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Journal of Inequalities and Applications

Erschienen in:

Open Access 01.12.2014 | Research

Extension of Jensen’s inequality to affine combinations

verfasst von: Zlatko Pavić

Erschienen in: Journal of Inequalities and Applications | Ausgabe 1/2014

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Abstract

The article provides the generalization of Jensen’s inequality for convex functions on the line segments. The main and preliminary inequalities are expressed in discrete form using affine combinations that can be reduced to convex combinations. The resulting quasi-arithmetic means are used to extend the two well-known inequalities.

MSC:26A51, 26D15.

Competing interests

The author declares that they have no competing interests.

1 Introduction

First of all, we recapitulate the basic notations of affinity and convexity relating to the binomial combinations. Let

X

be a real vector space.

A set

A \subseteq X

is affine if it contains all binomial affine combinations in

A

, that is, the combinations

α a + β b

of points

a, b \in A

and coefficients

α, β \in R

of the sum

α + β = 1

. The affine hull of a set

S \subseteq X

as the smallest affine set that contains

S

is denoted

aff S

. A function

f : A \to R

is affine if the equality

f (α a + β b) = α f (a) + β f (b)

(1.1)

holds for all binomial affine combinations

α a + β b

A

A set

C \subseteq X

is convex if it contains all binomial convex combinations in

C

, that is, the combinations

α a + β b

of points

a, b \in C

and non-negative coefficients

α, β \in R

of the sum

α + β = 1

. The convex hull of a set

S \subseteq X

as the smallest convex set that contains

S

is denoted

conv S

. A function

f : C \to R

is convex if the inequality

f (α a + β b) \leq α f (a) + β f (b)

(1.2)

holds for all binomial convex combinations

α a + β b

C

Using mathematical induction, it can be proved that everything said above is true for all n-membered affine or convex combinations. In this way, Jensen (see [1]) has extended the inequality in equation (1.2).

We will consider affine and convex combinations of the real numbers, the case where

X = R

. Take real numbers a and b such that

a \neq b

. Every number

x \in R

can be uniquely presented as the binomial affine combination

x = \frac{b - x}{b - a} a + \frac{x - a}{b - a} b,

(1.3)

which is convex if, and only if, the number x belongs to the interval

conv {a, b}

. Given the function

f : R \to R

, let

f_{{a, b}}^{line} : R \to R

be the function of the chord line passing through the points

A (a, f (a))

and

B (b, f (b))

of the graph of f. Applying the affinity of

f_{{a, b}}^{line}

to the combination in equation (1.3), we get

f_{{a, b}}^{line} (x) = \frac{b - x}{b - a} f (a) + \frac{x - a}{b - a} f (b) .

(1.4)

If the function f is convex, then we have the inequality

f (x) \leq f_{{a, b}}^{line} (x) if x \in conv {a, b},

(1.5)

and the reverse inequality

f (x) \geq f_{{a, b}}^{line} (x) if x \notin conv {a, b} ∖ {a, b} .

(1.6)

A wide area of convex analysis including convex functions and their inequalities is covered in [2]. The practical applications of convex analysis are presented in [3].

A brief scientific-historic background on Jensen’s inequality follows at the end of this introduction. Because of its attractiveness, Jensen’s and related inequalities were studied during the whole last century. So, there are Steffensen’s, Brunk’s, and Olkin’s inequality. In this century the research goes on, and we got Jensen-Mercer’s and Mercer-Steffensen’s inequality. For information as regards these inequalities, one may refer to papers [4‐7] and [8].

2 Generalization and reversal of Jensen-Mercer’s inequality

In order to generalize Jensen’s inequality, we firstly consider Jensen-Mercer’s inequality. A generalization of this inequality is given in Corollary 2.2, and the reverse inequality is specified in Corollary 2.4. The presentation of the section is based on the affine combinations in equation (2.1), and the simple inequality in equation (2.2).

Let

α, β, γ \in R

be coefficients such that

α + β - γ = 1

. Let

a, b, c \in R

be points. We consider the affine combination

α a + β b - γ c

. Including the affine combination

c = λ a + μ b

assuming that

λ + μ = 1

, we get the binomial form

α a + β b - γ c = (α - γ λ) a + (β - γ μ) b .

(2.1)

Lemma 2.1 Let

α, β, γ \in [0, 1]

be coefficients such that

α + β - γ = 1

. Let

a, b, c \in R

be points such that

c \in conv {a, b}

Then the affine combination

α a + β b - γ c

is in

conv {a, b}

, and every convex function

f : conv {a, b} \to R

satisfies the inequality

f (α a + β b - γ c) \leq α f (a) + β f (b) - γ f (c) .

(2.2)

Proof The condition

c = λ a + μ b \in conv {a, b}

involves

λ, μ \in [0, 1]

. Then the binomial combination of the right-hand side in (2.1) is convex since its coefficients

α - γ λ \geq α - γ = 1 - β \geq 0

and also

β - γ μ \geq 0

. So, the combination

α a + β b - γ c

belongs to

conv {a, b}

Since the inequality in equation (2.2) is trivially true for

a = b

, we suppose that

a \neq b

. Applying the inequality in equation (1.5), and the affinity of

f_{{a, b}}^{line}

, we get

f (α a + β b - γ c) \leq f_{{a, b}}^{line} (α a + β b - γ c)

(2.3)

= α f (a) + β f (b) - γ f_{{a, b}}^{line} (c)

(2.4)

\leq α f (a) + β f (b) - γ f (c)

(2.5)

because

f_{{a, b}}^{line} (c) \geq f (c)

. □

The inequality in equation (2.2) can also be proved applying the convexity of the function f to the right-hand side of the representation in equation (2.1), wherein we use the inequality

f (c) \leq λ f (a) + μ f (b)

Corollary 2.2 Let

α, β, γ \in [0, 1]

and

γ_{i} \in [0, 1]

be coefficients such that

α + β - γ = \sum_{i = 1}^{n} γ_{i} = 1

. Let

a, b, c_{i} \in R

be points such that all

c_{i} \in conv {a, b}

Then the affine combination

α a + β b - γ \sum_{i = 1}^{n} γ_{i} c_{i}

is in

conv {a, b}

, and every convex function

f : conv {a, b} \to R

satisfies the inequality

f (α a + β b - γ \sum_{i = 1}^{n} γ_{i} c_{i}) \leq α f (a) + β f (b) - γ \sum_{i = 1}^{n} γ_{i} f (c_{i}) .

(2.6)

As demonstrated above, to prove the inequality in equation (2.6) it is not necessary to use the Jensen inequality. Lemma 2.1 and Corollary 2.2 are also valid for

γ \in [- 1, 1]

because then the observed affine combinations with

γ \leq 0

become convex, and the associated inequalities follow from Jensen’s inequality. The combinations including

γ \in [- 1, 1]

were observed in [[9], Corollary 11 and Theorem 12] additionally using a monotone function g.

α = β = γ = 1

, then the inequality in equation (2.6) is reduced to Mercer’s variant of Jensen’s inequality obtained in [6]. Another generalization of Mercer’s result was achieved in [10] using the majorization assumptions. The importance and impact of the majorization theory to inequalities can be found in [11].

Lemma 2.3 Let

α, β, γ \in [1, \infty)

be coefficients such that

α + β - γ = 1

. Let

a, b, c \in R

be points such that

c \notin conv {a, b} ∖ {a, b}

Then the affine combination

α a + β b - γ c

is not in

conv {a, b} ∖ {a, b}

, and every convex function

f : conv {a, b, c} \to R

satisfies the inequality

f (α a + β b - γ c) \geq α f (a) + β f (b) - γ f (c) .

(2.7)

Proof The condition

c = λ a + μ b \notin conv {a, b} ∖ {a, b}

entails

λ \leq 0

λ \geq 1

, and the first coefficient of the binomial form of (2.1) satisfies

α - γ λ \geq α \geq 1

λ \leq 0

, or

α - γ λ \leq α - γ = 1 - β \leq 0

λ \geq 1

. So, the combination

α a + β b - γ

does not belong to

conv {a, b} ∖ {a, b}

. Applying the inequality in (1.6), we get the series of inequalities as in equations (2.3), (2.4), and (2.5), but with the reverse inequality signs. □

It is not necessary to require

γ \in [1, \infty)

in Lemma 2.3, it follows from the conditions

α, β \in [1, \infty)

and

α + β - γ = 1

Corollary 2.4 Let

α, β, γ \in [1, \infty)

and

γ_{i} \in [0, 1]

be coefficients such that

α + β - γ = \sum_{i = 1}^{n} γ_{i} = 1

. Let

a, b, c_{i} \in R

be points such that any

c_{i} \notin conv {a, b} ∖ {a, b}

and the convex combination

\sum_{i = 1}^{n} γ_{i} c_{i} \notin conv {a, b} ∖ {a, b}

Then the affine combination

α a + β b - γ \sum_{i = 1}^{n} γ_{i} c_{i}

is not in

conv {a, b} ∖ {a, b}

, and every convex function

f : conv {a, b, c_{i}} \to R

satisfies the inequality

f (α a + β b - γ \sum_{i = 1}^{n} γ_{i} c_{i}) \geq α f (a) + β f (b) - γ \sum_{i = 1}^{n} γ_{i} f (c_{i}) .

(2.8)

3 Main results

The aim of the paper is to generalize the inequality in equation (2.2). The first step is already done in Corollary 2.2 in the expression of the inequality in equation (2.6), presenting the point c as the convex combination. We are going to proceed in the same way with the points a and b. The main theorem follows.

Theorem 3.1 Let

α_{i}, β_{j}, γ_{k} \geq 0

be coefficients with the sums

α = \sum_{i = 1}^{n} α_{i}

β = \sum_{j = 1}^{m} β_{j}

, and

γ = \sum_{k = 1}^{l} γ_{k}

satisfying

0 < α, β \leq 1

and

α + β - γ = 1

. Let

a_{i}, b_{j}, c_{k} \in R

be points such that all

c_{k} \in conv {a, b}

where

a = \frac{1}{α} \sum_{i = 1}^{n} α_{i} a_{i}, b = \frac{1}{β} \sum_{j = 1}^{m} β_{j} b_{j} .

(3.1)

Then the affine combination

\sum_{i = 1}^{n} α_{i} a_{i} + \sum_{j = 1}^{m} β_{j} b_{j} - \sum_{k = 1}^{l} γ_{k} c_{k}

(3.2)

belongs to

conv {a, b}

, and every convex function

f : conv {a_{i}, b_{j}} \to R

satisfies the inequality

\begin{matrix} f (\sum_{i = 1}^{n} α_{i} a_{i} + \sum_{j = 1}^{m} β_{j} b_{j} - \sum_{k = 1}^{l} γ_{k} c_{k}) \\ \leq \sum_{i = 1}^{n} α_{i} f (a_{i}) + \sum_{j = 1}^{m} β_{j} f (b_{j}) - \sum_{k = 1}^{l} γ_{k} f (c_{k}) . \end{matrix}

(3.3)

If f is concave, then the reverse inequality is valid in equation (3.3). If f is affine, then the equality is valid in equation (3.3).

Proof Note that

γ \leq α, β

. Prove the theorem for the convex function f.

γ = 0

, the combination in equation (3.2) takes the convex form

α a + β b

belonging to

conv {a, b}

, and the inequality in equation (3.3) is reduced to Jensen’s inequality.

γ > 0

, then, including points a, b, and

c = \frac{1}{γ} \sum_{k = 1}^{l} γ_{k} c_{k}

(3.4)

in equation (3.2), we get the combination

α a + β b - γ c

which belongs to

conv {a, b}

by Lemma 2.1. The inequality in equation (3.3) is trivially true for

a = b

. So, we assume that

a \neq b

and use the function

f_{{a, b}}^{line}

. Applying the affinity of

f_{{a, b}}^{line}

to the convex combination in equation (3.4), and respecting the inequalities

f_{{a, b}}^{line} (c_{k}) \geq f (c_{k})

, we have

f_{{a, b}}^{line} (c) = \frac{1}{γ} \sum_{k = 1}^{l} γ_{k} f_{{a, b}}^{line} (c_{k}) \geq \frac{1}{γ} \sum_{k = 1}^{l} γ_{k} f (c_{k}) .

(3.5)

Using the inequality in equation (2.4), applying Jensen’s inequality to

f (a)

and

f (b)

, and finally using the inequality in equation (3.5) respecting the minus, we get

\begin{array}{rcl} f (\sum_{i = 1}^{n} α_{i} a_{i} + \sum_{j = 1}^{m} β_{j} b_{j} - \sum_{k = 1}^{l} γ_{k} c_{k}) & \leq & α f (a) + β f (b) - γ f_{{a, b}}^{line} (c) \\ \leq & \sum_{i = 1}^{n} α_{i} f (a_{i}) + \sum_{j = 1}^{m} β_{j} f (b_{j}) - \sum_{k = 1}^{l} γ_{k} f (c_{k}), \end{array}

completing the proof. □

Corollary 3.2 Let all the assumptions of Theorem 3.1 be fully satisfied. Let

\sum_{r = 1}^{s} λ_{r} x_{r}

x_{r} \in conv {a, b}

, and

λ a + μ b

be convex combinations having the common center with the affine combination in equation (3.2), that is,

\sum_{r = 1}^{s} λ_{r} x_{r} = λ a + μ b = \sum_{i = 1}^{n} α_{i} a_{i} + \sum_{j = 1}^{m} β_{j} b_{j} - \sum_{k = 1}^{l} γ_{k} c_{k} .

(3.6)

Then every convex function

f : conv {a_{i}, b_{j}} \to R

satisfies the double inequality

\sum_{r = 1}^{s} λ_{r} f (x_{r}) \leq λ f (a) + μ f (b) \leq \sum_{i = 1}^{n} α_{i} f (a_{i}) + \sum_{j = 1}^{m} β_{j} f (b_{j}) - \sum_{k = 1}^{l} γ_{k} f (c_{k}) .

(3.7)

If f is concave, then the reverse inequalities are valid in equation (3.7). If f is affine, then the equalities are valid in equation (3.7).

Proof Relying on the function

f_{{a, b}}^{line}

and the Jensen inequality, we derive the series of inequalities

\begin{array}{rcl} \sum_{r = 1}^{s} λ_{r} f (x_{r}) & \leq & \sum_{r = 1}^{s} λ_{r} f_{{a, b}}^{line} (x_{r}) = f_{{a, b}}^{line} (\sum_{r = 1}^{s} λ_{r} x_{r}) = f_{{a, b}}^{line} (λ a + μ b) = λ f (a) + μ f (b) \\ = & f_{{a, b}}^{line} (α a + β b - \sum_{k = 1}^{l} γ_{k} c_{k}) = α f (a) + β f (b) - \sum_{k = 1}^{l} γ_{k} f_{{a, b}}^{line} (c_{k}) \\ \leq & \sum_{i = 1}^{n} α_{i} f (a_{i}) + \sum_{j = 1}^{m} β_{j} f (b_{j}) - \sum_{k = 1}^{l} γ_{k} f (c_{k}) \end{array}

containing the double inequality in equation (3.7). □

Corollary 3.3 Let all the assumptions of Theorem 3.1 be fully satisfied. Let

\sum_{r = 1}^{s} λ_{r} x_{r}

be a convex combination with points

x_{r} \in conv {a, b}

such that

\sum_{r = 1}^{s} λ_{r} x_{r} + \sum_{k = 1}^{l} γ_{k} c_{k} = \sum_{i = 1}^{n} α_{i} a_{i} + \sum_{j = 1}^{m} β_{j} b_{j} .

(3.8)

Then every convex function

f : conv {a_{i}, b_{j}} \to R

satisfies the inequality

\sum_{r = 1}^{s} λ_{r} f (x_{r}) + \sum_{k = 1}^{l} γ_{k} f (c_{k}) \leq \sum_{i = 1}^{n} α_{i} f (a_{i}) + \sum_{j = 1}^{m} β_{j} f (b_{j}) .

(3.9)

We finish the section with the basic integral variant of the main theorem. In the corollary that follows we use segments with different endpoints.

Corollary 3.4 Let

α, β, γ \in [0, 1]

be coefficients such that

α + β - γ = 1

. Let

[a_{1}, a_{2}], [b_{1}, b_{2}], [c_{1}, c_{2}] \subset R

be segments such that

[c_{1}, c_{2}] \subseteq [\frac{a_{1} + a_{2}}{2}, \frac{b_{1} + b_{2}}{2}] .

Then the integral sum

\bar{x} = \frac{α}{a_{2} - a_{1}} \int_{a_{1}}^{a_{2}} x d x + \frac{β}{b_{2} - b_{1}} \int_{b_{1}}^{b_{2}} x d x - \frac{γ}{c_{2} - c_{1}} \int_{c_{1}}^{c_{2}} x d x

(3.10)

belongs to

[(a_{1} + a_{2}) / 2, (b_{1} + b_{2}) / 2]

, and every convex function

f : [a_{1}, b_{2}] \to R

satisfies the inequality

f (\bar{x}) \leq \frac{α}{a_{2} - a_{1}} \int_{a_{1}}^{a_{2}} f (x) d x + \frac{β}{b_{2} - b_{1}} \int_{b_{1}}^{b_{2}} f (x) d x - \frac{γ}{c_{2} - c_{1}} \int_{c_{1}}^{c_{2}} f (x) d x .

(3.11)

If f is concave, then the reverse inequality is valid in equation (3.11). If f is affine, then the equality is valid in equation (3.11).

Proof The sum of (3.10) is equal to the affine combination

\bar{x} = α \frac{a_{1} + a_{2}}{2} + β \frac{b_{1} + b_{1}}{2} - γ \frac{c_{1} + c_{2}}{2}

(3.12)

belonging to

[(a_{1} + a_{2}) / 2, (b_{1} + b_{2}) / 2]

by Lemma 2.1. Given the positive integer n, we can construct the affine combination

{\bar{x}}_{n} = \sum_{i = 1}^{n} α_{n i} a_{n i} + \sum_{j = 1}^{n} β_{n j} b_{n j} - \sum_{k = 1}^{n} γ_{n k} c_{n k},

(3.13)

which approaches the integral sum in equation (3.10) as n approaches infinity. Let us show how it looks in the example of the first member of the above combination.

We make the partition

[a_{1}, a_{2}] = ⋃_{i = 1}^{n} [{(a_{1})}_{n i}, {(a_{2})}_{n i}]

where all subsegments have the same length

(a_{2} - a_{1}) / n

, and the adjacent subsegments have a common endpoint. If we take points

a_{n i} \in [{(a_{1})}_{n i}, {(a_{2})}_{n i}]

, then we have

lim_{n \to \infty} \sum_{i = 1}^{n} \frac{1}{n} a_{n i} = lim_{n \to \infty} \sum_{i = 1}^{n} \frac{{(a_{2})}_{n i} - {(a_{1})}_{n i}}{a_{2} - a_{1}} a_{n i} = \frac{1}{a_{2} - a_{1}} \int_{a_{1}}^{a_{2}} x d x

(3.14)

suggesting to take

α_{n i} = α / n

Applying Theorem 3.1 to the affine combination in equation (3.13) and the convex function f, we get

f ({\bar{x}}_{n}) \leq \sum_{i = 1}^{n} α_{n i} f (a_{n i}) + \sum_{j = 1}^{n} β_{n j} f (b_{n j}) - \sum_{k = 1}^{n} γ_{n k} f (c_{n k}),

(3.15)

and letting n to infinity, we obtain the integral inequality in equation (3.11). □

4 Application to quasi-arithmetic means

Let

I \subseteq R

be an interval. Applying the Jensen inequality to the means, we use strictly monotone continuous functions

φ, ψ : I \to R

such that the function

ψ \circ φ^{- 1} : φ (I) \to R

is convex. In this case, we say that ψ is convex with respect to φ, or ψ is φ-convex. The same notation is used for the concavity. If the function ψ is φ-convex and φ-concave, we can say that it is φ-affine.

Suppose that

\bar{x} = \sum_{i = 1}^{n} α_{i} a_{i} + \sum_{j = 1}^{m} β_{j} b_{j} - \sum_{k = 1}^{l} γ_{k} c_{k}

(4.1)

is an affine combination as in Theorem 3.1, and

φ : I \to R

is a strictly monotone continuous function where

I = conv {a_{i}, b_{j}}

. The discrete φ-quasi-arithmetic mean of the combination

\bar{x}

can be defined as the point

M_{φ} (\bar{x}) = φ^{- 1} (\sum_{i = 1}^{n} α_{i} φ (a_{i}) + \sum_{j = 1}^{m} β_{j} φ (b_{j}) - \sum_{k = 1}^{l} γ_{k} φ (c_{k}))

(4.2)

belonging to

conv {a, b}

, because the affine combination enclosed in parentheses is located in

φ (conv {a, b})

. The quasi-arithmetic means defined in equation (4.2) are invariant with respect to the affinity, that is, the equality

M_{φ} (\bar{x}) = M_{u φ + v} (\bar{x})

(4.3)

holds for all pairs of real numbers

u \neq 0

and v.

The order of the pair of quasi-arithmetic means

M_{φ}

and

M_{ψ}

depends on convexity of the function

ψ \circ φ^{- 1}

and monotonicity of the function ψ, as follows.

Corollary 4.1 Let

\bar{x}

be an affine combination as in equation (4.1) satisfying all the assumptions of Theorem 3.1. Let

φ, ψ : I \to R

be strictly monotone continuous functions where

I = conv {a_{i}, b_{j}}

If ψ is either φ-convex and increasing or φ-concave and decreasing, then we have the inequality

M_{φ} (\bar{x}) \leq M_{ψ} (\bar{x}) .

(4.4)

If ψ is either φ-convex and decreasing or φ-concave and increasing, then we have the reverse inequality in equation (4.4).

If ψ is φ-affine, then the equality is valid in equation (4.4).

Proof We prove the case that the function ψ is φ-convex and increasing. Put the set

J = φ (I) = conv {φ (a_{i}), φ (b_{j})}

. Applying the inequality in equation (3.3) of Theorem 3.1 to the affine combination

{\bar{x}}_{φ} = \sum_{i = 1}^{n} α_{i} φ (a_{i}) + \sum_{j = 1}^{m} β_{j} φ (b_{j}) - \sum_{k = 1}^{l} γ_{k} φ (c_{k}),

(4.5)

which is in the set

φ (conv {a, b})

, and the convex function

f = ψ \circ φ^{- 1} : J \to R

, we get

ψ \circ φ^{- 1} ({\bar{x}}_{φ}) \leq {\bar{x}}_{ψ} .

Assigning the increasing function

ψ^{- 1}

to the above inequality, we attain

M_{φ} (\bar{x}) = φ^{- 1} ({\bar{x}}_{φ}) \leq ψ^{- 1} ({\bar{x}}_{ψ}) = M_{ψ} (\bar{x}),

which finishes the proof. □

The inequality in equation (4.4) may further be applied to the power means. The monotonicity of these power means is also valid. We will only derive the formula of the harmonic-geometric-arithmetic mean inequality.

Corollary 4.2 If

\bar{x}

is an affine combination as in equation (4.1) satisfying all the assumptions of Theorem 3.1 with the addition that all

a_{i}, b_{j} > 0

, then we have the harmonic-geometric-arithmetic mean inequality

\begin{array}{rcl} {(\sum_{i = 1}^{n} \frac{α_{i}}{a_{i}} + \sum_{j = 1}^{m} \frac{β_{j}}{b_{j}} - \sum_{k = 1}^{l} \frac{γ_{k}}{c_{k}})}^{- 1} & \leq & \prod_{i = 1}^{n} a_{i}^{α_{i}} \prod_{j = 1}^{m} b_{j}^{β_{j}} \prod_{k = 1}^{l} c_{k}^{- γ_{k}} \\ \leq & \sum_{i = 1}^{n} α_{i} a_{i} + \sum_{j = 1}^{m} β_{j} b_{j} - \sum_{k = 1}^{l} γ_{k} c_{k} . \end{array}

(4.6)

Proof Let

φ_{- 1} (x) = x^{- 1}

φ_{0} (x) = ln x

, and

φ_{1} (x) = x

. Applying the inequality in equation (4.4), first to the pair

φ_{- 1}

and

φ_{0}

, and then to the pair

φ_{0}

and

φ_{1}

, we get

M_{- 1} \leq M_{0} \leq M_{1}

representing the inequality in equation (4.6). □

The general approach to the theory of means and their inequalities was presented in [12]. Different forms of quasi-arithmetic means were considered in [13].

5 Application to other inequalities

The essential inequalities in equations (2.2) and (2.7) can be used to extend the Bernoulli and the Young inequality. Let us start with a simplified case of the mean inequality in equation (4.6).

Corollary 5.1 If numbers

α, β, γ \in [0, 1]

satisfy

α + β - γ = 1

, then the double inequality

{(\frac{α}{a} + \frac{β}{b} - \frac{γ}{c})}^{- 1} \leq \frac{a^{α} b^{β}}{c^{γ}} \leq α a + β b - γ c

(5.1)

holds for all numbers

a, b, c > 0

such that

c \in conv {a, b}

α, β, γ \in [1, \infty)

satisfy

α + β - γ = 1

, then the reverse double inequality is valid in equation (5.1) for all

a, b, c > 0

such that

c \notin conv {a, b} ∖ {a, b}

Proof The inequality in equation (5.1) follows from the inequality in equation (4.6) with

α_{i} = α

β_{j} = β

γ_{k} = γ

and

a_{i} = a

b_{j} = b

c_{k} = c

. So, it is actually a consequence of the inequality in equation (2.2).

The reverse inequality of equation (5.1) can be obtained from the inequality in equation (2.7) using the functions

φ_{- 1}

φ_{0}

, and

φ_{1}

. □

The previous corollary with a suitable choice of the coefficients and the points will be applied in the following two examples.

Example 5.2 If numbers

p, q, r \in [0, 1]

satisfy

p + q - r = 1

, then the double inequality

{(\frac{p}{1 + x} + \frac{q}{1 + y} - \frac{r}{1 + z})}^{- 1} \leq \frac{{(1 + x)}^{p} {(1 + y)}^{q}}{{(1 + z)}^{r}} \leq 1 + p x + q y - r z

(5.2)

holds for all numbers

x, y, z > - 1

such that

z \in conv {x, y}

p, q, r \in [1, \infty)

satisfy

p + q - r = 1

, then the reverse double inequality is valid in equation (5.2) for all

x, y, z > - 1

such that

z \notin conv {x, y} ∖ {x, y}

Proof Corollary 5.1 should be applied to the coefficients

α = p

β = q

γ = r

and the points

a = 1 + x

b = 1 + y

c = 1 + z

. □

The inequality

\frac{1 + t}{1 + t - p t} \leq {(1 + t)}^{p} \leq 1 + p t

(5.3)

applying to

p \in [0, 1]

and

t > - 1

follows from the inequality in equation (5.2) putting first

y = z = 0

and taking

t = x \in (- 1, 0]

, then putting

x = z = 0

and taking

t = y \in [0, \infty)

. In the same way, from the reverse inequality of equation (5.2) we can obtain the reverse inequality of equation (5.3) respecting

p \in [1, \infty)

and

t > - 1

. The right-hand sides of these inequalities are the well-known Bernoulli’s inequalities.

Example 5.3 If numbers

p, q \in [1, \infty)

and

r \in [0, 1]

satisfy

1 / p + 1 / q - r = 1

, then the double inequality

{(\frac{x^{- p}}{p} + \frac{y^{- q}}{q} - r z^{- 1})}^{- 1} \leq \frac{x y}{z^{r}} \leq \frac{x^{p}}{p} + \frac{y^{q}}{q} - r z

(5.4)

holds for all numbers

x, y, z > 0

such that

z \in conv {x^{p}, y^{q}}

p, q \in (0, 1]

and

r \in [1, \infty)

satisfy

1 / p + 1 / q - r = 1

, then the reverse double inequality is valid in equation (5.4) for all

x, y, z > 0

such that

z \notin conv {x^{p}, y^{q}} ∖ {x^{p}, y^{q}}

Proof Corollary 5.1 should be applied to the coefficients

α = 1 / p

β = 1 / q

γ = r

and the points

a = x^{p}

b = y^{q}

c = z

. □

Taking

r = 0

in equation (5.4), we have the inequality

{(\frac{x^{- p}}{p} + \frac{y^{- q}}{q})}^{- 1} \leq x y \leq \frac{x^{p}}{p} + \frac{y^{q}}{q},

(5.5)

which applies to numbers

p, q \in (1, \infty)

with the sum

1 / p + 1 / q = 1

, as well as to numbers

x, y > 0

. The right-hand side of the inequality in equation (5.5) represents the discrete form of Young’s inequality.

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Competing interests

The author declares that they have no competing interests.

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Titel: Extension of Jensen’s inequality to affine combinations
verfasst von: Zlatko Pavić
Publikationsdatum: 01.12.2014
Verlag: Springer International Publishing
Erschienen in: Journal of Inequalities and Applications / Ausgabe 1/2014
Elektronische ISSN: 1029-242X
DOI: https://doi.org/10.1186/1029-242X-2014-298

Springer Professional

Extension of Jensen’s inequality to affine combinations

Abstract

Competing interests

1 Introduction

2 Generalization and reversal of Jensen-Mercer’s inequality

3 Main results

4 Application to quasi-arithmetic means

5 Application to other inequalities

Competing interests

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Springer Professional

Abstract

Competing interests

1 Introduction

2 Generalization and reversal of Jensen-Mercer’s inequality

3 Main results

4 Application to quasi-arithmetic means

5 Application to other inequalities

Competing interests

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