(1)http://jipam.vu.edu.au/ Volume 7, Issue 2, Article 41, 2006 CERTAIN INEQUALITIES FOR CONVEX FUNCTIONS P.G

http://jipam.vu.edu.au/

Volume 7, Issue 2, Article 41, 2006

CERTAIN INEQUALITIES FOR CONVEX FUNCTIONS

P.G. POPESCU AND J.L. DÍAZ-BARRERO AUTOMATICS ANDCOMPUTERSCIENCEFACULTY

POLITEHNICAUNIVERSITY, BUCURE ¸STI, ROMANIA. pgpopescu@yahoo.com

APPLIEDMATHEMATICSIII

UNIVERSITATPOLITÈCNICA DECATALUNYA

JORDIGIRONA1-3, C2, 08034 BARCELONA, SPAIN

jose.luis.diaz@upc.edu

Received 12 July, 2005; accepted 09 December, 2005 Communicated by S.S. Dragomir

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

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

2000 Mathematics Subject Classification. 26D15.

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

P_n

n

X

k=1

p_ka_k

!

≤ 1 P_n

n

X

k=1

p_kf(a_k),

whereak ∈ I,1 ≤ k ≤ n are given data points andp1, p2, . . . , pn is a set of nonnegative real numbers constrained byPj

k=1pk =Pj.Iff 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_nare real numbers such thatp₁ >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 → R is a convex function in I and a_k ∈ I, 1 ≤ k ≤ n are such that

1 Pn

Pn

k=1p_kx_k∈I.Iffis concave (1.2) is reversed. Our aim in this paper is to use the preceding

ISSN (electronic): 1443-5756 c

2006 Victoria University. All rights reserved.

206-05

results to get new inequalities for convex functions. In addition, when thex_k’s are constrained some inequalities are obtained.

2. UNCONSTRAINEDINEQUALITIES

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

Theorem 2.1. Leta₀, a₁, . . . , a_nbe nonnegative real numbers. Then, the following inequality

exp

" _n X

k=0

n k

a_k 2ⁿ

#

≤ 1 8ⁿ

" _n X

k=0

n k

e^ak (1 +a_k)²

# " _n X

k=0

n k

(1 +a_k)

#2

holds.

Proof. Sincef(t) = _(1+t)^et 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

a_k

#−2

≤ 1 2ⁿ

n

X

k=0

n k

e^ak (1 +a_k)².

After rearranging terms, the inequality claimed immediately follows and the proof is complete.

Theorem 2.2. Letp₁, p₂, . . . , p_nbe a set of nonnegative real numbers constrained 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

holds.

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

1 +t²).Then, we have f⁰(t) = ^√1

1+t² > 0and f⁰⁰(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 P_n

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 thatf(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

ak+

q 1 +a²_k

1/n

≤ 1 n







n

X

k=1

ak+ v u u tn²+

n

X

k=1

ak

!2





holds.

LetT_nbe then^thtriangular number defined byT_n= ⁿ⁽ⁿ⁺¹⁾₂ .Then, settinga_k =T_k,1≤k≤ ninto the preceding result, we get

Corollary 2.4. For alln ≥1,

n

Y

k=1

T_k+

q 1 +T_k²

¹_n

≤ 1 3

T_n+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, F<sub>1</sub><sup>`</sup>+F₂^{`</sup>+...+F<sub>n</sub><sup>`}

1

F₁^`−4 + 1

F₂^{`−4</sup> +· · ·+ 1 F<sub>n</sub><sup>`−4}

≥F_n²F_n+1²

holds, whereFn is then^th Fibonacci number defined byF0 = 0, F1 = 1 and for all n ≥ 2, Fn =Fn−1+Fn−2.

Proof. Taking into account that F₁² +F₂² +· · ·+F_n² = F_nF_n+1, as is well known, and the fact that the function f : (0,∞) → R,defined by f(t) = 1/tis convex, we get after setting p_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_n²F_n+1 1

F₁^`−4 + 1

F₂^{`−4</sup> +· · ·+ 1 F<sub>n</sub><sup>`−4}

.

From the preceding expression immediately follows F₁^{`</sup>+F<sub>2</sub><sup>`}+· · ·+F_n^`

1

F₁^`−4 + 1

F₂^{`−4</sup> +· · ·+ 1 F<sub>n</sub><sup>`−4}

≥F_n²F_n+1² ,

and this completes the proof.

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

Theorem 2.6. Leta0, a1, . . . , anbe positive real numbers such thata0 ≥a1 ≥ · · · ≥anand let p0 =n(n+ 1)andpk =−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. Setting f(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.

3. CONSTRAINEDINEQUALITIES

In the sequel, imposing restrictions onx1, x2, . . . , xn,some inequalities with constraints are given. We begin with the following.

Theorem 3.1. Letp1, p2, . . . , pn∈[0,1)be a set of real numbers constrained byPj

k=1pk =Pj. Ifx₁, x₂, . . . , x_nare positive real numbers such that _x¹

1 +_x¹

2 +· · ·+_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 ≥ Pn

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 Pn

Pn k=1

pk

xk

≥ Pn

Pn k=1

1 x^pk_k

.

Then,

1 P_n

n

X

k=1

pkxk ≥ P_n Pn

k=1 1 x^pk_k

and the statement immediately follows.

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 Theorem 3.1 yields

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. Let a, b and c be positive real numbers such that a +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,then a² +b² +c² + 2(ab+bc+ca) = 1.Therefore, choosing p₁ =a², p₂ =b², p₃ = c², p₄ = 2ab, p₅ = 2bc, p₆ = 2caandx₁ = 1/a, x₂ = 1/b, x₃ = 1/c,

x₄ = 1/a, x₅ = 1/b, x₆ = 1/c, and applying 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. Leta, b, cbe positive numbers such thatab+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, choosing p₁ = _a¹, 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^1a+1c ·b^1b+a1 ·c^1c+1b.

Now, taking into account that _a¹ + ¹_b +¹_c = 1, we obtain:a+b+c≥a^1−1b ·b^1−1c ·c^1−1a, from which the statement immediately follows and the proof is complete.

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 Publishers, 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 ´CANDJ.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]