Open Access

An extension of Jensen's discrete inequality to half convex functions

Journal of Inequalities and Applications20112011:101

https://doi.org/10.1186/1029-242X-2011-101

Received: 17 April 2011

Accepted: 31 October 2011

Published: 31 October 2011

Abstract

We extend the right and left convex function theorems to weighted Jensen's type inequalities, and then combine the new theorems in a single one applicable to a half convex function f(u), defined on a real interval I and convex for us or us, where s I . The obtained results are applied for proving some open relevant inequalities.

Keywords

weighted Jensen's discrete inequalityright convex functionleft convex functionhalf convex function

1 Introduction

The right convex function theorem (RCF-Theorem) has the following statement (see [13]).

RCF-Theorem. Let f(u) be a function defined on a real interval I and convex for u s I . The inequality
f ( x 1 ) + f ( x 2 ) + + f ( x n ) n f ( x 1 + x 2 + + x n n )
(1)
holds for all x 1 , x 2 , , x n I satisfying x1 + x2 + ... + x n ns if and only if
f ( x ) + ( n - 1 ) f ( y ) n f ( s )
(2)

for all x , y I which satisfy xsy and x + (n - 1)y = ns.

Replacing f(u) by f(-u), s by -s, x by -x, y by -y, and each x i by -x i for i = 1, 2, ..., n, from RCF-Theorem we get the left convex function theorem (LCF-Theorem).

LCF-Theorem. Let f(u) be a function defined on a real interval I and convex for u s I . The inequality
f ( x 1 ) + f ( x 2 ) + + f ( x n ) n f ( x 1 + x 2 + + x n n )
(3)
holds for all x 1 , x 2 , , x n I satisfying x1 + x2 + ... + x n ns if and only if
f ( x ) + ( n - 1 ) f ( y ) n f ( s )
(4)

for all x , y I which satisfy xsy and × + (n - 1)y = ns.

Notice that from RCF- and LCF-Theorems, we get the following theorem, which we have called the half convex function theorem (HCF-Theorem).

HCF-Theorem. Let f(u) be a function defined on a real interval I and convex for us or us, where s I . The inequality
f ( x 1 ) + f ( x 2 ) + + f ( x n ) n f ( x 1 + x 2 + + x n n )
(5)
holds for all x 1 , x 2 , , x n I satisfying x1 + x2 + ... + x n = ns if and only if
f ( x ) + ( n - 1 ) f ( y ) n f ( s )
(6)

for all x , y I which satisfy x + (n - 1)y = ns.

Applying RCF-, LCF-, and HCF-Theorems to the function f(u) = g(e u ), and replacing s by ln r, x by ln x, y by ln y, and each x i by ln a i for i = 1, 2, ..., n, we get the following corollaries, respectively.

RCF-Corollary. Let g be a function defined on a positive interval I such that f(u) = g(e u ) is convex for e u r I . The inequality
g ( a 1 ) + g ( a 2 ) + + g ( a n ) n g ( a 1 a 2 a n n )
(7)
holds for all a 1 , a 2 , , a n I satisfying a1a2 ... a n r n if and only if
g ( a ) + ( n - 1 ) g ( b ) n g ( r )
(8)

for all a , b I which satisfy arb and abn-1= r n .

LCF-Corollary. Let g be a function defined on a positive interval I such that f(u) = g(e u ) is convex for e u r I . The inequality
g ( a 1 ) + g ( a 2 ) + + g ( a n ) n g ( a 1 a 2 a n n )
(9)
holds for all a 1 , a 2 , , a n I satisfying a1a2 ... a n r n if and only if
g ( a ) + ( n - 1 ) g ( b ) n g ( r )
(10)

for all a , b I which satisfy arb and abn-1= r n .

HCF-Corollary. Let g be a function defined on a positive interval I such that 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 )
(11)
holds for all a 1 , a 2 , , a n I satisfying a1a2 ... a n = r n if and only if
g ( a ) + ( n - 1 ) g ( b ) n g ( r )
(12)

for all a , b I which satisfy abn-1= r n .

2 Main results

In order to extend RCF-, LCF-, and HCF-Theorems to weighted Jensen's type inequalities, we need the following lemma.

Lemma 2.1 Let q1, q2and r1, r2, ..., r m be nonnegative real numbers such that
r 1 + r 2 + + r m = q 1 + q 2 ,
(13)
and let f be a convex function on I . If a , b I (ab) and x1, x2, ..., x m [a, b] such that
r 1 x 1 + r 2 x 2 + + r m x m = q 1 a + q 2 b ,
(14)
then
r 1 f ( x 1 ) + r 2 f ( x 2 ) + + r m f ( x m ) q 1 f ( a ) + q 2 f ( b ) .
(15)

The weighted right convex function theorem (WRCF-Theorem), weighted left convex function theorem (WLCF-Theorem), and weighted half convex function theorem (WHCF-Theorem) are the following.

WRCF-Theorem. Let f(u) be a function defined on a real interval I and convex for u s I , and let p1, p2, ..., p n be positive real numbers such that
p = min { p 1 , p 2 , , p n } , p 1 + p 2 + + p n = 1 .
(16)
The inequality
p 1 f ( x 1 ) + p 2 f ( x 2 ) + + p n f ( x n ) f ( p 1 x 1 + p 2 x 2 + + p n x n )
(17)
holds for all x 1 , x 2 , , x n I satisfying p1x1 + p2x2 + ... + p n x n s if and only if
p f ( x ) + ( 1 - p ) f ( y ) f ( s )
(18)

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

WLCF-Theorem. Let f(u) be a function defined on a real interval I and convex for u s I , and let p1, p2 , ..., p n be positive real numbers such that
p = min { p 1 , p 2 , , p n } , p 1 + p 2 + + p n = 1 .
(19)
The inequality
p 1 f ( x 1 ) + p 2 f ( x 2 ) + + p n f ( x n ) f ( p 1 x 1 + p 2 x 2 + + p n x n )
(20)
holds for all x 1 , x 2 , , x n I satisfying p1x1 + p2x2 + ... + p n x n s if and only if
p f ( x ) + ( 1 - p ) f ( y ) f ( s )
(21)

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

WHCF-Theorem. Let f(u) be a function defined on a real interval I and convex for us or us, where s I , and let p1, p2, ..., p n be positive real numbers such that
p = min { p 1 , p 2 , , p n } , p 1 + p 2 + + p n = 1 .
(22)
The inequality
p 1 f ( x 1 ) + p 2 f ( x 2 ) + + p n f ( x n ) f ( p 1 x 1 + p 2 x 2 + + p n x n )
(23)
holds for all x 1 , x 2 , , x n I satisfying p1x1 + p2x2 + ... + p n x n = s if and only if
p f ( x ) + ( 1 - p ) f ( y ) f ( s )
(24)

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

Notice that WLCF-Theorem can be obtained from WRCF-Theorem replacing f(u) by f(-u), s by -s, x by -x, y by -y, and each x i by -x i for i = 1,2, ..., n.

On the other hand, applying WRCF-, WLCF-, and WHCF-Theorems to the function f(u) = g(e u ) and replacing s by 1nr, x by 1nx, y by 1ny, and each x i by In a i for i = 1, 2, ... n, we get the following corollaries, respectively.

WRCF-Corollary Let g be a function defined on a positive interval I such that f(u) = g(e u ) is convex for ln u r I , and let p1, p2, ..., p n be positive real numbers such that
p = min { p 1 , p 2 , , p n } , p 1 + p 2 + + p n = 1 .
(25)
The inequality
p 1 g ( a 1 ) + p 2 g ( a 2 ) + + p n g ( a n ) g ( a 1 p 1 a 2 p 2 a n p n )
(26)
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 )
(27)

for all a , b I such that arb and a p b1-p= r.

WLCF-Corollary. Let g be a function defined on a positive interval I such that f(u) = g(e u ) is convex for e u r I , aand let p1, p2, ..., p n be positive real numbers such that
p = min { p 1 , p 2 , , p n } , p 1 + p 2 + + p n = 1 .
(28)
The inequality
p 1 g ( a 1 ) + p 2 g ( a 2 ) + + p n g ( a n ) g ( a 1 p 1 a 2 p 2 a n p n )
(29)
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 )
(30)

for all a , b I such that arb and a p b1-p= r.

WHCF-Corollary. Let g be a function defined on a positive interval I such that f(u) = g(e u ) is convex for e u r or e u r, where r I , and let p1, p2, ..., p n be positive real numbers such that
p = min { p 1 , p 2 , , p n } , p 1 + p 2 + + p n = 1 .
(31)
The inequality
p 1 g ( a 1 ) + p 2 g ( a 2 ) + + p n g ( a n ) g ( a 1 p 1 a 2 p 2 a n p n )
(32)
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 )
(33)

for all a , b I such that a p b1-p= r.

Remark 2.2. Let us denote
g ( u ) = f ( u ) - f ( s ) u - s , h ( x , y ) = g ( x ) - g ( y ) x - y .
(34)
In some applications, it is useful to replace the hypothesis
p f ( x ) + ( 1 - p ) f ( y ) f ( s )
(35)

in WRCF-, WLCF-, and WHCF-Theorems by the equivalent condition:

h(x, y) ≥ 0 for all x , y I such that px + (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 ) .

Remark 2.3. The required inequalities in WRCF-, WLCF-, and WHCF-Theorems turn into equalities for x1 = x2 = ... = x n . In addition, on the assumption that p1 = min{p1, p2, ..., p n }, equality holds for x1 = x and x2 = ... = x n = y if there exist x , y I , xy such that px + (1 - p)y = s and pf(x) + (1 - p)f(y) = f(s).

3 Proof of Lemma 2.1

Consider only the nontrivial case a < b. Since x1, x2, ..., x m [a, b] there exist λ1, λ2, ..., λ m [0, 1] such that
x i = λ i a + ( 1 - λ i ) b , i = 1 , 2 , , m .
From
λ i = x i - b a - b , i = 1 , 2 , , m ,
we have
i = 1 m r i λ i = 1 a - b ( i = 1 m r i x i - b i = 1 m r i ) = 1 a - b [ q 1 a + q 2 b - b ( q 1 + q 2 ) ] = q 1 .
Thus, according to Jensen's inequality, we get
i = 1 m r i f ( x i ) i = 1 m r i [ λ i f ( a ) + ( 1 - λ i ) f ( b ) ] = [ f ( a ) - f ( b ) ] i = 1 m r i λ i + f ( b ) i = 1 m r i = [ f ( a ) - f ( b ) ] q 1 + f ( b ) ( q 1 + q 2 ) = q 1 f ( a ) + q 2 f ( b ) .

4 Proof of WRCF-Theorem

Since the necessity is obvious, we prove further the sufficiency. Without loss of generality, assume that x1x2 ≤ ... ≤ x n . If x1s, then the required inequality follows by Jensen's inequality for convex functions. Otherwise, since
p 1 x 1 + p 2 x 2 + + p n x n ( p 1 + p 2 + + p n ) s ,
there exists k {1, 2, ..., n - 1} such that
x 1 x k < s x k + 1 x n .
Let us denote
q = p 1 + + p k ,
By Jensen's inequality, we have
i = k + 1 n p i f ( x i ) ) ( p k + 1 + + p n ) f ( z ) = ( 1 - q ) f ( z ) ,
where
z = p k + 1 x k + 1 + + p n x n p k + 1 + + p n , z s , z I .
Thus, it suffices to prove that
i = 1 k p i f ( x i ) + ( 1 - q ) f ( z ) f ( S ) ,
(36)
where
S = p 1 x 1 + p 2 x 2 + + p n x n = p 1 x 1 + p 2 x 2 + + p n x n p 1 + p 2 + + p n , s S z .
Let y i , i = 1, 2, ..., k, defined by
p x i + ( 1 - p ) y i = s .
We will show that
z y 1 y 2 y k > s .
We have
y 1 y 2 y k , y k - s = p ( s - x k ) 1 - p > 0 , y 1 = s - p x 1 1 - p S - p x 1 1 - p = ( p 1 - p ) x 1 + p 2 x 2 + + p n x n ( p 1 - p ) + p 2 + + p n .
Since p1 - p = p1 - min {p1, p2, ..., p n } ≥ 0, we get
( p 1 - p ) x 1 + p 2 x 2 + + p n x n ( p 1 - p ) + p 2 + + p n p 2 x 2 + + p n x n p 2 + + p n z ,
and hence y1z. Now, from zy1y2 ≥ ... ≥ yk > s, it follows that y 1 , y 2 , , y k I . Then, by hypothesis, we have
p f ( x i ) + ( 1 - p ) f ( y i ) f ( s )
for i = 1,2, ..., k. Summing all these inequalities multiplied by p i /p, respectively, we get
i = 1 k p i f ( x i ) + 1 - p p i = 1 k p i f ( y i ) q p f ( s ) .
Therefore, to prove (36), it suffices to show that
q p f ( s ) + ( 1 - q ) f ( z ) 1 - p p i = 1 k p i f ( y i ) + f ( S ) .
(37)
Since S [s, z], y i (s, z] for i = 1, 2, ..., k,
q p + 1 - q = 1 - p p i = 1 k p i + 1 , q p s + ( 1 - q ) z = 1 - p p i = 1 k p i y i + S ,
  1. (37)

    is a consequence of Lemma 2.1, where m = k + 1, a = s, b = z, q 1 = q/p, q 2 = 1 - q, r m = 1, x m = S, r i = (1 - p)p i /p and x i = y i for i = 1, 2, ..., k.

     

5 Applications

Proposition 5.1. Let a1, a2, ..., a n (n ≥ 3) be positive real numbers such that a1a2 ... a n = 1. If p and q are nonnegative real numbers such that p + qn - 1, then[4]
1 1 + p a 1 + q a 1 2 + 1 1 + p a 2 + q a 2 2 + + 1 1 + p a n + q a n 2 n 1 + p + q .
Proof. Write the desired inequality as
g ( a 1 ) + g ( a 2 ) + + g ( a n ) n g ( 1 ) ,
where
g ( t ) = 1 1 + p t + q t 2 , t > 0 .
To prove this inequality, we apply HCF-Corollary for r = 1. Let
f ( u ) = g ( e u ) = 1 1 + p e u + q e 2 u , u .
Using the second derivative,
f ( u ) = e u [ 4 q 2 e 3 u + 3 p q e 2 u + ( p 2 - 4 q ) e u - p ] ( 1 + p e u + q e 2 u ) 3 ,
we will show that f(u) is convex for e u r = 1. We need to show that
4 q 2 t 3 + 3 p q t 2 + ( p 2 - 4 q ) t - p 0
for t ≥ 1. Indeed,
4 q 2 t 3 + 3 p q t 2 + ( p 2 - 4 q ) t - p ( 4 q 2 + 3 p q + p 2 - 4 q - p ) t = [ ( p + 2 q ) ( p + q - 2 ) + 2 q 2 + p ] t > 0 ,

because p + qn - 1 ≥ 2.

By HCF-Corollary, it suffices to prove that g(a) + (n - 1)g(b) ≥ ng(1) for all a, b > 0 such that abn-1= 1. We write this inequality as
b 2 n - 2 b 2 n - 2 + p b n - 1 + q + n - 1 1 + p b + q b 2 n 1 + p + q .
Applying the Cauchy-Schwarz inequality, it suffices to prove that
( b n - 1 + n - 1 ) 2 ( b 2 n - 2 + p b n - 1 + q ) + ( n - 1 ) ( 1 + p b + q b 2 ) n 1 + p + q ,
which is equivalent to
p B + q C A ,
where
A = ( n - 1 ) ( b n - 1 - 1 ) 2 0 , B = ( b n - 1 - 1 ) 2 + n E = A n - 1 + n E , C = ( b n - 1 - 1 ) 2 + n F = A n - 1 + n F ,
with
E = b n - 1 + n - 2 - ( n - 1 ) b , F = 2 b n - 1 + n - 3 - ( n - 1 ) b 2 .
By the AM-GM inequality applied to n - 1 positive numbers, we have E ≥ 0 and F ≥ 0 for n ≥ 3. Since A ≥ 0 and p + qn - 1, we have
p B + q C - A p B + q C - ( p + q ) A n - 1 = n ( p E + q F ) 0 .

Equality holds for a1 = a2 = ... = a n = 1.

Remark 5.2. For p + q = n - 1 and n ≥ 3, by Proposition 5.1 we get the following beautiful inequality
1 1 + p a 1 + q a 1 2 + 1 1 + p a 2 + q a 2 2 + + 1 1 + p a n + q a n 2 1 .
If p = n - 1 and q = 0, then we get the well-known inequality
1 1 + ( n - 1 ) a 1 + 1 1 + ( n - 1 ) a 2 + + 1 1 + ( n - 1 ) a n 1 .
Remark 5.3. For p = q = 1 r , 0 < r 2 n - 1 and n ≥ 3, by Proposition 5.1 we obtain the inequality
1 r + a 1 + a 1 2 + 1 r + a 2 + a 2 2 + + 1 r + a n + a n 2 n r + 2 .
In addition, for r = 2 n - 1 , n ≥ 3, we get
i = 1 n 1 2 + ( n - 1 ) ( a i + a i 2 ) 1 2 .
Remark 5.4. For p = 2r, q = r2, r n - 1 and n ≥ 3, by Proposition 5.1 we obtain
1 ( 1 + r a 1 ) 2 + 1 ( 1 + r a 2 ) 2 + + 1 ( 1 + r a n ) 2 n ( 1 + r ) 2 .
Proposition 5.5. Let a1, a2, ..., a n (n ≥ 4) be positive real numbers such that a1a2 ... a n = 1. If p, q, r are nonnegative real numbers such that p + q + rn - 1, then[4]
i = 1 n 1 1 + p a i + q a i 2 + r a i 3 n 1 + p + q + r .
Proof. Write the required inequality as
g ( a 1 ) + g ( a 2 ) + + g ( a n ) n g ( 1 ) ,
where
g ( t ) = 1 1 + p t + q t 2 + r t 3 , t > 0 ,
and apply HCF-Corollary to g(t) for r = 1. Let
f ( u ) = g ( e u ) = 1 1 + p e u + q e 2 u + r e 3 u ,
defined on . For n ≥ 4, which implies p + q + r ≥ 3, we claim that f is convex for e u ≥ 1. Since
f ( u ) = t [ 9 r 2 t 5 + 1 1 q r t 4 + ( 2 p r + 4 q 2 ) t 3 + ( 3 p q - 9 r ) t 2 + ( p 2 - 4 q ) t - p ] ( 1 + p t + q t 2 + r t 3 ) 3 ,
where t = e u ≥ 1, we need to show that
9 r 2 t 5 + 1 1 q r t 4 + ( 2 p r + 4 q 2 ) t 3 + ( 3 p q - 9 r ) t 2 + ( p 2 - 4 q ) t - p 0
Since
9 r 2 t 5 + 1 1 q r t 4 + ( 2 p r + 4 q 2 ) t 3 - p ( 9 r 2 + 1 1 q r + 2 p r + 4 q 2 ) t 3 - p t ,
it suffices to show that
( 9 r 2 + 1 1 q r + 2 p r + 4 q 2 ) t 2 + ( 3 p q - 9 r ) t + p 2 - p - 4 q 0 .
Using the inequality t2 ≥ 2t - 1, we still have to prove that At + B ≥ 0, where
A = 1 8 r 2 + 2 2 q r + 4 p r + 8 q 2 + 3 p q - 9 r , B = - 9 r 2 - 1 1 q r - 2 p r - 4 q 2 + p 2 - p - 4 q .
Since p + q + r ≥ 3, we have
A 1 8 r 2 + 2 2 q r + 4 p r + 8 q 2 + 3 p q - 3 r ( p + q + r ) = 1 5 r 2 + 1 9 q r + p r + 8 q 2 + 3 p q 0 .
Therefore,
A t + B A + B = p 2 + 4 q 2 + 9 r 2 + 3 p q + 1 1 q r + 2 p r - ( p + 4 q + 9 r ) p 2 + 4 q 2 + 9 r 2 + 3 p q + 1 1 q r + 2 p r - ( p + 4 q + 9 r ) ( p + q + r ) 3 = 2 ( p - r ) 2 + 9 q 2 + 1 6 r 2 + 4 p q + 2 0 q r 3 > 0 .
According to HCF-Corollary, it suffices to prove that g(a) + (n - 1) g(b) ≥ ng(1) for all a, b > 0 such that abn-1= 1. We write this inequality as
b 3 n - 3 b 3 n - 3 + p b 2 n - 2 + q b n - 1 + r + n - 1 1 + p b + q b 2 + r b 3 n 1 + p + q + r ,
or
p 2 A 1 1 + q 2 A 2 2 + r 2 A 3 3 + p q A 1 2 + q r A 2 3 + r p A 3 1 A p + B q + C r ,
where
A 1 1 = b 2 n - 2 ( b n - n b + n - 1 ) , A 2 2 = b n - 1 ( b 2 n - n b 2 + n - 1 ) , A 3 3 = b 3 n - n b 3 + n - 1 ,
A 1 2 = b 3 n - 1 + b 3 n - 2 + ( n - 1 ) ( b 2 n - 2 + b n - 1 ) - n ( b 2 n + b n ) , A 2 3 = b 3 n + b 3 n - 1 + ( n - 1 ) ( b n - 1 + 1 ) - n ( b n + 2 + b 2 ) , A 3 1 = b 3 n + b 3 n - 2 + ( n - 1 ) ( b 2 n - 2 + 1 ) - n ( b 2 n + 1 + b ) ,
A = b 2 n - 2 [ ( n - 1 ) b n - n b n - 1 + 1 ] , B = b n - 1 [ ( n - 1 ) b 2 n - n b 2 n - 2 + 1 ] , C = ( n - 1 ) b 3 n - n b 3 n - 3 + 1 .
Since A,B,C ≥ 0 (from the AM-GM inequality applied to n positive numbers) and p + q + rn - 1, it suffices to show that
( n - 1 ) ( p 2 A 1 1 + q 2 A 2 2 + r 2 A 3 3 + p q A 1 2 + q r A 2 3 + r p A 3 1 )
which is equivalent to
p 2 B 1 1 + q 2 B 2 2 + r 2 B 3 3 + p q B 1 2 + q r B 2 3 + r p B 3 1 0 ,
(38)
where
B 1 1 = ( n - 1 ) A 1 1 - A = n b 2 n - 2 [ b n - 1 - ( n - 1 ) b + n - 2 ] , B 2 2 = ( n - 1 ) A 2 2 - B = n b n - 1 [ b 2 n - 2 - ( n - 1 ) b 2 + n - 2 ] , B 3 3 = ( n - 1 ) A 3 3 - C = n [ b 3 n - 3 - ( n - 1 ) b 3 + n - 2 ] ,
B 1 2 = ( n - 1 ) A 1 2 - A - B = n b n - 1 [ 2 b 2 n - 2 - ( n - 1 ) b n + 1 + ( n - 2 ) b n - 1 - ( n - 1 ) b + n - 2 ] = n b 2 n - 2 [ 2 b n - 1 - ( n - 1 ) b 2 + n - 3 ] + n b n - 1 [ b n - 1 - ( n - 1 ) b + n - 2 ] ,
B 2 3 = ( n - 1 ) A 2 3 - B - C = n [ 2 b 3 n - 3 - ( n - 1 ) b n + 2 + ( n - 2 ) b n - 1 - ( n - 1 ) b 2 + n - 2 ] ,
B 3 1 = ( n - 1 ) A 3 1 - C - A = n [ 2 b 3 n - 3 - ( n - 1 ) b 2 n + 1 + ( n - 2 ) b 2 n - 2 - ( n - 1 ) b + n - 2 ] .
We see that B11, B22, B33, B12 ≥ 0 (by the AM-GM inequality applied to n - 1 positive numbers). Also, we have
B 2 3 n = b n - 1 [ 3 b n - 1 - ( n - 1 ) b 3 + n - 4 ] + 2 b n - 1 ( b n - 1 - 1 ) 2 + b 2 n - 2 - ( n - 1 ) b 2 + n - 2 0 ,
since
3 b n - 1 - ( n - 1 ) b 3 + n - 4 0 , b 2 n - 2 - ( n - 1 ) b 2 + n - 2 0
(by the AM-GM inequality applied to n - 1 positive numbers). Using the inequality bn-1- (n - 1)b + n - 2 ≥ 0,we get B31D, where
D = n b n - 1 [ 2 b 2 n - 2 - ( n - 1 ) b n + 2 + ( n - 2 ) b n - 1 - 1 ] .
To prove (38), it suffices to show that p2B11 + r2B33 + prD ≥ 0. This is true if 4B11B33D2; that is,
4 [ b n 1 ( n 1 ) b + n 2 ] [ b 3 n 3 ( n 1 ) b 3 + n 2 ] [ 2 b 2 n 2 ( n 1 ) b n + 2 + ( n 2 ) b n 1 1 ] 2 .
(39)
In the case n = 4, (39) becomes in succession
4 ( b 3 3 b + 2 ) ( b 9 3 b 3 + 2 ) ( b 6 2 b 3 + 1 ) 2 , 4 ( b 1 ) 2 ( b + 2 ) ( b 3 1 ) 2 ( b 3 + 2 ) ( b 3 1 ) 4 , ( b 1 ) 2 ( b 3 1 ) 2 ( 3 b 4 + 5 b 3 3 b 2 + 6 b + 15 ) 0 .

Clearly, the last inequality is true. The inequality (39) also holds for n ≥ 5, but we leave this to the reader to prove. Equality occurs for a1 = a2 = ... = a n = 1.

Remark 5.6. For n = 4 and p + q + r = 3, by Proposition 5.5 we get the following beautiful inequality
i = 1 4 1 1 + p a i + q a i 2 + r a i 3 1 .
In addition, for p = q = r = 1, we get the known inequality ([2])
i = 1 4 1 1 + a i + a i 2 + a i 3 1 .
Conjecture 5.7. Let a1,a2, ..., a n be positive real numbers such that a1a2 ... a n = 1, and let k1,k2, ..., k m be nonnegative real numbers such that k1 + k2 + ... + k m n - 1. If mn - 1, then
i = 1 n 1 1 + k 1 a i + k 2 a i 2 + + k m a i m n 1 + k 1 + k 2 + + k m .
(40)
Remark 5.8. For m = n - 1 and k1 = k2 = ... = k m = 1, (40) turns into the known beautiful inequality ([2])
i = 1 n 1 1 + a i + a i 2 + + a i n - 1 1 .
Remark 5.9. For k 1 = ( m 1 ) r , k 2 = ( m 2 ) r 2 , , k m = ( m m ) r m , (40) turns into the known inequality [1, 2]
i = 1 n 1 ( 1 + r a i ) m n ( 1 + r ) m ,

which holds for 1 ≤ mn - 1 and r n m - 1 . □

Proposition 5.10. If x1, x2, ..., x n are nonnegative real numbers such that
x 1 + 2 x 2 + + n x n = n ( n + 1 ) 2 ,
then
( n - 1 ) ( n + 2 ) x 1 3 + 2 x 2 3 + + n x n 3 - n ( n + 1 ) 2 2 ( n 2 + n - 1 ) x 1 2 + 2 x 2 2 + + n x n 2 - n ( n + 1 ) 2 .
Proof. Since the inequality is trivial for n = 1, consider further that n ≥ 2. Write the inequality as
p 1 f ( x 1 ) + p 2 f ( x 2 ) + + p n f ( x n ) f ( p 1 x 1 + p 2 x 2 + + p n x n ) ,
where
p i = 2 i n ( n + 1 ) , i = 1 , 2 , , n , f ( u ) = ( n - 1 ) ( n + 2 ) u 3 - 2 ( n 2 + n - 1 ) u 2 , u 0 .
The function f(u) is convex for us = 1, since
f ( u ) = 6 ( n - 1 ) ( n + 2 ) u - 4 ( n 2 + n - 1 ) 6 ( n - 1 ) ( n + 2 ) - 4 ( n 2 + n - 1 ) = 2 ( n 2 + n - 4 ) > 0
for u ≥ 1. According to WHCF-Theorem and Remark 2.2, it suffices to prove that h(x, y) ≥ 0 for all x, y ≥ 0 such that px + (1 - p)y = 1, where
p = min { p 1 , p 2 , , p n } = 2 n ( n + 1 ) .
We have
g ( u ) = f ( u ) - f ( 1 ) u - 1 = ( n - 1 ) ( n + 2 ) ( u 2 + u + 1 ) - 2 ( n 2 + n - 1 ) ( u + 1 ) , h ( x , y ) = g ( x ) - g ( y ) x - y = ( n - 1 ) ( n + 2 ) ( x + y + 1 ) - 2 ( n 2 + n - 1 ) .
From px + (1 - p)y = 1, we get
x + y = x + 1 - p x 1 - p = 1 + ( 1 - 2 p ) x 1 - p = n ( n + 1 ) + ( n 2 + n - 4 ) x ( n - 1 ) ( n + 2 ) n ( n + 1 ) ( n - 1 ) ( n + 2 ) ,
and hence
h ( x , y ) ( n - 1 ) ( n + 2 ) n ( n + 1 ) ( n - 1 ) ( n + 2 ) + 1 - 2 ( n 2 + n - 1 ) = 0 .

This completes the proof. Equality holds for a1 = a2 = ... a n = 1, and also for a1 = 0 and a 2 = = a n = n ( n + 1 ) ( n - 1 ) ( n + 2 ) .

Declarations

Acknowledgements

The authors are grateful to the referees for their useful comments.

Authors’ Affiliations

(1)
Department of Automatic Control and Computers, University of Ploiesti

References

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Copyright

© Cirtoaje and Baiesu; licensee Springer. 2011

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