A Variant of Jensen-Steffensen's Inequality for

Journal of Inequalities in Pure and Applied Mathematics http://jipam.vu.edu.au/ Volume 7, Issue 2, Article 70, 2006

A VARIANT OF JENSEN-STEFFENSEN’S INEQUALITY FOR CONVEX AND SUPERQUADRATIC FUNCTIONS ˇ C´ BAKULA, AND S. BANIC´ S. ABRAMOVICH, M. KLARICI D EPARTMENT OF M ATHEMATICS U NIVERSITY OF H AIFA H AIFA , 31905, I SRAEL [email protected] D EPARTMENT OF M ATHEMATICS FACULTY OF NATURAL S CIENCES , M ATHEMATICS AND E DUCATION U NIVERSITY OF S PLIT T ESLINA 12, 21000 S PLIT C ROATIA [email protected] FACULTY OF C IVIL E NGINEERING AND A RCHITECTURE U NIVERSITY OF S PLIT M ATICE HRVATSKE 15, 21000 S PLIT C ROATIA [email protected]

Received 03 January, 2006; accepted 20 January, 2006 Communicated by C.P. Niculescu

A BSTRACT. A variant of Jensen-Steffensen’s inequality is considered for convex and for superquadratic functions. Consequently, inequalities for power means involving not only positive weights have been established. Key words and phrases: Jensen-Steffensen’s inequality, Monotonicity, Superquadraticity, Power means. 2000 Mathematics Subject Classification. 26D15, 26A51.

1. I NTRODUCTION . Let I be an interval in R and f : I → R a convex function on I. If ξ P = (ξ1 , . . . , ξm ) is any m-tuple in I m and p = (p1 , . . . , pm ) any nonnegative m-tuple such that m i=1 pi > 0, then the well known Jensen’s inequality (see for example [7, p. 43]) ! m m 1 X 1 X (1.1) f pi ξi ≤ pi f (ξi ) Pm i=1 Pm i=1 P holds, where Pm = m i=1 pi . ISSN (electronic): 1443-5756 c 2006 Victoria University. All rights reserved.

002-06

ˇ C´ BAKULA , S. A BRAMOVICH , M. K LARI CI

2

AND

S. BANI C´

If f is strictly convex, then (1.1) is strict unless ξi = c for all i ∈ {j : pj > 0}. It is well known that the assumption “p is a nonnegative m-tuple” can be relaxed at the expense of more restrictions on the m-tuple ξ. If p is a real m-tuple such that 0 ≤ Pj ≤ Pm , j = 1, . . . , m,

(1.2) where Pj = get

Pj

i=1

Pm > 0,

pi , then for any monotonic m-tuple ξ (increasing or decreasing) in I m we m 1 X ξ= pi ξi ∈ I, Pm i=1

and for any function f convex on I (1.1) still holds. Inequality (1.1) considered under conditions (1.2) is known as the Jensen-Steffensen’s inequality [7, p. 57] for convex functions. In his paper [5] A. McD. Mercer considered some monotonicity properties of power means. He proved the following theorem: Theorem A. Suppose that 0 < a < b and a ≤ x1 ≤ x2 ≤ · · · ≤ xn ≤ b hold withPat least one of the xk satisfying a < xk < b. If w = (w1 , . . . , wn ) is a positive n-tuple with ni=1 wi = 1 and −∞ < r < s < +∞, then a < Qr (a, b; x) < Qs (a, b; x) < b , where Qt (a, b; x) ≡

at + b t −

n X

! 1t wi xti

i=1

for all real t 6= 0, and ab Q0 (a, b; x) ≡ , G

G=

n Y

i xw i .

i=1

In his next paper [6], Mercer gave a variant of Jensen’s inequality for which Witkowski presented in [8] a shorter proof. This is stated in the following theorem: Theorem B. If f is a convex function on an interval containing an n-tuple x = (xP 1 , . . . , xn ) such that 0 < x1 ≤ x2 ≤ · · · ≤ xn and w = (w1 , . . . , wn ) is a positive n-tuple with ni=1 wi = 1, then ! n n X X f x1 + xn − wi xi ≤ f (x1 ) + f (xn ) − wi f (xi ) . i=1

i=1

This theorem is a special case of the following theorem proved in [4] by Abramovich, Klariˇcić Bakula, Matić and Peˇcarić: Theorem C ([4, Th. 2]). Let f : I → R, where I is an interval in R and let [a, b] ⊆ I, a < b. Let x = (x1 , . . . , xn ) be a monotonic n-tuple in [a, b]n and v = (v1 , . . . , vn ) a real n−tuple P such that vi 6= 0, i = 1, . . . , n, and 0 ≤ Vj ≤ Vn , j = 1, . . . , n, Vn > 0, where Vj = ji=1 vi . If f is convex on I, then ! n n 1 X 1 X vi xi ≤ f (a) + f (b) − vi f (xi ) . (1.3) f a+b− Vn i=1 Vn i=1 In case f is strictly convex, the equality holds in (1.3) iff one of the following two cases occurs: (1) either x = a or x = b,

J. Inequal. Pure and Appl. Math., 7(2) Art. 70, 2006

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

J ENSEN -S TEFFENSEN ’ S INEQUALITY - S UPERQUADRATIC FUNCTIONS

3

(2) there exists l ∈ {2, . . . , n − 1} such that x = x1 + xn − xl and    x1 = a, xn = b or x1 = b, xn = a, (1.4) V j (xj−1 − xj ) = 0, j = 2, . . . , l,   Vj (xj − xj+1 ) = 0, j = l, . . . , n − 1, Pn P where V j = i=j vi , j = 1, . . . , n and x = (1/Vn ) ni=1 vi xi . In the special case where v > 0 and f is strictly convex, the equality holds in (1.4) iff xi = a, i = 1, . . . , n, or xi = b, i = 1, . . . , n. Here, as in the rest of the paper, when we say that an n-tuple ξ is increasing (decreasing) we mean that ξ1 ≤ ξ2 ≤ · · · ≤ ξn (ξ1 ≥ ξ2 ≥ · · · ≥ ξn ). Similarly, when we say that a function f : I → R is increasing (decreasing) on I we mean that for all u, v ∈ I we have u < v ⇒ f (u) ≤ f (v) (u < v ⇒ f (u) ≥ f (v)). In Section 2 we refine Theorems A, B, and C. These refinements are achieved by superquadratic functions which were introduced in [1] and [2]. As Jensen’s inequality for convex functions is a generalization of Hölder’s inequality for f (x) = xp , p ≥ 1, so the inequalities satisfied by superquadratic functions are generalizations of the inequalities satisfied by the superquadratic functions f (x) = xp , p ≥ 2 (see [1], [2]). First we quote some definitions and state a list of basic properties of superquadratic functions. Definition 1.1. A function f : [0, ∞) → R is superquadratic provided that for all x ≥ 0 there exists a constant C(x) ∈ R such that (1.5)

f (y) − f (x) − f (|y − x|) ≥ C (x) (y − x)

for all y ≥ 0. Definition 1.2. A function f : [0, ∞) → R is said to be strictly superquadratic if (1.5) is strict for all x 6= y where xy 6= 0. LemmaP A ([2, Lemma 2.3]). Suppose that f is superquadratic. Let ξi ≥ 0, i = 1, . . . , m, and Pm let ξ = m p ξ , where p ≥ 0, i = 1, . . . , m, and p = 1. Then i i=1 i i=1 i i m m X X pi f (ξi ) − f ξ ≥ pi f ξi − ξ . i=1

i=1

Lemma B ([1, Lemma 2.2]). Let f be superquadratic function with C(x) as in Definition 1.1. Then: (i) f (0) ≤ 0, (ii) if f (0) = f 0 (0) = 0 then C(x) = f 0 (x) whenever f is differentiable at x > 0, (iii) if f ≥ 0, then f is convex and f (0) = f 0 (0) = 0. In [3] the following refinement of Jensen’s Steffensen’s type inequality for nonnegative superquadratic functions was proved: Theorem D ([3, Th. 1]). Let f : [0, ∞) → [0, ∞) be a differentiable and superquadratic function, let ξ be a nonnegative monotonic m-tuple in Rm and p a real m-tuple, m ≥ 3, satisfying 0 ≤ Pj ≤ Pm , j = 1, . . . , m, Pm > 0. Let ξ be defined as m 1 X ξ= pi ξi . Pm i=1




4

AND

S. BANI C´

Then (1.6)

m X

pi f (ξi ) − Pm f ξ ≥

i=1

k−1 X

Pi f (ξi+1 − ξi ) + Pk f ξ − ξk

i=1

+ P k+1 f ξk+1 − ξ +

m X

P i f (ξi − ξi−1 )

i=k+2

! p ξ − ξ i i ≥ Pi + P i f Pk i=1 Pm i=1 Pi + i=k+1 P i i=1 i=k+1 ! Pm i=1 pi ξ − ξi ≥ (m − 1) Pm f , (m − 1) Pm "

where P i =

Pm

j=i

k X

m X

#

Pm

pj and k ∈ {1, . . . , m − 1} satisfies ξk ≤ ξ ≤ ξk+1 .

In case f is also strictly superquadratic, inequality m X

! ξi − ξ p i i=1 (m − 1) Pm

Pm

pi f (ξi ) − Pm f ξ > (m − 1) Pm f

i=1

holds for ξ > 0 unless one of the following two cases occurs: (1) either ξ = ξ1 or ξ = ξm , (2) there exists k ∈ {3, . . . , m − 2} such that ξ = ξk and ( Pj (ξj − ξj+1 ) = 0, j = 1, . . . , k − 1 (1.7) P j (ξj − ξj−1 ) = 0, j = k + 1, . . . , m. In these two cases

m X

pi f (ξi ) − Pm f ξ = 0.

i=1

In Section 2 we refine Theorem B and Theorem C for functions which are superquadratic and positive. One of the refinements is derived easily from Theorem D. We use in Section 3 the following theorem [7, p. 323] to give an alternative proof of Theorem B. Theorem E. Let I be an interval in R, and ξ, η two decreasing m-tuples such that ξ, η ∈ I m . Let p be a real m-tuple such that k X

(1.8)

pi ξi ≤

i=1

k X

p i ηi

i=1

for k = 1, 2, . . . , m − 1, and m X

(1.9)

i=1

pi ξi =

m X

p i ηi .

i=1

Then for every continuous convex function f : I → R we have m m X X (1.10) pi f (ξi ) ≤ pi f (ηi ) . i=1


i=1



5

2. VARIANTS OF J ENSEN -S TEFFENSEN ’ S I NEQUALITY FOR P OSITIVE S UPERQUADRATIC F UNCTIONS In this section we refine in two ways Theorem C for functions which are superquadratic and positive. The refinement in Theorem 2.1 follows by showing that it is a special case of Theorem D for specific p. The refinement in Theorem 2.2 follows the steps in the proof of Theorem B given by Witkowski in [8]. Therefore the second refinement is confined only to the specific p given in Theorem B, which means that what we get is a variant of Jensen’s inequality and not of the more general Jensen-Steffensen’s inequality. Theorem 2.1. Let f : [0, ∞) → [0, ∞) and let [a, b] ⊆ [0, ∞) . Let x = (x1 , . . . , xn ) be a monotonic n−tuple in [a, b]n and v = (v1 , . . . , vn ) a real n-tuple such that vi 6= 0, i = 1, . . . , n, Pj 0 ≤ Vj ≤ Vn , j = 1, . . . , n, and Vn > 0, where Vj = i=1 vi . If f is differentiable and superquadratic, then ! n n 1 X 1 X (2.1) f (a) + f (b) − vi f (xi ) − f a + b − vi xi Vn i=1 Vn i=1   P P b − a − V1n ni=1 vi a + b − xi − V1n nj=1 vj xj . ≥ (n + 1) f  n+1 In case f is also strictly superquadratic and a > 0, inequality (2.1) is strict unless one of the following two cases occurs: (1) either x = a or x = b, (2) there exists l ∈ {2, . . . , n − 1} such that x = x1 + xn − xl and  x1 = a, xn = b or x1 = b, xn = a    (2.2) V j (xj−1 − xj ) = 0 , j = 2, . . . , l,    Vj (xj − xj+1 ) = 0 , j = l, . . . , n − 1, P P where V j = ni=j vi , j = 1, . . . , n, and x = V1n ni=1 vi xi . In these two cases we have n 1 X f (a) + f (b) − vi f (xi ) − f Vn i=1

n 1 X a+b− vi xi Vn i=1

! = 0.

In the special case where v > 0 and f is also strictly superquadratic, the equality holds in (2.1) iff xi = a, i = 1, . . . , n, or xi = b, i = 1, . . . , n. Proof. Suppose that x is an increasing n-tuple in [a, b]n . The proof of the theorem is an immediate result of Theorem D, by defining the (n + 2)-tuples ξ and p as ξ1 = a,

ξi+1 = xi , i = 1, . . . , n,

ξn+2 = b

p1 = 1,

pi+1 = −vi /Vn , i = 1, . . . , n,

pn+2 = 1.

Then we get (2.1) from the last inequality in (1.6) and from the fact that in our special case we have k n+2 X X Pi + P i ≤ n + 1, i=1


i=k+1



6

for Pi =

Pi

j=1

Pn+2

pj and P i =

j=i

AND

S. BANI C´

pj , and

m n 1 X 1 X pi ξi = a + b − vi xi = a + b − x. ξ= Pm i=1 Vn i=1

The proof of the equality case and the special case where v > 0 follows also from Theorem D. We have Vj , j = 1, . . . , n, Pn+1 = 0, Pn+2 = 1, Vn Vj−2 P 1 = 1, P 2 = 0, P j = , j = 3, . . . , n + 2. Vn Obviously, ξ = ξ1 is equivalent to x = b and ξ = ξn+2 is equivalent to x = a. Also, the existence of some k ∈ {3, . . . , m − 2} such that ξ = ξk and that (1.7) holds is equivalent to the existence of some l ∈ {2, . . . , n − 1} such that x = x1 + xn − xl = a + b − xl and that (2.2) holds. Therefore, applying Theorem D we get the desired conclusions. In the case e = (xn , . . . , x1 ) and v e = (vn , . . . , v1 ), when x is decreasing we simply replace x and v with x respectively, and then argue in the same manner. In the special case that v > 0 also, Vi > 0 and V j > 0, i = 1, . . . , n, and therefore according to (2.2) equality holds in (2.1) only when either x1 = · · · = xn = a or x1 = · · · = xn = b. Pj =

In the following P theorem we will prove a refinement of Theorem B. Without loss of generality we assume that ni=1 vi = 1. Theorem 2.2. Let f : [0, ∞) → [0, ∞) and let [a, b] ⊆ [0, ∞) , a < b. Let x = P(x1 , . . . , xn ) be an n-tuple in [a, b]n and v = (v1 , . . . , vn ) a real n-tuple such that v ≥ 0 and ni=1 vi = 1. If f is superquadratic we have ! n n X X f (a) + f (b) − vi f (xi ) − f a + b − vi xi i=1

≥

n X

vi f

i=1

(2.3)

≥

n X i=1

vi f

n ! i=1 n X X xi − a b − xi vj xj − xi + 2 vi f (b − xi ) + f (xi − a) b − a b − a j=1 i=1 n ! n X X 2 (xi − a) (b − xi ) vj xj − xi + 2 vi f . b−a j=1

i=1

If f is strictly superquadratic and v > 0 equality holds in (2.3) iff xi = a, i = 1, . . . , n, or xi = b, i = 1, . . . , n. Proof. The proof follows the technique in [8] and refines the result to positive superquadratic functions. From Lemma A we know that for any λ ∈ [0, 1] the following holds: λf (a) + (1 − λ) f (b) − f (λa + (1 − λ) b) ≥ λf (|a − λa − (1 − λ) b|) + (1 − λ) f (|b − λa − (1 − λ) b|) = λf (|(1 − λ) (a − b)|) + (1 − λ) f (|λ (b − a)|) (2.4)

= λf ((1 − λ) (b − a)) + (1 − λ) f (λ (b − a)) .

Also, for any xi ∈ [a, b] there exists a unique λi ∈ [0, 1] such that xi = λi a + (1 − λi ) b. We have n n X X (2.5) f (a) + f (b) − vi f (xi ) = f (a) + f (b) − vi f (λi a + (1 − λi ) b) . i=1


i=1



7

Applying (2.4) on every xi = λi a + (1 − λi ) b in (2.5) we obtain f (a) +f (b) −

n X

vi f (xi )

i=1 n X vi − λi f (a) − (1 − λi ) f (b) ≥ f (a) + f (b) + i=1

(2.6)

+ λi f ((1 − λi ) (b − a)) + (1 − λi ) f (λi (b − a)) n X = vi [(1 − λi ) f (a) + λi f (b)]

i=1

+

n X

vi [λi f ((1 − λi ) (b − a)) + (1 − λi ) f (λi (b − a))] .

i=1

Applying again (2.4) on (2.6) we get (2.7) f (a) + f (b) −

n X

vi f (xi ) ≥

n X

i=1

vi f ((1 − λi ) a + λi b)

i=1

+2

n X

vi [λi f ((1 − λi ) (b − a)) + (1 − λi ) f (λi (b − a))] .

i=1

Applying again Lemma A on (2.7) we obtain f (a) +f (b) − ≥f

n X

i=1 n X

+

vi f (xi ) !

vi [(1 − λi ) a + λi b]

i=1 n X

! n X vj [(1 − λj ) a + λj b] (1 − λi ) a + λi b −

vi f

i=1 n X

+2

j=1

vi [(1 − λi ) f (λi (b − a)) + λi f ((1 − λi ) (b − a))]

i=1

! n X =f a+b− vi xi + vi f vj xj − xi i=1 i=1 j=1 n X xi − a b − xi +2 vi f (b − xi ) + f (xi − a) , b − a b − a i=1 n X

(2.8)

!

n X

and this is the first inequality in (2.3). Since f is a nonnegative superquadratic function, from Lemma B we know that it is also convex, so from (2.8) we have X n n X xi − a b − xi 2 (b − xi ) (xi − a) vi f (b − xi ) + f (xi − a) ≥ vi f , b−a b−a b−a i=1 i=1 hence, the second inequality in (2.3) is proved.




8

AND

S. BANI C´

For the case when f is strictly superquadratic and v > 0 we may deduce that inequalities (2.6) and (2.7) become equalities iff each of the λi , i = 1, . . . , n, is either equal to 1 or equal to 0, which means that xi ∈ {a, b} , i = 1, . . . , n. However, since we also have n X

vj xj − xi = 0,

i = 1, . . . , n,

j=1

we deduce that xi = a, i = 1, . . . , n, or xi = b, i = 1, . . . , n. This completes the proof of the theorem.

P Corollary 2.3. Let v = (v1 , . . . , vn )be a real n-tuple such that v ≥ 0, ni=1 vi = 1 and let x = (x1 , . . . , xn ) be an n-tuple in [a, b]n , 0 < a < b. Then for any real numbers r and s such that rs ≥ 2 we have s Qs (a, b; x) −1 Qr (a, b; x)  n s n r X X 1 r r  vi ≥ vj xj − xi Qr (a, b; x)s i=1 j=1 r # n r r X s s x i − ar r b − x i (2.9) +2 vi (b − xri ) r + r (xri − ar ) r r r r b −a b −a i=1   s rs n n n X r r r r r X X 2 (x − a ) (b − x ) 1 i i  , ≥ vi vj xrj − xri + 2 vi s r r Qr (a, b; x) i=1 b − a j=1 i=1 1 P where Qp (a, b; x) = (ap + bp − ni=1 vi xi ) p , p ∈ R \ {0} . If rs > 2 and v > 0, the equalities hold in (2.9) iff xi = a, i = 1, . . . , n or xi = b, i = 1, . . . , n. s

Proof. We define a function f : (0, ∞) → (0, ∞) as f (x) = x r . It can be easily checked that for any real numbers r and s such that rs ≥ 2 the function f is superquadratic. We define a new positive n-tuple ξ in [ar , br ] as ξi = xri , i = 1, . . . , n. From Theorem 2.2 we have ! rs n n X X as + b s − vi xsi − ar + br − vi xri i=1

(2.10)

i=1

s r n n n X r X X s s br − xri r x i − ar r r r r r r r ≥ vi vj xj − xi + 2 vi r (b − xi ) + r (xi − a ) r r b − a b − a i=1 j=1 i=1 s s n n n X r X X 2 (xri − ar ) (br − xri ) r r r vi ≥ vi vj xj − xi + 2 ≥ 0. b r − ar i=1

i=1

j=1

We have as + b s −

n X

vi xsi −

ar + b r −

i=1

n X

! rs vi xri

= Qs (a, b; x)s − Qr (a, b; x)s

i=1

so from (2.10) the inequalities in (2.9) follow. s The equality case follows from the equality case in Theorem 2.2, as the function f (x) = x r is strictly superquadratic for rs > 2.




9

Remark 2.4. It is an immediate result of Corollary 2.3 that if rs > 2 and there is at least one j ∈ {1, . . . , n} such that vj xrj − ar br − xrj > 0, then for this j we have ! rs s 2 xrj − ar br − xrj Qs (a, b; x) 2vj > 0. −1> Qr (a, b; x) Qr (a, b; x)s b r − ar 3. A N A LTERNATIVE P ROOF OF T HEOREM B In this section we give an interesting alternative proof of Theorem B based on Theorem E. To carry out that proof we need the following technical lemma. Lemma 3.1. Let y = (y1 , . . P . , ym ) be a decreasing real m-tuple and p = (p1 , . . . , pm ) a nonnegative real m-tuple with m i=1 pi = 1 . We define m X y= pi yi i=1

and the m-tuple y = (y, y, . . . , y) . Then the m-tuples η = y, ξ = y and p satisfy conditions (1.8) and (1.9). Proof. Note that y is a convex combination of y1 , y2 , . . . , ym , so we know that ym ≤ y ≤ y1 . From the definitions of the m-tuples ξ and η we have m m m m X X X X pi = y = pi yi = p i ηi . pi ξi = y i=1

i=1

i=1

i=1

Hence, condition (1.9) is satisfied. Furthermore, for k = 1, 2, . . . , m − 1 we have k X

p i ηi −

k X

i=1

pi ξi =

k X

i=1

pi yi − y

i=1

= Since

k X i=1

i=1

pi

i=1

k X

pi yi −

i=1

Pm

k X

m X

p j yj

j=1

k X

pi .

i=1

pi = 1 , we can write

p i ηi −

k X

pi ξi =

i=1

k X

pj +

j=1

=

=

m X

pj

k X

j=k+1 k X

m X

i=1

pj

pi yi −

i=1

k X

!

pi

m X

k X

pi yi −

pj yi −

k X j=1

m X

pi

i=1

j=k+1

pi

k X i=1

j=k+1

i=1

=

m X

pj yj +

m X j=k+1

! pj yj

k X

pi

i=1

pj yj

j=k+1

m X

! pj yj

j=k+1

pj (yi − yj ) .

j=k+1




10

AND

S. BANI C´

Since p is nonnegative and y is decreasing, we obtain k X

p i ηi −

i=1

k X

pi ξi ≥ 0,

k = 1, 2, . . . , m − 1,

i=1

which means that condition (1.8) is satisfied as well.

Now we can give an alternative proof of Theorem B which is mainly based on Theorem E. P Proof of Theorem B. Since x = ni=1 wi xi is a convex combination of x1 , x2 , . . . , xn it is clear that there is an s ∈ {1, 2, . . . , n − 1} such that x1 ≤ · · · ≤ xs ≤ x ≤ xs+1 ≤ · · · ≤ xn , that is, −x1 ≥ · · · ≥ −xs ≥ −x ≥ −xs+1 ≥ · · · ≥ −xn .

(3.1)

Adding x1 + xn to all the inequalities in (3.1) we obtain xn ≥ · · · ≥ x1 + xn − xs ≥ x1 + xn − x ≥ x1 + xn − xs+1 ≥ · · · ≥ x1 , which gives us (3.2)

x1 + xn − x = x1 + xn −

n X

wi xi ∈ [x1 , xn ] .

i=1

We use (1.10) to prove the theorem. For this, we define the (n + 2)-tuples ξ, η and p as follows: η1 = xn , η2 = xn , η3 = xn−1 , . . . , ηn = x2 , ηn+1 = x1 , ηn+2 = x1 , p1 = 1, p2 = −wn , p3 = −wn−1 , . . . , pn = −w2 , pn+1 = −w1 , pn+2 = 1, ξ1 = ξ2 = · · · = ξn+2 = η,

η=

n+2 X

p i ηi = x 1 + x n −

i=1

wj xj .

j=1

It is easily verified that ξ and η are decreasing and that η and p satisfy conditions (1.8) and (1.9). Condition (1.9) is trivially fulfilled since n+2 X

n X

pi ξi = η

n+2 X

pi = η =

i=1

i=1

Pn+2 i=1

n+2 X

pi = 1. It remains to see that ξ,

p i ηi .

i=1

Further, we have ξi = η, i = 1, 2, . . . , n + 2 . To prove (1.8) , we need to demonstrate that (3.3)

η

k X

pi ≤

i=1

k X

pi ηi , k = 1, 2, . . . , n + 1.

i=1

For k = 1, (3.3) becomes η ≤ xn , and this holds because of (3.2) . On the other hand, for k = n + 1, (3.3) becomes ! n n X X η 1− wi ≤ xn − wi xi , i=1

i=1

that is, 0 ≤ xn − x, and this holds because of (3.2) .




11

If k ∈ {2, . . . , n} , (3.3) can be rewritten and in its stead we have to prove that ! n n X X wi xi . (3.4) η 1− wi ≤ xn − i=n+2−k

i=n+2−k

Let us consider the decreasing n-tuple y, where yi = x1 + xn − xi , i = 1, 2, . . . , n. We have y=

n X

wi yi

i=1

=

n X

wi (x1 + xn − xi )

i=1

= x1 + xn −

n X

wi xi = x1 + xn − x = η.

i=1

If we apply Lemma 3.1 to the n-tuple y and to the weights w, then m = n and for all l ∈ {1, 2, . . . , n − 1} the inequality l X

y

wi ≤

i=1

l X

wi (x1 + xn − xi )

i=1

P P holds. Taking into consideration that y = η, li=1 wi = 1 − ni=l+1 wi and changing indices as l = n + 1 − k, we deduce that ! n+1−k n X X (3.5) η 1− wi ≤ wi (x1 + xn − xi ) , i=1

i=n+2−k

for all k ∈ {2, . . . , n}. The difference between the right side of (3.4) and the right side of (3.5) is n n+1−k X X xn − wi xi − wi (x1 + xn − xi ) i=1

i=n+2−k

n X

= xn −

wi xi − xn

= xn 1 −

! wi

= xn

i=n+2−k n X

=

i=n+2−k

wi −

wi xi −

wi xi −

wi (x1 − xi )

n+1−k X

wi (x1 − xi )

i=1

i=n+2−k

wi (xn − xi ) +

n+1−k X i=1

i=n+2−k n X

wi (x1 − xi )

i=1

n X

−

i=1 n X

wi −

n+1−k X

i=1

i=n+2−k n+1−k X

n+1−k X

n+1−k X

wi (xi − x1 ) ≥ 0,

i=1

since w is nonnegative and x is increasing. Therefore, the inequality (3.6)

n+1−k X

wi (x1 + xn − xi ) ≤ xn −

i=1


n X

wi xi

i=n+2−k



12

AND

S. BANI C´

holds for all k ∈ {2, . . . , n}. From (3.5) and (3.6) we obtain (3.4) . This completes the proof that the m-tuples ξ, η and p satisfy conditions (1.8) and (1.9) and we can apply Theorem E to obtain n n+2 X X wi f (xi ) + f (x1 ) . pi f (η) ≤ f (xn ) − i=1

i=1

Taking into consideration that f

Pn+2

x1 + xn −

i=1

n X

pi = 1 and η = x1 + xn − !

wi xi

≤ f (x1 ) + f (xn ) −

i=1

Pn

j=1

n X

wj xj we finally get

wi f (xi ) .

i=1

R EFERENCES [1] S. ABRAMOVICH, G. JAMESON AND G. SINNAMON, Refining Jensen’s inequality, Bull. Math. Soc. Math. Roum., 47 (2004), 3–14. [2] S. ABRAMOVICH, G. JAMESON AND G. SINNAMON, Inequalities for averages of convex and superquadratic functions, J. Inequal. in Pure and Appl. Math., 5(4) (2004), Art. 91. [ONLINE: http://jipam.vu.edu.au/article.php?sid=444]. ´ M. MATIC´ AND J. PECARI ˇ ´ Jensen-Steffensen’s and related in[3] S. ABRAMOVICH, S. BANIC, C, equalities for superquadratic functions, submitted for publication. ˇ C´ BAKULA, M. MATIC´ AND J. PECARI ˇ ´ A variant of Jensen[4] S. ABRAMOVICH, M. KLARICI C, Steffensen’s inequality and quasi-arithmetic means, J. Math. Anal. Applics., 307 (2005), 370–385. [5] A. McD. MERCER, A monotonicity property of power means, J. Inequal. in Pure and Appl. Math., 3(3) (2002), Art. 40. [ONLINE: http://jipam.vu.edu.au/article.php?sid=192]. [6] A. McD. MERCER, A variant of Jensen’s inequality, J. Inequal. in Pure and Appl. Math., 4(4) (2003), Art. 73. [ONLINE: http://jipam.vu.edu.au/article.php?sid=314]. ˇ ´ F. PROSCHAN AND Y.L. TONG, Convex Functions, Partial Orderings, and Statis[7] J.E. PECARI C, tical Applications, Academic Press, Inc. (1992). [8] A. WITKOWSKI, A new proof of the monotonicity property of power means, J. Inequal. in Pure and Appl. Math., 5(3) (2004), Art. 73. [ONLINE: http://jipam.vu.edu.au/article. php?sid=425].

