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

Journal of Inequalities and Applications20132013:54

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

  • Received: 16 October 2012
  • Accepted: 30 January 2013
  • Published:

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 ] I , decreasing for u c and increasing for u c , where c [ 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 , , x n } be a sequence of real numbers belonging to a given real interval I , and let p = { p 1 , p 2 , , p n } be a sequence of given positive weights associated to x and satisfying p 1 + p 2 + + p n = 1 . If f is a convex function on I , then
i = 1 n p i f ( x i ) f ( 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 s or u s , where s I .

WHCF-Theorem Let f be a function defined on a real interval I and convex for u s or u s , where s I , and let p 1 , p 2 , , p n be positive real numbers such that
p = min { p 1 , p 2 , , p n } , p 1 + p 2 + + p n = 1 .
The inequality
p 1 f ( x 1 ) + p 2 f ( x 2 ) + + p n f ( x n ) f ( s )
holds for all x 1 , x 2 , , x n I satisfying p 1 x 1 + p 2 x 2 + + p n x n = s if and only if
p f ( x ) + ( 1 p ) f ( y ) f ( s )

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

For the particular case p 1 = p 2 = = 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 s or u s , where s I . The inequality
f ( x 1 ) + f ( x 2 ) + + f ( x n ) n f ( s )
holds for all x 1 , x 2 , , x n I satisfying x 1 + x 2 + + x n = n s if and only if
f ( x ) + ( n 1 ) f ( y ) n f ( s )

for all x , y 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 , , 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 r or e u r , where r I . The inequality
g ( a 1 ) + g ( a 2 ) + + g ( a n ) n g ( r )
holds for all a 1 , a 2 , , a n I satisfying a 1 a 2 a n = r n if and only if
g ( a ) + ( n 1 ) g ( b ) n g ( r )

for all a , b 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 r or e u r , where r I , and let p 1 , p 2 , , p n be positive real numbers such that
p = min { p 1 , p 2 , , p n } , p 1 + p 2 + + p n = 1 .
The inequality
p 1 g ( a 1 ) + p 2 g ( a 2 ) + + p n g ( a n ) g ( r )
holds for all a 1 , a 2 , , a n I satisfying a 1 p 1 a 2 p 2 a n p n = r if and only if
p g ( a ) + ( 1 p ) g ( b ) g ( r )

for all a , b 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 ] 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 s y
when f is convex for u s , and
x s y
when f is convex for u s (see [3]). Also, in HCF-Corollary and WHCF-Corollary, it suffices to consider that
a r b
when f is convex for e u r , and
a r b

when f is convex for e u 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 s 0 and increasing for u s 0 , where s 0 I . In addition, assume that f is convex on [ s 0 , s ] or [ s , s 0 ] , where s I . The inequality
f ( x 1 ) + f ( x 2 ) + + f ( x n ) n f ( s )
holds for all x 1 , x 2 , , x n I satisfying x 1 + x 2 + + x n = n s if and only if
f ( x ) + ( n 1 ) f ( y ) n f ( s )

for all x , y 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 s 0 and increasing for u s 0 , where s 0 I , and let p 1 , p 2 , , p n be positive real numbers such that
p = min { p 1 , p 2 , , p n } , p 1 + p 2 + + p n = 1 .
In addition, assume that f is convex on [ s 0 , s ] or [ s , s 0 ] , where s I . The inequality
p 1 f ( x 1 ) + p 2 f ( x 2 ) + + p n f ( x n ) f ( s )
holds for all x 1 , x 2 , , x n I satisfying p 1 x 1 + p 2 x 2 + + p n x n = s if and only if
p f ( x ) + ( 1 p ) f ( y ) f ( s )

for all x , y 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 , , n , we get the following corollaries, respectively.

PCF-Corollary Let g be a function defined on a positive interval I , decreasing for t r 0 and increasing for t r 0 , where r 0 I . In addition, assume that the function f defined by f ( u ) = g ( e u ) is convex for r 0 e u r or r e u r 0 , where r I . The inequality
g ( a 1 ) + g ( a 2 ) + + g ( a n ) n g ( r )
holds for all a 1 , a 2 , , a n I satisfying a 1 a 2 a n = r n if and only if
g ( a ) + ( n 1 ) g ( b ) n g ( r )

for all a , b 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 r 0 and increasing for t r 0 , where r 0 I , and let p 1 , p 2 , , p n be positive real numbers such that
p = min { p 1 , p 2 , , p n } , p 1 + p 2 + + p n = 1 .
In addition, assume that the function f defined by f ( u ) = g ( e u ) is convex for r 0 e u r or r e u r 0 , where r I . The inequality
p 1 g ( a 1 ) + p 2 g ( a 2 ) + + p n g ( a n ) g ( r )
holds for all a 1 , a 2 , , a n I satisfying a 1 p 1 a 2 p 2 a n p n = r if and only if
p g ( a ) + ( 1 p ) g ( b ) g ( r )

for all a , b 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 s 0 and increasing for u s 0 , where s 0 I , and let p 1 , p 2 , , p n be positive real numbers such that
p 1 + p 2 + + p n = 1 .
For s I , s s 0 , if the inequality
p 1 f ( x 1 ) + p 2 f ( x 2 ) + + p n f ( x n ) f ( s )
holds for all x 1 , x 2 , , x n I such that
x 1 , x 2 , , x n s 0 , p 1 x 1 + p 2 x 2 + + p n x n = s ,
then it holds for all x 1 , x 2 , , x n I such that
p 1 x 1 + p 2 x 2 + + p n x n = s .
Lemma 2.2 Let f be a function defined on a real interval I , decreasing for u s 0 and increasing for u s 0 , where s 0 I , and let p 1 , p 2 , , p n be positive real numbers such that
p 1 + p 2 + + p n = 1 .
For s I , s s 0 , if the inequality
p 1 f ( x 1 ) + p 2 f ( x 2 ) + + p n f ( x n ) f ( s )
holds for all x 1 , x 2 , , x n I such that
x 1 , x 2 , , x n s 0 , p 1 x 1 + p 2 x 2 + + p n x n = s ,
then it holds for all x 1 , x 2 , , x n I such that
p 1 x 1 + p 2 x 2 + + p n x n = s .
Notice that in the case s s 0 , WPCF-Theorem follows immediately from Lemma 2.1 and WHCF-Theorem applied to the interval
I 0 = { u I u s 0 } ,
because f is convex for u I 0 , u s . Also, in the case s s 0 , WPCF-Theorem follows immediately from Lemma 2.2 and WHCF-Theorem applied to the interval
I 0 = { u I u s 0 } ,

because f is convex for u I 0 , u s .

Remark 2.3

According to Remark 1.2, it suffices to consider in PCF-Theorem and WPCF-Theorem that
x s y
when f is convex on [ s 0 , s ] , and
x s y
when f is convex on [ s , s 0 ] . Also, it suffices to consider in PCF-Corollary and WPCF-Corollary that
a r b
when f is convex for r 0 e u r , and
a r b

when f is convex for r e u r 0 .

Remark 2.4

Let us denote
g ( u ) = f ( u ) f ( s ) u s , h ( x , y ) = g ( x ) g ( y ) x y .
In many applications, it is useful to replace the hypothesis
p f ( x ) + ( 1 p ) f ( y ) f ( s )
in WHCF-Theorem and WPCF-Theorem by the equivalent condition:
h ( x , y ) 0 x , y I , p x + ( 1 p ) y = s .
This equivalence is true since
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 ) .
In the particular case p 1 = p 2 = = p n = 1 / n , this condition becomes
h ( x , y ) 0 x , y 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 = = x n = s . In addition, on the assumption that
p = min { p 1 , p 2 , , p n } ,
the equality also holds for x 1 = x and x 2 = = x n = y if there exist x , y I , x 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 , , n , define the numbers y i I as
y i = { s 0 , x i s 0 , x i , x i > s 0 .
Since y i x i for i = 1 , 2 , , n , we have
p 1 y 1 + p 2 y 2 + + p n y n p 1 x 1 + p 2 x 2 + + p n x n = s .
In addition, since f ( y i ) f ( x i ) for x i s 0 and f ( y i ) = f ( x i ) for x i > s 0 , we have f ( y i ) f ( x i ) for i = 1 , 2 , , n , and hence
p 1 f ( y 1 ) + p 2 f ( y 2 ) + + p n f ( y n ) p 1 f ( x 1 ) + p 2 f ( x 2 ) + + p n f ( x n ) .
Thus, it suffices to show that
p 1 f ( y 1 ) + p 2 f ( y 2 ) + + p n f ( y n ) f ( s )

for all y 1 , y 2 , , y n s 0 such that p 1 y 1 + p 2 y 2 + + p n y n s . By hypothesis, this inequality is true for y 1 , y 2 , , y n s 0 and p 1 y 1 + p 2 y 2 + + p n y n = s . Since f is increasing for u I , u s 0 , the more we have p 1 f ( y 1 ) + p 2 f ( y 2 ) + + p n f ( y n ) f ( s ) for y 1 , y 2 , , y n s 0 and p 1 y 1 + p 2 y 2 + + p n y n s . □

Proof of Lemma 2.2 For i = 1 , 2 , , n , define the numbers y i I as follows:
y i = { x i , x i s 0 , s 0 , x i > s 0 .
We have y i s 0 , y i x i and f ( y i ) f ( x i ) for i = 1 , 2 , , n . Therefore,
p 1 y 1 + p 2 y 2 + + p n y n p 1 x 1 + p 2 x 2 + + p n x n = s
and
p 1 f ( y 1 ) + p 2 f ( y 2 ) + + p n f ( y n ) p 1 f ( x 1 ) + p 2 f ( x 2 ) + + p n f ( x n ) .
Thus, it suffices to show that
p 1 f ( y 1 ) + p 2 f ( y 2 ) + + p n f ( y n ) f ( s )

for all y 1 , y 2 , , y n s 0 such that p 1 y 1 + p 2 y 2 + + p n y n s . By hypothesis, this inequality is true for y 1 , y 2 , , y n s 0 and p 1 y 1 + p 2 y 2 + + p n y n = s . Since f is decreasing for u I , u s 0 , we have also p 1 f ( y 1 ) + p 2 f ( y 2 ) + + p n f ( y n ) f ( s ) for y 1 , y 2 , , y n s 0 and p 1 y 1 + p 2 y 2 + + p n y n s . □

4 Applications

Application 4.1 Let x 1 , x 2 , , x n n n 2 ( n 3 ) such that
x 1 + x 2 + + x n = n .
If k n ( 3 n 4 ) ( n 2 ) 2 , then
1 x 1 k + x 1 2 + 1 x 2 k + x 2 2 + + 1 x n k + x n 2 0 ,

with equality for x 1 = x 2 = = x n = 1 , and also for x 1 = n n 2 and x 2 = = x n = n n 2 (or any cyclic permutation).

Proof

Rewrite the desired inequality as
f ( x 1 ) + f ( x 2 ) + + f ( x n ) n f ( s ) ,
where
s = 1 , f ( u ) = 1 u k + u 2 , u I = [ n n 2 , n ( 2 n 3 ) n 2 ] .
We have
f ( u ) = u 2 2 u k ( u 2 + k ) 2 , f ( u ) = 2 f 1 ( u ) ( u 2 + k ) 3 ,
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: k + 1 2 ( n 1 ) 2 n 2 . For u I , u 1 , we have
f 1 ( u ) > ( k + 1 ) ( u 1 ) ( u 1 ) 3 = ( u 1 ) [ k + 1 ( u 1 ) 2 ] 0 ,
since
u 1 n ( 2 n 3 ) n 2 1 = 2 ( n 1 ) 2 n 2 k + 1 .
Therefore, f is convex for u I , u 1 . By HCF-Theorem, we only need to show that f ( x ) + ( n 1 ) f ( y ) n f ( 1 ) for all x , y I which satisfy x + ( n 1 ) y = n . According to Remark 2.4, this is true if h ( x , y ) 0 for x , y I and x + ( n 1 ) y = n , where
h ( x , y ) = g ( x ) g ( y ) x y , g ( u ) = f ( u ) f ( 1 ) u 1 .
Indeed, we have
g ( u ) = 1 u 2 + k
and
h ( x , y ) = x + y ( x 2 + k ) ( y 2 + k ) = n + ( n 2 ) x ( n 1 ) ( x 2 + k ) ( y 2 + k ) 0 .
Case 2: 2 ( n 1 ) n 2 k + 1 < 2 ( n 1 ) 2 n 2 . Since
1 1 + k n n 2
and
1 + 1 + k < 1 + 2 ( n 1 ) 2 n 2 = n ( 2 n 3 ) n 2 ,
from the expression of f it follows that f is decreasing on [ n n 2 , s 0 ] and increasing on [ s 0 , n ( 2 n 3 ) n 2 ] , where
s 0 = 1 + k + 1 > 1 .
On the other hand, for u [ 1 , s 0 ] , we have
f 1 ( u ) > ( k + 1 ) ( u 1 ) ( u 1 ) 3 = ( u 1 ) [ k + 1 ( u 1 ) 2 ] 0 ,
since
u 1 s 0 1 = 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 ) n f ( 1 ) for all x , y I such that x + ( n 1 ) y = n . We have proved this before (at Case 1). □

Application 4.2 If x 1 , x 2 , , x n ( n 3 ) are real numbers such that
x 1 + x 2 + + x n = n ,
then [6]
i = 1 n n ( n + 1 ) 2 x i n 2 + ( n 2 ) x i 2 n ,

with equality for x 1 = x 2 = = x n = 1 , and also for x 1 = n and x 2 = = x n = 0 (or any cyclic permutation).

Proof The desired inequality is true for x 1 > n ( n + 1 ) 2 since
n ( n + 1 ) 2 x 1 n 2 + ( n 2 ) x 1 2 < 0
and
n ( n + 1 ) 2 x i n 2 + ( n 2 ) x i 2 < n n 1 , i = 2 , 3 , , n .
Consider further that x 1 , x 2 , , x n n ( n + 1 ) 2 and rewrite the desired inequality as
f ( x 1 ) + f ( x 2 ) + + f ( x n ) n f ( s ) ,
where
s = 1 , f ( u ) = 2 u n ( n + 1 ) ( n 2 ) u 2 + n 2 , u I = [ n ( 3 n 2 ) 2 , n ( n + 1 ) 2 ] .
We have
f ( u ) 2 ( n 2 ) = n 2 + n ( n + 1 ) u u 2 [ ( n 2 ) u 2 + n 2 ] 2
and
f ( u ) 2 ( n 2 ) = 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 [ n ( 3 n 2 ) 2 , s 0 ] and increasing on [ s 0 , n ( n + 1 ) 2 ] , where
s 0 = n 2 ( n + 1 n 2 + 2 n + 5 ) ( 1 , 0 ) .
On the other hand, for 1 u 1 , we have
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 ,
and hence f ( u ) > 0 . Since [ s 0 , s ] [ 1 , 1 ] , f is convex on [ s 0 , s ] . By PCF-Theorem, we only need to show that f ( x ) + ( n 1 ) f ( y ) n f ( 1 ) for all x , y I which satisfy x + ( n 1 ) y = n . According to Remark 2.4, this is true if h ( x , y ) 0 for x , y I and x + ( n 1 ) y = n , where
h ( x , y ) = g ( x ) g ( y ) x y , g ( u ) = f ( u ) f ( 1 ) u 1 .
Indeed, we have
g ( u ) = ( n 2 ) u + n ( n 2 ) u 2 + n 2
and
h ( x , y ) n 2 = n 2 n ( x + y ) ( n 2 ) x y [ ( n 2 ) x 2 + n 2 ] [ ( n 2 ) y 2 + n 2 ] = ( n 1 ) ( n 2 ) y 2 [ ( n 2 ) x 2 + n 2 ] [ ( n 2 ) y 2 + n 2 ] 0 .

 □

Application 4.3 Let x 1 , x 2 , , x n ( n 2 ) be positive real numbers such that
x 1 + x 2 + + x n n .
If k > 1 , then [6]
x 1 x 1 k + x 2 + + x n + x 2 x 1 + x 2 k + + x n + + x n x 1 + x 2 + + x n k 1 ,

with equality for x 1 = x 2 = = x n = 1 .

Proof

Using the substitutions
s = x 1 + x 2 + + x n n ,
and
y 1 = x 1 s , y 2 = x 2 s , , y n = x n s ,
the desired inequality becomes
y 1 s k 1 y 1 k + y 2 + + y n + + y n y 1 + y 2 + + s k 1 y n k 1 ,
where s 1 and y 1 + y 2 + + y n = n . Clearly, if this inequality holds for s = 1 , then it holds for any s 1 . Therefore, we need only to consider the case s = 1 , when x 1 + x 2 + + x n = n , and the desired inequality is equivalent to
x 1 x 1 k x 1 + n + x 2 x 2 k x 2 + n + + x n x n k x n + n 1 .

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

Case 1: 1 < k n + 1 . By Bernoulli’s inequality, we have
x 1 k 1 + k ( x 1 1 ) ,
and hence
x 1 k x 1 + n n k + 1 + ( k 1 ) x 1 0 .
Consequently, it suffices to prove that
i = 1 n x i n k + 1 + ( k 1 ) x i 1 .
For k = n + 1 , this inequality is an equality. Otherwise, for 1 < k < n + 1 , we rewrite the inequality as
i = 1 n 1 n k + 1 + ( k 1 ) x i 1 ,
which follows from the AM-HM inequality as follows:
i = 1 n 1 n k + 1 + ( k 1 ) x i n 2 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 ) + + f ( x n ) n f ( s ) ,
where
s = 1 , f ( u ) = u u k u + n , u I = ( 0 , n ) .
We have
f ( u ) = ( k 1 ) u k n ( u k u + n ) 2
and
f ( u ) = 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 = ( n k 1 ) 1 / k < 1 .
On the other hand, for u [ s 0 , s ] , we have
( k 1 ) u k n ( k 1 ) s 0 k n = 0 ,
and hence
f 1 ( u ) 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 ) ] k u k 1 [ 2 ( k 1 ) + 2 ( k + 1 ) ] = 4 k u k 1 > 0 .
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 ) n f ( 1 ) for all positive x, y which satisfy x 1 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 ) n f ( 1 ) as follows:
x x k x n + ( n 1 ) y y k y + n 1 , x k x + n x ( y k y + n ) y k n y + n , x k x ( n 1 ) y ( y y k ) y k n y + n .
Since y < 1 and y k n y + n > y k y + 1 > 0 , it suffices to show that
x k x ( n 1 ) ( y y k ) y k y + 1 ,
which is equivalent to
h ( x ) y y k ( 1 y ) ( y k y + 1 ) ,
where
h ( x ) = x k x x 1 .
By the weighted AM-GM inequality, we have
h ( x ) = ( 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 ) 1 y , we get
h ( x ) h ( 2 y ) = ( 2 y ) k 2 + y 1 y .
Thus, it suffices to show that
( 2 y ) k 2 + y y y k y k y + 1 .
Putting 1 y = t , 0 < t < 1 , we write this inequality as
( 2 y ) k 1 + y 1 y k y + 1 , ( 1 + t ) k t 1 ( 1 t ) k + t , ( 1 t 2 ) k + t ( 1 + t ) k 1 + t 2 + t ( 1 t ) k .
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 ) t ( 1 t ) k , which is clearly true. □

Application 4.4 Let x 1 , x 2 , , x n ( n 2 ) be positive real numbers such that
x 1 + x 2 + + x n = n .
If 0 < k n n 1 , then [6]
x 1 k / x 1 + x 2 k / x 2 + + x n k / x n n

with equality for x 1 = x 2 = = x n = 1 .

Proof

According to the power mean inequality, we have
( x 1 p / x 1 + x 2 p / x 2 + + x n p / x n n ) 1 / p ( x 1 q / x 1 + x 2 q / x 2 + + x n q / x n n ) 1 / q
for all p q > 0 . Thus, it suffices to prove the desired inequality for
k = n n 1 , 1 < k 2 .
Rewrite the desired inequality as
f ( x 1 ) + f ( x 2 ) + + f ( x n ) n f ( s ) ,
where
s = 1 , f ( u ) = u k / u , u I = ( 0 , n ) .
We have
f ( u ) = k u k u 2 ( ln u 1 ) , f ( u ) = k u k u 4 [ u + ( 1 ln u ) ( 2 u k + k ln u ) ] .
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 0 and
2 u k + k ln u 2 k 0 ,
we have f > 0 for u [ 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 n
for 0 < x 1 y and x + ( n 1 ) y = n . We have
k x k > 1 .
Also, from
k y = n ( n 1 ) y > n x + ( n 1 ) y = 1 , k y 2 y 2 ,
we get
0 < k y 1 1 .
Therefore, by Bernoulli’s inequality, we have
x k / x + ( n 1 ) y k / y n = 1 ( 1 x ) k / x + ( n 1 ) y y k / y 1 n 1 1 + k x ( 1 x 1 ) + ( n 1 ) y [ 1 + ( k y 1 ) ( y 1 ) ] n = x 2 x 2 k x + k ( k 1 ) x 2 ( 2 k ) x = ( x 1 ) 2 [ ( k 1 ) x + k ( 2 k ) ] x 2 k x + k 0 .

 □

Application 4.5 If a, b, c are positive real numbers such that a b c = 1 , then
1 a 17 + 4 a + 6 a 2 + 1 b 17 + 4 b + 6 b 2 + 1 c 17 + 4 c + 6 c 2 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 ) 3 g ( r ) ,
where r = 1 and
g ( t ) = 1 t 1 + p t + q t 2 , t I = ( 0 , ) ,
where p = 4 17 , q = 6 17 . From
g ( t ) = 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 , ) , where
r 0 = 1 + 1 + p + 1 q > 1 .
We have
f ( u ) = g ( e u ) = 1 e u 1 + p e u + q e 2 u , f ( u ) = t h ( t ) ( 1 + p t + q t 2 ) 3 ,
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 [ r , r 0 ] , and hence f is convex for
e u [ r , r 0 ] = [ 1 , 1 + 1 + p + 1 q ] .
We have
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 ] .
Since
h ( t ) = 6 q [ 2 ( q t 2 + 2 q t + p + 1 ) + p ( t 1 ) ] 12 q ( q t 2 + 2 q t + p + 1 ) 0 ,
h ( t ) is increasing,
h ( t ) h ( 1 ) = p 2 + 9 p q + 8 q 2 + p + 8 q > 0 ,
h is increasing, and hence
h ( t ) 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 .
By PCF-Corollary, we only need to prove that g ( a ) + 2 g ( b ) 3 g ( 1 ) for a b 2 = 1 ; that is,
1 a 1 + p a + q a 2 + 2 ( 1 b ) 1 + p b + q b 2 0 ,
b 2 ( b 2 1 ) b 4 + p b 2 + q + 2 ( 1 b ) 1 + p b + q b 2 0 ,
p A + q B C ,
where
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 ) .
Indeed, we have
17 ( p A + q B C ) = 3 ( b 1 ) 2 ( b 2 ) 2 ( 2 b 2 + 2 b + 1 ) 0 .

 □

Application 4.6 If a, b, c are positive real numbers such that a b c = 1 , then [6]
7 6 a 2 + a 2 + 7 6 b 2 + b 2 + 7 6 c 2 + c 2 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 ) 3 g ( r ) ,
where r = 1 and
g ( t ) = 7 6 t 2 + t 2 , t I = ( 0 , ) .
From
g ( t ) = 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 , ) , where
r 0 = 3 .
We have
f ( u ) = g ( e u ) = 7 6 e u 2 + e 2 u , f ( u ) = 2 t h ( t ) ( 2 + t 2 ) 3 ,
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 [ r , r 0 ] , and hence f is convex for
e u [ r , r 0 ] = [ 1 , 3 ] .
We have
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 .
By PCF-Corollary, we only need to prove that g ( a ) + 2 g ( b ) 3 g ( 1 ) for a b 2 = 1 ; that is,
7 6 a 2 + a 2 + 2 ( 7 6 b ) 2 + b 2 1 , b 2 ( 7 b 2 6 ) 2 b 4 + 1 + 2 ( 7 6 b ) 2 + b 2 1 , ( b 1 ) 2 ( b 2 ) 2 ( 5 b 2 + 6 b + 3 ) 0 .

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 [ 13 3 3 , 13 3 3 ] ,
then [6]
a + k a 2 + 1 + b + k b 2 + 1 + c + k c 2 + 1 3 ( 1 + k ) 2 ,

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

Proof

The desired inequality is equivalent to
( a 1 ) 2 a 2 + 1 k ( 2 a 2 + 1 3 ) .
Thus, it suffices to prove this inequality for only | k | = 13 3 3 . On the other hand, replacing a, b, c by 1 / a , 1 / b , 1 / c , the inequality becomes
( a 1 ) 2 a 2 + 1 k ( 3 2 a 2 + 1 ) .
Thus, we only need to prove the desired inequality for k = 13 3 3 . Write this inequality as
g ( a ) + g ( b ) + g ( c ) 3 g ( r ) ,
where r = 1 and
g ( t ) = t k t 2 + 1 , t I = ( 0 , ) .
From
g ( t ) = 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 , ) , where
r 0 = 3 9 .
We have
f ( u ) = g ( e u ) = e u k e 2 u + 1 , f ( u ) = t h ( t ) ( t 2 + 1 ) 3 ,
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 [ r 0 , r ] , and hence f is convex for
e u [ r 0 , r ] = [ 3 9 , 1 ] .
Indeed, since
4 k t = 52 t 3 3 = 52 27 > 1 ,
we have
h ( t ) = t 4 + 6 t 2 1 + 4 k t ( 1 t 2 ) 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 ) 3 g ( 1 ) for a b 2 = 1 ; that is,
a + k a 2 + 1 + 2 ( b + k ) b 2 + 1 3 ( 1 + k ) 2 , b 2 ( k b 2 + 1 ) b 4 + 1 + 2 ( b + k ) b 2 + 1 3 ( 1 + k ) 2 , 3 b 6 4 b 5 + b 4 + b 2 4 b + 3 k ( 1 b 2 ) 3 0 , ( b 1 ) 2 [ ( 3 + k ) b 4 + 2 ( 1 + k ) b 3 + 2 b 2 + 2 ( 1 k ) b + 3 k ] 0 , ( b 1 ) 2 ( b 2 + 3 ) 2 [ ( 27 + 13 3 ) b 2 + 24 ( 2 + 3 ) b + 33 + 17 3 ] 0 .

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 2 + 2 2 , then [6]
a 3 k a 2 + b c + b 3 k b 2 + c a + c 3 k c 2 + a b a + b + c k + 1 ,

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

Proof For k < 2 + 2 2 , the proof is similar to the one of the main case k = 2 + 2 2 . For this reason, we consider further only the case where
k = 2 + 2 2 .
Due to homogeneity, we may assume that a b c = 1 . On this hypothesis,
a 3 k a 2 + b c 1 k + 1 a = ( a 4 k a 3 + 1 a k + 1 ) = 1 k + 1 a 4 a k a 3 + 1 .
Thus, we can write the inequality as
g ( a ) + g ( b ) + g ( c ) 3 g ( r ) ,
where r = 1 and
g ( t ) = t 4 t k t 3 + 1 , t I = ( 0 , ) .
From
g ( t ) = 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 , ) , where
r 0 = k 2 + ( k + 1 ) ( k + 4 ) k 3 0.4149 .
We have
f ( u ) = g ( e u ) = e 4 u e u k e 3 u + 1 , f ( u ) = t h ( t ) ( k t 3 + 1 ) 3 ,
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 ) 0 for t [ t 1 , t 2 ] , where t 1 0.2345 and t 2 1.02 . Since
[ r 0 , r ] [ t 1 , t 2 ] ,
f is convex for e u [ r 0 , r ] . Then, by PCF-Corollary, it suffices to show that g ( a ) + 2 g ( b ) 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
a 3 k a 2 + 1 + 2 a + k a + 2 k + 1 ,
which is equivalent to the obvious inequality
( a 1 ) 2 ( a 2 ) 2 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]
1 a 1 1 + a 1 2 + 1 a 2 1 + a 2 2 + 1 a 3 1 + a 3 2 + 1 a 4 1 + a 4 2 + 1 a 5 1 + a 5 2 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 ) 3 g ( r ) ,
where r = 1 and
g ( t ) = 1 t 1 + t 2 , t I = ( 0 , ) .
From
g ( t ) = t 2 2 t 1 ( t 2 + 1 ) 2 ,
it follows that g is decreasing on ( 0 , r 0 ] and increasing on [ r 0 , ) , where
r 0 = 1 + 2 .
We have
f ( u ) = g ( e u ) = 1 e u 1 + e 2 u , f ( u ) = t h ( t ) ( t 2 + 1 ) 3 ,
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 [ r , r 0 ] , and hence f is convex for
e u [ r , r 0 ] = [ 1 , 1 + 2 ] .
Indeed,
h ( t ) t 4 + 6 t 2 1 = 8 ( 3 t 2 ) 2 4 .
By PCF-Corollary, we only need to prove that g ( a ) + 4 g ( b ) 5 g ( 1 ) for a b 4 = 1 ; that is,
1 a 1 + a 2 + 4 ( 1 b ) 1 + b 2 0 , b 4 ( b 4 1 ) 1 + b 8 + 4 ( 1 b ) 1 + b 2 0 , 1 + 4 ( 1 b ) 1 + b 2 1 + b 4 1 + b 8 .
Since
1 + b 4 1 + b 8 2 1 + b 4 4 ( 1 + b 2 ) 2 ,
it suffices to show that
1 + 4 ( 1 b ) 1 + b 2 4 ( 1 + b 2 ) 2 .

This inequality is equivalent to ( 1 b ) 4 0 , and the proof is completed. □

Remark 4.1

The inequality
1 a 1 1 + a 1 2 + 1 a 2 1 + a 2 2 + 1 a 3 1 + a 3 2 + 1 a 4 1 + a 4 2 + 1 a 5 1 + a 5 2 + 1 a 6 1 + a 6 2 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
a 1 ( a 1 + 1 ) 1 + a 1 2 0 ,
which is false for
a 1 = 1 a 2 a 3 a 4 a 5 a 6 = 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

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Copyright

© Cirtoaje; licensee Springer 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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