View of New inequalities of Hermite-Hadamard type for HA-convex functions

J. Numer. Anal. Approx. Theory, vol. 47 (2018) no. 1, pp. 26–41 ictp.acad.ro/jnaat

NEW INEQUALITIES OF HERMITE-HADAMARD TYPE FOR HA-CONVEX FUNCTIONS

SILVESTRU SEVER DRAGOMIR^∗

Abstract. Some new inequalities of Hermite-Hadamard type for HA-convex functions defined on positive intervals are given.

MSC 2010. 26D15; 25D10.

Keywords. Convex functions, Integral inequalities,HA-Convex functions.

1. INTRODUCTION

Following [1] (see also [41]) we say that the functionf :I ⊂R\ {0} →Ris HA-convex orharmonically convex if

(1) f _tx+(1−t)y^xy ≤(1−t)f(x) +tf(y)

for allx, y∈I and t∈[0,1]. If the inequality in (1) is reversed, thenf is said to beHA-concave or harmonically concave.

In order to avoid any confusion with the class of AH-convex functions, namely the functions satisfying the condition

(2) f((1−t)x+ty)≤ _(1−t)f^f(x)f(y)_(y)+tf(x),

we call the class of functions satisfying (1) asHA-convex functions.

If I ⊂ (0,∞) and f is convex and nondecreasing function then f is HA- convex and if f isHA-convex and nonincreasing function thenf is convex.

The following simple but important fact is as follows:

Criterion 1. If [a, b] ⊂ I ⊂ (0,∞) and if we consider the function g : ₁

b,_a¹→R, defined byg(t) =f ¹_t,thenf is HA-convex on [a, b]if and only if g is convex in the usual sense on ¹_b,¹_a.

∗Mathematics, College of Engineering & Science Victoria University, PO Box 14428 Mel- bourne City, MC 8001, Australia., DST-NRF Centre of Excellence in the Mathematical and Statistical Sciences, School of Computer Science and Applied Mathematics,University of the Witwatersrand, Johannesburg, Private Bag 3, Wits 2050, South Africa, e-mail:

[email protected].

For a convex functionh : [c, d]→R, the following inequality is well known in the literature as the Hermite-Hadamard inequality

(3) h^c+d₂ ≤ _d−c¹

Z d c

h(t)dt≤ ^h(c)+h(d)₂ for any convex function h: [c, d]→R.

For related results, see [1]–[18], [21]–[26], [27]–[37] and [38]–[49].

If we write the Hermite-Hadamard inequality for the convex functiong(t) = f ¹_ton the closed interval¹_b,_a¹,then we have

(4) f_a+b^2ab≤ _b−a^ab Z ¹

a

1 b

f ¹_tdt≤ ^f(b)+f₂ ^(a). Using the change of variables= ¹_t,we have

Z ¹

a

1 b

f ¹_tdt= Z b

a f(s)

s² ds and by (4) we get

(5) f _a+b^2ab≤ _b−a^ab Z b

a f(s)

s² ds≤ ^f(b)+f₂ ^(a).

The inequality (5) has been obtained in a different manner in [41] by I. I¸scan.

The identric mean I(a, b) is defined by I(a, b) := ¹_{e a}^bba

_b−a¹ while the logarithmic mean is defined by

L(a, b) := _ln^b−a_b−lna.

In the recent paper [25] we established the following inequalities for HA- convex functions:

Theorem 2. Let f : [a, b]⊂(0,∞) →R be an HA-convex function on the interval [a, b].Then

(6) f(L(a, b))≤ _b−a¹ Z b

a

f(x)dx≤ (L(a,b)−a)bf(b)+(b−L(a,b))af(a)

(b−a)L(a,b) ,

and

Theorem 3. Let f : [a, b]⊂ (0,∞) → R be a HA-convex function on the interval [a, b].Then

(7) f^a+b₂ ^a+b₂ ≤ _b−a¹ Z b

a

xf(x)dx≤ bf(b)+af(a)

2 .

Motivated by the above results, we establish in this paper some new inequal- ities of Hermite-Hadamard type for HA-convex functions. Some applications for special means are also given.

2. FURTHER RESULTS

We start with the following characterization ofHA-convex functions.

Theorem 4. Let f, h : [a, b] ⊂ (0,∞) → R be so that h(t) = tf(t) for t∈[a, b].Thenf is HA-convex on the interval[a, b]if and only if h is convex on[a, b].

Proof. Assume thatf isHA-convex on the interval [a, b].Then the function g:¹_b,_a¹→R,g(t) =f ¹_tis convex on¹_b,_a¹.By replacingtwith ¹_t we have f(t) =g ¹_t.

Ifλ∈[0,1] andx, y∈[a, b] then, by the convexity ofg on¹_b,¹_a,we have h((1−λ)x+λy) = [(1−λ)x+λy]f((1−λ)x+λy)

= [(1−λ)x+λy]g_{(1−λ)x+λy}¹

= [(1−λ)x+λy]g^(1−λ)x

1 x+λy¹_y (1−λ)x+λy

≤[(1−λ)x+λy]^(1−λ)xg(¹_x)+λyg ¹_y

(1−λ)x+λy

= (1−λ)xg_x¹+λyg¹_y

= (1−λ)xf(x) +λyf(y) = (1−λ)h(x) +λh(y), which shows thath is convex on [a, b].

We havef(t) = ^h(t)_t fort∈[a, b].If λ∈[0,1] andx, y∈[a, b] then, by the convexity ofh on [a, b],we have

f_{λx+(1−λ)y}^xy =

h

_xy

λx+(1−λ)y

xy λx+(1−λ)y

= ^{λx+(1−λ)y}_xy h_{λx+(1−λ)y}^xy

= ^{λx+(1−λ)y}_xy h

1 (1−λ)_x¹+λ_y¹

= ^{λx+(1−λ)y}_xy h

_(1−λ)1 xx+λ¹_yy (1−λ)_x¹+λ_y¹

≤ ^{λx+(1−λ)y}_xy ^(1−λ)

1

xh(x)+λ¹_yh(y) (1−λ)¹

x+λ_y¹

= (1−λ)¹_xh(x) +λ¹_yh(y) = (1−λ)f(x) +λf(y), which shows thatf isHA-convex on the interval [a, b].

Remark5. Iff is HA-convex on the interval [a, b],then by Theorem 4 the functionh(t) =tf(t) is convex on [a, b] and by Hermite-Hadamard inequality (3) we get the inequality (7). This gives a direct proof of (7) and it is simpler

than in [25].

In 1994, [11] (see also [32, p. 22]) we proved the following refinement of Hermite-Hadamard inequality. For a direct proof that is different from the one in [11], see the recent paper [24].

Lemma 6. Let p : [c, d]→ R be a convex function on [c, d]. Then for any division c=y₀< y₁ < ... < yn−1 < y_n=dwithn≥1 we have the inequalities

p^c+d₂ ≤ _d−c¹

n−1

X

i=0

(y_i+1−y_i)p^yi+1₂^+yi (8)

≤ _d−c¹ Z d

c

p(y)dy≤ _d−c¹

n−1

X

i=0

(yi+1−yi)^p(yi)+p(y₂ ⁱ⁺¹⁾

≤ ¹₂[p(c) +p(d)]. We can state the following result:

Theorem 7. Let f : [a, b]⊂ (0,∞) → R be a HA-convex function on the interval [a, b]. Then for any division a=x₀ < x₁ < ... < xn−1 < x_n=b with n≥1 we have the inequalities

a+b

2 f^a+b₂ ≤ _2(b−a)¹

n−1

X

i=0

x²_i+1−x²_if^xi+1₂^+xi (9)

≤ _b−a¹ Z b

a

xf(x)dx

≤ _b−a¹

n−1

X

i=0

(xi+1−xi)^xif(xi)+x₂^i+1f(xi+1)

≤ ¹₂[af(a) +bf(b)].

Follows by Lemma 6 for the convex function p(x) =xf(x), x∈[a, b]. If we take n= 2 and x∈[a, b],then by (9) we have

a+b

2 f ^a+b₂ ≤ _2(b−a)^{1 h} x²−a²f ^x+a₂ + b²−x²f ^x+b₂ ⁱ (10)

≤ _b−a¹ Z b

a

tf(t)dt

≤ _2(b−a)¹ [(b−a)xf(x) + (x−a)af(a) + (b−x)bf(b)]

≤ ¹₂[af(a) +bf(b)].

If in this inequality we choosex= ^a+b₂ ,then we get the inequality

a+b

2 f ^a+b₂ ≤ _2(b−a)^{1 hb+3a}₄ f ^b+3a₄ + ^a+3b₄ f ^a+3b₄ ⁱ (11)

≤ _b−a¹ Z b

a

tf(t)dt

≤ ¹₂

"

a+b

2 f ^a+b₂ +^{af a} +bf(b)

2

#

≤ ¹₂[af(a) +bf(b)].

If we take in (10) x= _a+b^2ab,then we get

a+b

2 f ^a+b₂ ≤ ¹

4(a+b)²

ha²(a+ 3b)f^a(a+3b)_2(a+b)+b²(3a+b)f^b(3a+b)_2(a+b)ⁱ (12)

≤ _b−a¹ Z b

a

tf(t)dt

≤ _a+b^{1 h}abf _a+b^2ab+ ^a2f(a)+b₂ ^2f(b)i≤ ¹₂[af(a) +bf(b)]. We also have:

Theorem 8. Let f : [a, b]⊂ (0,∞) → R be a HA-convex function on the interval [a, b]. Then for any division a=x₀ < x₁ < ... < xn−1 < x_n=b with n≥1 we have the inequalities

f_a+b^2ab≤ _b−a^ab

n−1

X

j=0

_x

j+1−x_j xj+1xj

f_x^2xj+1xj

j+1+xj

(13)

≤ _b−a^ab Z b

a f(x)

x² dx

≤ _b−a^ab

n−1

X

i=0

_x

j+1−x_j xj+1xj

_f(x

j)+f(xj+1)

2 ≤ ^f(b)+f(a)₂ .

Proof. Consider the convex function p(x) = f ¹_x that is convex on the interval ¹_b,¹_a. The division a =x0 < x1 < ... < xn−1 < xn =b with n≥ 1 produces the division y_i= _x¹

n−i, i∈ {0, ..., n}of the interval ¹_b,_a¹. Using the inequality (8) we get

f 1¹ b+ 1a

2

!

≤ ₁¹

a−¹_b n−1

X

i=0

t_x ¹

n−i−1 − _x¹

n−i

f



 ₁ ¹

xn−i−1+ 1 xn−i 2

 (14) 

≤ ₁¹

a−¹_b

Z ¹

a

1 b

f ¹_tdt

≤ ₁¹

a−¹

b

n−1

X

i=0

1

xn−i−1 −_x¹

n−i

f ¹1 xn−i−1

!

+f ¹1 xn−i

!

2

≤ ¹₂^hf¹1 b

+f¹1 a

i

that is equivalent to

f_a+b^2ab≤ _b−a^ab

n−1

X

i=0

_x

n−i−xn−i−1

xn−i−1xn−i

f_x^{2xn−i−1xn−i}

n−i+xn−i−1

(15)

≤ _b−a^ab Z ¹

a

1 b

f¹_tdt≤

≤ _b−a^ab

n−1

X

i=0

_x

n−i−xn−i−1

xn−i−1xn−i

_f(x

n−i−1)+f(xn−i) 2

≤ ¹₂[f(b) +f(a)].

By re-indexing the sums and taking into account that Z ¹

a

1 b

f ¹_tdt= Z b

a f(x)

x² dx

we obtain the desired result (13).

Remark 9. If we take n = 2 and x ∈ [a, b], then by (13) we have, after appropriate calculations, that

f_a+b^2ab≤ _x¹

_(x−a)bf

(_a+x^2ax)^+(b−x)af(_x+b^2xb)

b−a

(16)

≤ _b−a^ab Z b

a f(x)

x² dx

≤ ¹₂^hf(x) + (x−a)bf(a)+(b−x)af(b) x(b−a)

i

≤ ^f(b)+f(a)₂ .

If we take in (16) x= _a+b^2ab ∈[a, b],then we get f _a+b^2ab≤ ¹₂^hf _a+3b^4ab +f _3a+b^{4ab i} (17)

≤ _b−a^ab Z b

a f(x)

x² dx

≤ ¹₂^hf_a+b^2ab+ ^f(a)+f(b)₂ ⁱ≤ ^f(a)+f(b)₂ . If we take in (16) x= ^a+b₂ ∈[a, b],then we get

f _a+b^2ab≤ bf^a(a+b)_3a+b +af^b(a+b)_a+3b a+b

(18)

≤ _b−a^ab Z b

a f(x)

x² dx

≤ ¹₂^hf ^a+b₂ +^bf(a)+af_a+b ^(b)i≤ ^f(b)+f(a)₂ .

3. RELATED RESULTS

We recall some facts on the lateral derivatives of a convex function.

Suppose thatI is an interval of real numbers with interior ˚I and f :I →R is a convex function on I. Then f is continuous on ˚I and has finite left and right derivatives at each point of ˚I. Moreover, if x, y ∈ ˚I and x < y, then f₋⁰ (x) ≤ f₊⁰ (x) ≤ f₋⁰ (y) ≤ f₊⁰ (y) which shows that both f₋⁰ and f₊⁰ are

nondecreasing function on ˚I. It is also known that a convex function must be differentiable except for at most countably many points.

For a convex function f :I →R, the subdifferential of f denoted by ∂f is the set of all functions ϕ:I →[−∞,∞] such that ϕ ˚I⊂R and

(19) f(x)≥f(a) + (x−a)ϕ(a) for any x, a∈I.

It is also well known that if f is convex on I, then ∂f is nonempty, f₋⁰ , f₊⁰ ∈∂f and ifϕ∈∂f, then

f₋⁰ (x)≤ϕ(x)≤f₊⁰ (x) for anyx∈˚I.

In particular, ϕ is a nondecreasing function. Iff is differentiable and convex on ˚I, then ∂f={f⁰}.

Lemma 10. Let f : [a, b] ⊂ (0,∞) → R be an HA-convex function on the interval [a, b].Then f has lateral derivatives in every point of (a, b) and (20) f(t)−f(s)≥sf_±⁰ (s) 1− ^s_t

for any s∈(a, b) and t∈[a, b]. Also, we have

(21) f(t)−f(a)≥af₊⁰ (a) 1− ^a_t and

(22) f(t)−f(b)≥bf₋⁰ (b)1−^b_t

for any t∈[a, b]provided the lateral derivatives f₊⁰ (a) andf₋⁰ (b) are finite.

Proof. If f is HA-convex function on the interval [a, b], then the function h(t) =tf(t) is convex on [a, b], therefore the functionf has lateral derivatives in each point of (a, b) and

h⁰_±(t) =f(t) +tf_±⁰ (t)

for anyt∈(a, b).Also, if f₊⁰ (a) and f₋⁰ (b) are finite then

h⁰₊(a) =f(a) +af₊⁰ (a) andh⁰₋(b) =f(b) +bf₋⁰ (b). Writing the gradient inequality for the convex functionh, namely

h(t)−h(s)≥h⁰_±(s) (t−s) for anys∈(a, b) and t∈[a, b],we have

tf(t)−sf(s)≥f(s) +sf_±⁰ (s)(t−s) =f(s) (t−s) +sf_±⁰ (s) (t−s) that is equivalent to

tf(t)−tf(s)≥sf_±⁰ (s) (t−s) for anys∈(a, b) and t∈[a, b].

Now, by dividing witht >0 we get the desired result (20).

The rest follows by the corresponding properties of convex function h.

We use the following results obtained by the author in [19] and [20]

Lemma11. Let h: [α, β]→Rbe a convex function on[α, β].Then we have the inequalities

1 8

hh⁰₊^α+β₂ −h⁰₋ ^α+β₂ ⁱ(β−α)≤ ^h(α)+h(β)₂ −_β−α¹ Z β

α

h(t)dt (23)

≤ ¹₈h⁰₋(β)−h⁰₊(α)(β−α) and

1 8

h

h⁰₊ ^α+β₂ −h⁰₋ ^α+β₂ ⁱ(β−α)≤ _β−α¹ Z β

α

h(t)dt−h ^α+β₂ (24)

≤ ¹₈h⁰₋(β)−h⁰₊(α)(β−α). The constant ¹₈ is best possible in (23) and (24).

The following result holds:

Theorem12. Let f : [a, b]⊂(0,∞)→Rbe an HA-convex function on the interval [a, b].Then we have

1 16

hf₊^{0 a+b}₂ −f₋^{0 a+b}₂ ⁱ b²−a²≤ (25)

≤ af(a)+bf(b) 2 −_b−a¹

Z b a

tf(t)dt

≤ ¹₈[f(b)−f(a)] (b−a) +¹₈bf₋⁰ (b)−af₊⁰ (a)(b−a) and

1 16

hf₊^{0 a+b}₂ −f₋^{0 a+b}₂ ⁱ b²−a²≤ (26)

≤ _b−a¹ Z b

a

tf(t)dt− ^a+b₂ f^a+b₂

≤ ¹₈[f(b)−f(a)] (b−a) +¹₈bf₋⁰ (b)−af₊⁰ (a)(b−a).

Proof. Making use of inequality (23) in Lemma 11 for the convex function h(t) =tf(t) we have

1 8

ha+b

2 f₊^{0 a+b}₂ −^a+b₂ f₋^{0 a+b}₂ ⁱ(b−a)≤

≤ af(a)+bf(b) 2 −_b−a¹

Z b a

tf(t)dt

≤ ¹₈f(b) +bf₋⁰ (b)−f(a)−af₊⁰ (a)(b−a), which proves the inequality (25).

The inequality (26) follows by (24).

Corollary 13. Let f : [a, b]⊂ (0,∞)→ R be a differentiable HA-convex function on the interval [a, b].Then we have

0≤ af(a)+bf(b) 2 −_b−a¹

Z b a

tf(t)dt (27)

≤ ¹₈[f(b)−f(a)] (b−a) +¹₈bf₋⁰ (b)−af₊⁰ (a)(b−a)

and

0≤ _b−a¹ Z b

a

tf(t)dt− ^a+b₂ f^a+b₂ (28)

≤ ¹₈[f(b)−f(a)] (b−a) +¹₈bf₋⁰ (b)−af₊⁰ (a)(b−a). We remark that from (27) we have

(3a+b)f(a)+(a+3b)f(b)

8 −¹₈bf₋⁰ (b)−af₊⁰ (a)(b−a)≤ (29)

≤ _b−a¹ Z b

a

tf(t)dt≤ ^af(a)+bf(b)₂ and from (28) we have

a+b

2 f^a+b₂ ≤_b−a¹ Z b

a

tf(t)dt (30)

≤^a+b₂ f^a+b₂ +¹₈[f(b)−f(a)] (b−a) +¹₈bf₋⁰ (b)−af₊⁰ (a)(b−a). The identric mean I(a, b) is defined by

I(a, b) := ¹_ea^bba

_b−a¹

while the logarithmic mean is defined by

L(a, b) := _ln^b−a_b−lna. The following result also holds:

Theorem14. Let f : [a, b]⊂(0,∞)→Rbe an HA-convex function on the interval [a, b].

(i) Ifbf(b)−af(a)6=^R_a^bf(s)ds and

(31) αf :=

Rb

as²f⁰(s)ds

Rb

asf⁰(s)ds = ^b

2f(b)−a²f(a)−2Rb asf(s)ds bf(b)−af(a)−Rb

af(s)ds ∈[a, b]

then

(32) f(α_f)≥ _b−a¹

Z b a

f(s)ds.

(ii) Iff(b)6=f(a) and

(33) βf =

Rb asf⁰(s)ds

Rb

af⁰(s)ds = ^{bf(b)−af(a)−}

Rb af(s)ds

f(b)−f(a) ∈[a, b]

then

(34) f(β_f)≥ _lnb−lna¹

Z b a

f(s)ds.

(iii) Ifaf(b)6=bf(a) and

(35) γf := ^(f_af^(b)−f_(b)−bf(a)^(a))ab ∈[a, b]

then

(36) f(γ_f)≥ _b−a^2ab

Z b a

f(s) s² ds.

Proof. We know that if f : [a, b] ⊂(0,∞) → R is an HA-convex function on the interval [a, b] then the functions is differentiable except for at most countably many points. Then, from (20) we have

(37) f(t)−f(s)≥sf⁰(s) 1−^s_t for anyt∈[a, b] and almost everys∈(a, b).

(i) If we take the Lebesgue integral mean in (37), then we get (38) f(t)−_b−a¹

Z b a

f(s)ds≥ _b−a¹ Z b

a

sf⁰(s)ds−¹_tb−a¹ Z b

a

s²f⁰(s)ds for anyt∈[a, b].

If we take t=α_f in (38) then we get the desired inequality (32).

(ii) If we divide the inequality (37) by sthen we get (39) ¹_sf(t)− ^f(s)_s ≥f⁰(s)−¹_tsf⁰(s) for anyt∈[a, b] and almost everys∈(a, b).

If we take the Lebesgue integral mean in (39), then we get f(t)_b−a¹

Z b a

1

sds−_b−a¹ Z b

a f(s)

s ds_b−a¹ Z b

a

f⁰(s)ds−¹_tb−a¹ Z b

a

sf⁰(s)ds that is equivalent to

f(t)

L(a,b) −_b−a¹ Z b

a f(s)

s ds≥ ^f(b)−f_b−a^(a) −¹_tbf(b)−af(a)−Rb af(s)ds

(40) b−a

for anyt∈[a, b]

If we take t=β_f in (40) then we get the desired result (34).

(iii) If we divide the inequality (37) bys² then we get (41) _s¹₂f(t)−^f_s^(s)₂ ≥ ^f0(s)_s −¹_tf⁰(s) for anyt∈[a, b] and almost everys∈(a, b).

If we take the Lebesgue integral mean in (41), then we get f(t)_b−a¹

Z b a

1

s²ds−_b−a¹ Z b

a f(s)

s² ds≥ _b−a¹ Z b

a f⁰(s)

s ds−¹_tb−a¹ Z b

a

f⁰(s)ds, which is equivalent to

f(t)_ab¹ −_b−a¹ Z b

a f(s)

s² ds≥ _b−a^{1 hf(b)}_b −^f(a)_a + Z b

a f(s)

s² dsⁱ−¹_t^f(b)−f_b−a^(a) or, to

f(t)_ab¹ −_b−a² Z b

a f(s)

s² ds≥ _b−a¹ af(b)−bf(a)

ba −¹_t^f(b)−f(a)_b−a .

Remark 15. We observe that a sufficient condition for (31) and (33) to hold is that f is increasing on [a, b]. Iff(a)<0< f(b),then the inequality (35) also holds.

We also have the following result:

Theorem16. Let f : [a, b]⊂(0,∞)→Rbe an HA-convex function on the interval [a, b].Then we have

(42) f ^a+b₂ ≤ _lnb−ln¹ _a Z b

a f(t)

a+b−tdt≤ af(a)+bf(b)

a+b .

Proof. Since the functionh(t) =tf(t) is convex, then we have

x+y

2 f ^x+y₂ ≤ ^xf(x)+yf(y)₂ for anyx, y∈[a, b].

If we divide this inequality by xy >0 we get

(43) ¹₂ ¹_x+_y¹f ^x+y₂ ≤ ¹₂^f(x)_y +^f(y)_x , for anyx, y∈[a, b].

If we replace xby (1−t)a+tband y by ta+ (1−t)b in (43), then we get

1 2

1

(1−t)a+tb+_ta+(1−t)b¹ f ^a+b₂ ≤ ¹₂^f((1−t)a+tb)

ta+(1−t)b + f(ta+(1−t)b) (1−t)a+tb

, (44)

for anyt∈[0,1].

Integrating (44) on [0,1] overt we get

1 2

Z ₁

0 1

(1−t)a+tbdt+ Z 1

0 1 ta+(1−t)bdt

f ^a+b₂ ≤ (45)

≤ ¹₂ Z ₁

0

f((1−t)a+tb) ta+(1−t)b dt+

Z 1 0

f(ta+(1−t)b) (1−t)a+tb dt

. Observe that, by the appropriate change of variable,

Z 1 0

1

(1−t)a+tbdt= Z 1

0 1

ta+(1−t)bdt= _b−a¹ Z b

a du

u = ^lnb−ln_b−a ^a and

Z 1 0

f((1−t)a+tb) ta+(1−t)b dt=

Z 1 0

f(ta+(1−t)b) (1−t)a+tb = _b−a¹

Z b a

f(u) a+b−udu and by (45) we get the first inequality in (42).

From the convexity of hwe also have

((1−t)a+tb)f((1−t)a+tb)≤(1−t)af(a) +tbf(b) and

(ta+ (1−t)b)f(ta+ (1−t)b)≤taf(a) + (1−t)bf(b) for anyt∈[0,1].

Add these inequalities to get

((1−t)a+tb)f((1−t)a+tb) + (ta+ (1−t)b)f(ta+ (1−t)b)≤

≤af(a) +bf(b) for anyt∈[0,1].

If we divide this inequality by ((1−t)a+tb) (ta+ (1−t)b),then we get

(46) ^f((1−t)a+tb)

ta+(1−t)b +^f(ta+(1−t)b)

(1−t)a+tb ≤ af(a)+bf(b) ((1−t)a+tb)(ta+(1−t)b)

for anyt∈[0,1].

If we integrate the inequality (46) over ton [0,1],then we obtain Z 1

0

f((1−t)a+tb) ta+(1−t)b dt+

Z 1 0

f(ta+(1−t)b) (1−t)a+tb dt≤ (47)

≤[af(a) +bf(b)]

Z 1 0

dt

((1−t)a+tb)(ta+(1−t)b). Since

Z 1 0

dt

((1−t)a+tb)(ta+(1−t)b) = _b−a¹ Z b

a du u(a+b−u)

and

1

u(a+b−u) = _a+b^{1 1}_u +_a+b−u¹ , then

Z b a

du

u(a+b−u) = _a+b¹ Z b

a 1

u +_a+b−u¹ du= _a+b² (lnb−lna). By (47) we then have

2 b−a

Z b a

f(u)

a+b−udu≤2^haf(a)+bf(b) a+b

i_lnb−lna

b−a ,

which proves the second inequality in (42).

4. APPLICATIONS

We consider thearithmetic meanA(a, b) = ^a+b₂ ,thegeometric mean G(a, b)

=√

aband harmonic mean H(a, b) = _a+b^2ab for the positive numbers a, b >0.

If we use the inequalities (13) for the HA-convex function f(t) = t on the interval [a, b]⊂(0,∞) then for any divisiona=x₀< x₁ < ... < xn−1 < x_n=b withn≥1 we have the inequalities

2ab a+b ≤ _b−a^2ab

n−1

X

j=0

xj+1−x_j

xj+1+xj ≤ ^G_L(a,b)^2(a,b) ≤ _2(b−a)^ab

n−1

X

i=0

x²_j+1−x²_j

xj+1xj ≤A(a, b). (48)

In particular, we have

H(a, b)≤2ab _a+3b¹ +_3a+b¹ ≤ ^G_L(a,b)^2(a,b) ≤ H(a,b)+A(a,b)

2 (≤A(a, b)). (49)

Consider the functionf : (0,∞)→R,f(t) = ^ln_t^t.Observe thatg(t) =f ¹_t=

−tlnt, which shows thatf is HA-concave on (0,∞).

If we write the inequality (11) for the HA-concave function f(t) = ^ln_t^t on (0,∞),then we have for any divisiona=x₀< x₁ < ... < xn−1 < x_n=bwith n≥1 that

A(a, b)≥

n−1

Y

i=0

xi+1+xi

2

^xi+1

−xi

b−a ≥I(a, b)≥

n−1

Y

i=0

(x_ix_i+1)^xi+1

−xi

2(b−a) ≥G(a, b). (50)

In particular, we have

A(a, b)≥ ^b+3a₄ ^{2(b−a)1 a+3b}₄ ^2(b−a)1 (51)

≥I(a, b)≥^qA(a, b)G(a, b) (≥G(a, b)).

The interested reader may apply the above inequalities for other HA-convex functions such as f(t) = ^h(t)_t , t >0 withh any convex function on an interval I ⊂(0,∞) etc. The details are omitted.

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