Volume 2010, Article ID 128258, 12 pagesdoi:10.1155/2010/128258 Research Article The Best Lower Bound Depended on Two Fixed Variables for Jensen’s Inequality with Ordered Variables Vasil
Trang 1Volume 2010, Article ID 128258, 12 pages
doi:10.1155/2010/128258
Research Article
The Best Lower Bound Depended on
Two Fixed Variables for Jensen’s Inequality with Ordered Variables
Vasile Cirtoaje
Department of Automatic Control and Computers, University of Ploiesti, 100680 Ploiesti, Romania
Correspondence should be addressed to Vasile Cirtoaje,vcirtoaje@upg-ploiesti.ro
Received 16 June 2010; Accepted 4 November 2010
Academic Editor: R N Mohapatra
Copyrightq 2010 Vasile Cirtoaje This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited
We give the best lower bound for the weighted Jensen’s discrete inequality with ordered variables
applied to a convex function f, in the case when the lower bound depends on f, weights, and two
given variables Furthermore, under the same conditions, we give some sharp lower bounds for the weighted AM-GM inequality and AM-HM inequality
1 Introduction
Let x {x1, x2, , x n } be a sequence of real numbers belonging to an interval I, and let
p {p1, p2, , p n } be a sequence of given positive weights associated to x and satisfying
p1 p2 · · · p n 1 If f is a convex function on I, then the well-known discrete Jensen’s
inequality1 states that
Δf, p, x≥ 0, 1.1 where
Δf, p, x p1f x1 p2f x2 · · · p n f x n − fp1x1 p2x2 · · · p n x n
1.2
is the so-called Jensen’s difference The next refinement of Jensen’s inequality was proven in
2, as a consequence of its Theorem2.1, partii
Δf, p, x≥ max
1≤i<k≤n
p i f x i p k f x k −p i p k
f
p
i x i p k x k
p i p k
≥ 0. 1.3
Trang 2By1.3, for fixed x i and x k, we get
Δf, p, x≥ p i f x i p k f x k −p i p k
f
p
i x i p k x k
p i p k
: Sp,f x i , x k . 1.4
In this paper, we will establish that the best lower bound L p,f x i , x k of Jensen’s difference
Δf, p, x for
x1≤ · · · ≤ x i ≤ · · · ≤ x k ≤ · · · ≤ x n 1.5 has the expression
L p,f x i , x k Q i f x i R k f x k − Q i R k f
Q i x i R k x k
Q i R k
where
Q i p1 p2 · · · p i , R k p k p k1 · · · p n 1.7 Logically, we need to have
L p,f x i , x k ≥ S p,f x i , x k . 1.8 Indeed, this inequality is equivalent to Jensen’s inequality
Q i − p i
f x i R k − p k
f x k p i p k
f
p
i x i p k x k
p i p k
≥ Q i R k f
Q i x i R k x k
Q i R k
.
1.9
2 Main Results
Theorem 2.1 Let f be a convex function on I, and let x1 , x2, , x n ∈ I n ≥ 3 such that
x1≤ x2≤ · · · ≤ x n 2.1
For fixed x i and x k (1 ≤ i < k ≤ n), Jensen’s difference Δf, p, x is minimal when
x1 x2 · · · x i−1 x i , x k1 x k2 · · · x n x k ,
x i1 x i2 · · · x k−1 Q i x i R k x k
Q i R k
,
2.2
Trang 3that is,
Δf, p, x≥ Q i f x i R k f x k − Q i R k f
Q i x i R k x k
Q i R k
: Lp,f x i , x k .
2.3
For proving Theorem2.1, we will need the following three lemmas
Lemma 2.2 Let p, q be nonnegative real numbers, and let f be a convex function on I If a, b, c, d ∈ I
such that c, d ∈ a, b and
then
pf a qfb ≥ pfc qfd. 2.5
Lemma 2.3 Let f be a convex function on I, and let x1 , x2, , x n ∈ I (n ≥ 3) such that
x1≤ x2≤ · · · ≤ x n 2.6
For fixed x i , x i1, , x n , where i ∈ {2, 3, , n − 1}, Jensen’s difference Δf, p, x is minimal when
x1 x2 · · · x i−1 x i 2.7
Lemma 2.4 Let f be a convex function on I, and let x1 , x2, , x n ∈ I (n ≥ 3) such that
x1≤ x2≤ · · · ≤ x n 2.8
For fixed x1, x2, , x k , where k ∈ {2, 3, , n − 1}, Jensen’s difference Δf, p, x is minimal when
x k1 x k2 · · · x n x k 2.9
Applying Theorem 2.1 for f x e x and using the substitutions a1 e x1, a2
e x2, , a n e xn, we obtain
Corollary 2.5 Let
0 < a1≤ · · · ≤ a i ≤ · · · ≤ a k ≤ · · · ≤ a n , 2.10
and let p1, p2, , p n be positive real numbers such that p1 p2 · · · p n 1 Then,
p1a1 p2a2 · · · p n a n − a p1
1a p2
2 · · · a pn n
≥ Q i a i R k a k − Q i R k a Qi/ Q i R k
2.11
Trang 4with equality for
a1 a2 · · · a i , a k a k1 · · · a n ,
a i1 a i2 · · · a k−1 a Qi/ Q i R k
2.12
Using Corollary2.5, we can prove the propositions below
Proposition 2.6 Let
0 < a1≤ · · · ≤ a i ≤ · · · ≤ a k ≤ · · · ≤ a n , 2.13
and let p1, p2, , p n be positive real numbers such that p1 p2 · · · p n 1 If
P
⎧
⎪
⎪
2Q i R k
Q i R k
, Q i ≤ R k ,
R k , Q i ≥ R k ,
2.14
then
p1a1 p2a2 · · · p n a n − a p1
1 a p2
2 · · · a pn
with equality for a1 a2 · · · a n When Q i R k , equality holds again for a1 a2 · · · a i ,
a i1 · · · a k−1 √a i a k , a k a k1 · · · a n
Proposition 2.7 Let
0 < a1≤ · · · ≤ a i ≤ · · · ≤ a k ≤ · · · ≤ a n , 2.16
and let p1, p2, , p n be positive real numbers such that p1 p2 · · · p n 1 Then,
p1a1 p2a2 · · · p n a n − a p1
1a p2
2 · · · a pn
4Q i 2R k a k 2Q i 4R k a i
, 2.17
with equality if and only if a1 a2 · · · a n
Remark 2.8 For p1 p2 · · · p n 1/n, from Proposition2.6we get the inequality
a1 a2 · · · a n − n√n
a1a2· · · a n ≥ P√a k−√a i2, 2.18 where
P
⎧
⎪
⎪
2in − k 1
n i − k 1 , i k ≤ n 1,
n − k 1, i k ≥ n 1.
2.19
Trang 5Equality in2.18 holds for a1 a2 · · · a n If i k n 1, then equality holds again for
a1 a2 · · · a i , a i1 · · · a k−1 √a i a k , a k a k1 · · · a n
Remark 2.9 For p1 p2 · · · p n 1/n, from Proposition2.7, we get the inequality
a1 a2 · · · a n − n√n
a1a2· · · a n≥ 3in − k 1a k − a i2
2n 2i − k 1ak 22n i − 2k 2a i , 2.20
with equality if and only if a1 a2 · · · a n
Applying Theorem2.1for fx − ln x, we obtain
Corollary 2.10 Let
0 < a1≤ · · · ≤ a i ≤ · · · ≤ a k ≤ · · · ≤ a n , 2.21
and let p1, p2, , p n be positive real numbers such that p1 p2 · · · p n 1 Then,
p1a1 p2a2 · · · p n a n
a p1
1 a p2
2 · · · a pn n
≥ Q i a i R k a k /Q i R kQi R k
a Qi i a Rk k , 2.22
with equality for
a1 a2 · · · a i , a k a k1 · · · a n ,
a i1 a i2 · · · a k−1 Q i a i R k a k
Q i R k
2.23
Remark 2.11 For p1 p2 · · · p n 1/n, from Corollary2.10, we get the inequality
a1 a2 · · · a n
n√n
a1a2· · · a n ≥ n
ia i n − k 1a k /n i − k 1 n i−k1
a i
i a n −k1
k
, 2.24
with equality for
a1 a2 · · · a i , a k a k1 · · · a n ,
a i1 a i2 · · · a k−1 ia i n − k 1a k
n i − k 1 .
2.25
If i ≤ n/2 and k n − i 1, then 2.24 becomes
a1 a2 · · · a n
n√n
a1a2· · · a n ≥
a i /a n −i1a n −i1 /a i
2
2i/n
Trang 6with equality for
a1 a2 · · · a i , a n −i1 a n −i2 · · · a n ,
a i1 a i2 · · · a n −i a i a n −i1
2 .
2.27
In the case i 1, from 2.26, we get
a1 a2 · · · a n
n√n
a1a2· · · a n ≥
a1/a na n /a1
2
2/n
with equality for
a2 a3 · · · a n−1 a1 a n
Applying Theorem2.1for fx 1/x, we obtain the following.
Corollary 2.12 Let
0 < a1≤ · · · ≤ a i ≤ · · · ≤ a k ≤ · · · ≤ a n , 2.30
and let p1, p2, , p n be positive real numbers such that p1 p2 · · · p n 1 Then,
p1
a1 p2
a2 · · · p n
a n − 1
p1a1 p2a2 · · · p n a n ≥ Q i R k a k − a i2
a i a k Q i a i R k a k, 2.31
with equality for
a1 a2 · · · a i , a k a k1 · · · a n ,
a i1 a i2 · · · a k−1 Q i a i R k a k
Q i R k
.
2.32
Using Corollary2.12, we can prove the following proposition
Proposition 2.13 Let
0 < a1≤ · · · ≤ a i ≤ · · · ≤ a k ≤ · · · ≤ a n , 2.33
and let p1, p2, , p n be positive real numbers such that p1 p2 · · · p n 1 If
P
⎧
⎪
⎪
Q i , Q i ≤ 3R k ,
4Q i R k
Q i R k
Trang 7p1
a1 p2
a2 · · · p n
a n − 1
p1a1 p2a2 · · · p n a n ≥ P
1
√
a i −√1
a k
2
, 2.35
with equality for a1 a2 · · · a n
3 Proof of Lemmas
Proof of Lemma 2.2 Since c, d ∈ a, b, there exist λ1, λ2∈ 0, 1 such that
c λ1a 1 − λ1b, d λ2a 1 − λ2b. 3.1
In addition, from pa qb pc qd, we get
Applying Jensen’s inequality twice, we obtain
f c fλ1a 1 − λ1b ≤ λ1f a 1 − λ1fb,
f d fλ2a 1 − λ2b ≤ λ2f a 1 − λ2fb,
3.3
and hence
pf c qfd ≤ pλ1f a 1 − λ1fb qλ2f a 1 − λ2fb
pfa qfb.
3.4
Proof of Lemma 2.3 We need to show that
p1f x1 p2f x2 · · · p i f x i − fp1x1 p2x2 · · · p n x n
≥ Q i f x i − fQ i x i p i1x i1 · · · p n x n
.
3.5
Using Jensen’s inequality
p1f x1 p2f x2 · · · p i f x i ≥ Q i f
p
1x1 p2x2 · · · p i x i
Q i
, 3.6
Trang 8it suffices to prove that
Q i f
p
1x1 p2x2 · · · p i x i
Q i
fQ i x i p i1x i1 · · · p n x n
≥ Q i f x i fp1x1 p2x2 · · · p n x n
,
3.7
which can be written as
Q i f X i fY i ≥ Q i f x i fX, 3.8 where
X i p1x1 p2x2 · · · p i x i
Q i
,
Y i Q i x i p i1x i1 · · · p n x n ,
X p1x1 p2x2 · · · p n x n
3.9
Since x i , X ∈ X i , Y i and
Q i X i Y i Q i x i X, 3.10
by Lemma2.2, the conclusion follows
Proof of Lemma 2.4 We need to prove that
p k f x k p k1f x k1 · · · p n f x n − fp1x1 p2x2 · · · p n x n
≥ R k f x k − fp1x1 · · · p k−1x k−1 R k x k
.
3.11
By Jensen’s inequality, we have
p k f x k p k1f x k1 · · · p n f x n ≥ R k f
p
k x k p k1x k1 · · · p n x n
R k
. 3.12 Therefore, it suffices to prove that
R k f
p k x k p k1x k1 · · · p n x n
R k
fp1x1 · · · p k−1x k−1 R k x k
≥ R k f x k fp1x1 p2x2 · · · p n x n
,
3.13
Trang 9or, equivalently,
R k f X k fY k ≥ R k f x k fX, 3.14 where
X k p k x k p k1x k1 · · · p n x n
Y k p1x1 · · · p k−1x k−1 R k x k ,
X p1x1 p2x2 · · · p n x n
3.15
The inequality3.14 follows from Lemma2.2, since x k , X ∈ Y k , X k and
R k X k Y k R k x k X. 3.16
4 Proof of Theorem
Proof By Lemmas 2.3 and 2.4, it follows that for fixed x i , x i1, , x k, Jensen’s difference
Δf, p, x is minimal when x1 x2 · · · x i−1 x i and x k1 x k2 · · · x n x k; that is,
Δf, p, x≥ Q i f x i p i1f x i1 · · · p k−1f x k−1 R k f x k
− fQ i x i p i1x i1 · · · p k−1x k−1 R k x k
.
4.1
Therefore, towards proving2.3, we only need to show that
p i1f x i1 · · · p k−1f x k−1 Q i R k f
Q i x i R k x k
Q i R k
≥ fQ i x i p i1x i1 · · · p k−1x k−1 R k x k
.
4.2
Since
Q i p i1 · · · p k−1 R k 1, 4.3 this inequality is a consequence of Jensen’s inequality Thus, the proof is completed
Trang 105 Proof of Propositions
Proof of Proposition 2.6 Using Corollary2.5, we need to prove that
Q i a i R k a k − Q i R k a Qi/ Q i R k
k ≥ P√a k−√a i2. 5.1
Since this inequality is homogeneous in a i and a k , and also in Q i and R k, without loss of
generality, assume that a i 1 and Q i 1 Using the notations a k x2and R k p, where
x ≥ 1 and p > 0, the inequality is equivalent to gx ≥ 0, where
g x 1 px2−1 px 2p/1p − Px − 12, 5.2 with
P
⎧
⎪
⎪
2p
p 1, p ≥ 1,
p, p ≤ 1.
5.3
We have
gx 2px − x p−1/p1
− 2Px − 1,
gx 2p − P−2p
p− 1
p 1 x −2/p1 .
5.4
If p≥ 1, then
gx 2p
p− 1
p 1
1− x −2/p1
and if p≤ 1, then
gx 2p
1− p
Since gx ≥ 0 for x ≥ 1, and gx is increasing, gx ≥ g1 0, gx is increasing, and hence gx ≥ g1 0 for x ≥ 1 This concludes the proof.
Proof of Proposition 2.7 Using Corollary2.5, we need to prove that
Q i a i R k a k − Q i R k a Qi/ Q i R k
4Q i 2R k a k 2Q i 4R k a i
. 5.7
Trang 11Since this inequality is homogeneous in a i and a k , and also in Q i and R k , we may set a i 1
and Q i 1 Using the notations a k x and R k p, where x ≥ 1 and p > 0, the inequality is equivalent to gx ≥ 0, where
g x 4 2px 2 4p1 px −1 px p/ 1p
− 3px − 12. 5.8
We have
1
2
1 2p gx p
x − x −1/1p
−2 px p/ 1p− 1,
1 p 2p
1 2p gx 1 p x −2−p/1p−
2 px −1/1p ,
1 p2
2p
1 2p2 p gx x − 1x −3−2p/1p .
5.9
Since gx ≥ 0 for x ≥ 1, gx is strictly increasing, gx ≥ g1 0, and gx is strictly increasing, gx ≥ g1 0, gx is strictly increasing, and hence gx ≥ g1 0 for
x≥ 1
Proof of Proposition 2.13 Using Corollary2.12, we need to prove that
Q i R k a k − a i2
a i a k Q i a i R k a k≥ P
1
√
a i −√1
a k
2
This inequality is true if
Q i R k√a i√a k2≥ PQ i a i R k a k . 5.11
For Q i ≤ 3R k, we have
Q i R k√a i√a k2
− PQ i a i R k a k Q i a i
2R k
a k
a i R k − Q i
≥ Q i a i 3R k − Q i ≥ 0.
5.12
Also, for Q i ≥ 3R k, we get
Q i R k√a i√a k2− PQ i a i R k a k
Q i R k
Q i R k R k − 3Q i a i Q i − 3R k a k 2Q i R k√a i a k
≥ Q i R k
Q i R k R k − 3Q i a i Q i − 3R k a i 2Q i R k a i 0.
5.13
The proposition is proved
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