An extension of Jensen’s discrete inequality to partially convex functions

Vasile Cirtoaje

doi:10.1186/1029-242X-2013-54

Research
Open Access

An extension of Jensen’s discrete inequality to partially convex functions

Vasile Cirtoaje¹Email author

Journal of Inequalities and Applications20132013:54

https://doi.org/10.1186/1029-242X-2013-54

Received: 16 October 2012
Accepted: 30 January 2013
Published: 18 February 2013

Abstract

This paper deals with a new extension of Jensen’s discrete inequality to a partially convex function f, which is defined on a real interval $I$ , convex on a subinterval $[a, b] \subset I$ , decreasing for $u \leq c$ and increasing for $u \geq c$ , where $c \in [a, b]$ . Several relevant applications are given to show the effectiveness of the proposed partially convex function theorem.

MSC:26D07, 26D10, 41A44.

Keywords

Jensen’s discrete inequality
partially convex function
increasing/decreasing function

1 Introduction

Let

x = {x_{1}, x_{2}, \dots, x_{n}}

be a sequence of real numbers belonging to a given real interval

I

, and let

p = {p_{1}, p_{2}, \dots, p_{n}}

be a sequence of given positive weights associated to x and satisfying

p_{1} + p_{2} + \dots + p_{n} = 1

. If f is a convex function on

I

, then

\sum_{i = 1}^{n} p_{i} f (x_{i}) \geq f (\sum_{i = 1}^{n} p_{i} x_{i})

is the classical Jensen discrete inequality (see [1, 2]).

In [3], we extended the weighted Jensen discrete inequality to a half convex function f, defined on a real interval $I$ and convex for $u \leq s$ or $u \geq s$ , where $s \in I$ .

WHCF-Theorem Let f be a function defined on a real interval

I

and convex for

u \leq s

or

u \geq s

, where

s \in I

, and let

p_{1}, p_{2}, \dots, p_{n}

be positive real numbers such that

p = min {p_{1}, p_{2}, \dots, p_{n}}, p_{1} + p_{2} + \dots + p_{n} = 1 .

The inequality

p_{1} f (x_{1}) + p_{2} f (x_{2}) + \dots + p_{n} f (x_{n}) \geq f (s)

holds for all

x_{1}, x_{2}, \dots, x_{n} \in I

satisfying

p_{1} x_{1} + p_{2} x_{2} + \dots + p_{n} x_{n} = s

if and only if

p f (x) + (1 - p) f (y) \geq f (s)

for all $x, y \in I$ such that $p x + (1 - p) y = s$ .

For the particular case $p_{1} = p_{2} = \dots = p_{n} = 1 / n$ , from the weighted half convex function theorem, we get the half convex function theorem (see [4, 5]).

HCF-Theorem Let f be a function defined on a real interval

I

and convex for

u \leq s

or

u \geq s

, where

s \in I

. The inequality

f (x_{1}) + f (x_{2}) + \dots + f (x_{n}) \geq n f (s)

holds for all

x_{1}, x_{2}, \dots, x_{n} \in I

satisfying

x_{1} + x_{2} + \dots + x_{n} = n s

if and only if

f (x) + (n - 1) f (y) \geq n f (s)

for all $x, y \in I$ which satisfy $x + (n - 1) y = n s$ .

Applying HCF-Theorem and WHCF-Theorem to the function f defined by $f (u) = g (e^{u})$ and replacing s by lnr, x by lnx, y by lny, and each $x_{i}$ by $ln a_{i}$ for $i = 1, 2, \dots, n$ , we get the following corollaries, respectively.

HCF-Corollary Let g be a function defined on a positive interval

I

such that the function f defined by

f (u) = g (e^{u})

is convex for

e^{u} \leq r

or

e^{u} \geq r

, where

r \in I

. The inequality

g (a_{1}) + g (a_{2}) + \dots + g (a_{n}) \geq n g (r)

holds for all

a_{1}, a_{2}, \dots, a_{n} \in I

satisfying

a_{1} a_{2} \dots a_{n} = r^{n}

if and only if

g (a) + (n - 1) g (b) \geq n g (r)

for all $a, b \in I$ which satisfy $a b^{n - 1} = r^{n}$ .

WHCF-Corollary Let g be a function defined on a positive interval

I

such that the function f defined by

f (u) = g (e^{u})

is convex for

e^{u} \leq r

or

e^{u} \geq r

, where

r \in I

, and let

p_{1}, p_{2}, \dots, p_{n}

be positive real numbers such that

p = min {p_{1}, p_{2}, \dots, p_{n}}, p_{1} + p_{2} + \dots + p_{n} = 1 .

The inequality

p_{1} g (a_{1}) + p_{2} g (a_{2}) + \dots + p_{n} g (a_{n}) \geq g (r)

holds for all

a_{1}, a_{2}, \dots, a_{n} \in I

satisfying

a_{1}^{p_{1}} a_{2}^{p_{2}} \dots a_{n}^{p_{n}} = r

if and only if

p g (a) + (1 - p) g (b) \geq g (r)

for all $a, b \in I$ such that $a^{p} b^{1 - p} = r$ .

In this paper, we will use HCF-Theorem and WHCF-Theorem to extend Jensen’s inequality to partially convex functions, which are defined on a real interval $I$ and convex only on a subinterval $[a, b] \subset I$ .

Remark 1.1 Clearly, HCF-Theorem is a particular case of WHCF-Theorem. However, we posted here the both theorems because HCF-Theorem is much more useful to prove many inequalities of extended Jensen type.

Remark 1.2

Actually, in HCF-Theorem and WHCF-Theorem, it suffices to consider that

x \geq s \geq y

when f is convex for

u \leq s

, and

x \leq s \leq y

when f is convex for

u \geq s

(see [3]). Also, in HCF-Corollary and WHCF-Corollary, it suffices to consider that

a \geq r \geq b

when f is convex for

e^{u} \leq r

, and

a \leq r \leq b

when f is convex for $e^{u} \geq r$ .

2 Main results

The main results of the paper are given by the following two theorems: partially convex function theorem (PCF-Theorem) and weighted partially convex function theorem (WPCF-Theorem).

PCF-Theorem Let f be a function defined on a real interval

I

, decreasing for

u \leq s_{0}

and increasing for

u \geq s_{0}

, where

s_{0} \in I

. In addition, assume that f is convex on

[s_{0}, s]

or

[s, s_{0}]

, where

s \in I

. The inequality

f (x_{1}) + f (x_{2}) + \dots + f (x_{n}) \geq n f (s)

holds for all

x_{1}, x_{2}, \dots, x_{n} \in I

satisfying

x_{1} + x_{2} + \dots + x_{n} = n s

if and only if

f (x) + (n - 1) f (y) \geq n f (s)

for all $x, y \in I$ which satisfy $x + (n - 1) y = n s$ .

WPCF-Theorem Let f be a function defined on a real interval

I

, decreasing for

u \leq s_{0}

and increasing for

u \geq s_{0}

, where

s_{0} \in I

, and let

p_{1}, p_{2}, \dots, p_{n}

be positive real numbers such that

p = min {p_{1}, p_{2}, \dots, p_{n}}, p_{1} + p_{2} + \dots + p_{n} = 1 .

In addition, assume that f is convex on

[s_{0}, s]

or

[s, s_{0}]

, where

s \in I

. The inequality

p_{1} f (x_{1}) + p_{2} f (x_{2}) + \dots + p_{n} f (x_{n}) \geq f (s)

holds for all

x_{1}, x_{2}, \dots, x_{n} \in I

satisfying

p_{1} x_{1} + p_{2} x_{2} + \dots + p_{n} x_{n} = s

if and only if

p f (x) + (1 - p) f (y) \geq f (s)

for all $x, y \in I$ such that $p x + (1 - p) y = s$ .

Applying PCF-Theorem and WPCF-Theorem to the function f defined by $f (u) = g (e^{u})$ and replacing $s_{0}$ by $ln r_{0}$ , s by lnr, x by lnx, y by lny, and each $x_{i}$ by $ln a_{i}$ for $i = 1, 2, \dots, n$ , we get the following corollaries, respectively.

PCF-Corollary Let g be a function defined on a positive interval

I

, decreasing for

t \leq r_{0}

and increasing for

t \geq r_{0}

, where

r_{0} \in I

. In addition, assume that the function f defined by

f (u) = g (e^{u})

is convex for

r_{0} \leq e^{u} \leq r

or

r \leq e^{u} \leq r_{0}

, where

r \in I

. The inequality

g (a_{1}) + g (a_{2}) + \dots + g (a_{n}) \geq n g (r)

holds for all

a_{1}, a_{2}, \dots, a_{n} \in I

satisfying

a_{1} a_{2} \dots a_{n} = r^{n}

if and only if

g (a) + (n - 1) g (b) \geq n g (r)

for all $a, b \in I$ which satisfy $a b^{n - 1} = r^{n}$ .

WPCF-Corollary Let g be a function defined on a positive interval

I

, decreasing for

t \leq r_{0}

and increasing for

t \geq r_{0}

, where

r_{0} \in I

, and let

p_{1}, p_{2}, \dots, p_{n}

be positive real numbers such that

p = min {p_{1}, p_{2}, \dots, p_{n}}, p_{1} + p_{2} + \dots + p_{n} = 1 .

In addition, assume that the function f defined by

f (u) = g (e^{u})

is convex for

r_{0} \leq e^{u} \leq r

or

r \leq e^{u} \leq r_{0}

, where

r \in I

. The inequality

p_{1} g (a_{1}) + p_{2} g (a_{2}) + \dots + p_{n} g (a_{n}) \geq g (r)

holds for all

a_{1}, a_{2}, \dots, a_{n} \in I

satisfying

a_{1}^{p_{1}} a_{2}^{p_{2}} \dots a_{n}^{p_{n}} = r

if and only if

p g (a) + (1 - p) g (b) \geq g (r)

for all $a, b \in I$ such that $a^{p} b^{1 - p} = r$ .

In order to prove WPCF-Theorem, we need the following lemmas.

Lemma 2.1 Let f be a function defined on a real interval

I

, decreasing for

u \leq s_{0}

and increasing for

u \geq s_{0}

, where

s_{0} \in I

, and let

p_{1}, p_{2}, \dots, p_{n}

be positive real numbers such that

p_{1} + p_{2} + \dots + p_{n} = 1 .

For

s \in I

,

s \geq s_{0}

, if the inequality

p_{1} f (x_{1}) + p_{2} f (x_{2}) + \dots + p_{n} f (x_{n}) \geq f (s)

holds for all

x_{1}, x_{2}, \dots, x_{n} \in I

such that

x_{1}, x_{2}, \dots, x_{n} \geq s_{0}, p_{1} x_{1} + p_{2} x_{2} + \dots + p_{n} x_{n} = s,

then it holds for all

x_{1}, x_{2}, \dots, x_{n} \in I

such that

p_{1} x_{1} + p_{2} x_{2} + \dots + p_{n} x_{n} = s .

Lemma 2.2 Let f be a function defined on a real interval

I

, decreasing for

u \leq s_{0}

and increasing for

u \geq s_{0}

, where

s_{0} \in I

, and let

p_{1}, p_{2}, \dots, p_{n}

be positive real numbers such that

p_{1} + p_{2} + \dots + p_{n} = 1 .

For

s \in I

,

s \leq s_{0}

, if the inequality

p_{1} f (x_{1}) + p_{2} f (x_{2}) + \dots + p_{n} f (x_{n}) \geq f (s)

holds for all

x_{1}, x_{2}, \dots, x_{n} \in I

such that

x_{1}, x_{2}, \dots, x_{n} \leq s_{0}, p_{1} x_{1} + p_{2} x_{2} + \dots + p_{n} x_{n} = s,

then it holds for all

x_{1}, x_{2}, \dots, x_{n} \in I

such that

p_{1} x_{1} + p_{2} x_{2} + \dots + p_{n} x_{n} = s .

Notice that in the case

s \geq s_{0}

, WPCF-Theorem follows immediately from Lemma 2.1 and WHCF-Theorem applied to the interval

I_{0} = {u \in I ∣ u \geq s_{0}},

because f is convex for

u \in I_{0}

,

u \leq s

. Also, in the case

s \leq s_{0}

, WPCF-Theorem follows immediately from Lemma 2.2 and WHCF-Theorem applied to the interval

I_{0} = {u \in I ∣ u \leq s_{0}},

because f is convex for $u \in I_{0}$ , $u \geq s$ .

Remark 2.3

According to Remark 1.2, it suffices to consider in PCF-Theorem and WPCF-Theorem that

x \geq s \geq y

when f is convex on

[s_{0}, s]

, and

x \leq s \leq y

when f is convex on

[s, s_{0}]

. Also, it suffices to consider in PCF-Corollary and WPCF-Corollary that

a \geq r \geq b

when f is convex for

r_{0} \leq e^{u} \leq r

, and

a \leq r \leq b

when f is convex for $r \leq e^{u} \leq r_{0}$ .

Remark 2.4

Let us denote

g (u) = \frac{f (u) - f (s)}{u - s}, h (x, y) = \frac{g (x) - g (y)}{x - y} .

In many applications, it is useful to replace the hypothesis

p f (x) + (1 - p) f (y) \geq f (s)

in WHCF-Theorem and WPCF-Theorem by the equivalent condition:

h (x, y) \geq 0 \forall x, y \in I, p x + (1 - p) y = s .

This equivalence is true since

\begin{aligned} p f (x) + (1 - p) f (y) - f (s) & = p [f (x) - f (s)] + (1 - p) [f (y) - f (s)] \\ = p (x - s) g (x) + (1 - p) (y - s) g (y) \\ = p (1 - p) (x - y) [g (x) - g (y)] \\ = p (1 - p) {(x - y)}^{2} h (x, y) . \end{aligned}

In the particular case

p_{1} = p_{2} = \dots = p_{n} = 1 / n

, this condition becomes

h (x, y) \geq 0 \forall x, y \in I, x + (n - 1) y = n s .

Remark 2.5 The required inequalities in WHCF-Theorem and WPCF-Theorem turn into equalities for

x_{1} = x_{2} = \dots = x_{n} = s

. In addition, on the assumption that

p = min {p_{1}, p_{2}, \dots, p_{n}},

the equality also holds for

x_{1} = x

and

x_{2} = \dots = x_{n} = y

if there exist

x, y \in I

,

x \neq y

, such that

p x + (1 - p) y = s, p f (x) + (1 - p) f (y) = f (s) .

3 Proof of lemmas

Proof of Lemma 2.1 For

i = 1, 2, \dots, n

, define the numbers

y_{i} \in I

as

y_{i} = {\begin{cases} s_{0}, & x_{i} \leq s_{0}, \\ x_{i}, & x_{i} > s_{0} . \end{cases}

Since

y_{i} \geq x_{i}

for

i = 1, 2, \dots, n

, we have

p_{1} y_{1} + p_{2} y_{2} + \dots + p_{n} y_{n} \geq p_{1} x_{1} + p_{2} x_{2} + \dots + p_{n} x_{n} = s .

In addition, since

f (y_{i}) \leq f (x_{i})

for

x_{i} \leq s_{0}

and

f (y_{i}) = f (x_{i})

for

x_{i} > s_{0}

, we have

f (y_{i}) \leq f (x_{i})

for

i = 1, 2, \dots, n

, and hence

p_{1} f (y_{1}) + p_{2} f (y_{2}) + \dots + p_{n} f (y_{n}) \leq p_{1} f (x_{1}) + p_{2} f (x_{2}) + \dots + p_{n} f (x_{n}) .

Thus, it suffices to show that

p_{1} f (y_{1}) + p_{2} f (y_{2}) + \dots + p_{n} f (y_{n}) \geq f (s)

for all $y_{1}, y_{2}, \dots, y_{n} \geq s_{0}$ such that $p_{1} y_{1} + p_{2} y_{2} + \dots + p_{n} y_{n} \geq s$ . By hypothesis, this inequality is true for $y_{1}, y_{2}, \dots, y_{n} \geq s_{0}$ and $p_{1} y_{1} + p_{2} y_{2} + \dots + p_{n} y_{n} = s$ . Since f is increasing for $u \in I$ , $u \geq s_{0}$ , the more we have $p_{1} f (y_{1}) + p_{2} f (y_{2}) + \dots + p_{n} f (y_{n}) \geq f (s)$ for $y_{1}, y_{2}, \dots, y_{n} \geq s_{0}$ and $p_{1} y_{1} + p_{2} y_{2} + \dots + p_{n} y_{n} \geq s$ . □

Proof of Lemma 2.2 For

i = 1, 2, \dots, n

, define the numbers

y_{i} \in I

as follows:

y_{i} = {\begin{cases} x_{i}, & x_{i} \leq s_{0}, \\ s_{0}, & x_{i} > s_{0} . \end{cases}

We have

y_{i} \leq s_{0}

,

y_{i} \leq x_{i}

and

f (y_{i}) \leq f (x_{i})

for

i = 1, 2, \dots, n

. Therefore,

p_{1} y_{1} + p_{2} y_{2} + \dots + p_{n} y_{n} \leq p_{1} x_{1} + p_{2} x_{2} + \dots + p_{n} x_{n} = s

and

p_{1} f (y_{1}) + p_{2} f (y_{2}) + \dots + p_{n} f (y_{n}) \leq p_{1} f (x_{1}) + p_{2} f (x_{2}) + \dots + p_{n} f (x_{n}) .

Thus, it suffices to show that

p_{1} f (y_{1}) + p_{2} f (y_{2}) + \dots + p_{n} f (y_{n}) \geq f (s)

for all $y_{1}, y_{2}, \dots, y_{n} \leq s_{0}$ such that $p_{1} y_{1} + p_{2} y_{2} + \dots + p_{n} y_{n} \leq s$ . By hypothesis, this inequality is true for $y_{1}, y_{2}, \dots, y_{n} \leq s_{0}$ and $p_{1} y_{1} + p_{2} y_{2} + \dots + p_{n} y_{n} = s$ . Since f is decreasing for $u \in I$ , $u \leq s_{0}$ , we have also $p_{1} f (y_{1}) + p_{2} f (y_{2}) + \dots + p_{n} f (y_{n}) \geq f (s)$ for $y_{1}, y_{2}, \dots, y_{n} \leq s_{0}$ and $p_{1} y_{1} + p_{2} y_{2} + \dots + p_{n} y_{n} \leq s$ . □

4 Applications

Application 4.1 Let

x_{1}, x_{2}, \dots, x_{n} \geq \frac{- n}{n - 2}

(

n \geq 3

) such that

x_{1} + x_{2} + \dots + x_{n} = n .

If

k \geq \frac{n (3 n - 4)}{{(n - 2)}^{2}}

, then

\frac{1 - x_{1}}{k + x_{1}^{2}} + \frac{1 - x_{2}}{k + x_{2}^{2}} + \dots + \frac{1 - x_{n}}{k + x_{n}^{2}} \geq 0,

with equality for $x_{1} = x_{2} = \dots = x_{n} = 1$ , and also for $x_{1} = \frac{- n}{n - 2}$ and $x_{2} = \dots = x_{n} = \frac{n}{n - 2}$ (or any cyclic permutation).

Proof

Rewrite the desired inequality as

f (x_{1}) + f (x_{2}) + \dots + f (x_{n}) \geq n f (s),

where

\begin{matrix} s = 1, \\ f (u) = \frac{1 - u}{k + u^{2}}, u \in I = [\frac{- n}{n - 2}, \frac{n (2 n - 3)}{n - 2}] . \end{matrix}

We have

\begin{matrix} f^{'} (u) = \frac{u^{2} - 2 u - k}{{(u^{2} + k)}^{2}}, \\ f^{″} (u) = \frac{2 f_{1} (u)}{{(u^{2} + k)}^{3}}, \end{matrix}

where

f_{1} (u) = - u^{3} + 3 u^{2} + 3 k u - k = (k + 1) (3 u - 1) - {(u - 1)}^{3} .

There are two cases to consider.

Case 1:

\sqrt{k + 1} \geq \frac{2 {(n - 1)}^{2}}{n - 2}

. For

u \in I

,

u \geq 1

, we have

f_{1} (u) > (k + 1) (u - 1) - {(u - 1)}^{3} = (u - 1) [k + 1 - {(u - 1)}^{2}] \geq 0,

since

u - 1 \leq \frac{n (2 n - 3)}{n - 2} - 1 = \frac{2 {(n - 1)}^{2}}{n - 2} \leq \sqrt{k + 1} .

Therefore, f is convex for

u \in I

,

u \geq 1

. By HCF-Theorem, we only need to show that

f (x) + (n - 1) f (y) \geq n f (1)

for all

x, y \in I

which satisfy

x + (n - 1) y = n

. According to Remark 2.4, this is true if

h (x, y) \geq 0

for

x, y \in I

and

x + (n - 1) y = n

, where

h (x, y) = \frac{g (x) - g (y)}{x - y}, g (u) = \frac{f (u) - f (1)}{u - 1} .

Indeed, we have

g (u) = \frac{- 1}{u^{2} + k}

and

h (x, y) = \frac{x + y}{(x^{2} + k) (y^{2} + k)} = \frac{n + (n - 2) x}{(n - 1) (x^{2} + k) (y^{2} + k)} \geq 0 .

Case 2:

\frac{2 (n - 1)}{n - 2} \leq \sqrt{k + 1} < \frac{2 {(n - 1)}^{2}}{n - 2}

. Since

1 - \sqrt{1 + k} \leq \frac{- n}{n - 2}

and

1 + \sqrt{1 + k} < 1 + \frac{2 {(n - 1)}^{2}}{n - 2} = \frac{n (2 n - 3)}{n - 2},

from the expression of

f^{'}

it follows that f is decreasing on

[\frac{- n}{n - 2}, s_{0}]

and increasing on

[s_{0}, \frac{n (2 n - 3)}{n - 2}]

, where

s_{0} = 1 + \sqrt{k + 1} > 1 .

On the other hand, for

u \in [1, s_{0}]

, we have

f_{1} (u) > (k + 1) (u - 1) - {(u - 1)}^{3} = (u - 1) [k + 1 - {(u - 1)}^{2}] \geq 0,

since

u - 1 \leq s_{0} - 1 = \sqrt{k + 1} .

Thus, f is convex on $[1, s_{0}]$ . By PCF-Theorem, we only need to show that $f (x) + (n - 1) f (y) \geq n f (1)$ for all $x, y \in I$ such that $x + (n - 1) y = n$ . We have proved this before (at Case 1). □

Application 4.2 If

x_{1}, x_{2}, \dots, x_{n}

(

n \geq 3

) are real numbers such that

x_{1} + x_{2} + \dots + x_{n} = n,

then [6]

\sum_{i = 1}^{n} \frac{n (n + 1) - 2 x_{i}}{n^{2} + (n - 2) x_{i}^{2}} \leq n,

with equality for $x_{1} = x_{2} = \dots = x_{n} = 1$ , and also for $x_{1} = n$ and $x_{2} = \dots = x_{n} = 0$ (or any cyclic permutation).

Proof The desired inequality is true for

x_{1} > \frac{n (n + 1)}{2}

since

\frac{n (n + 1) - 2 x_{1}}{n^{2} + (n - 2) x_{1}^{2}} < 0

and

\frac{n (n + 1) - 2 x_{i}}{n^{2} + (n - 2) x_{i}^{2}} < \frac{n}{n - 1}, i = 2, 3, \dots, n .

Consider further that

x_{1}, x_{2}, \dots, x_{n} \leq \frac{n (n + 1)}{2}

and rewrite the desired inequality as

f (x_{1}) + f (x_{2}) + \dots + f (x_{n}) \geq n f (s),

where

\begin{matrix} s = 1, \\ f (u) = \frac{2 u - n (n + 1)}{(n - 2) u^{2} + n^{2}}, u \in I = [\frac{n (3 - n^{2})}{2}, \frac{n (n + 1)}{2}] . \end{matrix}

We have

\frac{f^{'} (u)}{2 (n - 2)} = \frac{n^{2} + n (n + 1) u - u^{2}}{{[(n - 2) u^{2} + n^{2}]}^{2}}

and

\frac{f^{″} (u)}{2 (n - 2)} = \frac{f_{1} (u)}{{[(n - 2) u^{2} + n^{2}]}^{3}},

where

f_{1} (u) = 2 (n - 2) u^{3} - 3 n (n + 1) (n - 2) u^{2} - 2 n^{2} (2 n - 3) u + n^{3} (n + 1) .

From the expression of

f^{'}

, it follows that f is decreasing on

[\frac{n (3 - n^{2})}{2}, s_{0}]

and increasing on

[s_{0}, \frac{n (n + 1)}{2}]

, where

s_{0} = \frac{n}{2} (n + 1 - \sqrt{n^{2} + 2 n + 5}) \in (- 1, 0) .

On the other hand, for

- 1 \leq u \leq 1

, we have

\begin{aligned} f_{1} (u) & > - 2 (n - 2) - 3 n (n + 1) (n - 2) - 2 n^{2} (2 n - 3) + n^{3} (n + 1) \\ = n^{2} {(n - 3)}^{2} + 4 (n + 1) > 0, \end{aligned}

and hence

f^{″} (u) > 0

. Since

[s_{0}, s] \subset [- 1, 1]

, f is convex on

[s_{0}, s]

. By PCF-Theorem, we only need to show that

f (x) + (n - 1) f (y) \geq n f (1)

for all

x, y \in I

which satisfy

x + (n - 1) y = n

. According to Remark 2.4, this is true if

h (x, y) \geq 0

for

x, y \in I

and

x + (n - 1) y = n

, where

h (x, y) = \frac{g (x) - g (y)}{x - y}, g (u) = \frac{f (u) - f (1)}{u - 1} .

Indeed, we have

g (u) = \frac{(n - 2) u + n}{(n - 2) u^{2} + n^{2}}

and

\begin{aligned} \frac{h (x, y)}{n - 2} & = \frac{n^{2} - n (x + y) - (n - 2) x y}{[(n - 2) x^{2} + n^{2}] [(n - 2) y^{2} + n^{2}]} \\ = \frac{(n - 1) (n - 2) y^{2}}{[(n - 2) x^{2} + n^{2}] [(n - 2) y^{2} + n^{2}]} \geq 0 . \end{aligned}

□

Application 4.3 Let

x_{1}, x_{2}, \dots, x_{n}

(

n \geq 2

) be positive real numbers such that

x_{1} + x_{2} + \dots + x_{n} \geq n .

If

k > 1

, then [6]

\frac{x_{1}}{x_{1}^{k} + x_{2} + \dots + x_{n}} + \frac{x_{2}}{x_{1} + x_{2}^{k} + \dots + x_{n}} + \dots + \frac{x_{n}}{x_{1} + x_{2} + \dots + x_{n}^{k}} \leq 1,

with equality for $x_{1} = x_{2} = \dots = x_{n} = 1$ .

Proof

Using the substitutions

s = \frac{x_{1} + x_{2} + \dots + x_{n}}{n},

and

y_{1} = \frac{x_{1}}{s}, y_{2} = \frac{x_{2}}{s}, \dots, y_{n} = \frac{x_{n}}{s},

the desired inequality becomes

\frac{y_{1}}{s^{k - 1} y_{1}^{k} + y_{2} + \dots + y_{n}} + \dots + \frac{y_{n}}{y_{1} + y_{2} + \dots + s^{k - 1} y_{n}^{k}} \leq 1,

where

s \geq 1

and

y_{1} + y_{2} + \dots + y_{n} = n

. Clearly, if this inequality holds for

s = 1

, then it holds for any

s \geq 1

. Therefore, we need only to consider the case

s = 1

, when

x_{1} + x_{2} + \dots + x_{n} = n

, and the desired inequality is equivalent to

\frac{x_{1}}{x_{1}^{k} - x_{1} + n} + \frac{x_{2}}{x_{2}^{k} - x_{2} + n} + \dots + \frac{x_{n}}{x_{n}^{k} - x_{n} + n} \leq 1 .

There are two cases to consider: $1 < k \leq n + 1$ and $k > n + 1$ .

Case 1:

1 < k \leq n + 1

. By Bernoulli’s inequality, we have

x_{1}^{k} \geq 1 + k (x_{1} - 1),

and hence

x_{1}^{k} - x_{1} + n \geq n - k + 1 + (k - 1) x_{1} \geq 0 .

Consequently, it suffices to prove that

\sum_{i = 1}^{n} \frac{x_{i}}{n - k + 1 + (k - 1) x_{i}} \leq 1 .

For

k = n + 1

, this inequality is an equality. Otherwise, for

1 < k < n + 1

, we rewrite the inequality as

\sum_{i = 1}^{n} \frac{1}{n - k + 1 + (k - 1) x_{i}} \geq 1,

which follows from the AM-HM inequality as follows:

\sum_{i = 1}^{n} \frac{1}{n - k + 1 + (k - 1) x_{i}} \geq \frac{n^{2}}{\sum_{i = 1}^{n} [n - k + 1 + (k - 1) x_{i}]} = 1 .

Case 2:

k > n + 1

. Write the desired inequality as

f (x_{1}) + f (x_{2}) + \dots + f (x_{n}) \geq n f (s),

where

\begin{matrix} s = 1, \\ f (u) = \frac{- u}{u^{k} - u + n}, u \in I = (0, n) . \end{matrix}

We have

f^{'} (u) = \frac{(k - 1) u^{k} - n}{{(u^{k} - u + n)}^{2}}

and

f^{″} (u) = \frac{f_{1} (u)}{{(u^{k} - u + n)}^{3}},

where

f_{1} (u) = k (k - 1) u^{k - 1} (u^{k} - u + n) - 2 (k u^{k - 1} - 1) [(k - 1) u^{k} - n] .

From the expression of

f^{'}

, it follows that f is decreasing on

(0, s_{0}]

and increasing on

[s_{0}, n)

, where

s_{0} = {(\frac{n}{k - 1})}^{1 / k} < 1 .

On the other hand, for

u \in [s_{0}, s]

, we have

(k - 1) u^{k} - n \geq (k - 1) s_{0}^{k} - n = 0,

and hence

\begin{aligned} f_{1} (u) & \geq k (k - 1) u^{k - 1} (u^{k} - u + n) - 2 k u^{k - 1} [(k - 1) u^{k} - n] \\ = k u^{k - 1} [- (k - 1) (u^{k} + u) + n (k + 1)] \\ \geq k u^{k - 1} [- 2 (k - 1) + 2 (k + 1)] = 4 k u^{k - 1} > 0 . \end{aligned}

Thus,

f^{″} (u) > 0

, and hence f is convex on

[s_{0}, s]

. By PCF-Theorem and Remark 2.3, we need to show that

f (x) + (n - 1) f (y) \geq n f (1)

for all positive x, y which satisfy

x \geq 1 \geq y > 0

and

x + (n - 1) y = n

. Consider the nontrivial case where

x > 1 > y > 0

and write the inequality

f (x) + (n - 1) f (y) \geq n f (1)

as follows:

\begin{matrix} \frac{x}{x^{k} - x - n} + \frac{(n - 1) y}{y^{k} - y + n} \leq 1, \\ x^{k} - x + n \geq \frac{x (y^{k} - y + n)}{y^{k} - n y + n}, \\ x^{k} - x \geq \frac{(n - 1) y (y - y^{k})}{y^{k} - n y + n} . \end{matrix}

Since

y < 1

and

y^{k} - n y + n > y^{k} - y + 1 > 0

, it suffices to show that

x^{k} - x \geq \frac{(n - 1) (y - y^{k})}{y^{k} - y + 1},

which is equivalent to

h (x) \geq \frac{y - y^{k}}{(1 - y) (y^{k} - y + 1)},

where

h (x) = \frac{x^{k} - x}{x - 1} .

By the weighted AM-GM inequality, we have

h^{'} (x) = \frac{(k - 1) x^{k} + 1 - k x^{k - 1}}{{(x - 1)}^{2}} > 0,

and hence h is strictly increasing. Since

x - 1 = (n - 1) (1 - y) \geq 1 - y

, we get

h (x) \geq h (2 - y) = \frac{{(2 - y)}^{k} - 2 + y}{1 - y} .

Thus, it suffices to show that

{(2 - y)}^{k} - 2 + y \geq \frac{y - y^{k}}{y^{k} - y + 1} .

Putting

1 - y = t

,

0 < t < 1

, we write this inequality as

\begin{matrix} {(2 - y)}^{k} - 1 + y \geq \frac{1}{y^{k} - y + 1}, \\ {(1 + t)}^{k} - t \geq \frac{1}{{(1 - t)}^{k} + t}, \\ {(1 - t^{2})}^{k} + t {(1 + t)}^{k} \geq 1 + t^{2} + t {(1 - t)}^{k} . \end{matrix}

By Bernoulli’s inequality,

{(1 - t^{2})}^{k} + t {(1 + t)}^{k} > 1 - k t^{2} + t (1 + k t) = 1 + t .

So, we only need to show that $t (1 - t) \geq t {(1 - t)}^{k}$ , which is clearly true. □

Application 4.4 Let

x_{1}, x_{2}, \dots, x_{n}

(

n \geq 2

) be positive real numbers such that

x_{1} + x_{2} + \dots + x_{n} = n .

If

0 < k \leq \frac{n}{n - 1}

, then [6]

x_{1}^{k / x_{1}} + x_{2}^{k / x_{2}} + \dots + x_{n}^{k / x_{n}} \leq n

with equality for $x_{1} = x_{2} = \dots = x_{n} = 1$ .

Proof

According to the power mean inequality, we have

{(\frac{x_{1}^{p / x_{1}} + x_{2}^{p / x_{2}} + \dots + x_{n}^{p / x_{n}}}{n})}^{1 / p} \geq {(\frac{x_{1}^{q / x_{1}} + x_{2}^{q / x_{2}} + \dots + x_{n}^{q / x_{n}}}{n})}^{1 / q}

for all

p \geq q > 0

. Thus, it suffices to prove the desired inequality for

k = \frac{n}{n - 1}, 1 < k \leq 2 .

Rewrite the desired inequality as

f (x_{1}) + f (x_{2}) + \dots + f (x_{n}) \geq n f (s),

where

\begin{matrix} s = 1, \\ f (u) = - u^{k / u}, u \in I = (0, n) . \end{matrix}

We have

\begin{matrix} f^{'} (u) = k u^{\frac{k}{u} - 2} (ln u - 1), \\ f^{″} (u) = k u^{\frac{k}{u} - 4} [u + (1 - ln u) (2 u - k + k ln u)] . \end{matrix}

From the expression of

f^{'}

, it follows that f is decreasing on

(0, s_{0}]

and increasing on

[s_{0}, n)

, where

s_{0} = e .

In addition, we claim that f is convex on

[s, s_{0}]

. Indeed, since

1 - ln u \geq 0

and

2 u - k + k ln u \geq 2 - k \geq 0,

we have

f^{″} > 0

for

u \in [s, s_{0}]

. Therefore, by PCF-Theorem and Remark 2.3, we only need to show that

x^{k / x} + (n - 1) y^{k / y} \leq n

for

0 < x \leq 1 \leq y

and

x + (n - 1) y = n

. We have

\frac{k}{x} \geq k > 1 .

Also, from

\frac{k}{y} = \frac{n}{(n - 1) y} > \frac{n}{x + (n - 1) y} = 1, \frac{k}{y} \leq \frac{2}{y} \leq 2,

we get

0 < \frac{k}{y} - 1 \leq 1 .

Therefore, by Bernoulli’s inequality, we have

\begin{aligned} x^{k / x} + (n - 1) y^{k / y} - n \\ = \frac{1}{{(\frac{1}{x})}^{k / x}} + (n - 1) y \cdot y^{k / y - 1} - n \\ \leq \frac{1}{1 + \frac{k}{x} (\frac{1}{x} - 1)} + (n - 1) y [1 + (\frac{k}{y} - 1) (y - 1)] - n \\ = \frac{x^{2}}{x^{2} - k x + k} - (k - 1) x^{2} - (2 - k) x \\ = \frac{- {(x - 1)}^{2} [(k - 1) x + k (2 - k)]}{x^{2} - k x + k} \leq 0 . \end{aligned}

□

Application 4.5 If a, b, c are positive real numbers such that

a b c = 1

, then

\frac{1 - a}{17 + 4 a + 6 a^{2}} + \frac{1 - b}{17 + 4 b + 6 b^{2}} + \frac{1 - c}{17 + 4 c + 6 c^{2}} \geq 0,

with equality for $a = b = c = 1$ and also for $8 a = b = c = 2$ (or any cyclic permutation).

Proof

Write the desired inequality as

g (a) + g (b) + g (c) \geq 3 g (r),

where

r = 1

and

g (t) = \frac{1 - t}{1 + p t + q t^{2}}, t \in I = (0, \infty),

where

p = \frac{4}{17}

,

q = \frac{6}{17}

. From

g^{'} (t) = \frac{q t^{2} - 2 q t - p - 1}{{(1 + p t + q t^{2})}^{2}},

it follows that g is decreasing on

(0, r_{0}]

and increasing on

[r_{0}, \infty)

, where

r_{0} = 1 + \sqrt{1 + \frac{p + 1}{q}} > 1 .

We have

\begin{matrix} f (u) = g (e^{u}) = \frac{1 - e^{u}}{1 + p e^{u} + q e^{2 u}}, \\ f^{″} (u) = \frac{t \cdot h (t)}{{(1 + p t + q t^{2})}^{3}}, \end{matrix}

where

t = e^{u}, h (t) = - q^{2} t^{4} + q (p + 4 q) t^{3} + 3 q (p + 2) t^{2} + (p - 4 q + p^{2}) t - p - 1 .

We will show that

h (t) > 0

for

t \in [r, r_{0}]

, and hence f is convex for

e^{u} \in [r, r_{0}] = [1, 1 + \sqrt{1 + \frac{p + 1}{q}}] .

We have

\begin{matrix} h^{'} (t) = - 4 q^{2} t^{3} + 3 q (p + 4 q) t^{2} + 6 q (p + 2) t + p - 4 q + p^{2}, \\ h^{″} (t) = 6 q [- 2 q t^{2} + (p + 4 q) t + p + 2] . \end{matrix}

Since

h^{″} (t) = 6 q [2 (- q t^{2} + 2 q t + p + 1) + p (t - 1)] \geq 12 q (- q t^{2} + 2 q t + p + 1) \geq 0,

h^{'} (t)

is increasing,

h^{'} (t) \geq h^{'} (1) = p^{2} + 9 p q + 8 q^{2} + p + 8 q > 0,

h is increasing, and hence

\begin{aligned} h (t) \geq h (1) & = p^{2} + 4 p q + 3 q^{2} + 2 q - 1 = {(p + 2 q)}^{2} - {(q - 1)}^{2} \\ = (p + q + 1) (p + 3 q - 1) > 0 . \end{aligned}

By PCF-Corollary, we only need to prove that

g (a) + 2 g (b) \geq 3 g (1)

for

a b^{2} = 1

; that is,

\frac{1 - a}{1 + p a + q a^{2}} + \frac{2 (1 - b)}{1 + p b + q b^{2}} \geq 0,

\frac{b^{2} (b^{2} - 1)}{b^{4} + p b^{2} + q} + \frac{2 (1 - b)}{1 + p b + q b^{2}} \geq 0,

p A + q B \geq C,

where

\begin{matrix} A = b^{2} {(b - 1)}^{2} (b + 2), \\ B = {(b - 1)}^{2} (b^{4} + 2 b^{3} + 2 b^{2} + 2 b + 2), \\ C = b^{2} {(b - 1)}^{2} (2 b + 1) . \end{matrix}

Indeed, we have

17 (p A + q B - C) = 3 {(b - 1)}^{2} {(b - 2)}^{2} (2 b^{2} + 2 b + 1) \geq 0 .

□

Application 4.6 If a, b, c are positive real numbers such that

a b c = 1

, then [6]

\frac{7 - 6 a}{2 + a^{2}} + \frac{7 - 6 b}{2 + b^{2}} + \frac{7 - 6 c}{2 + c^{2}} \geq 1,

with equality for $a = b = c = 1$ and also for $8 a = b = c = 2$ (or any cyclic permutation).

Proof

Write the desired inequality as

g (a) + g (b) + g (c) \geq 3 g (r),

where

r = 1

and

g (t) = \frac{7 - 6 t}{2 + t^{2}}, t \in I = (0, \infty) .

From

g^{'} (t) = \frac{2 (3 t + 2) (t - 3)}{{(2 + t^{2})}^{2}},

it follows that g is decreasing on

(0, r_{0}]

and increasing on

[r_{0}, \infty)

, where

r_{0} = 3 .

We have

\begin{matrix} f (u) = g (e^{u}) = \frac{7 - 6 e^{u}}{2 + e^{2 u}}, \\ f^{″} (u) = \frac{2 t \cdot h (t)}{{(2 + t^{2})}^{3}}, \end{matrix}

where

t = e^{u}, h (t) = - 3 t^{4} + 14 t^{3} + 36 t^{2} - 28 t - 12 .

We will show that

h (t) > 0

for

t \in [r, r_{0}]

, and hence f is convex for

e^{u} \in [r, r_{0}] = [1, 3] .

We have

\begin{aligned} h (t) & = 3 (t^{2} - 1) (9 - t^{2}) + 14 t^{3} + 6 t^{2} - 28 t + 15 \\ = 3 (t^{2} - 1) (9 - t^{2}) + 14 t^{2} (t - 1) + 14 {(t - 1)}^{2} + 6 t^{2} + 1 > 0 . \end{aligned}

By PCF-Corollary, we only need to prove that

g (a) + 2 g (b) \geq 3 g (1)

for

a b^{2} = 1

; that is,

\begin{matrix} \frac{7 - 6 a}{2 + a^{2}} + \frac{2 (7 - 6 b)}{2 + b^{2}} \geq 1, \\ \frac{b^{2} (7 b^{2} - 6)}{2 b^{4} + 1} + \frac{2 (7 - 6 b)}{2 + b^{2}} \geq 1, \\ {(b - 1)}^{2} {(b - 2)}^{2} (5 b^{2} + 6 b + 3) \geq 0 . \end{matrix}

Since the last inequality is true, the proof is completed. □

Application 4.7 Let a, b, c be positive real numbers such that

a b c = 1

. If

k \in [\frac{- 13}{3 \sqrt{3}}, \frac{13}{3 \sqrt{3}}],

then [6]

\frac{a + k}{a^{2} + 1} + \frac{b + k}{b^{2} + 1} + \frac{c + k}{c^{2} + 1} \leq \frac{3 (1 + k)}{2},

with equality for $a = b = c = 1$ . If $k = \frac{13}{3 \sqrt{3}}$ , then the equality holds also for $a = 7 + 4 \sqrt{3}$ and $b = c = 2 - \sqrt{3}$ (or any cyclic permutation). If $k = \frac{- 13}{3 \sqrt{3}}$ , then the equality holds also for $a = 7 - 4 \sqrt{3}$ and $b = c = 2 + \sqrt{3}$ (or any cyclic permutation).

Proof

The desired inequality is equivalent to

\sum \frac{{(a - 1)}^{2}}{a^{2} + 1} \geq k (\sum \frac{2}{a^{2} + 1} - 3) .

Thus, it suffices to prove this inequality for only

| k | = \frac{13}{3 \sqrt{3}}

. On the other hand, replacing a, b, c by

1 / a

,

1 / b

,

1 / c

, the inequality becomes

\sum \frac{{(a - 1)}^{2}}{a^{2} + 1} \geq k (3 - \sum \frac{2}{a^{2} + 1}) .

Thus, we only need to prove the desired inequality for

k = \frac{13}{3 \sqrt{3}}

. Write this inequality as

g (a) + g (b) + g (c) \geq 3 g (r),

where

r = 1

and

g (t) = \frac{- t - k}{t^{2} + 1}, t \in I = (0, \infty) .

From

g^{'} (t) = \frac{t^{2} + 2 k t - 1}{{(t^{2} + 1)}^{2}},

it follows that g is decreasing on

(0, r_{0}]

and increasing on

[r_{0}, \infty)

, where

r_{0} = \frac{\sqrt{3}}{9} .

We have

\begin{matrix} f (u) = g (e^{u}) = \frac{- e^{u} - k}{e^{2 u} + 1}, \\ f^{″} (u) = \frac{t \cdot h (t)}{{(t^{2} + 1)}^{3}}, \end{matrix}

where

t = e^{u}, h (t) = - t^{4} - 4 k t^{3} + 6 t^{2} + 4 k t - 1 .

We will show that

h (t) > 0

for

t \in [r_{0}, r]

, and hence f is convex for

e^{u} \in [r_{0}, r] = [\frac{\sqrt{3}}{9}, 1] .

Indeed, since

4 k t = \frac{52 t}{3 \sqrt{3}} = \frac{52}{27} > 1,

we have

h (t) = - t^{4} + 6 t^{2} - 1 + 4 k t (1 - t^{2}) \geq - t^{4} + 6 t^{2} - 1 + (1 - t^{2}) = t^{2} (5 - t^{2}) > 0 .

By PCF-Corollary, we only need to prove that

g (a) + 2 g (b) \geq 3 g (1)

for

a b^{2} = 1

; that is,

\begin{matrix} \frac{a + k}{a^{2} + 1} + \frac{2 (b + k)}{b^{2} + 1} \leq \frac{3 (1 + k)}{2}, \\ \frac{b^{2} (k b^{2} + 1)}{b^{4} + 1} + \frac{2 (b + k)}{b^{2} + 1} \leq \frac{3 (1 + k)}{2}, \\ 3 b^{6} - 4 b^{5} + b^{4} + b^{2} - 4 b + 3 - k {(1 - b^{2})}^{3} \geq 0, \\ {(b - 1)}^{2} [(3 + k) b^{4} + 2 (1 + k) b^{3} + 2 b^{2} + 2 (1 - k) b + 3 - k] \geq 0, \\ {(b - 1)}^{2} {(b - 2 + \sqrt{3})}^{2} [(27 + 13 \sqrt{3}) b^{2} + 24 (2 + \sqrt{3}) b + 33 + 17 \sqrt{3}] \geq 0 . \end{matrix}

The last inequality is clearly true, and the proof is completed. □

Application 4.8 If a, b, c are positive real numbers and

0 < k \leq 2 + 2 \sqrt{2}

, then [6]

\frac{a^{3}}{k a^{2} + b c} + \frac{b^{3}}{k b^{2} + c a} + \frac{c^{3}}{k c^{2} + a b} \geq \frac{a + b + c}{k + 1},

with equality for $a = b = c = 1$ . If $k = 2 + 2 \sqrt{2}$ , then the equality holds also for $\frac{a}{\sqrt{2}} = b = c$ (or any cyclic permutation).

Proof For

k < 2 + 2 \sqrt{2}

, the proof is similar to the one of the main case

k = 2 + 2 \sqrt{2}

. For this reason, we consider further only the case where

k = 2 + 2 \sqrt{2} .

Due to homogeneity, we may assume that

a b c = 1

. On this hypothesis,

\sum \frac{a^{3}}{k a^{2} + b c} - \frac{1}{k + 1} \sum a = \sum (\frac{a^{4}}{k a^{3} + 1} - \frac{a}{k + 1}) = \frac{1}{k + 1} \sum \frac{a^{4} - a}{k a^{3} + 1} .

Thus, we can write the inequality as

g (a) + g (b) + g (c) \geq 3 g (r),

where

r = 1

and

g (t) = \frac{t^{4} - t}{k t^{3} + 1}, t \in I = (0, \infty) .

From

g^{'} (t) = \frac{k t^{6} + 2 (k + 2) t^{3} - 1}{{(k t^{3} + 1)}^{2}},

it follows that g is decreasing on

(0, r_{0}]

and increasing on

[r_{0}, \infty)

, where

r_{0} = \sqrt[3]{\frac{- k - 2 + \sqrt{(k + 1) (k + 4)}}{k}} \approx 0.4149 .

We have

\begin{matrix} f (u) = g (e^{u}) = \frac{e^{4 u} - e^{u}}{k e^{3 u} + 1}, \\ f^{″} (u) = \frac{t \cdot h (t)}{{(k t^{3} + 1)}^{3}}, \end{matrix}

where

t = e^{u}, h (t) = k^{2} t^{9} - k (4 k + 1) t^{6} + (13 k + 16) t^{3} - 1 .

We have

h (t) \geq 0

for

t \in [t_{1}, t_{2}]

, where

t_{1} \approx 0.2345

and

t_{2} \approx 1.02

. Since

[r_{0}, r] \subset [t_{1}, t_{2}],

f is convex for

e^{u} \in [r_{0}, r]

. Then, by PCF-Corollary, it suffices to show that

g (a) + 2 g (b) \geq 3 g (1)

for

a b^{2} = 1

. This is true if the original inequality holds for

b = c = 1

. Thus, we need to show that

\frac{a^{3}}{k a^{2} + 1} + \frac{2}{a + k} \geq \frac{a + 2}{k + 1},

which is equivalent to the obvious inequality

{(a - 1)}^{2} {(a - \sqrt{2})}^{2} \geq 0 .

□

Application 4.9 If

a_{1}

,

a_{2}

,

a_{3}

,

a_{4}

,

a_{5}

are positive real numbers such that

a_{1} a_{2} a_{3} a_{4} a_{5} = 1,

then [6]

\frac{1 - a_{1}}{1 + a_{1}^{2}} + \frac{1 - a_{2}}{1 + a_{2}^{2}} + \frac{1 - a_{3}}{1 + a_{3}^{2}} + \frac{1 - a_{4}}{1 + a_{4}^{2}} + \frac{1 - a_{5}}{1 + a_{5}^{2}} \geq 0,

with equality for $a_{1} = a_{2} = a_{3} = a_{4} = a_{5} = 1$ .

Proof

Write the inequality as

g (a_{1}) + g (a_{2}) + g (a_{3}) + g (a_{4}) + g (a_{5}) \geq 3 g (r),

where

r = 1

and

g (t) = \frac{1 - t}{1 + t^{2}}, t \in I = (0, \infty) .

From

g^{'} (t) = \frac{t^{2} - 2 t - 1}{{(t^{2} + 1)}^{2}},

it follows that g is decreasing on

(0, r_{0}]

and increasing on

[r_{0}, \infty)

, where

r_{0} = 1 + \sqrt{2} .

We have

\begin{matrix} f (u) = g (e^{u}) = \frac{1 - e^{u}}{1 + e^{2 u}}, \\ f^{″} (u) = \frac{t \cdot h (t)}{{(t^{2} + 1)}^{3}}, \end{matrix}

where

t = e^{u}, h (t) = - t^{4} + 4 t^{3} + 6 t^{2} - 4 t - 1 .

We will show that

h (t) > 0

for

t \in [r, r_{0}]

, and hence f is convex for

e^{u} \in [r, r_{0}] = [1, 1 + \sqrt{2}] .

Indeed,

h (t) \geq - t^{4} + 6 t^{2} - 1 = 8 - {(3 - t^{2})}^{2} \geq 4 .

By PCF-Corollary, we only need to prove that

g (a) + 4 g (b) \geq 5 g (1)

for

a b^{4} = 1

; that is,

\begin{matrix} \frac{1 - a}{1 + a^{2}} + \frac{4 (1 - b)}{1 + b^{2}} \geq 0, \\ \frac{b^{4} (b^{4} - 1)}{1 + b^{8}} + \frac{4 (1 - b)}{1 + b^{2}} \geq 0, \\ 1 + \frac{4 (1 - b)}{1 + b^{2}} \geq \frac{1 + b^{4}}{1 + b^{8}} . \end{matrix}

Since

\frac{1 + b^{4}}{1 + b^{8}} \leq \frac{2}{1 + b^{4}} \leq \frac{4}{{(1 + b^{2})}^{2}},

it suffices to show that

1 + \frac{4 (1 - b)}{1 + b^{2}} \geq \frac{4}{{(1 + b^{2})}^{2}} .

This inequality is equivalent to ${(1 - b)}^{4} \geq 0$ , and the proof is completed. □

Remark 4.1

The inequality

\frac{1 - a_{1}}{1 + a_{1}^{2}} + \frac{1 - a_{2}}{1 + a_{2}^{2}} + \frac{1 - a_{3}}{1 + a_{3}^{2}} + \frac{1 - a_{4}}{1 + a_{4}^{2}} + \frac{1 - a_{5}}{1 + a_{5}^{2}} + \frac{1 - a_{6}}{1 + a_{6}^{2}} \geq 0

is not true for any positive numbers

a_{1}

,

a_{2}

,

a_{3}

,

a_{4}

,

a_{5}

,

a_{6}

satisfying

a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} = 1

. Indeed, for

a_{2} = a_{3} = a_{4} = a_{5} = a_{6} = 2

, the inequality becomes

\frac{- a_{1} (a_{1} + 1)}{1 + a_{1}^{2}} \geq 0,

which is false for

a_{1} = \frac{1}{a_{2} a_{3} a_{4} a_{5} a_{6}} = \frac{1}{32} .

Declarations

Acknowledgements

The author is grateful to the referees for their useful comments.

Authors’ Affiliations

(1)

Department of Automatic Control and Computers, University of Ploiesti, Ploiesti, 100680, Romania

References

Jensen JLWV: Sur les fonctions convexes et les inegalites entre les valeurs moyennes. Acta Math. 1906, 30(1):175–193. 10.1007/BF02418571MathSciNetView ArticleGoogle Scholar
Mitrinović DS, Pečarić JE, Fink AM: Classical and New Inequalities in Analysis. Kluwer Academic, Dordrecht; 1993.View ArticleGoogle Scholar
Cirtoaje V, Baiesu A: An extension of Jensen’s discrete inequality to half convex functions. J. Inequal. Appl. 2011., 2011: Article ID 101Google Scholar
Cirtoaje V: A generalization of Jensen’s inequality. Gaz. Mat., Ser. A 2005, 2: 124–138.Google Scholar
Cirtoaje V: Algebraic Inequalities-Old and New Methods. GIL Publishing House, Zalau; 2006.Google Scholar
Art of Problem Solving. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45516/115797/220970/242881/246265/453476/465869Google Scholar

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