ
≤
1
.To prove the main results, we first establish some lemmas.Lemma 20.
Let the functionf
:
I
⊂
R
+
+
→
R
+
+ be continuous on Ω and differentiable in the interior of Ω. For m
≤
0, if f is increasing and multiplicatively convex, then x
1
-
m
f
′
(
x
)
/
f
(
x
) is increasing.Proof.
Sincef is multiplicatively convex, and by using Lemma 15, we can easily see that x
f
′
(
x
)
/
f
(
x
) is increasing. Further, by applying f
(
x
)
≥
0 and the monotonicity of f, it follows that x
1
-
m
f
′
(
x
)
/
f
(
x
) is also increasing for x
>
0 and m
≤
0.Lemma 21.
Ifr
>
0, c
>
0 and m
≤
0, the function g
(
x
)
=
(
x
r
-
c
)
(
r
-
m
)
/
r
/
x
r is increasing.Proof.
We can easily derive that(22)
g
′
(
x
)
=
1
x
2
r
[
r
-
m
r
(
x
r
-
c
)
-
m
/
r
nnn
·
r
x
r
-
1
·
x
r
-
r
x
r
-
1
(
x
r
-
c
)
(
r
-
m
)
/
r
]
=
1
x
2
r
r
x
r
-
1
(
x
r
-
c
)
-
m
/
r
[
-
m
r
x
r
+
c
]
≥
0
.
So the functiong
(
x
) is increasing.Lemma 22.
Letf
:
Ω
⊂
R
+
+
n
→
R
+
+ be continuous on Ω and differentiable in the interior of Ω. For m
≤
0 and r
>
0, if f is increasing and multiplicatively convex, then
(23)
I
=
F
n
,
2
(
x
,
r
)
(
x
1
-
x
2
)
(
x
1
r
-
m
-
x
2
r
-
m
)
×
f
′
[
(
x
1
r
+
x
2
r
)
1
/
r
]
f
[
(
x
1
r
+
x
2
r
)
1
/
r
]
(
x
1
r
+
x
2
r
)
1
/
r
-
1
≥
0
,
(24)
I
I
=
F
n
,
2
(
x
,
r
)
(
x
1
-
x
2
)
×
∑
j
=
3
n
‍
{
f
′
[
(
x
1
r
+
x
j
r
)
1
/
r
]
f
[
(
x
1
r
+
x
j
r
)
1
/
r
]
·
(
x
1
r
+
x
j
r
)
1
/
r
-
1
·
x
1
r
-
m
vvv
-
f
′
[
(
x
2
r
+
x
j
r
)
1
/
r
]
f
[
(
x
2
r
+
x
j
r
)
1
/
r
]
v
v
v
·
(
x
2
r
-
m
+
x
j
r
)
1
/
r
-
1
·
x
2
r
f
′
[
(
x
1
r
+
x
j
r
)
1
/
r
]
f
[
(
x
1
r
+
x
j
r
)
1
/
r
]
}
≥
0
.Proof.
Firstly, we prove that (23) holds. Since m
≤
0 and r
>
0, then
(25)
(
x
1
-
x
2
)
(
x
1
r
-
m
-
x
2
r
-
m
)
≥
0
.
So, we deduce that I
≥
0.
Secondly, we prove that (24) holds. Set u
=
(
x
1
r
+
x
j
r
)
1
/
r,v
=
(
x
2
r
+
x
j
r
)
1
/
r, obviously, u
,
v
∈
Ω. One can easily find that
(26)
I
I
=
F
n
,
2
(
x
,
r
)
x
1
-
x
2
u
-
v
(
u
-
v
)
×
∑
j
=
3
n
‍
{
u
f
′
(
u
)
f
(
u
)
·
(
u
r
-
x
j
r
)
(
r
-
m
)
/
r
u
r
v
v
i
i
-
v
f
′
(
v
)
f
(
v
)
·
(
u
r
-
x
j
r
)
(
r
-
m
)
/
r
v
r
}
.
Because f is multiplicatively convex, and by Lemma 15, we get
(27)
(
u
-
v
)
(
u
f
′
(
u
)
f
(
u
)
-
v
f
′
(
v
)
f
(
v
)
)
≥
0
.
On the other hand, forr
>
0 and m
≤
0, we easily know that the functions x
r and (
x
r
+
x
j
r
)
1
/
r are increasing about x. By applying Lemma 21, we have
(28)
(
x
1
-
x
2
)
(
x
1
r
-
x
2
r
)
≥
0
;
x
1
-
x
2
u
-
v
=
(
x
1
-
x
2
)
2
(
x
1
-
x
2
)
(
u
-
v
)
≥
0
;
(
u
-
v
)
{
u
f
′
(
u
)
f
(
u
)
·
(
u
r
-
x
j
r
)
(
r
-
m
)
/
r
u
r
-
v
f
′
(
v
)
f
(
v
)
·
(
u
r
-
x
j
r
)
(
r
-
m
)
/
r
v
r
}
≥
0
.
Because the functionf is increasing and f
≥
0, and applying (27)-(28), we obtain I
I
≥
0.Lemma 23.
Letf
:
Ω
⊂
R
+
+
n
→
R
+
+ be continuous on Ω and differentiable in the interior of Ω. If f is increasing and multiplicatively convex, then for m
≤
0 and r
>
0, one has
(29)
I
I
I
=
F
n
,
k
(
x
,
r
)
(
x
1
-
x
2
)
(
x
1
r
-
m
-
x
2
r
-
m
)
×
∑
2
≤
i
1
<
i
2
<
⋯
<
i
k
-
2
≤
n
‍
f
′
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
f
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
≥
0
,
I
V
=
F
n
,
k
(
x
,
r
)
(
x
1
-
x
2
)
×
∑
3
≤
i
1
<
i
2
<
…
<
i
k
-
1
≤
n
‍
{
f
′
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
f
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
vvvvvvvvmmmii
·
(
x
1
r
+
∑
j
=
1
k
-
1
‍
x
i
j
r
)
1
/
r
-
1
vvvvvvvvvvmmmik
·
x
1
r
-
m
-
f
′
[
(
x
2
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
f
[
(
x
2
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
k
k
k
k
k
k
k
k
k
k
k
k
k
i
·
(
x
2
r
+
∑
j
=
1
k
-
1
‍
x
i
j
r
)
1
/
r
-
1
·
x
2
r
-
m
f
′
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
f
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
}
≥
0
.Proof.
Setu
=
(
x
1
r
+
∑
j
=
1
k
-
1
‍
x
i
j
r
)
1
/
r, v
=
(
x
1
r
+
∑
j
=
1
k
-
1
‍
x
i
j
r
)
1
/
r; it is easy that u
,
v
∈
Ω. Then
(30)
I
V
=
F
n
,
k
(
x
,
r
)
x
1
-
x
2
u
-
v
(
u
-
v
)
×
∑
3
≤
i
1
<
i
2
<
⋯
<
i
k
-
1
≤
n
{
u
f
′
(
u
)
f
(
u
)
·
(
u
r
-
w
)
(
r
-
m
)
/
r
u
r
vvvvvvvvvvvvvvv
-
v
f
′
(
v
)
f
(
v
)
·
(
v
r
-
w
)
(
r
-
m
)
/
r
v
r
}
,
where w
=
∑
j
=
1
k
-
1
‍
x
i
j
r.
By (27)-(28) and the monotonicity property of the nonnegative function f, we get that I
I
I
≥
0
,
I
V
≥
0.Proof of Theorem16.
By Lemma10 and Remark 12, we only need to prove that
(31)
(
x
1
-
x
2
)
(
x
1
1
-
m
∂
F
n
,
k
(
x
,
r
)
x
1
-
x
2
1
-
m
∂
F
n
,
k
(
x
,
r
)
x
2
)
≥
0
.
To prove the above inequality, we consider the following three cases fork.
Case
1.For k
=
1. It is clear that F
n
,
1
(
x
,
r
)
=
∏
i
=
1
n
‍
f
(
x
i
). From (31), it follows that
(32)
Δ
1
=
(
x
1
-
x
2
)
(
x
1
1
-
m
∂
F
n
,
1
(
x
,
r
)
x
1
-
x
2
1
-
m
∂
F
n
,
1
(
x
,
r
)
x
2
)
=
(
x
1
-
x
2
)
(
x
1
1
-
m
f
′
(
x
1
)
f
(
x
1
)
-
x
2
1
-
m
f
′
(
x
2
)
f
(
x
2
)
)
∏
i
=
1
n
‍
f
(
x
i
)
.
By Lemma20, and f
(
x
)
≥
0, it follows that Δ
1
≥
0.
Case
2.For k
=
2
,
r
>
0, we have F
n
,
2
(
x
,
r
)
=
∏
1
≤
i
<
j
≤
n
‍
f
[
(
x
i
r
+
x
j
r
)
1
/
r
]. We can easily derive that
(33)
ln
⁡
F
n
,
2
(
x
,
r
)
=
∑
1
≤
i
<
j
≤
n
‍
ln
⁡
f
[
(
x
i
r
+
x
j
r
)
1
/
r
]
=
∑
j
=
2
n
‍
ln
⁡
f
[
(
x
1
r
+
x
j
r
)
1
/
r
]
+
∑
2
≤
i
<
j
≤
n
‍
ln
⁡
f
[
(
x
i
r
+
x
j
r
)
1
/
r
]
.
By differentiating the above equation with respect to x
1, we obtain
(34)
∂
F
n
,
2
(
x
,
r
)
x
1
=
F
n
,
2
(
x
,
r
)
·
∂
ln
⁡
F
n
,
2
(
x
,
r
)
x
1
=
F
n
,
2
(
x
,
r
)
·
{
∑
j
=
2
n
‍
f
′
[
(
x
1
r
+
x
j
r
)
1
/
r
]
f
[
(
x
1
r
+
x
j
r
)
1
/
r
]
x
x
x
x
x
x
x
·
(
x
1
r
+
x
j
r
)
1
/
r
-
1
·
x
1
r
-
1
∑
j
=
2
n
‍
f
′
[
(
x
1
r
+
x
j
r
)
1
/
r
]
f
[
(
x
1
r
+
x
j
r
)
1
/
r
]
∑
j
=
2
n
‍
}
=
F
n
,
2
(
x
,
r
)
{
f
′
[
(
x
1
r
+
x
2
r
)
1
/
r
]
f
[
(
x
1
r
+
x
2
r
)
1
/
r
]
vvvvvv
·
(
x
1
r
+
x
2
r
)
1
/
r
-
1
·
x
1
r
-
1
vvvvvv
+
∑
j
=
3
n
‍
f
′
[
(
x
1
r
+
x
j
r
)
1
/
r
]
f
[
(
x
1
r
+
x
j
r
)
1
/
r
]
vvvvvv
·
(
x
1
r
+
x
j
r
)
1
/
r
-
1
·
x
1
r
-
1
f
′
[
(
x
1
r
+
x
2
r
)
1
/
r
]
f
[
(
x
1
r
+
x
2
r
)
1
/
r
]
}
.
Similarly, we have(35)
∂
F
n
,
2
(
x
,
r
)
x
2
=
F
n
,
2
(
x
,
r
)
×
{
(
x
2
r
+
x
j
r
)
1
/
r
-
1
·
x
2
r
-
1
f
′
[
(
x
1
r
+
x
2
r
)
1
/
r
]
f
[
(
x
1
r
+
x
2
r
)
1
/
r
]
·
(
x
1
r
+
x
2
r
)
1
/
r
-
1
·
x
2
r
-
1
i
i
iii
+
∑
j
=
3
n
‍
f
′
[
(
x
2
r
+
x
j
r
)
1
/
r
]
f
[
(
x
2
r
+
x
j
r
)
1
/
r
]
iiiii
·
(
x
2
r
+
x
j
r
)
1
/
r
-
1
·
x
2
r
-
1
f
′
[
(
x
1
r
+
x
2
r
)
1
/
r
]
f
[
(
x
1
r
+
x
2
r
)
1
/
r
]
}
.
So, from (34) and (35), and by applying Lemma 22, we have
(36)
Δ
2
=
(
x
1
-
x
2
)
(
x
1
1
-
m
∂
F
n
,
2
(
x
,
r
)
x
1
-
x
2
1
-
m
∂
F
n
,
2
(
x
,
r
)
x
2
)
=
F
n
,
2
(
x
,
r
)
(
x
1
-
x
2
)
(
x
1
r
-
m
-
x
2
r
-
m
)
f
′
[
(
x
1
r
+
x
2
r
)
1
/
r
]
f
[
(
x
1
r
+
x
2
r
)
1
/
r
]
×
(
x
1
r
+
x
2
r
)
1
/
r
-
1
+
F
n
,
2
(
x
,
r
)
(
x
1
-
x
2
)
·
∑
j
=
3
n
‍
{
f
′
[
(
x
1
r
+
x
j
r
)
1
/
r
]
f
[
(
x
1
r
+
x
j
r
)
1
/
r
]
·
(
x
1
r
+
x
j
r
)
1
/
r
-
1
·
x
1
r
-
m
n
n
n
-
f
′
[
(
x
2
r
+
x
j
r
)
1
/
r
]
f
[
(
x
2
r
+
x
j
r
)
1
/
r
]
·
(
x
2
r
+
x
j
r
)
1
/
r
-
1
·
x
2
r
-
m
}
=
I
+
I
I
≥
0
.
So we get thatΔ
2
≥
0.
Case
3. For 3
≤
k
≤
n
,
r
>
0, similarly to the discussion of Case 2, we have
(37)
ln
⁡
F
n
,
2
(
x
,
r
)
=
∑
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
ln
⁡
f
[
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
]
=
∑
2
≤
i
1
<
i
2
<
⋯
<
i
k
-
1
≤
n
‍
ln
⁡
f
[
(
x
1
r
+
∑
j
=
1
k
-
1
‍
x
i
j
r
)
1
/
r
]
+
∑
2
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
ln
⁡
f
[
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
]
.
By differentiating to the above with respect tox
1, we have
(38)
∂
F
n
,
k
(
x
,
r
)
x
1
=
F
n
,
k
(
x
,
r
)
·
∂
ln
⁡
F
n
,
k
(
x
,
r
)
x
1
=
F
n
,
k
(
x
,
r
)
·
∑
2
≤
i
1
<
i
2
<
⋯
<
i
k
-
1
≤
n
‍
f
′
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
f
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
·
(
x
1
r
+
∑
j
=
1
k
-
1
‍
x
i
j
r
)
1
/
r
-
1
·
x
1
r
-
1
=
F
n
,
k
(
x
,
r
)
{
∑
3
≤
i
1
<
i
2
<
⋯
<
i
k
-
2
≤
n
‍
f
′
[
(
x
1
r
+
x
2
r
+
∑
j
=
1
k
-
2
x
i
j
r
)
1
/
r
]
f
[
(
x
1
r
+
x
2
r
+
∑
j
=
1
k
-
2
x
i
j
r
)
1
/
r
]
yyyyyyy
·
(
x
1
r
+
x
2
r
+
∑
j
=
1
k
-
2
‍
x
i
j
r
)
1
/
r
-
1
·
x
1
r
-
1
yyyyyyyi
+
∑
3
≤
i
1
<
i
2
<
⋯
<
i
k
-
1
≤
n
f
′
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
f
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
‍
y
y
y
y
y
y
y
i
·
(
x
1
r
+
∑
j
=
1
k
-
1
‍
x
i
j
r
)
1
/
r
-
1
·
x
1
r
-
1
∑
3
≤
i
1
<
i
2
<
⋯
<
i
k
-
2
≤
n
‍
f
′
[
(
x
1
r
+
x
2
r
+
∑
j
=
1
k
-
2
x
i
j
r
)
1
/
r
]
f
[
(
x
1
r
+
x
2
r
+
∑
j
=
1
k
-
2
x
i
j
r
)
1
/
r
]
}
.
Similarly, we can have(39)
∂
F
n
,
k
(
x
,
r
)
x
2
=
F
n
,
k
(
x
,
r
)
·
∂
ln
⁡
F
n
,
k
(
x
,
r
)
x
2
=
F
n
,
k
(
x
,
r
)
{
∑
3
≤
i
1
<
i
2
<
⋯
<
i
k
-
2
≤
n
‍
f
′
[
(
x
1
r
+
x
2
r
+
∑
j
=
1
k
-
2
x
i
j
r
)
1
/
r
]
f
[
(
x
1
r
+
x
2
r
+
∑
j
=
1
k
-
2
x
i
j
r
)
1
/
r
]
mmmm
·
(
x
1
r
+
x
2
r
+
∑
j
=
1
k
-
2
‍
x
i
j
r
)
1
/
r
-
1
·
x
1
r
-
1
mmmm
+
∑
3
≤
i
1
<
i
2
<
⋯
<
i
k
-
1
≤
n
‍
f
′
[
(
x
2
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
f
[
(
x
2
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
·
(
x
2
r
+
∑
j
=
1
k
-
1
‍
x
i
j
r
)
1
/
r
-
1
·
x
2
r
-
1
∑
3
≤
i
1
<
i
2
<
⋯
<
i
k
-
2
≤
n
‍
f
′
[
(
x
1
r
+
x
2
r
+
∑
j
=
1
k
-
2
x
i
j
r
)
1
/
r
]
f
[
(
x
1
r
+
x
2
r
+
∑
j
=
1
k
-
2
x
i
j
r
)
1
/
r
]
}
.
From (38) and (39), we have
(40)
Δ
3
=
(
x
1
-
x
2
)
(
x
1
1
-
m
∂
F
n
,
k
(
x
,
r
)
x
1
1
-
m
-
x
2
1
-
m
∂
F
n
,
k
(
x
,
r
)
x
2
)
=
F
n
,
k
(
x
,
r
)
(
x
1
-
x
2
)
(
x
1
r
-
m
-
x
2
r
-
m
)
×
∑
2
≤
i
1
<
i
2
<
⋯
<
i
k
-
2
≤
n
‍
f
′
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
f
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
+
F
n
,
k
(
x
,
r
)
(
x
1
-
x
2
)
×
∑
3
≤
i
1
<
i
2
<
⋯
<
i
k
-
1
≤
n
‍
{
f
′
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
f
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
vvvvvvvvvvvvv
·
(
x
1
r
+
∑
j
=
1
k
-
1
‍
x
i
j
r
)
1
/
r
-
1
·
x
1
r
-
m
v
v
v
v
v
v
v
v
v
v
v
v
v
i
-
f
′
[
(
x
2
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
f
[
(
x
2
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
m
m
m
m
m
m
m
m
·
(
x
2
r
+
∑
j
=
1
k
-
1
‍
x
i
j
r
)
1
/
r
-
1
·
x
2
r
-
m
f
′
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
f
[
(
x
1
r
+
∑
j
=
1
k
-
1
x
i
j
r
)
1
/
r
]
}
=
I
I
I
+
I
V
≥
0
.
So we get thatΔ
2
≥
0. So the proof of Theorem 16 is complete.Proof of Theorem19.
Letx
=
(
x
1
,
x
2
,
…
,
x
n
)
∈
R
+
+
n, and ∑
i
=
1
n
‍
x
i
=
s. From [13, 14], we have
(41)
s
-
μ
x
n
-
μ
=
(
s
-
μ
x
1
n
-
μ
,
s
-
μ
x
2
n
-
μ
,
…
,
s
-
μ
x
n
n
-
μ
)
≺
(
x
1
,
x
2
,
…
,
x
n
)
=
x
,
s
+
μ
x
n
+
μ
=
(
s
+
μ
x
1
n
+
μ
,
s
+
μ
x
2
n
+
μ
,
…
,
s
+
μ
x
n
n
+
μ
)
≺
(
x
1
,
x
2
,
…
,
x
n
)
=
x
.
From (41), and contacting Corollaries 17 and 18, we get (21).


## 4. Applications

Letx
i
>
0
,
i
=
1,2
,
…
,
n be n positive real numbers and set x
=
{
x
1
,
x
2
,
…
,
x
n
}. The pth power mean of of order p
∈
R of x
i is defined by
(42)
M
p
(
x
)
=
{
(
x
1
p
+
x
2
p
+
⋯
+
x
n
p
n
)
1
/
p
,
p
≠
0
;
x
1
x
2
⋯
x
n
,
p
=
0
.In particular, forp
=
1
,
p
=
0, and p
=
1 we, respectively, get the arithmetic, the geometric, and the harmonic means of x
i, and set A
n
(
x
)
=
M
1
(
x
)
=
(
1
/
n
)
∑
i
=
1
n
‍
x
i, G
n
(
x
)
=
M
0
(
x
)
=
(
∏
i
=
1
n
‍
x
i
)
1
/
n, H
n
(
x
)
=
M
-
1
(
x
)
=
n
/
∑
i
=
1
n
(
1
/
x
i
)
.In this section, some applications of the results in Section3 are given. Some analytic inequalities are established. In particular, several inequalities involving the pth power mean and the arithmetic, the geometric, or the harmonic means are presented.


### 4.1. Analytic Inequalities

To establish some analytic inequalities, we first give a lemma.Lemma 24 (see [20]).
Letx
i
>
0
,
i
=
1,2
,
…
,
n. Then the following relations is known:
(43)
(
A
n
(
x
)
,
A
n
(
x
)
,
…
,
A
n
(
x
)
)
≺
(
x
1
,
x
2
,
…
,
x
n
)
;
ln
⁡
(
G
n
(
x
)
,
G
n
(
x
)
,
…
,
G
n
(
x
)
)
≺
ln
⁡
(
x
1
,
x
2
,
…
,
x
n
)
;
(
1
H
n
(
x
)
,
1
H
n
(
x
)
,
…
,
1
H
n
(
x
)
)
≺
(
1
x
1
,
1
x
2
,
…
,
1
x
n
)
.By applying Lemma24 and Theorem 16, Corollaries 17 and 18, we can get the following several theorems and corollaries.Theorem 25.
Forr
>
0 and k
=
1,2
,
…
,
n, the symmetric functions
(44)
U
k
(
x
,
r
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
tan
⁡
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
are Schur m-power convex in R
+
+
n when m
≤
0. In particular, U
k
(
x
,
r
) are Schur geometrically convex and Schur harmonically convex in R
+
+
n.Proof.
Forx
∈
R
+
+, let f
(
x
)
=
tan
x. Then f
′
(
x
)
=
1
/
cos
⁡
⁡
2
x
=
sec
2
x
≥
0 and
(45)
x
[
f
(
x
)
f
′′
(
x
)
-
(
f
′
(
x
)
)
2
]
+
f
(
x
)
f
′
(
x
)
=
sec
2
x
(
x
sec
2
x
-
2
x
+
tan
x
)
.
Let g
(
x
)
=
x
sec
2
x
-
2
x
+
tan
x; then g
′
(
x
)
=
2
(
1
+
x
tan
x
)
sec
2
x
-
2
≥
0, and g
(
x
) is increasing. From the monotonicity of g
(
x
), we get that
(46)
x
[
f
(
x
)
f
′′
(
x
)
-
(
f
′
(
x
)
)
2
]
+
f
(
x
)
f
′
(
x
)
=
sec
2
x
g
(
x
)
≥
g
(
0
)
=
0
.
Therefore, f
(
x
) is increasing and multiplicatively convex in R
+
+
n. By applying Theorem 16, we obtain the result.By using Theorem25 and Lemma 24, we get the following inequalities.Corollary 26.
Forx
∈
R
+
+
n, then
(47)
∏
i
=
1
n
‍
(
tan
1
x
i
)
1
/
n
≥
tan
(
1
A
(
x
)
)
;
∏
i
=
1
n
‍
(
tan
x
i
)
1
/
n
≥
tan
G
(
x
)
;
∏
i
=
1
n
‍
(
tan
x
i
)
1
/
n
≥
tan
H
(
x
)
.Theorem 27.
Forr
>
0 and k
=
1,2
,
…
,
n, the symmetric functions
(48)
V
k
(
x
,
r
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
arcsin
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
are Schur m-power convex in (
0,1
)
n when m
≤
0. In particular, V
k
(
x
,
r
) are Schur geometrically convex and Schur harmonically convex in (
0,1
)
n.Proof.
Forx
∈
(
0,1
), let f
(
x
)
=
arcsin
x. Then f
′
(
x
)
=
1
/
1
-
x
2
≥
0 and
(49)
x
[
f
(
x
)
f
′′
(
x
)
-
(
f
′
(
x
)
)
2
]
+
f
(
x
)
f
′
(
x
)
=
arcsin
x
-
x
1
-
x
2
(
1
-
x
2
)
1
-
x
2
.
Let h
(
x
)
=
arctan
x
-
x
2
arctan
x
-
x; then h
′
(
x
)
=
2
x
2
/
1
-
x
2
≥
0, and h
(
x
) is increasing. From the monotonicity of h
(
x
), we get that
(50)
x
[
f
(
x
)
f
′′
(
x
)
-
(
f
′
(
x
)
)
2
]
+
f
(
x
)
f
′
(
x
)
=
h
(
x
)
(
1
-
x
2
)
1
-
x
2
≥
h
(
0
)
=
0
.
Therefore, f
(
x
) is increasing and multiplicatively convex in R
+
+
n. By applying Theorem 16, we obtain the result.By using Theorem27 and Lemma 24, we get the following inequalities.Corollary 28.
Forx
∈
(
0,1
)
n, then
(51)
∏
i
=
1
n
‍
(
arcsin
1
x
i
)
1
/
n
≥
arcsin
(
1
A
(
x
)
)
;
∏
i
=
1
n
‍
(
arcsin
x
i
)
1
/
n
≥
arcsin
G
(
x
)
;
∏
i
=
1
n
‍
(
arcsin
x
i
)
1
/
n
≥
arcsin
H
(
x
)
.Theorem 29.
Forr
>
0 and k
=
1,2
,
…
,
n, the symmetric functions
(52)
M
k
(
x
,
r
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
are Schur m-power convex in R
+
+
n when m
≤
0. In particular, M
k
(
x
,
r
) are Schur geometrically convex in R
+
+
n and Schur harmonically convex in R
+
+
n.Proof.
Letf
(
x
)
=
x
,
x
∈
R
+
+. We can easily see that f
(
x
) is increasing and multiplicatively convex in R
+
+. By applying Theorem 16, we obtain the result.By using Theorem29 and Lemma 24, we obtain the following results.Corollary 30.
Letx
∈
R
+
+, for r
>
0 and k
=
1,2
,
…
,
n; one has
(53)
(
k
1
/
r
1
A
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
1
x
i
j
r
)
1
/
r
,
(54)
(
k
1
/
r
G
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
,
(55)
(
k
1
/
r
H
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
.In particular, takek
=
1 in (53); we get the known A-G-H inequality. And take k
=
n in (53); we get the following inequalities:
(56)
(
∑
i
=
1
n
‍
1
x
i
r
)
1
/
r
·
∑
i
=
1
n
‍
x
i
≥
n
1
/
r
+
1
.Theorem 31.
Forr
>
0 and k
=
1,2
,
…
,
n, the symmetric functions
(57)
N
k
(
x
,
r
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
1
-
(
∑
j
=
1
k
x
i
j
k
)
1
/
r
are Schur m-power convex on (
0,1
)
n when m
≤
0. In particular, N
k
(
x
,
r
) are Schur geometrically convex on (
0,1
)
n and Schur harmonically convex on (
0,1
)
n.Proof.
Letf
(
x
)
=
x
/
(
1
-
x
)
,
x
∈
(
0,1
). Then f
′
(
x
)
=
(
1
/
(
1
-
x
)
2
)
≥
0, and
(58)
x
[
f
(
x
)
f
′′
(
x
)
-
f
′
2
(
x
)
]
+
f
(
x
)
f
′
(
x
)
=
x
2
(
1
-
x
)
4
≥
0
.
So f
(
x
) is increasing and multiplicatively convex in (
0,1
). By applying Theorem 16, we obtain the result.By using Theorem31 and Lemma 24, we obtain the following results.Corollary 32.
Letx
∈
(
0,1
)
n, for r
>
0 and k
=
1,2
,
…
,
n, one has
(59)
(
k
1
/
r
G
n
(
x
)
1
-
k
1
/
r
G
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
1
-
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
,
(
k
1
/
r
H
n
(
x
)
1
-
k
1
/
r
H
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
1
-
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
.In particular, takek
=
1 or k
=
n in (59), we get the following inequalities:
(60)
(
G
n
(
x
)
1
-
G
n
(
x
)
)
n
≤
∏
i
=
1
n
‍
x
i
1
-
x
i
,
(
H
n
(
x
)
1
-
H
n
(
x
)
)
n
≤
∏
i
=
1
n
‍
x
i
1
-
x
i
,
G
n
(
x
)
1
-
n
1
/
r
G
n
(
x
)
≤
M
r
(
x
)
1
-
n
1
/
r
M
r
(
x
)
,
H
n
(
x
)
1
-
n
1
/
r
H
n
(
x
)
≤
M
r
(
x
)
1
-
n
1
/
r
M
r
(
x
)
.Theorem 33.
Forr
>
0 and k
=
1,2
,
…
,
n, the symmetric functions
(61)
I
k
(
x
,
r
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
1
+
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
1
-
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
are Schur m-power convex on (
0,1
)
n when m
≤
0. In particular, I
k
(
x
,
r
) are Schur geometrically convex and Schur harmonically convex on (
0,1
)
n.Proof.
Letf
(
x
)
=
x
/
(
1
-
x
)
,
x
∈
(
0,1
). Then f
′
(
x
)
=
(
2
/
(
1
-
x
)
2
)
≥
0, and
(62)
x
[
f
(
x
)
f
′′
(
x
)
-
f
′
2
(
x
)
]
+
f
(
x
)
f
′
(
x
)
=
2
-
x
2
(
1
-
x
)
4
≥
0
.
So f
(
x
) is increasing and multiplicatively convex in (
0,1
). By applying Theorem 16, we obtain the result.By using Theorem33 and Lemma 24, we get the following inequalities.Corollary 34.
Letx
∈
(
0,1
)
n, for r
>
0 and k
=
1,2
,
…
,
n; one has
(63)
(
1
+
k
1
/
r
G
n
(
x
)
1
-
k
1
/
r
G
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
1
+
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
1
-
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
,
(
1
+
k
1
/
r
H
n
(
x
)
1
-
k
1
/
r
H
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
1
+
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
1
-
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
.In particular, takek
=
1 or k
=
n in (63); we get the following inequalities:
(64)
(
1
+
G
n
(
x
)
1
-
G
n
(
x
)
)
n
≤
∏
i
=
1
n
‍
1
+
x
i
1
-
x
i
,
(
1
+
H
n
(
x
)
1
-
H
n
(
x
)
)
n
≤
∏
i
=
1
n
‍
1
+
x
i
1
-
x
i
,
1
+
n
1
/
r
G
n
(
x
)
1
-
n
1
/
r
G
n
(
x
)
≤
1
+
n
1
/
r
M
r
(
x
)
1
-
n
1
/
r
M
r
(
x
)
,
1
+
n
1
/
r
H
n
(
x
)
1
-
n
1
/
r
H
n
(
x
)
≤
1
+
n
1
/
r
M
r
(
x
)
1
-
n
1
/
r
M
r
(
x
)
.Theorem 35.
Forr
>
0 and k
=
1,2
,
…
,
n, the symmetric functions
(65)
T
k
(
x
,
r
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
[
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
]
(
∑
j
=
1
r
x
i
j
r
)
1
/
r
are Schur m-power convex on (
e
-
1
,
+
∞
)
n when m
≤
0. In particular, T
k
(
x
,
r
) are Schur geometrically convex and Schur harmonically convex on (
e
-
1
,
+
∞
)
n.Proof.
Letf
(
x
)
=
x
x
,
x
∈
(
e
-
1
,
+
∞
). Then f
′
(
x
)
=
x
x
(
ln
⁡
x
+
1
)
≥
0, and
(66)
x
[
f
(
x
)
f
′′
(
x
)
-
f
′
2
(
x
)
]
+
f
(
x
)
f
′
(
x
)
=
f
2
(
x
)
(
ln
⁡
x
+
2
)
≥
0
.
So f
(
x
) is increasing and multiplicatively convex on (
e
-
1
,
+
∞
). By applying Theorem 16, we obtain the result.By using Theorem35 and Lemma 24, we get the following inequalities.Corollary 36.
Letx
∈
(
e
-
1
,
+
∞
)
n, for r
>
0 and k
=
1,2
,
…
,
n; one has
(67)
(
k
1
/
r
1
A
n
(
x
)
)
k
1
/
r
(
1
/
A
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
1
x
i
j
r
)
(
1
/
r
)
(
∑
j
=
1
k
(
1
/
x
i
j
r
)
)
1
/
r
,
(
k
1
/
r
G
n
(
x
)
)
k
1
/
r
G
n
(
x
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
x
i
j
r
)
(
1
/
r
)
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
,
(
k
1
/
r
H
n
(
x
)
)
k
1
/
r
H
n
(
x
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
x
i
j
r
)
(
1
/
r
)
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
.In particular, takek
=
1 or k
=
n in (67); we get the following inequalities:
(68)
(
1
A
n
(
x
)
)
n
/
A
n
(
x
)
≤
∏
i
=
1
n
‍
(
1
x
i
)
1
/
x
i
,
(
G
n
(
x
)
)
n
·
G
n
(
x
)
≤
∏
i
=
1
n
‍
x
i
x
i
,
(
H
n
(
x
)
)
n
·
H
n
(
x
)
≤
∏
i
=
1
n
‍
x
i
x
i
,
(
n
1
/
r
1
A
n
(
x
)
)
n
1
/
r
(
1
/
A
n
(
x
)
)
≤
(
n
1
/
r
M
r
(
1
x
)
)
n
1
/
r
M
r
(
1
/
x
)
,
(
n
1
/
r
G
n
(
x
)
)
n
1
/
r
G
n
(
x
)
≤
(
n
1
/
r
M
r
(
x
)
)
n
1
/
r
M
r
(
x
)
,
(
n
1
/
r
H
n
(
x
)
)
n
1
/
r
H
n
(
x
)
≤
(
n
1
/
r
M
r
(
x
)
)
n
1
/
r
M
r
(
x
)
.


### 4.2. Geometric Inequalities

In this section, some geometric inequalities ofn-dimensional simplex are established by use of the results of Theorem 16.Lots of geometric inequalities for ann-dimensional simplex are established (see [37–42]). In this section, applying the above Lemma and the main results in Section 2, we establish some interesting geometric inequalities on n-dimensional simplex in n-dimensional Euclidean space E
n.In what follows, LetΩ
=
{
A
1
,
A
2
,
…
,
A
n
+
1
} be an n-dimensional simplex in n-dimensional Euclidean space E
n  (
n
≥
2
) with V the volume. We denote by h
i, r
i, F
i
(
i
=
1,2
,
…
,
n
+
1
), and r the altitudes, the radii of excircles, the areas of lateral surfaces, and the inradius of Ω, respectively. For a given point P in Ω, let B
i stand for the intersection point of straight line A
i
P and hyperplane a
i
=
{
A
1
⋯
A
i
-
1
A
i
+
1
⋯
A
n
+
1
}.We first give some lemmas.Lemma 37.
LetΩ
=
{
A
1
,
A
2
,
…
,
A
n
+
1
} be an n-dimensional simplex in n-dimensional Euclidean space E
n  (
n
≥
2
). The following relation is
(69)
(
1
)
(
1
n
+
1
,
1
n
+
1
,
…
,
1
n
+
1
)
≺
(
r
h
1
,
r
h
2
,
…
,
r
h
n
+
1
)
;
(70)
(
2
)
(
n
-
1
n
+
1
,
n
-
1
n
+
1
,
…
,
n
-
1
n
+
1
)
≺
(
r
r
1
,
r
r
2
,
…
,
r
r
n
+
1
)
;
(71)
(
3
)
(
1
n
+
1
,
1
n
+
1
,
…
,
1
n
+
1
)
≺
(
r
1
h
1
+
2
r
1
,
r
2
h
2
+
2
r
2
,
…
,
r
n
+
1
h
n
+
1
+
2
r
n
+
1
)
;
(72)
(
4
)
(
1
n
+
1
,
1
n
+
1
,
…
,
1
n
+
1
)
≺
(
|
P
B
1
|
|
A
1
B
1
|
,
|
P
B
2
|
|
A
2
B
2
|
,
…
,
|
P
B
n
+
1
|
|
A
n
+
1
B
n
+
1
|
)
,
where F
=
∑
i
=
1
n
+
1
‍
F
i.Proof.
By the formula [37, 38]
(73)
F
i
h
i
=
n
V
,
∑
i
=
1
n
+
1
‍
F
i
r
=
n
V
,
∑
i
=
1
n
+
1
‍
(
F
-
2
F
i
)
r
i
=
n
V
,
where F
=
∑
i
=
1
n
+
1
‍
F
i. From these we get
(74)
∑
i
=
1
n
+
1
r
h
i
=
∑
i
=
1
n
+
1
‍
F
i
r
F
i
h
i
=
∑
i
=
1
n
+
1
F
i
r
n
V
=
n
V
n
V
=
1
,
(75)
∑
i
=
1
n
+
1
‍
r
r
i
=
∑
i
=
1
n
+
1
‍
(
F
-
2
F
i
)
r
(
F
-
2
F
i
)
r
i
=
∑
i
=
1
n
+
1
(
F
-
2
F
i
)
r
n
V
=
(
n
-
1
)
n
V
n
V
=
n
-
1
,
(76)
∑
i
=
1
n
+
1
‍
r
i
h
i
+
2
r
i
=
∑
i
=
1
n
+
1
‍
(
F
-
2
F
i
)
F
i
r
i
(
F
-
2
F
i
)
F
i
(
h
i
+
2
r
i
)
=
∑
i
=
1
n
+
1
‍
n
V
F
i
n
V
(
F
-
2
F
i
)
+
2
n
V
F
i
=
∑
i
=
1
n
+
1
F
i
F
=
1
.
By Definition3, and using (74), (75), and (76), respectively, we obtain (69), (70), and (71). In addition, (72) follows from Definition 3 and the fact that ∑
i
=
1
n
+
1
‍
|
P
B
i
|
/
|
A
i
B
i
|
=
1. So the proof is complete.Applying Lemma37 and Theorem 29, we establish some interesting geometric inequalities on n-dimensional simplex in n-dimensional Euclidean space E
n.Theorem 38.
LetΩ
=
{
A
1
,
A
2
,
…
,
A
n
+
1
} be an n-dimensional simplex in n-dimensional Euclidean space E
n  (
n
≥
2
). For t
>
0 and k
=
1,2
,
…
,
n, one has
(77)
(
r
·
k
1
/
t
(
n
+
1
)
)
C
n
+
1
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
h
i
j
t
)
1
/
t
;
(
r
·
k
1
/
t
n
+
1
n
-
1
)
C
n
+
1
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
r
i
j
t
)
1
/
t
;
(
r
·
k
1
/
t
(
n
+
1
)
)
C
n
+
1
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
[
∑
j
=
1
k
‍
(
h
i
j
+
2
r
i
j
r
i
j
)
t
]
1
/
t
;
(
r
·
k
1
/
t
(
n
+
1
)
)
C
n
+
1
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
[
∑
j
=
1
k
‍
(
|
P
A
i
j
|
|
P
B
i
j
|
)
t
]
1
/
t
.In particular, takek
=
1 in (77), we get the following inequality
(78)
r
(
n
+
1
)
≤
∏
i
=
1
n
+
1
‍
h
i
1
/
(
n
+
1
)
;
r
(
n
+
1
n
-
1
)
≤
∏
i
=
1
n
+
1
‍
r
i
1
/
(
n
+
1
)
;
(
n
+
1
)
≤
∏
i
=
1
n
+
1
‍
(
h
i
r
i
+
2
)
1
/
(
n
+
1
)
;
i
(
n
+
1
)
≤
∏
i
=
1
n
+
1
‍
(
|
P
B
i
|
P
A
i
|
)
1
/
(
n
+
1
)
.


## 4.1. Analytic Inequalities

To establish some analytic inequalities, we first give a lemma.Lemma 24 (see [20]).
Letx
i
>
0
,
i
=
1,2
,
…
,
n. Then the following relations is known:
(43)
(
A
n
(
x
)
,
A
n
(
x
)
,
…
,
A
n
(
x
)
)
≺
(
x
1
,
x
2
,
…
,
x
n
)
;
ln
⁡
(
G
n
(
x
)
,
G
n
(
x
)
,
…
,
G
n
(
x
)
)
≺
ln
⁡
(
x
1
,
x
2
,
…
,
x
n
)
;
(
1
H
n
(
x
)
,
1
H
n
(
x
)
,
…
,
1
H
n
(
x
)
)
≺
(
1
x
1
,
1
x
2
,
…
,
1
x
n
)
.By applying Lemma24 and Theorem 16, Corollaries 17 and 18, we can get the following several theorems and corollaries.Theorem 25.
Forr
>
0 and k
=
1,2
,
…
,
n, the symmetric functions
(44)
U
k
(
x
,
r
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
tan
⁡
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
are Schur m-power convex in R
+
+
n when m
≤
0. In particular, U
k
(
x
,
r
) are Schur geometrically convex and Schur harmonically convex in R
+
+
n.Proof.
Forx
∈
R
+
+, let f
(
x
)
=
tan
x. Then f
′
(
x
)
=
1
/
cos
⁡
⁡
2
x
=
sec
2
x
≥
0 and
(45)
x
[
f
(
x
)
f
′′
(
x
)
-
(
f
′
(
x
)
)
2
]
+
f
(
x
)
f
′
(
x
)
=
sec
2
x
(
x
sec
2
x
-
2
x
+
tan
x
)
.
Let g
(
x
)
=
x
sec
2
x
-
2
x
+
tan
x; then g
′
(
x
)
=
2
(
1
+
x
tan
x
)
sec
2
x
-
2
≥
0, and g
(
x
) is increasing. From the monotonicity of g
(
x
), we get that
(46)
x
[
f
(
x
)
f
′′
(
x
)
-
(
f
′
(
x
)
)
2
]
+
f
(
x
)
f
′
(
x
)
=
sec
2
x
g
(
x
)
≥
g
(
0
)
=
0
.
Therefore, f
(
x
) is increasing and multiplicatively convex in R
+
+
n. By applying Theorem 16, we obtain the result.By using Theorem25 and Lemma 24, we get the following inequalities.Corollary 26.
Forx
∈
R
+
+
n, then
(47)
∏
i
=
1
n
‍
(
tan
1
x
i
)
1
/
n
≥
tan
(
1
A
(
x
)
)
;
∏
i
=
1
n
‍
(
tan
x
i
)
1
/
n
≥
tan
G
(
x
)
;
∏
i
=
1
n
‍
(
tan
x
i
)
1
/
n
≥
tan
H
(
x
)
.Theorem 27.
Forr
>
0 and k
=
1,2
,
…
,
n, the symmetric functions
(48)
V
k
(
x
,
r
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
arcsin
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
are Schur m-power convex in (
0,1
)
n when m
≤
0. In particular, V
k
(
x
,
r
) are Schur geometrically convex and Schur harmonically convex in (
0,1
)
n.Proof.
Forx
∈
(
0,1
), let f
(
x
)
=
arcsin
x. Then f
′
(
x
)
=
1
/
1
-
x
2
≥
0 and
(49)
x
[
f
(
x
)
f
′′
(
x
)
-
(
f
′
(
x
)
)
2
]
+
f
(
x
)
f
′
(
x
)
=
arcsin
x
-
x
1
-
x
2
(
1
-
x
2
)
1
-
x
2
.
Let h
(
x
)
=
arctan
x
-
x
2
arctan
x
-
x; then h
′
(
x
)
=
2
x
2
/
1
-
x
2
≥
0, and h
(
x
) is increasing. From the monotonicity of h
(
x
), we get that
(50)
x
[
f
(
x
)
f
′′
(
x
)
-
(
f
′
(
x
)
)
2
]
+
f
(
x
)
f
′
(
x
)
=
h
(
x
)
(
1
-
x
2
)
1
-
x
2
≥
h
(
0
)
=
0
.
Therefore, f
(
x
) is increasing and multiplicatively convex in R
+
+
n. By applying Theorem 16, we obtain the result.By using Theorem27 and Lemma 24, we get the following inequalities.Corollary 28.
Forx
∈
(
0,1
)
n, then
(51)
∏
i
=
1
n
‍
(
arcsin
1
x
i
)
1
/
n
≥
arcsin
(
1
A
(
x
)
)
;
∏
i
=
1
n
‍
(
arcsin
x
i
)
1
/
n
≥
arcsin
G
(
x
)
;
∏
i
=
1
n
‍
(
arcsin
x
i
)
1
/
n
≥
arcsin
H
(
x
)
.Theorem 29.
Forr
>
0 and k
=
1,2
,
…
,
n, the symmetric functions
(52)
M
k
(
x
,
r
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
are Schur m-power convex in R
+
+
n when m
≤
0. In particular, M
k
(
x
,
r
) are Schur geometrically convex in R
+
+
n and Schur harmonically convex in R
+
+
n.Proof.
Letf
(
x
)
=
x
,
x
∈
R
+
+. We can easily see that f
(
x
) is increasing and multiplicatively convex in R
+
+. By applying Theorem 16, we obtain the result.By using Theorem29 and Lemma 24, we obtain the following results.Corollary 30.
Letx
∈
R
+
+, for r
>
0 and k
=
1,2
,
…
,
n; one has
(53)
(
k
1
/
r
1
A
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
1
x
i
j
r
)
1
/
r
,
(54)
(
k
1
/
r
G
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
,
(55)
(
k
1
/
r
H
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
.In particular, takek
=
1 in (53); we get the known A-G-H inequality. And take k
=
n in (53); we get the following inequalities:
(56)
(
∑
i
=
1
n
‍
1
x
i
r
)
1
/
r
·
∑
i
=
1
n
‍
x
i
≥
n
1
/
r
+
1
.Theorem 31.
Forr
>
0 and k
=
1,2
,
…
,
n, the symmetric functions
(57)
N
k
(
x
,
r
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
1
-
(
∑
j
=
1
k
x
i
j
k
)
1
/
r
are Schur m-power convex on (
0,1
)
n when m
≤
0. In particular, N
k
(
x
,
r
) are Schur geometrically convex on (
0,1
)
n and Schur harmonically convex on (
0,1
)
n.Proof.
Letf
(
x
)
=
x
/
(
1
-
x
)
,
x
∈
(
0,1
). Then f
′
(
x
)
=
(
1
/
(
1
-
x
)
2
)
≥
0, and
(58)
x
[
f
(
x
)
f
′′
(
x
)
-
f
′
2
(
x
)
]
+
f
(
x
)
f
′
(
x
)
=
x
2
(
1
-
x
)
4
≥
0
.
So f
(
x
) is increasing and multiplicatively convex in (
0,1
). By applying Theorem 16, we obtain the result.By using Theorem31 and Lemma 24, we obtain the following results.Corollary 32.
Letx
∈
(
0,1
)
n, for r
>
0 and k
=
1,2
,
…
,
n, one has
(59)
(
k
1
/
r
G
n
(
x
)
1
-
k
1
/
r
G
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
1
-
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
,
(
k
1
/
r
H
n
(
x
)
1
-
k
1
/
r
H
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
1
-
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
.In particular, takek
=
1 or k
=
n in (59), we get the following inequalities:
(60)
(
G
n
(
x
)
1
-
G
n
(
x
)
)
n
≤
∏
i
=
1
n
‍
x
i
1
-
x
i
,
(
H
n
(
x
)
1
-
H
n
(
x
)
)
n
≤
∏
i
=
1
n
‍
x
i
1
-
x
i
,
G
n
(
x
)
1
-
n
1
/
r
G
n
(
x
)
≤
M
r
(
x
)
1
-
n
1
/
r
M
r
(
x
)
,
H
n
(
x
)
1
-
n
1
/
r
H
n
(
x
)
≤
M
r
(
x
)
1
-
n
1
/
r
M
r
(
x
)
.Theorem 33.
Forr
>
0 and k
=
1,2
,
…
,
n, the symmetric functions
(61)
I
k
(
x
,
r
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
1
+
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
1
-
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
are Schur m-power convex on (
0,1
)
n when m
≤
0. In particular, I
k
(
x
,
r
) are Schur geometrically convex and Schur harmonically convex on (
0,1
)
n.Proof.
Letf
(
x
)
=
x
/
(
1
-
x
)
,
x
∈
(
0,1
). Then f
′
(
x
)
=
(
2
/
(
1
-
x
)
2
)
≥
0, and
(62)
x
[
f
(
x
)
f
′′
(
x
)
-
f
′
2
(
x
)
]
+
f
(
x
)
f
′
(
x
)
=
2
-
x
2
(
1
-
x
)
4
≥
0
.
So f
(
x
) is increasing and multiplicatively convex in (
0,1
). By applying Theorem 16, we obtain the result.By using Theorem33 and Lemma 24, we get the following inequalities.Corollary 34.
Letx
∈
(
0,1
)
n, for r
>
0 and k
=
1,2
,
…
,
n; one has
(63)
(
1
+
k
1
/
r
G
n
(
x
)
1
-
k
1
/
r
G
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
1
+
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
1
-
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
,
(
1
+
k
1
/
r
H
n
(
x
)
1
-
k
1
/
r
H
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
1
+
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
1
-
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
.In particular, takek
=
1 or k
=
n in (63); we get the following inequalities:
(64)
(
1
+
G
n
(
x
)
1
-
G
n
(
x
)
)
n
≤
∏
i
=
1
n
‍
1
+
x
i
1
-
x
i
,
(
1
+
H
n
(
x
)
1
-
H
n
(
x
)
)
n
≤
∏
i
=
1
n
‍
1
+
x
i
1
-
x
i
,
1
+
n
1
/
r
G
n
(
x
)
1
-
n
1
/
r
G
n
(
x
)
≤
1
+
n
1
/
r
M
r
(
x
)
1
-
n
1
/
r
M
r
(
x
)
,
1
+
n
1
/
r
H
n
(
x
)
1
-
n
1
/
r
H
n
(
x
)
≤
1
+
n
1
/
r
M
r
(
x
)
1
-
n
1
/
r
M
r
(
x
)
.Theorem 35.
Forr
>
0 and k
=
1,2
,
…
,
n, the symmetric functions
(65)
T
k
(
x
,
r
)
=
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
[
(
∑
j
=
1
k
‍
x
i
j
r
)
1
/
r
]
(
∑
j
=
1
r
x
i
j
r
)
1
/
r
are Schur m-power convex on (
e
-
1
,
+
∞
)
n when m
≤
0. In particular, T
k
(
x
,
r
) are Schur geometrically convex and Schur harmonically convex on (
e
-
1
,
+
∞
)
n.Proof.
Letf
(
x
)
=
x
x
,
x
∈
(
e
-
1
,
+
∞
). Then f
′
(
x
)
=
x
x
(
ln
⁡
x
+
1
)
≥
0, and
(66)
x
[
f
(
x
)
f
′′
(
x
)
-
f
′
2
(
x
)
]
+
f
(
x
)
f
′
(
x
)
=
f
2
(
x
)
(
ln
⁡
x
+
2
)
≥
0
.
So f
(
x
) is increasing and multiplicatively convex on (
e
-
1
,
+
∞
). By applying Theorem 16, we obtain the result.By using Theorem35 and Lemma 24, we get the following inequalities.Corollary 36.
Letx
∈
(
e
-
1
,
+
∞
)
n, for r
>
0 and k
=
1,2
,
…
,
n; one has
(67)
(
k
1
/
r
1
A
n
(
x
)
)
k
1
/
r
(
1
/
A
n
(
x
)
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
1
x
i
j
r
)
(
1
/
r
)
(
∑
j
=
1
k
(
1
/
x
i
j
r
)
)
1
/
r
,
(
k
1
/
r
G
n
(
x
)
)
k
1
/
r
G
n
(
x
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
x
i
j
r
)
(
1
/
r
)
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
,
(
k
1
/
r
H
n
(
x
)
)
k
1
/
r
H
n
(
x
)
C
n
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
x
i
j
r
)
(
1
/
r
)
(
∑
j
=
1
k
x
i
j
r
)
1
/
r
.In particular, takek
=
1 or k
=
n in (67); we get the following inequalities:
(68)
(
1
A
n
(
x
)
)
n
/
A
n
(
x
)
≤
∏
i
=
1
n
‍
(
1
x
i
)
1
/
x
i
,
(
G
n
(
x
)
)
n
·
G
n
(
x
)
≤
∏
i
=
1
n
‍
x
i
x
i
,
(
H
n
(
x
)
)
n
·
H
n
(
x
)
≤
∏
i
=
1
n
‍
x
i
x
i
,
(
n
1
/
r
1
A
n
(
x
)
)
n
1
/
r
(
1
/
A
n
(
x
)
)
≤
(
n
1
/
r
M
r
(
1
x
)
)
n
1
/
r
M
r
(
1
/
x
)
,
(
n
1
/
r
G
n
(
x
)
)
n
1
/
r
G
n
(
x
)
≤
(
n
1
/
r
M
r
(
x
)
)
n
1
/
r
M
r
(
x
)
,
(
n
1
/
r
H
n
(
x
)
)
n
1
/
r
H
n
(
x
)
≤
(
n
1
/
r
M
r
(
x
)
)
n
1
/
r
M
r
(
x
)
.


## 4.2. Geometric Inequalities

In this section, some geometric inequalities ofn-dimensional simplex are established by use of the results of Theorem 16.Lots of geometric inequalities for ann-dimensional simplex are established (see [37–42]). In this section, applying the above Lemma and the main results in Section 2, we establish some interesting geometric inequalities on n-dimensional simplex in n-dimensional Euclidean space E
n.In what follows, LetΩ
=
{
A
1
,
A
2
,
…
,
A
n
+
1
} be an n-dimensional simplex in n-dimensional Euclidean space E
n  (
n
≥
2
) with V the volume. We denote by h
i, r
i, F
i
(
i
=
1,2
,
…
,
n
+
1
), and r the altitudes, the radii of excircles, the areas of lateral surfaces, and the inradius of Ω, respectively. For a given point P in Ω, let B
i stand for the intersection point of straight line A
i
P and hyperplane a
i
=
{
A
1
⋯
A
i
-
1
A
i
+
1
⋯
A
n
+
1
}.We first give some lemmas.Lemma 37.
LetΩ
=
{
A
1
,
A
2
,
…
,
A
n
+
1
} be an n-dimensional simplex in n-dimensional Euclidean space E
n  (
n
≥
2
). The following relation is
(69)
(
1
)
(
1
n
+
1
,
1
n
+
1
,
…
,
1
n
+
1
)
≺
(
r
h
1
,
r
h
2
,
…
,
r
h
n
+
1
)
;
(70)
(
2
)
(
n
-
1
n
+
1
,
n
-
1
n
+
1
,
…
,
n
-
1
n
+
1
)
≺
(
r
r
1
,
r
r
2
,
…
,
r
r
n
+
1
)
;
(71)
(
3
)
(
1
n
+
1
,
1
n
+
1
,
…
,
1
n
+
1
)
≺
(
r
1
h
1
+
2
r
1
,
r
2
h
2
+
2
r
2
,
…
,
r
n
+
1
h
n
+
1
+
2
r
n
+
1
)
;
(72)
(
4
)
(
1
n
+
1
,
1
n
+
1
,
…
,
1
n
+
1
)
≺
(
|
P
B
1
|
|
A
1
B
1
|
,
|
P
B
2
|
|
A
2
B
2
|
,
…
,
|
P
B
n
+
1
|
|
A
n
+
1
B
n
+
1
|
)
,
where F
=
∑
i
=
1
n
+
1
‍
F
i.Proof.
By the formula [37, 38]
(73)
F
i
h
i
=
n
V
,
∑
i
=
1
n
+
1
‍
F
i
r
=
n
V
,
∑
i
=
1
n
+
1
‍
(
F
-
2
F
i
)
r
i
=
n
V
,
where F
=
∑
i
=
1
n
+
1
‍
F
i. From these we get
(74)
∑
i
=
1
n
+
1
r
h
i
=
∑
i
=
1
n
+
1
‍
F
i
r
F
i
h
i
=
∑
i
=
1
n
+
1
F
i
r
n
V
=
n
V
n
V
=
1
,
(75)
∑
i
=
1
n
+
1
‍
r
r
i
=
∑
i
=
1
n
+
1
‍
(
F
-
2
F
i
)
r
(
F
-
2
F
i
)
r
i
=
∑
i
=
1
n
+
1
(
F
-
2
F
i
)
r
n
V
=
(
n
-
1
)
n
V
n
V
=
n
-
1
,
(76)
∑
i
=
1
n
+
1
‍
r
i
h
i
+
2
r
i
=
∑
i
=
1
n
+
1
‍
(
F
-
2
F
i
)
F
i
r
i
(
F
-
2
F
i
)
F
i
(
h
i
+
2
r
i
)
=
∑
i
=
1
n
+
1
‍
n
V
F
i
n
V
(
F
-
2
F
i
)
+
2
n
V
F
i
=
∑
i
=
1
n
+
1
F
i
F
=
1
.
By Definition3, and using (74), (75), and (76), respectively, we obtain (69), (70), and (71). In addition, (72) follows from Definition 3 and the fact that ∑
i
=
1
n
+
1
‍
|
P
B
i
|
/
|
A
i
B
i
|
=
1. So the proof is complete.Applying Lemma37 and Theorem 29, we establish some interesting geometric inequalities on n-dimensional simplex in n-dimensional Euclidean space E
n.Theorem 38.
LetΩ
=
{
A
1
,
A
2
,
…
,
A
n
+
1
} be an n-dimensional simplex in n-dimensional Euclidean space E
n  (
n
≥
2
). For t
>
0 and k
=
1,2
,
…
,
n, one has
(77)
(
r
·
k
1
/
t
(
n
+
1
)
)
C
n
+
1
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
h
i
j
t
)
1
/
t
;
(
r
·
k
1
/
t
n
+
1
n
-
1
)
C
n
+
1
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
(
∑
j
=
1
k
‍
r
i
j
t
)
1
/
t
;
(
r
·
k
1
/
t
(
n
+
1
)
)
C
n
+
1
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
[
∑
j
=
1
k
‍
(
h
i
j
+
2
r
i
j
r
i
j
)
t
]
1
/
t
;
(
r
·
k
1
/
t
(
n
+
1
)
)
C
n
+
1
k
≤
∏
1
≤
i
1
<
i
2
<
⋯
<
i
k
≤
n
‍
[
∑
j
=
1
k
‍
(
|
P
A
i
j
|
|
P
B
i
j
|
)
t
]
1
/
t
.In particular, takek
=
1 in (77), we get the following inequality
(78)
r
(
n
+
1
)
≤
∏
i
=
1
n
+
1
‍
h
i
1
/
(
n
+
1
)
;
r
(
n
+
1
n
-
1
)
≤
∏
i
=
1
n
+
1
‍
r
i
1
/
(
n
+
1
)
;
(
n
+
1
)
≤
∏
i
=
1
n
+
1
‍
(
h
i
r
i
+
2
)
1
/
(
n
+
1
)
;
i
(
n
+
1
)
≤
∏
i
=
1
n
+
1
‍
(
|
P
B
i
|
P
A
i
|
)
1
/
(
n
+
1
)
.



---

*Source: 258108-2014-05-20.xml*