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An extension of Jensen's discrete inequality to half convex functions
Journal of Inequalities and Applications volume 2011, Article number: 101 (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 and convex for u ≤ s or u ≥ s, where . The obtained results are applied for proving some open relevant inequalities.
1 Introduction
The right convex function theorem (RCF-Theorem) has the following statement (see [1–3]).
RCF-Theorem. Let f(u) be a function defined on a real interval and convex for . The inequality
holds for allsatisfying x1 + x2 + ... + x n ≥ ns if and only if
for allwhich satisfy x ≤ s ≤ y 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 and convex for . The inequality
holds for allsatisfying x1 + x2 + ... + x n ≤ ns if and only if
for allwhich satisfy x ≥ s ≥ y 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 and convex for u≤ s or u ≥ s, where . The inequality
holds for allsatisfying x1 + x2 + ... + x n = ns if and only if
for allwhich satisfy x + (n - 1)y = ns.
Applying RCF-, LCF-, and HCF-Theorems to the function f(u) = g(eu ), 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 such that f(u) = g(eu ) is convex for . The inequality
holds for allsatisfying a1a2 ... a n ≥ rn if and only if
for allwhich satisfy a ≤ r ≤ b and abn-1= rn .
LCF-Corollary. Let g be a function defined on a positive interval such that f(u) = g(eu ) is convex for . The inequality
holds for allsatisfying a1a2 ... a n ≤ rn if and only if
for allwhich satisfy a ≥ r ≥ b and abn-1= rn .
HCF-Corollary. Let g be a function defined on a positive interval such that f(u) = g(eu ) is convex for eu ≤ r or eu ≥ r, where . The inequality
holds for allsatisfying a1a2 ... a n = rn if and only if
for allwhich satisfy abn-1= rn .
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
and let f be a convex function on. If(a ≤ b) and x1, x2, ..., x m ∈ [a, b] such that
then
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 and convex for , and let p1, p2, ..., p n be positive real numbers such that
The inequality
holds for allsatisfying p1x1 + p2x2 + ... + p n x n ≥ s if and only if
for allsuch that x ≤ s ≤ y and px + (1 - p)y = s.
WLCF-Theorem. Let f(u) be a function defined on a real interval and convex for , and let p1, p2 , ..., p n be positive real numbers such that
The inequality
holds for allsatisfying p1x1 + p2x2 + ... + p n x n ≤ s if and only if
for allsuch that x ≥ s ≥ y and px + (1 - p)y = s.
WHCF-Theorem. Let f(u) be a function defined on a real interval and convex for u≤ s or u ≥ s, where , and let p1, p2, ..., p n be positive real numbers such that
The inequality
holds for allsatisfying p1x1 + p2x2 + ... + p n x n = s if and only if
for allsuch 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(eu ) 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 such that f(u) = g(eu ) is convex for , and let p1, p2, ..., p n be positive real numbers such that
The inequality
holds for all satisfying if and only if
for allsuch that a ≤ r ≤ b and apb1-p= r.
WLCF-Corollary. Let g be a function defined on a positive interval such that f(u) = g(eu ) is convex for , aand let p1, p2, ..., p n be positive real numbers such that
The inequality
holds for all satisfying if and only if
for allsuch that a ≥ r ≥ b and apb1-p= r.
WHCF-Corollary. Let g be a function defined on a positive interval such that f(u) = g(eu ) is convex for eu ≤ r or eu ≥ r, where , and let p1, p2, ..., p n be positive real numbers such that
The inequality
holds for all satisfying if and only if
for allsuch that apb1-p= r.
Remark 2.2. Let us denote
In some applications, it is useful to replace the hypothesis
in WRCF-, WLCF-, and WHCF-Theorems by the equivalent condition:
h(x, y) ≥ 0 for all such that px + (1 - p)y = s.
This equivalence is true since
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 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
From
we have
Thus, according to Jensen's inequality, we get
4 Proof of WRCF-Theorem
Since the necessity is obvious, we prove further the sufficiency. Without loss of generality, assume that x1 ≤ x2 ≤ ... ≤ x n . If x1 ≥ s, then the required inequality follows by Jensen's inequality for convex functions. Otherwise, since
there exists k ∈ {1, 2, ..., n - 1} such that
Let us denote
By Jensen's inequality, we have
where
Thus, it suffices to prove that
where
Let y i , i = 1, 2, ..., k, defined by
We will show that
We have
Since p1 - p = p1 - min {p1, p2, ..., p n } ≥ 0, we get
and hence y1 ≤ z. Now, from z ≥ y1 ≥ y2 ≥ ... ≥ yk > s, it follows that . Then, by hypothesis, we have
for i = 1,2, ..., k. Summing all these inequalities multiplied by p i /p, respectively, we get
Therefore, to prove (36), it suffices to show that
Since S ∈ [s, z], y i ∈ (s, z] for i = 1, 2, ..., k,
-
(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 + q ≥ n - 1, then[4]
Proof. Write the desired inequality as
where
To prove this inequality, we apply HCF-Corollary for r = 1. Let
Using the second derivative,
we will show that f(u) is convex for eu ≥ r = 1. We need to show that
for t ≥ 1. Indeed,
because p + q ≥ n - 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
Applying the Cauchy-Schwarz inequality, it suffices to prove that
which is equivalent to
where
with
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 + q ≥ n - 1, we have
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
If p = n - 1 and q = 0, then we get the well-known inequality
Remark 5.3. For , and n ≥ 3, by Proposition 5.1 we obtain the inequality
In addition, for , n ≥ 3, we get
Remark 5.4. For p = 2r, q = r2, and n ≥ 3, by Proposition 5.1 we obtain
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 + r ≥ n - 1, then[4]
Proof. Write the required inequality as
where
and apply HCF-Corollary to g(t) for r = 1. Let
defined on ℝ. For n ≥ 4, which implies p + q + r ≥ 3, we claim that f is convex for eu ≥ 1. Since
where t = eu ≥ 1, we need to show that
Since
it suffices to show that
Using the inequality t2 ≥ 2t - 1, we still have to prove that At + B ≥ 0, where
Since p + q + r ≥ 3, we have
Therefore,
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
or
where
Since A,B,C ≥ 0 (from the AM-GM inequality applied to n positive numbers) and p + q + r ≥ n - 1, it suffices to show that
which is equivalent to
where
We see that B11, B22, B33, B12 ≥ 0 (by the AM-GM inequality applied to n - 1 positive numbers). Also, we have
since
(by the AM-GM inequality applied to n - 1 positive numbers). Using the inequality bn-1- (n - 1)b + n - 2 ≥ 0,we get B31 ≥ D, where
To prove (38), it suffices to show that p2B11 + r2B33 + prD ≥ 0. This is true if 4B11B33 ≥ D2; that is,
In the case n = 4, (39) becomes in succession
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
In addition, for p = q = r = 1, we get the known inequality ([2])
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 m ≤ n - 1, then
Remark 5.8. For m = n - 1 and k1 = k2 = ... = k m = 1, (40) turns into the known beautiful inequality ([2])
Remark 5.9. For ,, (40) turns into the known inequality [1, 2]
which holds for 1 ≤ m ≤ n - 1 and . □
Proposition 5.10. If x1, x2, ..., x n are nonnegative real numbers such that
then
Proof. Since the inequality is trivial for n = 1, consider further that n ≥ 2. Write the inequality as
where
The function f(u) is convex for u ≥ s = 1, since
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
We have
From px + (1 - p)y = 1, we get
and hence
This completes the proof. Equality holds for a1 = a2 = ... a n = 1, and also for a1 = 0 and
References
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Cirtoaje V: Algebraic Inequalities--Old and New Methods. GIL Publishing House, Zalau 2006.
Tetiva M: A new proof for the right convex function theorem. Gazeta Matematica Seria A 2006, 2: 126–133.
Art of Problem Solving[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=223266]
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The authors are grateful to the referees for their useful comments.
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Authors' contributions
VC conceived and proved the main results and their applications. AB performed numerical verification for all applications and prepared the first manuscript. Both authors read and approved the final manuscript.
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Cirtoaje, V., Baiesu, A. An extension of Jensen's discrete inequality to half convex functions. J Inequal Appl 2011, 101 (2011). https://doi.org/10.1186/1029-242X-2011-101
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DOI: https://doi.org/10.1186/1029-242X-2011-101