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

. Some new inequalities of Hermite-Hadamard type for HA -convex functions deﬁned on positive intervals are given. MSC 2010. 26D15; 25D10.


INTRODUCTION
Following [1] (see also [41]) we say that the function f : for all x, y ∈ I and t ∈ [0, 1]. If the inequality in (1) is reversed, then f is said to be HA-concave or harmonically concave. In order to avoid any confusion with the class of AH-convex functions, namely the functions satisfying the condition we call the class of functions satisfying (1) as HA-convex functions.
If I ⊂ (0, ∞) and f is convex and nondecreasing function then f is HAconvex and if f is HA-convex and nonincreasing function then f is convex.
If we write the Hermite-Hadamard inequality for the convex function g (t) = f 1 t on the closed interval 1 b , 1 a , then we have .
Using the change of variable s = 1 t , we have and by (4) we get The inequality (5) has been obtained in a different manner in [41] by I. Işcan. The identric mean I (a, b) is defined by In the recent paper [25] we established the following inequalities for HAconvex functions: and Motivated by the above results, we establish in this paper some new inequalities of Hermite-Hadamard type for HA-convex functions. Some applications for special means are also given.

FURTHER RESULTS
We start with the following characterization of HA-convex functions.
Proof. Assume that f is HA-convex on the interval [a, b] . Then the function which shows that f is HA-convex on the interval [a, b] .
Remark 5. If f is HA-convex on the interval [a, b] , then by Theorem 4 the function h (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].
Then for any division c = y 0 < y 1 < ... < y n−1 < y n = d with n ≥ 1 we have the inequalities . We can state the following result:

Follows by Lemma 6 for the convex function
If we take n = 2 and x ∈ [a, b] , then by (9) we have If we take in (10) x = 2ab a+b , then we get . We also have: Using the inequality (8) we get . By re-indexing the sums and taking into account that 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 If we take in (16) If we take in (16)

RELATED RESULTS
We recall some facts on the lateral derivatives of a convex function. Suppose that I is an interval of real numbers with interiorI and f : I → R is a convex function on I. Then f is continuous onI and has finite left and right derivatives at each point ofI. Moreover, if x, y ∈I and x < y, then shows that both f − and f + are nondecreasing function onI. 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 − , f + ∈ ∂f and if ϕ ∈ ∂f , then In particular, ϕ is a nondecreasing function. If f is differentiable and convex onI, then ∂f = {f } .  Writing the gradient inequality for the convex function h, namely ∈ (a, b) and t ∈ [a, b] . Now, by dividing with t > 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] Lemma 11. Let h : [α, β] → R be a convex function on [α, β] . Then we have the inequalities  a) . Proof. Making use of inequality (23) in Lemma 11 for the convex function h (t) = tf (t) we have , which proves the inequality (25).
and a) . We remark that from (27) we have and from (28) we have then then (34) f (β f ) ≥ 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 ∈ [a, b] and almost every s ∈ (a, b) .
(i) If we take the Lebesgue integral mean in (37), then we get If we take t = α f in (38) then we get the desired inequality (32).
(ii) If we divide the inequality (37) by s then we get ∈ [a, b] and almost every s ∈ (a, b) .
If we take the Lebesgue integral mean in (39), then we get (40) for any t ∈ [a, b] If we take t = β f in (40) then we get the desired result (34).
(iii) If we divide the inequality (37) by s 2 then we get and almost every s ∈ (a, b) .
If we take the Lebesgue integral mean in (41), then we get which is equivalent to Remark 15. We observe that a sufficient condition for (31) and (33) to hold is that f is increasing on [a, b] . If f (a) < 0 < f (b) , then the inequality (35) also holds.
We also have the following result: Proof. Since the function h (t) = tf (t) is convex, then we have If we divide this inequality by xy > 0 we get for any x, y ∈ [a, b] . If we replace x by (1 − t) a + tb and y by ta (43), then we get for any t ∈ [0, 1] .
Integrating (44) Observe that, by the appropriate change of variable, a+b−u du and by (45) we get the first inequality in (42).
From the convexity of h we also have for any t ∈ [0, 1] .
Add these inequalities to get for any t ∈ [0, 1] .

APPLICATIONS
We consider the arithmetic mean A (a, b) = a+b 2 , the geometric mean G (a, b) = √ ab and harmonic mean H (a, b) = 2ab a+b 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 division a = x 0 < x 1 < ... < x n−1 < x n = b with n ≥ 1 we have the inequalities In particular, we have (a, b)) .
If we write the inequality (11) for the HA-concave function f (t) = ln t t on (0, ∞) , then we have for any division a = x 0 < x 1 < ... < x n−1 < x n = b with n ≥ 1 that 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 with h any convex function on an interval I ⊂ (0, ∞) etc. The details are omitted.