# Some Hermite-Hadamard Type Inequalities for Harmonicallys-Convex Functions

Authors: Feixiang Chen; Shanhe Wu
Journal: The Scientific World Journal
 (2014)

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## Abstract

We establish some estimates of the right-hand side of Hermite-Hadamard type inequalities for functions whose derivatives absolute values are harmonicallys-convex. Several Hermite-Hadamard type inequalities for products of two harmonically s-convex functions are also considered.


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## Body

## 1. Introduction

Letf:I⊆R→R be a convex function and a,b∈I with a<b; then
(1)f(a+b2)≤1b-a∫ab‍f(t)dt≤f(a)+f(b)2.
Inequality (1) is known as the Hermite-Hadamard inequality.In [1], Hudzik and Maligranda considered the class of functions which are s-convex in the second sense. This class of functions is defined as follows.A functionf:[0,∞)→[0,∞) is said to be s-convex in the second sense if the inequality
(2)f(αx+(1-α)y)≤αsf(x)+(1-α)sf(y)
holds for all x,y∈[0,∞), α∈[0,1] and for some fixed s∈(0,1].It can be easily seen that, fors=1, s-convexity reduces to ordinary convexity of functions defined on [0,∞).In [2], Dragomir and Fitzpatrick established a variant of Hermite-Hadamard inequality which holds for the s-convex functions in the second sense.Theorem 1 (see [2]).
Suppose thatf:[0,∞)→[0,∞) is an s-convex function in the second sense, where s∈(0,1] and let a,b∈[0,∞), a<b. If f∈L[a,b], then the following inequalities hold:
(3)2s-1f(a+b2)≤1b-a∫ab‍f(t)dt≤f(a)+f(b)s+1.Some generalizations, improvements, and extensions of inequalities (1) and (3) can be found in the recent papers [2–18].In [16], İşcan investigated the Hermite-Hadamard type inequalities for harmonically convex functions.Definition 2 (see [16]).
LetI⊆R∖{0} be a real interval. A function f:I→R is said to be harmonically convex, if
(4)f(xytx+(1-t)y)≤tf(y)+(1-t)f(x),
for all x,y∈I and t∈[0,1]. If the inequality in (4) is reversed, then f is said to be harmonically concave.Theorem 3 (see [16]).
Letf:I⊆R∖{0}→R be a harmonically convex function and a,b∈I with a<b. If f∈L(a,b), then one has
(5)f(2aba+b)≤abb-a∫ab‍f(x)x2dx≤f(a)+f(b)2.Theorem 4 (see [16]).
Letf:I⊆R∖{0}→R be a differentiable function on I0 (I0 is the interior of I), a,b∈I with a<b, and f′∈L[a,b]; then
(6)f(a)+f(b)2-abb-a∫ab‍f(x)x2dx=ab(b-a)2∫01‍1-2t(tb+(1-t)a)2f′(abtb+(1-t)a)dt.In [19], İşcan investigated the Hermite-Hadamard type inequalities for harmonically s-convex functions.Definition 5 (see [19]).
LetI⊆R∖{0} be a real interval. A function f:I⊆(0,∞)→R is said to be harmonically s-convex, if
(7)f(xytx+(1-t)y)≤tsf(y)+(1-t)sf(x),
for all x,y∈I, t∈[0,1] and for some fixed s∈(0,1]. If the inequality in (7) is reversed, then f is said to be harmonically s-concave.Theorem 6 (see [19]).
Letf:I⊆(0,∞)→R be a harmonically s-convex function and a,b∈I with a<b. If f∈L(a,b), then one has
(8)2s-1f(2aba+b)≤abb-a∫ab‍f(x)x2dx≤f(a)+f(b)s+1.In [20], Pachpatte established two new Hermite-Hadamard type inequalities for products of convex functions asserted by Theorem 7.Theorem 7 (see [20]).
Letf and g be real-valued, nonnegative, and convex functions on [a,b]. Then
(9)1b-a∫ab‍f(x)g(x)dx≤13M(a,b)+16N(a,b),2f(a+b2)g(a+b2)≤1b-a∫ab‍f(x)g(x)dx+16M(a,b)+13N(a,b),
where M(a,b)=f(a)g(a)+f(b)g(b) and N(a,b)=f(a)g(b)+f(b)g(a).For more results concerning the Hermite-Hadamard inequality, we refer the reader to [21–25] and the references cited therein.In this paper, we establish some estimates of the right-hand side of Hermite-Hadamard type inequalities for functions whose derivatives absolute values are harmonicallys-convex. Moreover, we provide several Hermite-Hadamard type inequalities for products of two harmonically s-convex functions.


## 2. Inequalities for Harmonicallys-Convex Functions

We recall the following special functions.The gamma function is as follows:(10)Γ(x)=∫0+∞‍e-xtx-1dt,x>0;
the beta function is as follows:
(11)β(x,y)=∫01‍(1-t)y-1tx-1dt,x>0,y>0,β(x,y)=Γ(x)Γ(y)Γ(x+y);
the hypergeometric function is as follows:
(12)2F1(x,y;c;z)=1β(y,c-y)∫01‍ty-1(1-t)c-y-1(1-zt)-xdt,hhhhhhhhhhhhhhhhhhhhhh|z|<1,c>y>0.Our main results are given in the following theorems.Theorem 8.
Letf:I⊆(0,∞)→R be a differentiable function on I0 such that f′∈L[a,b], where a,b∈I0 with a<b. If |f′|q is harmonically s-convex on [a,b] for some fixed s∈(0,1], q≥1, then
(13)|f(a)+f(b)2-abb-a∫ab‍f(x)x2dx|≤ab(b-a)2C11-1/q(a,b)[C2(s;a,b)|f′(b)|qhhhhhhhhhhhhhhhhhhhhh+C3(s;a,b)|f′(a)|q]1/q,
where
(14)C1(a,b)=b-2(+122F1(2,1;3;12(1-ab))2F1(2,2;3;1-ab)-2F1(2,1;2;1-ab)hhhhhhhhhhhhh+122F1(2,1;3;12(1-ab))),C2(s;a,b)=b-2(12s(s+1)(s+2)2F1(2,s+1;s+3;12(1-ab))2s+22F1(2,s+2;s+3;1-ab)hhihhhhh-1s+12F1(2,s+1;s+2;1-ab)hhihhhhh+12s(s+1)(s+2)2F1hihhhhhh×(2,s+1;s+3;12(1-ab))),C3(s;a,b)=b-2(122F1(2,1;3;12(1-ab))2(s+1)(s+2)2F1(2,2;s+3;1-ab)hhhhhhhhhhhhhh-1s+12F1(2,1;s+2;1-ab)hhhhhhhhhhhhhh+122F1(2,1;3;12(1-ab))).Proof.
LetAt=ta+(1-t)b. Using Theorem 4, the power mean inequality, and the harmonically s-convexity of |f′|q, we have
(15)|f(a)+f(b)2-abb-a∫ab‍f(x)x2dx|≤ab(b-a)2∫01‍|1-2t|At2|f′(abAt)|dt≤ab(b-a)2(∫01‍|1-2t|At2dt)1-1/q×(∫01‍|1-2t|At2|f′(abAt)|qdt)1/q≤ab(b-a)2K11-1/q×(∫01‍|1-2t|At2[(1-t)s|f′(a)|q+ts|f′(b)|q]dt)1/q≤ab(b-a)2K11-1/q(K2|f′(b)|q+K3|f′(a)|q)1/q,
where
(16)K1=∫01‍|1-2t|At2dt,K2=∫01‍|1-2t|At2tsdt,K3=∫01‍|1-2t|At2(1-t)sdt.
Calculating K1, K2, and K3, we find
(17)K1=∫01‍|1-2t|At2dt=∫01/2‍1-2tAt2dt+∫1/21‍2t-1At2dt=∫01‍2t-1At2dt+2∫01/2‍1-2tAt2dt=2∫01‍tAt-2dt-∫01‍At-2dt+∫01‍(1-u)b-2(1-u12(1-ab))-2du=b-2(2F1(2,1;2;1-ab)+122F1(2,1;3;12(1-ab))2F1(2,2;3;1-ab)hhhhhh-2F1(2,1;2;1-ab)hhhhhh+122F1(2,1;3;12(1-ab)))=C1(a,b).
Similarly, we get
(18)K2=∫01‍|1-2t|tsAt2dt=∫01‍2t-1At2tsdt+2∫01/2‍1-2tAt2tsdt=2∫01‍ts+1At-2dt-∫01‍tsAt-2dt+12s∫01‍(1-u)usb-2(1-u12(1-ab))-2du=2b-2s+22F1(2,s+2;s+3;1-ab)-b-2s+12F1(2,s+1;s+2;1-ab)+b-22s(s+1)(s+2)2F1(2,s+1;s+3;12(1-ab))=C2(s;a,b),(19)K3=∫01‍|1-2t|(1-t)sAt2dt=∫01‍2t-1At2(1-t)sdt+2∫01/2‍1-2tAt2(1-t)sdt≤2∫01‍t(1-t)sAt-2dt-∫01‍(1-t)sAt-2dt+2∫01/2‍1-2tAt2dt=b-2(122F1(2,1;3;12(1-ab))2(s+1)(s+2)2F1(2,2;s+3;1-ab)hhhhhh-1s+12F1(2,1;s+2;1-ab)hhhhhh+122F1(2,1;3;12(1-ab)))=C3(s;a,b).
This completes the proof of Theorem 8.Theorem 9.
Letf:I⊆(0,∞)→R be a differentiable function on I0 such that f′∈L[a,b], where a,b∈I0 with a<b. If |f′|q is harmonically s-convex on [a,b] for some fixed s∈(0,1], q>1, then
(20)|f(a)+f(b)2-abb-a∫ab‍f(x)x2dx|≤a(b-a)2b(1p+1)1/p×((2F1(2q,s+1;s+2;1-ab)|f′(b)|qhhhhhhh+2F1(2q,1;s+2;1-ab)|f′(a)|q)×(s+1)-1(2F1(2q,s+1;s+2;1-ab)|f′(b)|q)1/q,
where (1/p)+(1/q)=1.Proof.
LetAt=ta+(1-t)b. Utilizing Theorem 4, the Hölder inequality, and the harmonically s-convexity of |f′|q, we have
(21)|f(a)+f(b)2-abb-a∫ab‍f(x)x2dx|≤ab(b-a)2∫01‍|1-2t|At2|f′(abAt)|dt≤ab(b-a)2(∫01‍|1-2t|pdt)1/p(∫01‍1At2q|f′(abAt)|qdt)1/q≤ab(b-a)2K41/p×(∫01‍1At2q[(1-t)s|f′(a)|q+ts|f′(b)|q]dt)1/q=ab(b-a)2K41/p[K5|f′(b)|q+K6|f′(a)|q]1/q,
where
(22)K4=∫01‍|1-2t|pdt=1p+1,K5=∫01‍At-2qtsdti=b-2q∫01‍ts(1-t(1-ab))-2qdti=1(s+1)b2q2F1(2q,s+1;s+2;1-ab),K6=∫01‍At-2q(1-t)sdti=b-2q∫01‍(1-t)s(1-t(1-ab))-2qdti=1(s+1)b2q2F1(2q,1;s+2;1-ab).
The proof of Theorem 9 is completed.


## 3. Inequalities for Products of Harmonicallys-Convex Functions

Theorem 10.
Letf,g:[a,b]→[0,∞), a,b∈(0,∞), a<b, be functions such that f,g,fg∈L[a,b]. If f is harmonically s1-convex and g is harmonically s2-convex on [a,b] for some fixed s1,s2∈(0,1], then
(23)abb-a∫ab‍f(x)g(x)x2dx≤11+s1+s2M(a,b)+Γ(1+s1)Γ(s2+1)Γ(s1+s2+2)N(a,b),
where M(a,b)=f(a)g(a)+f(b)g(b) and N(a,b)=f(a)g(b)+f(b)g(a).Proof.
Sincef is harmonically s1-convex and g is harmonically s2-convex on [a,b], then for t∈[0,1] we get
(24)f(abta+(1-t)b)≤ts1f(b)+(1-t)s1f(a),g(abta+(1-t)b)≤ts2g(b)+(1-t)s2g(a).
From (24), we get
(25)f(abta+(1-t)b)g(abta+(1-t)b)≤ts1+s2f(b)g(b)+(1-t)s1+s2f(a)g(a)+ts1(1-t)s2f(b)g(a)+(1-t)s1ts2f(a)g(b).
Integrating both sides of the above inequality with respect to t over [0,1], we obtain
(26)∫01‍f(abta+(1-t)b)g(abta+(1-t)b)dt=abb-a∫ab‍f(x)g(x)x2dx≤1s1+s2+1[f(a)g(a)+f(b)g(b)]+f(a)g(b)∫01‍(1-t)s1ts2dt+f(b)g(a)∫01‍ts1(1-t)s2dt=11+s1+s2M(a,b)+β(1+s1,s2+1)N(a,b).
The proof of Theorem 10 is completed.Remark 11.
Takings1=s2=1 in Theorem 10, we obtain
(27)abb-a∫ab‍f(x)g(x)x2dx≤13M(a,b)+16N(a,b).Remark 12.
Choosings1=s2=1 and g≡1 in Theorem 10 gives
(28)abb-a∫ab‍f(x)x2dx≤f(a)+f(b)2,
which is the right-hand side inequality of (5).Theorem 13.
Letf,g:[a,b]→[0,∞), a,b∈(0,∞), a<b, be functions such that f,g,fg∈L[a,b]. If f is harmonically s1-convex and g is harmonically s2-convex on [a,b] for some fixed s1,s2∈(0,1], then
(29)2s1+s2-1f(2aba+b)g(2aba+b)≤abb-a∫ab‍f(x)g(x)x2dx+M(a,b)Γ(1+s1)Γ(s2+1)Γ(s1+s2+2)+1s2+s1+1N(a,b),
where M(a,b)=f(a)g(a)+f(b)g(b) and N(a,b)=f(a)g(b)+f(a)g(b).Proof.
Using the harmonicallys-convexity of f and g, we have for all x,y∈[a,b](30)f(2xyx+y)≤f(y)+f(x)2s1,g(2xyx+y)≤g(y)+g(x)2s2.
Choosing x=ab/(tb+(1-t)a) and y=ab/(tb+(1-t)a), we have
(31)f(2aba+b)g(2aba+b)≤f(ab/(tb+(1-t)a))+f(ab/(ta+(1-t)b))2s1×g(ab/(tb+(1-t)a))+g(ab/(ta+(1-t)b))2s2=12s1+s2[f(abtb+(1-t)a)g(abtb+(1-t)a)hhhhhhhh+f(abtb+(1-t)a)g(abta+(1-t)b)hhhhhhhh+f(abta+(1-t)b)g(abtb+(1-t)a)hhhhhhhh+f(abta+(1-t)b)g(abta+(1-t)b)]≤12s1+s2[f(abtb+(1-t)a)g(abtb+(1-t)a)hhhhhhhh+f(abta+(1-t)b)g(abta+(1-t)b)]+12s1+s2{[ts1f(a)+(1-t)s1f(b)]hhhhhhhhhh×[(1-t)s2g(a)+ts2g(b)]hhhhhhhhhh+[(1-t)s1f(a)+ts1f(b)]hhhhhhhhhh×[ts2g(a)+(1-t)s2g(b)]}=12s1+s2[f(abtb+(1-t)a)g(abtb+(1-t)a)hhhhhhhhh+f(abta+(1-t)b)g(abta+(1-t)b)]+12s1+s2{[ts1(1-t)s2+(1-t)s1ts2]M(a,b)hhhhhhhhhhh+[(1-t)s2+s1+ts2+s1]N(a,b)}.
Integrating the resulting inequality with respect to t over [0,1], we get
(32)f(2aba+b)g(2aba+b)≤12s1+s2[∫01‍f(abtb+(1-t)a)g(abtb+(1-t)a)dthhhhhhhhh+∫01‍f(abta+(1-t)b)g(abta+(1-t)b)dt]+12s1+s2{M(a,b)∫01‍[ts1(1-t)s2+(1-t)s1ts2]dthihhhhhhhhh+N(a,b)∫01‍[(1-t)s2+s1+ts2+s1]dt}.
That is,
(33)f(2aba+b)g(2aba+b)≤12s1+s2-1abb-a∫ab‍f(x)g(x)x2dx+12s1+s2{M(a,b)∫01‍[ts1(1-t)s2+(1-t)s1ts2]dthhhhhhhhhhh+N(a,b)∫01‍[(1-t)s2+s1+ts2+s1]dt}.
From
(34)∫01‍[ts1(1-t)s2+(1-t)s1ts2]dt=2β(s1+1,s2+1),∫01‍[(1-t)s2+s1+ts2+s1]dt=2s2+s1+1,
we get
(35)2s1+s2-1f(2aba+b)g(2aba+b)≤abb-a∫ab‍f(x)g(x)x2dx+M(a,b)β(s1+1,s2+1)+N(a,b)1s2+s1+1.
This completes the proof of Theorem 13.Remark 14.
Puttings1=s2=1 in Theorem 13 gives
(36)2f(2aba+b)g(2aba+b)≤abb-a∫ab‍f(x)g(x)x2dx+16M(a,b)+13N(a,b).Remark 15.
If we takes1=s2=1 and g≡1 in Theorem 13, then we obtain
(37)2f(2aba+b)≤abb-a∫ab‍f(x)x2dx+f(a)+f(b)2.



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