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volume 7, issue 2, article 41, 2006.

Received 12 July, 2005;

accepted 09 December, 2005.

Communicated by:S.S. Dragomir

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Journal of Inequalities in Pure and Applied Mathematics

CERTAIN INEQUALITIES FOR CONVEX FUNCTIONS

P.G. POPESCU AND J.L. DI AZ-BARRERO

Automatics and Computer Science Faculty Politehnica University, Bucure¸sti, Romania.

EMail:pgpopescu@yahoo.com Applied Mathematics III

Universitat Politècnica de Catalunya

Jordi Girona 1-3, C2, 08034 Barcelona, Spain EMail:jose.luis.diaz@upc.edu

c

2000Victoria University ISSN (electronic): 1443-5756 206-05

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Certain Inequalities for Convex Functions

P.G. Popescu and J.L. Díaz-Barrero

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J. Ineq. Pure and Appl. Math. 7(2) Art. 41, 2006

http://jipam.vu.edu.au

Abstract

Classical inequalities like Jensen and its reverse are used to obtain some elementary numerical inequalities for convex functions. Furthermore, imposing restrictions on the data points several new constrained inequalities are given.

2000 Mathematics Subject Classification:26D15.

Key words: Convex functions, Numerical Inequalities, Inequalities with constraints.

1. Introduction

It is well known ([1], [2]) that a continous function,f,convex in a real interval I ⊆Rhas the property

(1.1) f 1

Pn n

X

k=1

p_ka_k

!

≤ 1 Pn

n

X

k=1

p_kf(a_k),

where a_k ∈ I, 1 ≤ k ≤ n are given data points and p₁, p₂, . . . , p_n is a set of nonnegative real numbers constrained by Pj

k=1p_k = P_j. If f is concave the preceding inequality is reversed.

A broad consideration of inequalities for convex functions can be found, among others, in ([3], [4]). Furthermore, in [5] a reverse of Jensen’s inequality is presented. It states that if p₁, p₂, . . . , p_n are real numbers such that p₁ >

0, p_k ≤0for2≤k ≤nandP_n >0,then

(1.2) f 1

P_n

n

X

k=1

p_ka_k

!

≥ 1 P_n

n

X

k=1

p_kf(a_k)

holds, where f : I → Ris a convex function inI anda_k ∈ I,1 ≤ k ≤ n are such that _P¹

n

Pn

k=1p_kx_k ∈ I.Iff is concave (1.2) is reversed. Our aim in this paper is to use the preceding results to get new inequalities for convex functions.

In addition, when thex_k’s are constrained some inequalities are obtained.

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2. Unconstrained Inequalities

In the sequel, applying the preceding results and some numerical identities, some elementary inequalities are obtained. We begin with:

Theorem 2.1. Let a₀, a₁, . . . , a_n be nonnegative real numbers. Then, the fol- lowing inequality

exp

" _n X

k=0

n k

a_k 2ⁿ

#

≤ 1 8ⁿ

" _n X

k=0

n k

e^a^k (1 +a_k)²

# " _n X

k=0

n k

(1 +a_k)

#2

holds.

Proof. Sincef(t) = _(1+t)^e^t 2 is convex in[0,+∞), then settingp_k = ⁿ_k

2ⁿ,0≤ k ≤ n, into (1.1) and taking into account the well known identityPn

k=0 n k

= 2ⁿ,we have

exp

n

X

k=0

n k

a_k 2ⁿ

! "

1 + 1 2ⁿ

n

X

k=0

n k

ak

#⁻²

≤ 1 2ⁿ

n

X

k=0

n k

e^a^k (1 +a_k)². After rearranging terms, the inequality claimed immediately follows and the proof is complete.

Theorem 2.2. Let p1, p2, . . . , pn be a set of nonnegative real numbers con- strained byPj

k=1p_k =P_j.Ifa₁, a₂, . . . , a_nare positive real numbers, then

" _n Y

k=1

a_k+

q 1 +a²_k

pk#_Pn¹

≤ 1 P_n

n

X

k=1

p_ka_k+ v u u

t1 + 1 P_n

n

X

k=1

p_ka_k

!2

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holds.

Proof. Letf : (0,+∞)→Rbe the function defined byf(t) = ln(t+√

1 +t²).

Then, we havef⁰(t) = ^√¹

1+t² >0andf⁰⁰(t) = −√ ^t

(1+t²)³ ≤0.Therefore,f is concave and applying (1.1) yields

ln





 1 P_n

n

X

k=1

p_ka_k+ v u u

t1 + 1 P_n

n

X

k=1

p_ka_k

!2





≥ 1 Pn

n

X

k=1

p_kln

a_k+ q

1 +a²_k

= ln

" _n Y

k=1

a_k+

q 1 +a²_k

pk#_Pn¹ .

Taking into account that f(t) = log(t) is injective, the statement immediately follows and this completes the proof.

Settingp_k = 1

n, 1≤k ≤ninto the preceding result we get

Corollary 2.3. Leta₁, a₂, . . . , a_nbe a set of positive real numbers. Then

n

Y

k=1

a_k+

q 1 +a²_k

1/n

≤ 1 n







n

X

k=1

a_k+ v u u tn²+

n

X

k=1

a_k

!2





holds.

LetT_n be the n^th triangular number defined byT_n = ⁿ⁽ⁿ⁺¹⁾₂ . Then, setting ak=Tk,1≤k ≤ninto the preceding result, we get

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Corollary 2.4. For alln≥1,

n

Y

k=1

Tk+

q 1 +T_k²

_n¹

≤ 1 3

Tn+1+ q

9 +T_n+1²

holds.

An interesting result involving Fibonacci numbers that can be proved using convex functions is the following

Theorem 2.5. Letnbe a positive integer and`be a whole number. Then, F₁^`+F₂^`+...+F_n^`

1

F₁^`−4 + 1

F₂^`−4 +· · ·+ 1 F_n^`−4

≥F_n²F_n+1²

holds, whereF_nis then^thFibonacci number defined byF₀ = 0, F₁ = 1and for alln≥2, F_n=Fn−1+Fn−2.

Proof. Taking into account thatF₁²+F₂²+· · ·+F_n² =F_nF_n+1,as is well known, and the fact that the functionf : (0,∞)→R,defined byf(t) = 1/tis convex, we get after settingp_i = _F ^Fⁱ²

nFn+1,1≤i≤nanda_i =F_nF_i^`−2,1≤i≤n: 1

F₁^`

Fn+1 +_F^F²^`

n+1 +· · ·+ _F^Fⁿ^`

n+1

≤ 1

F_n²Fn+1

1

F₁^`−4 + 1

F₂^`−4 +· · ·+ 1 F_n^`−4

.

From the preceding expression immediately follows F₁^`+F₂^`+· · ·+F_n^`

1

F₁^`−4 + 1

F₂^`−4 +· · ·+ 1 F_n^`−4

≥F_n²F_n+1² , and this completes the proof.

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Finally, using the reverse Jensen’s inequality, we state and prove:

Theorem 2.6. Leta0, a1, . . . , anbe positive real numbers such thata0 ≥ a1 ≥

· · · ≥a_nand letp₀ =n(n+ 1)andp_k =−k, k= 1,2, . . . , n.Then

(2.1)

n

X

k=0

p_ka_k

! _n X

k=0

p_k a_k

!

≤

n+ 1 2

2

.

Proof. Settingf(t) = ¹_t,that is convex in(0,+∞),and taking into account that Pn

k=1k = ⁿ⁽ⁿ⁺¹⁾₂ from (1.2) we have

f 2

n(n+ 1)

n

X

k=0

p_ka_k

!

≥ 2

n(n+ 1)

n

X

k=0

p_kf(a_k)

or

2 n(n+ 1)

n

X

k=0

p_ka_k

!−1

≥ 2

n(n+ 1)

n

X

k=0

p_k a_k

from which, after rearranging terms, (2.1) immediately follows and the proof is complete.

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3. Constrained Inequalities

In the sequel, imposing restrictions on x₁, x₂, . . . , x_n, some inequalities with constraints are given. We begin with the following.

Theorem 3.1. Let p₁, p₂, . . . , p_n ∈ [0,1)be a set of real numbers constrained by Pj

k=1p_k = P_j. If x₁, x₂, . . . , x_n are positive real numbers such that _x¹

1 +

1

x2 +· · ·+_x¹

n = 1, then

n

X

k=1

p_kx_k

! _n X

k=1

1 x^p_k^k

!

≥P_n²

holds.

Proof. Taking into account the weighted AM-HM inequality, we have 1

P_n

n

X

k=1

p_kx_k ≥ P_n Pn

k=1

p_k x_k

.

Since0≤p_k <1for1≤k ≤n,then ^p_x^k

k ≤ ¹

x^pk_k .From which, we get P_n

Pn k=1

p_k x_k

≥ P_n Pn

k=1 1 x^pk_k

.

Then,

1 P_n

n

X

k=1

p_kx_k ≥ P_n Pn

k=1 1 x^pk_k

and the statement immediately follows.

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Corollary 3.2. Ifx₁, x₂, . . . , x_n are positive real numbers such that _x¹

1 +_x¹

2 +

· · ·+ _x¹

n = 1,then

1 n ≤

n

X

k=1

1 x^1/x_k ^k.

Proof. Settingp_k = 1/x_k, 1≤k ≤ninto Theorem3.1yields

n

X

k=1

p_kx_k

! _n X

k=1

1 x^p_k^k

!

=n

n

X

k=1

1 x^1/x_k ^k

!

≥

n

X

k=1

1 x_k

!2

= 1

completing the proof.

Finally, we give two inequalities similar to the ones obtained in [6] for the triangle.

Theorem 3.3. Leta, bandcbe positive real numbers such thata+b+c= 1.

Then, the following inequality

a^a(a+2b)·b^b(b+2c)·c^c(c+2a) ≥ 1 3 holds.

Proof. Sincea+b+c= 1,thena²+b²+c²+ 2(ab+bc+ca) = 1.Therefore, choosing p₁ = a², p₂ = b², p₃ = c², p₄ = 2ab, p₅ = 2bc, p₆ = 2ca and x₁ = 1/a, x₂ = 1/b, x₃ = 1/c, x₄ = 1/a, x₅ = 1/b, x₆ = 1/c, and applying

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Jensen’s inequality to the functionf(t) = lntthat is concave for allt ≥ 0,we obtain

ln

a²1 a +b²1

b +c²1

c+ 2ab1

a + 2bc1

b + 2ca1 c

≥a²ln1

a +b²ln1

b +c²ln1

c+ 2abln1

a + 2bcln1

b + 2caln1 c, from which, we get

ln 3≥ln

1

a^a(a+2b)·b^b(b+2c)·c^c(c+2a)

and this completes the proof.

Theorem 3.4. Let a, b, c be positive numbers such that ab+bc+ca = abc.

Then,

√b

a^c

√ b√^a

c(a+b+c)≥abc holds.

Proof. Sinceab+bc+ca = abc, then ¹_a + ¹_b + ¹_c = 1. So, choosingp₁ = ¹_a, p₂ = ¹_b, p₃ = ¹_c andx₁ =ab, x₂ =bc, x₃ =ca, and applying Jensen’s inequality tof(t) = lntagain, we get

ln (a+b+c)≥ 1

alnab+1

b lnbc+ 1 clnca or

a+b+c≥a^a¹⁺¹^c ·b¹^b⁺¹^a ·c¹^c⁺¹^b.

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Now, taking into account that ¹_a + ¹_b + ¹_c = 1, we obtain: a+b+c ≥ a¹⁻¹^b · b¹⁻¹^c ·c¹⁻¹^a, from which the statement immediately follows and the proof is complete.

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References

[1] J.L.W.V. JENSEN, Sur les fonctions convexes et les inégalités entre les valeurs moyennes, Acta Math., 30 (1906), 175–193.

[2] Th.M. RASSIAS, Survey on Classical Inequalities, Kluwer Academic Pub- lishers, Dordrecht, 2000.

[3] M. BENCZE, Octogon Mathematical Magazine Collection, 1993-2004.

[4] J.F. STEFFENSEN, On certain inequalities and methods of approximation, J. Institute Actuaries, 51 (1919), 279–297.

[5] P.M. VASI ´C AND J.E. PE ˇCARI ´C, On Jensen inequality, Univ. Beograd.

Publ. Elektrotehn Fak. Ser. Mat. Fis., 639-677 (1979), 50–54.

[6] J.L. DÍAZ-BARRERO, Some cyclical inequalities for the triangle, J. Ineq.

Pure and Appl. Math., 6(1) (2005), Art. 20. [ONLINE:http://jipam.

vu.edu.au/article.php?sid=489]