In recent years, the concept of convex function has been extended by some scholars. For example, Breckner^[1] introduced the concept of s-convexity, and Varošanec^[2] defined h-convex functions. Some results for Hadamard-type inequalities related to the extended convex functions have been obtained^[3-7].

1 Background knowledge

In 2011, Özdemir et al.^[8] presented the (h, m)-convex function as follows.

Definition 1.1 Let h: J⊆ℝ→ℝ be a non-negative function. We say that f:[0, b]→ℝ is an (h, m)-convex function with m∈[0, 1], if f is non-negative and for all x, y∈[0, b] and α∈(0, 1), we have

$ \begin{array}{*{20}{c}} {f\left( {\alpha x + m\left( {1 - \alpha } \right)y} \right) \le h\left( \alpha \right)f\left( x \right) + }\\ {mh\left( {1 - \alpha } \right)f\left( y \right).} \end{array} $

If the above inequality is reversed, f is said to be (h, m)-concave function on [0, b].

Remark 1.1

1) if we choose m=1, we have h-convex functions;

2) if we choose m=1 and h(α)=α, we obtain non-negative ordinary convex functions;

3) if we choose m=1 and h(α)=α^s, we have s-convex functions in the second sense;

4) if we choose h(α)=α, we have m-convex functions.

One of important applications of the concept of convex function is the Hadamard's inequality as follows.

Let f:I⊆ℝ→ℝ be a convex function and a, b ∈ I with a < b, then the inequality

$ f\left( {\frac{{a + b}}{2}} \right) \le \frac{1}{{b - a}} \le \int_a^b {f\left( x \right){\rm{d}}x} \le \frac{{f\left( a \right) + f\left( b \right)}}{2} $

(1)

holds, which is well known as Hadamard's inequality.

In Ref. [9], Dragomir and Fitzpatrick established the Hadamard's type inequalities for s-convex function as follows.

Theorem 1.1 Suppose that f: [0, ∞)→[0, ∞) is an s-convex function in the second sense, where s∈(0, 1], and let a, b∈[0, ∞), a < b. If f∈L¹([a, b]), the inequalities

$ {2^{s - 1}}f\left( {\frac{{a + b}}{2}} \right) \le \frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} \le \frac{{f\left( a \right) + f\left( b \right)}}{{s + 1}} $

(2)

hold. The constant $ k = \frac{1}{{s + 1}} $ is the best possible in the second inequality in (2).

In Ref.[3], Sarkaya proved the Hadamard's type inequalities for class of h-convex functions as follows.

Theorem 1.2 Let f:I⊆ℝ→ℝ be an h-convex function, a, b∈I, a < b and f∈L¹[a, b].Then

$ \begin{array}{*{20}{c}} {\frac{1}{{2h\left( {\frac{1}{2}} \right)}}f\left( {\frac{{a + b}}{2}} \right) \le \frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} }\\ { \le \left[ {f\left( a \right) + f\left( b \right)} \right]\int_0^1 {h\left( t \right){\rm{d}}t} .} \end{array} $

(3)

In Ref.[5], Dragomir and Toader proved the inequality for m-convex function as follows.

Theorem 1.3 Let f:[0, ∞)→ℝ be an m-convex function with m∈(0, 1]. If 0≤a < b < ∞ and f∈L¹[a, b], one has the inequality

$ \begin{array}{*{20}{c}} {\frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} \le }\\ {\min \left\{ {\frac{{f\left( a \right) + mf\left( {\frac{b}{m}} \right)}}{2},\frac{{f\left( b \right) + mf\left( {\frac{a}{m}} \right)}}{2}} \right\}.} \end{array} $

(4)

Some inequalities of Hadamard-type related to this new class of (h, m)-convex functions are given^[8].

Theorem 1.4 Let f:[0, ∞)→ℝ be an (h, m)-convex function with m∈(0, 1] and t∈[0, 1]. If 0≤a < b < ∞ and f∈L¹[a, b], the inequality

$ \begin{array}{*{20}{c}} {\frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} \le }\\ {\min \left\{ {f\left( a \right)\int_0^1 {h\left( t \right){\rm{d}}t} + mf\left( {\frac{b}{m}} \right)\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} ,} \right.}\\ {\left. {f\left( b \right)\int_0^1 {h\left( t \right){\rm{d}}t} + mf\left( {\frac{a}{m}} \right)\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} } \right\}} \end{array} $

(5)

holds.

In Ref. [10], Pachpatte established some Hadamard's type inequalities for products of convex functions as follows.

Theorem 1.5 Let f, g : [a, b] →[0, ∞) be convex functions on [a, b] ∈ℝ, a < b, then

$ \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} \le \frac{1}{3}M\left( {a,b} \right) + \frac{1}{6}N\left( {a,b} \right), $

(6)

and

$ \begin{array}{*{20}{c}} {2f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right) \le }\\ {\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} + \frac{1}{6}M\left( {a,b} \right) + \frac{1}{3}N\left( {a,b} \right),} \end{array} $

(7)

where $ M\left( {a, b} \right) = f\left( a \right)g\left( a \right) + f\left( b \right)g\left( b \right) $, $ N\left( {a, b} \right) = f\left( a \right)g\left( b \right) + f\left( b \right)g\left( a \right) $.

In Ref.[11], some Hadamard's type inequalities for products of s-convex functions in the second sense are constructed by Kirmaci et al. as follows.

Theorem 1.6  Let f, g:[a, b]→ℝ, a, b∈[0, ∞), a < b, be functions such that f and fg are in L¹([a, b]). If f is convex and nonnegative on [a, b] and if g is s-convex on [a, b] for some fixed s∈(0, 1), then

$ \begin{array}{*{20}{c}} {\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} \le }\\ {\frac{1}{{s + 2}}M\left( {a,b} \right) + \frac{1}{{\left( {s + 1} \right)\left( {s + 2} \right)}}N\left( {a,b} \right),} \end{array} $

(8)

where $ M\left( {a, b} \right) = f\left( a \right)g\left( a \right) + f\left( b \right)g(b) $, $ N\left( {a, b} \right) = f\left( a \right)g\left( b \right) + f\left( b \right)g(a) $.

Theorem 1.7 Let f, g:[a, b]→ℝ, a, b∈[0, ∞), a < b, be functions such that f and fg in L¹([a, b]). If f is s₁-convex and g is s₂-convex connegative on [a, b] for some fixed s₁, s₂∈(0, 1), then

$ \begin{array}{*{20}{c}} {\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} \le }\\ {\frac{1}{{{s_1} + {s_2} + 1}}M\left( {a,b} \right) + B\left( {{s_1} + 1,{s_2} + 1} \right)N\left( {a,b} \right) = }\\ {\frac{1}{{{s_1} + {s_2} + 1}}\left[ {M\left( {a,b} \right) + {s_1}{s_2}\frac{{\Gamma \left( {{s_1}} \right)\Gamma \left( {{s_2}} \right)}}{{\Gamma \left( {{s_1} + {s_2} + 1} \right)}}N\left( {a,b} \right)} \right],} \end{array} $

(9)

where $ B\left( {x, y} \right) = \smallint _{_0}^{^1}{t^{x-1}}{\left( {1-t} \right)^{y-1}}{\rm{d}}t = \frac{{\Gamma \left( x \right)\Gamma \left( y \right)}}{{\Gamma (x + y)}} $.

Theorem 1.8  Let f, g:[a, b]→ℝ, a, b∈[0, ∞), a < b, be functions such that f and fg are in L¹ ([a, b]). If f is convex and nonnegative on [a, b] and if g is s-convex on [a, b] for some fixed s∈(0, 1), then

$ \begin{array}{*{20}{c}} {{2^s}f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right) - \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} \le }\\ {\frac{1}{{\left( {s + 1} \right)\left( {s + 2} \right)}}M\left( {a,b} \right) + \frac{1}{{s + 2}}N\left( {a,b} \right).} \end{array} $

(10)

The main purpose of this work is to establish some new Hadamard-type inequalities, similar to the above inequalities, for products of convex functions and (h, m)-convex functions, which are generalizations of the above inequalities.

2 Main results and applications

Theorem 2.1 Let h:[0, 1]→(0, ∞), f, g:[0, ∞)→ℝ, be functions such that h∈L¹([0, 1]), f g∈L¹([a, b]) with a, b∈[0, ∞), a < b. If f is convex and nonnegative on [0, ∞), and if g is (h, m)-convex and nonnegative on [0, ∞) with m∈(0, 1] and t∈[0, 1], then the following inequality holds,

$ \begin{array}{*{20}{c}} {\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} }\\ { \le \min \left\{ {\left[ {f\left( a \right) - f\left( b \right)} \right]P\left( {t,h\left( t \right),h\left( {1 - t} \right)} \right) + } \right.}\\ {f\left( b \right)P\left( {1,h\left( t \right),h\left( {1 - t} \right)} \right),}\\ {\left[ {f\left( b \right) - f\left( a \right)} \right]Q\left( {t,h\left( t \right),h\left( {1 - t} \right)} \right) + }\\ {\left. {f\left( a \right)Q\left( {1,h\left( t \right),h\left( {1 - t} \right)} \right)} \right\},} \end{array} $

(11)

where

$ \begin{array}{l} \;\;\;\;\;P\left( {t,h\left( t \right),h\left( {1 - t} \right)} \right)\\ = g\left( a \right)\int_0^1 {th\left( t \right){\rm{d}}t} + mg\left( {\frac{b}{m}} \right)\int_0^1 {th\left( {1 - t} \right){\rm{d}}t} , \end{array} $

$ \begin{array}{l} \;\;\;\;\;P\left( {1,h\left( t \right),h\left( {1 - t} \right)} \right)\\ = g\left( a \right)\int_0^1 {h\left( t \right){\rm{d}}t} + mg\left( {\frac{b}{m}} \right)\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} , \end{array} $

$ \begin{array}{l} \;\;\;\;\;Q\left( {t,h\left( t \right),h\left( {1 - t} \right)} \right)\\ = g\left( b \right)\int_0^1 {th\left( t \right){\rm{d}}t} + mg\left( {\frac{a}{m}} \right)\int_0^1 {th\left( {1 - t} \right){\rm{d}}t} , \end{array} $

$ \begin{array}{l} \;\;\;\;\;Q\left( {t,h\left( t \right),h\left( {1 - t} \right)} \right)\\ = g\left( b \right)\int_0^1 {h\left( t \right){\rm{d}}t} + mg\left( {\frac{a}{m}} \right)\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} . \end{array} $

Proof  Since f is convex and nonnegative on [0, ∞), we have

$ f\left( {ta + \left( {1 - t} \right)b} \right) \le tf\left( a \right) + \left( {1 - t} \right)f\left( b \right) $

and

$ f\left( {tb + \left( {1 - t} \right)a} \right) \le tf\left( b \right) + \left( {1 - t} \right)f\left( a \right). $

From g is (h, m)-convex and nonnegative on [0, ∞), that is

$ g\left( {tx + m\left( {1 - t} \right)y} \right) \le h\left( t \right)g\left( x \right) + mh\left( {1 - t} \right)g\left( y \right), $

for all x, y∈[0, ∞). It follows that, for all t∈[0, 1],

$ g\left( {ta + \left( {1 - t} \right)b} \right) \le h\left( t \right)g\left( a \right) + mh\left( {1 - t} \right)g\left( {\frac{b}{m}} \right), $

and

$ g\left( {tb + \left( {1 - t} \right)a} \right) \le h\left( t \right)g\left( b \right) + mh\left( {1 - t} \right)g\left( {\frac{a}{m}} \right). $

By the nonnegativeness of f and g, we obtain

$ \begin{array}{l} f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right)\\ \le \left[ {tf\left( a \right) + \left( {1 - t} \right)f\left( b \right)} \right]\left[ {h\left( t \right)g\left( a \right) + } \right.\\ \left. {mh\left( {1 - t} \right)g\left( {\frac{b}{m}} \right)} \right]\\ = \left[ {f\left( a \right) - f\left( b \right)} \right]\left[ {g\left( a \right)th\left( t \right) + } \right.\\ \left. {mg\left( {\frac{b}{m}} \right)th\left( {1 - t} \right)} \right] + \\ f\left( b \right)\left[ {g\left( a \right)h\left( t \right) + mg\left( {\frac{b}{m}} \right)h\left( {1 - t} \right)} \right], \end{array} $

and

$ \begin{array}{l} f\left( {tb + \left( {1 - t} \right)b} \right)g\left( {tb + \left( {1 - t} \right)a} \right)\\ \le \left[ {tf\left( b \right) + \left( {1 - t} \right)f\left( a \right)} \right]\left[ {h\left( t \right)g\left( b \right) + } \right.\\ \left. {mh\left( {1 - t} \right)g\left( {\frac{a}{m}} \right)} \right]\\ = \left[ {f\left( b \right) - f\left( a \right)} \right]\left[ {g\left( b \right)th\left( t \right) + } \right.\\ \left. {mg\left( {\frac{a}{m}} \right)th\left( {1 - t} \right)} \right] + \\ f\left( a \right)\left[ {g\left( b \right)h\left( t \right) + mg\left( {\frac{a}{m}} \right)h\left( {1 - t} \right)} \right]. \end{array} $

Integrating the above two inequalities on [0, 1], with respect to t, we obtain

$ \begin{array}{l} \int_0^1 {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} \\ \le \left[ {f\left( a \right) - f\left( b \right)} \right]\left[ {g\left( a \right)\int_0^1 {th\left( t \right){\rm{d}}t} + } \right.\\ \left. {mg\left( {\frac{b}{m}} \right)\int_0^1 {th\left( {1 - t} \right){\rm{d}}t} } \right] + \\ f\left( b \right)\left[ {g\left( a \right)\int_0^1 {\left( {h\left( t \right){\rm{d}}t + mg\left( {\frac{b}{m}} \right)\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} } \right)} } \right], \end{array} $

and

$ \begin{array}{l} \int_0^1 {f\left( {tb + \left( {1 - t} \right)a} \right)g\left( {tb + \left( {1 - t} \right)a} \right){\rm{d}}t} \\ \le \left[ {f\left( b \right) - f\left( a \right)} \right]\left[ {g\left( b \right)\int_0^1 {th\left( t \right){\rm{d}}t} + } \right.\\ \left. {mg\left( {\frac{b}{m}} \right)\int_0^1 {th\left( {1 - t} \right){\rm{d}}t} } \right] + \\ f\left( a \right)\left[ {g\left( b \right)\int_0^1 {\left( {h\left( t \right){\rm{d}}t + mg\left( {\frac{b}{m}} \right)\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} } \right)} } \right], \end{array} $

It is easy to see

$ \begin{array}{l} \int_0^1 {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} \\ = \int_0^1 {f\left( {tb + \left( {1 - t} \right)a} \right)g\left( {tb + \left( {1 - t} \right)a} \right){\rm{d}}t} \\ = \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} . \end{array} $

Using the above inequalities and equality, we obtain the required result.

Remark 2.1 If we choose f(x)=1 in (11) for all x∈[a, b], we obtain the inequality (5).

Remark 2.2 If we choose f(x)=1 and h(t)=t in (11) for x∈[a, b], we obtain the inequality (4).

Remark 2.3 If we choose f(x)=1, m=1 and h(t)=t in (11), we obtain the right hand side of the Hadamard's inequality (1). If we choose f(x)=1, m=1 and h(t)=t^s in (11), we obtain the right hand side of the inequality (2).

Theorem 2.2 Let h₁, h₂:[0, 1]→(0, ∞), f, g:[0, ∞)→ℝ, be functions such that h₁h₂∈L¹ ([0, 1]), fg∈L¹([a, b]) with a, b∈[0, ∞), a < b. If f is (h₁, m₁)-convex and nonnegative on [0, ∞), and if g is (h₂, m₂)-convex and nonnegative on [0, ∞) with m₁, m₂∈(0, 1] and t∈[0, 1], then the following inequality holds,

$ \begin{array}{l} \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} \\ \le \min \left\{ {f\left( a \right)P\left( {{h_1}\left( t \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right) + } \right.\\ {m_1}f\left( {\frac{b}{{{m_1}}}} \right)P\left( {{h_1}\left( {1 - t} \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right),\\ f\left( b \right)Q\left( {{h_1}\left( t \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right) + \\ \left. {{m_1}f\left( {\frac{a}{{{m_1}}}} \right)Q\left( {{h_1}\left( {1 - t} \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right)} \right\}, \end{array} $

(12)

where

$ \begin{array}{*{20}{c}} {P\left( {{h_1}\left( t \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right)}\\ { = g\left( a \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( t \right){\rm{d}}t} + }\\ {{m_2}g\left( {\frac{b}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} ,} \end{array} $

$ \begin{array}{*{20}{c}} {P\left( {{h_1}\left( {1 - t} \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right)}\\ { = g\left( a \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( t \right){\rm{d}}t} + }\\ {{m_2}g\left( {\frac{b}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right){\rm{d}}t} ,} \end{array} $

$ \begin{array}{*{20}{c}} {Q\left( {{h_1}\left( t \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right)}\\ { = g\left( b \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( t \right){\rm{d}}t} + }\\ {{m_2}g\left( {\frac{a}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} ,} \end{array} $

$ \begin{array}{*{20}{c}} {Q\left( {{h_1}\left( {1 - t} \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right)}\\ { = g\left( b \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( t \right){\rm{d}}t} + }\\ {{m_2}g\left( {\frac{a}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right){\rm{d}}t} .} \end{array} $

Proof Since f is (h₁, m₁)-convex and nonnegative on [0, ∞), that is

$ \begin{array}{*{20}{c}} {f\left( {tx + {m_1}\left( {1 - t} \right)y} \right) \le {h_1}\left( t \right)f\left( x \right) + }\\ {{m_1}{h_1}\left( {1 - t} \right)f\left( y \right)} \end{array} $

for all x, y∈[0, ∞), we have

$ f\left( {ta + \left( {1 - t} \right)b} \right) \le {h_1}\left( t \right)f\left( a \right) + {m_1}{h_1}\left( {1 - t} \right)f\left( {\frac{b}{{{m_1}}}} \right), $

and

$ f\left( {tb + \left( {1 - t} \right)a} \right) \le {h_1}\left( t \right)f\left( b \right) + {m_1}{h_1}\left( {1 - t} \right)f\left( {\frac{a}{{{m_1}}}} \right). $

From g is (h₂, m₂)-convex and nonnegative on [0, ∞), that is

$ g\left( {tx + {m_2}\left( {1 - t} \right)y} \right) \le {h_2}\left( t \right)g\left( x \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( y \right), $

for all x, y∈[0, ∞). It follows that, for all t∈[0, 1],

$ g\left( {ta + \left( {1 - t} \right)b} \right) \le {h_2}\left( t \right)g\left( a \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( {\frac{b}{{{m_2}}}} \right) $

and

$ g\left( {tb + \left( {1 - t} \right)a} \right) \le {h_2}\left( t \right)g\left( b \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( {\frac{a}{{{m_2}}}} \right). $

By the nonnegativeness of f and g, we obtain

$ \begin{array}{*{20}{c}} {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right)}\\ { \le \left[ {{h_1}\left( t \right)f\left( a \right) + {m_1}{h_1}\left( {1 - t} \right)f\left( {\frac{b}{{{m_1}}}} \right)} \right] \times }\\ {\left[ {{h_2}\left( t \right)g\left( a \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right]}\\ { = f\left( a \right)\left[ {g\left( a \right){h_1}\left( t \right){h_2}\left( t \right) + } \right.}\\ {\left. {{m_2}g\left( {\frac{b}{{{m_2}}}} \right){h_1}\left( t \right){h_2}\left( {1 - t} \right)} \right] + }\\ {{m_1}f\left( {\frac{b}{{{m_1}}}} \right)\left[ {g\left( a \right){h_1}\left( {1 - t} \right){h_2}\left( t \right) + } \right.}\\ {\left. {{m_2}g\left( {\frac{b}{{{m_2}}}} \right){h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right)} \right],} \end{array} $

and

$ \begin{array}{*{20}{c}} {f\left( {tb + \left( {1 - t} \right)a} \right)g\left( {tb + \left( {1 - t} \right)a} \right)}\\ { \le \left[ {{h_1}\left( t \right)f\left( b \right) + {m_1}{h_1}\left( {1 - t} \right)f\left( {\frac{a}{{{m_1}}}} \right)} \right] \times }\\ {\left[ {{h_2}\left( t \right)g\left( b \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( {\frac{a}{{{m_2}}}} \right)} \right]}\\ { = f\left( b \right)\left[ {g\left( b \right){h_1}\left( t \right){h_2}\left( t \right) + } \right.}\\ {\left. {{m_2}g\left( {\frac{a}{{{m_2}}}} \right){h_1}\left( t \right){h_2}\left( {1 - t} \right)} \right] + }\\ {{m_1}f\left( {\frac{a}{{{m_1}}}} \right)\left[ {g\left( b \right){h_1}\left( {1 - t} \right){h_2}\left( t \right) + } \right.}\\ {\left. {{m_2}g\left( {\frac{a}{{{m_2}}}} \right){h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right)} \right],} \end{array} $

Integrating the above two inequalities on [0, 1], with respect to t, we obtain

$ \begin{array}{*{20}{c}} {\int_0^1 {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} }\\ { \le f\left( a \right)\left[ {g\left( a \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( t \right){\rm{d}}t} } \right] + }\\ {{m_2}g\left( {\frac{b}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} + }\\ {{m_1}f\left( {\frac{b}{{{m_1}}}} \right)\left[ {g\left( a \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} } \right] + }\\ {\left. {{m_2}g\left( {\frac{b}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right){\rm{d}}t} } \right],} \end{array} $

and

$ \begin{array}{*{20}{c}} {\int_0^1 {f\left( {tb + \left( {1 - t} \right)a} \right)g\left( {tb + \left( {1 - t} \right)a} \right){\rm{d}}t} }\\ { \le f\left( b \right)\left[ {g\left( a \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( t \right){\rm{d}}t} } \right] + }\\ {{m_2}g\left( {\frac{a}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} + }\\ {{m_1}f\left( {\frac{a}{{{m_1}}}} \right)\left[ {g\left( b \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( t \right){\rm{d}}t} } \right] + }\\ {\left. {{m_2}g\left( {\frac{a}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right){\rm{d}}t} } \right],} \end{array} $

It is easy to see that

$ \begin{array}{l} \int_0^1 {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} \\ = \int_0^1 {f\left( {tb + \left( {1 - t} \right)a} \right)g\left( {tb + \left( {1 - t} \right)a} \right){\rm{d}}t} \\ = \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} . \end{array} $

Using the above three inequalities and equality, we obtain the required result.

Remark 2.4 If we choose h₁(t)=t, m₁=1 in(12), we can obtain the inequality (11).

Corollary 2.1 Let h:[0, 1] →(0, ∞), f, g : [0, ∞)→ℝ, be functions such that h∈L¹ ([0, 1]), fg∈L¹([a, b]) with a, b∈[0, ∞), a < b. If f and g is (h, m)-convex and nonnegative on [0, ∞) with m∈(0, 1] and t∈[0, 1], then the following inequality holds,

$ \begin{array}{l} \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} \\ \le \min \left\{ {f\left( a \right)P\left( {h\left( t \right),h\left( t \right),h\left( {1 - t} \right)} \right) + } \right.\\ mf\left( {\frac{b}{m}} \right)P\left( {h\left( {1 - t} \right),h\left( t \right),h\left( {1 - t} \right)} \right),\\ f\left( b \right)Q\left( {h\left( t \right),h\left( t \right),h\left( {1 - t} \right)} \right) + \\ \left. {mf\left( {\frac{a}{m}} \right)Q\left( {h\left( {1 - t} \right),h\left( t \right),h\left( {1 - t} \right)} \right)} \right\}, \end{array} $

(13)

where

$ \begin{array}{*{20}{c}} {P\left( {h\left( t \right),h\left( t \right),h\left( {1 - t} \right)} \right)}\\ { = g\left( a \right)\int_0^1 {{h^2}\left( t \right){\rm{d}}t} + mg\left( {\frac{b}{m}} \right)\int_0^1 {h\left( t \right)h\left( {1 - t} \right){\rm{d}}t} ,} \end{array} $

$ \begin{array}{*{20}{c}} {P\left( {h\left( {1 - t} \right),h\left( t \right),h\left( {1 - t} \right)} \right)}\\ { = g\left( a \right)\int_0^1 {h\left( {1 - t} \right)h\left( t \right){\rm{d}}t} + mg\left( {\frac{b}{m}} \right)\int_0^1 {h{{\left( {1 - t} \right)}^2}{\rm{d}}t} ,} \end{array} $

$ \begin{array}{*{20}{c}} {Q\left( {h\left( t \right),h\left( t \right),h\left( {1 - t} \right)} \right)}\\ { = g\left( b \right)\int_0^1 {{h^2}\left( t \right){\rm{d}}t} + mg\left( {\frac{a}{m}} \right)\int_0^1 {h\left( t \right)h\left( {1 - t} \right){\rm{d}}t} ,} \end{array} $

$ \begin{array}{*{20}{c}} {Q\left( {h\left( {1 - t} \right),h\left( t \right),h\left( {1 - t} \right)} \right)}\\ { = g\left( b \right)\int_0^1 {h\left( {1 - t} \right)h\left( t \right){\rm{d}}t} + mg\left( {\frac{a}{m}} \right)\int_0^1 {{h^2}\left( {1 - t} \right){\rm{d}}t} .} \end{array} $

Proof From Theorem 2.2 let h₁=h₂=h and m₁=m₂=m, so Corollary 2.1 immediately holds.

Theorem 2.3 Let h:[0, 1]→(0, ∞), f, g:[0, ∞)→ℝ, be functions such that h∈L¹([0, 1]), f g∈L¹([a, b]) with a, b∈[0, ∞), a < b. If f is convex and nonnegative on [0, ∞), and g is (h, m)-convex and nonnegative on [0, ∞) with m∈(0, 1] and t∈[0, 1], then the following inequality holds,

$ \begin{array}{l} \frac{2}{{h\left( {\frac{1}{2}} \right)}}f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right) - \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} \\ \le \left[ {mg\left( {\frac{b}{m}} \right)f\left( a \right) + mg\left( {\frac{b}{m}} \right)f\left( b \right) + } \right.\\ \;\;\;\;\;\left. {f\left( a \right)g\left( a \right)} \right]\int_0^1 {h\left( t \right){\rm{d}}t} + \\ \left[ {{m^2}g\left( {\frac{a}{{{m^2}}}} \right)f\left( a \right) + {m^2}g\left( {\frac{a}{{{m^2}}}} \right)f\left( b \right) + } \right.\\ \left. {mg\left( {\frac{b}{m}} \right)f\left( a \right)} \right]\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} + \left[ {f\left( b \right) - f\left( a \right)} \right] \times \\ \int_0^1 {\left[ {g\left( a \right)th\left( t \right) + mg\left( {\frac{b}{m}} \right)th\left( {1 - t} \right)} \right]{\rm{d}}t} . \end{array} $

(14)

Proof We can write

$ \frac{{a + b}}{2} = \frac{{ta + \left( {1 - t} \right)b}}{2} + \frac{{\left( {1 - t} \right)a + tb}}{2}. $

Since f is convex and nonnegative on [0, ∞), and g is (h, m)-convex and nonnegative on [0, ∞), so we have

$ \begin{array}{l} f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)\\ = f\left( {\frac{{ta + \left( {1 - t} \right)b}}{2} + \frac{{\left( {1 - t} \right)a + tb}}{2}} \right) \times \\ g\left( {\frac{{ta + \left( {1 - t} \right)b}}{2} + \frac{{\left( {1 - t} \right)a + tb}}{2}} \right)\\ \le \frac{{h\left( {\frac{1}{2}} \right)}}{2}\left[ {f\left( {ta + \left( {1 - t} \right)b} \right) + f\left( {\left( {1 - t} \right)a + tb} \right)} \right] \times \\ \left[ {g\left( {ta + \left( {1 - t} \right)b} \right) + mg\left( {\frac{{\left( {1 - t} \right)a}}{m} + \frac{{tb}}{m}} \right)} \right]\\ \le \frac{{h\left( {\frac{1}{2}} \right)}}{2}\{ f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right) + \\ m\left[ {tf\left( a \right) + \left( {1 - t} \right)f\left( b \right)} \right]g\left( {\frac{{\left( {1 - t} \right)a}}{m} + \frac{{tb}}{m}} \right) + \\ \left[ {\left( {1 - t} \right)f\left( a \right) + tf\left( b \right)} \right]\left[ {h\left( t \right)g\left( a \right) + mh\left( {1 - t} \right)g\left( {\frac{b}{m}} \right)} \right] + \\ \left. {m\left[ {\left( {1 - t} \right)f\left( a \right) + tf\left( b \right)} \right]g\left( {\frac{{\left( {1 - t} \right)a}}{m} + \frac{{tb}}{m}} \right)} \right\}\\ \le \frac{{h\left( {\frac{1}{2}} \right)}}{2}\left\{ {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right) + } \right.\\ m\left[ {f\left( a \right) + f\left( b \right)} \right] \times \\ \left[ {h\left( t \right)g\left( {\frac{b}{m}} \right) + mh\left( {1 - t} \right)g\left( {\frac{a}{{{m^2}}}} \right)} \right] + \\ f\left( a \right)\left[ {g\left( a \right)h\left( t \right) + mg\left( {\frac{b}{m}} \right)h\left( {1 - t} \right)} \right] + \\ \left. {\left[ {f\left( b \right) - f\left( a \right)} \right]\left[ {g\left( a \right)th\left( t \right) + mg\left( {\frac{b}{m}} \right)th\left( {1 - t} \right)} \right]} \right\}\\ = \frac{{h\left( {\frac{1}{2}} \right)}}{2}\left\{ {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right) + } \right.\\ \left[ {mg\left( {\frac{b}{m}} \right)f\left( a \right) + mf\left( {\frac{b}{m}} \right)f\left( b \right) + f\left( a \right)g\left( a \right)} \right]h\left( t \right) + \\ \left[ {{m^2}g\left( {\frac{a}{{{m^2}}}} \right)f\left( a \right) + {m^2}g\left( {\frac{a}{{{m^2}}}} \right)f\left( b \right)} \right] + \\ \left. {mg\left( {\frac{b}{m}} \right)f\left( a \right)} \right]h\left( {1 - t} \right) + \\ \left. {\left[ {f\left( b \right) - f\left( a \right)} \right]\left[ {g\left( a \right)th\left( t \right) + mg\left( {\frac{b}{m}} \right)th\left( {1 - t} \right)} \right]} \right\}. \end{array} $

Integrating both side of the above inequality on [0, 1], with respect to t, and by the fact that

$ \begin{array}{*{20}{c}} {\int_0^1 {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} }\\ { = \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} ,} \end{array} $

we obtain

$ \begin{array}{l} f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)\\ \le \frac{{h\left( {\frac{1}{2}} \right)}}{2}\left\{ {\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} + } \right.\\ \left[ {mg\left( {\frac{b}{m}} \right)f\left( a \right) + mg\left( {\frac{b}{m}} \right)f\left( b \right) + f\left( a \right)g\left( a \right)} \right]\int_0^1 {h\left( t \right){\rm{d}}t} + \\ \left[ {{m^2}g\left( {\frac{a}{{{m^2}}}} \right)f\left( a \right) + {m^2}g\left( {\frac{a}{{{m^2}}}} \right)f\left( b \right)} \right] + \\ \left. {mg\left( {\frac{b}{m}} \right)f\left( a \right)} \right]\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} + \\ \left[ {f\left( b \right) - f\left( a \right)} \right]\int_0^1 {\left[ {g\left( a \right)th\left( t \right) + mg\left( {\frac{b}{m}} \right)th\left( {1 - t} \right){\rm{d}}t} \right]} , \end{array} $

which completes the proof.

Corollary 2.2 Let f, g:[0, ∞)→[0, ∞), be convex functions such that fg∈L¹([a, b]) with a, b∈[0, ∞), a < b, then the following inequality holds,

$ \begin{array}{*{20}{c}} {4f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)}\\ { \le \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} + \frac{2}{3}M\left( {a,b} \right) + \frac{5}{6}N\left( {a,b} \right)} \end{array} $

(15)

where $ M\left( {a, b} \right) = f\left( a \right)g\left( a \right) + f\left( b \right)g(b)$, $ N\left( {a, b} \right) = f\left( a \right)g\left( b \right) + f\left( b \right)g(a) $.

Proof From Theorem 2.3 let h(t)=t, m=1, so Corollary 2.2 immediately holds.

Theorem 2.4 Let h₁, h₂:[0, 1]→(0, ∞), f, g:[0, ∞)→ℝ, be functions such that h₁h₂∈L¹([0, 1]), $f\;g \in ([\min \{ \frac{a}{{{m_1}}},\frac{a}{{{m_2}}}\} ,\max\{ \frac{b}{{{m_1}}},\frac{b}{{{m_2}}}\} ]) $ with a, b∈[0, ∞), a < b and m₁, m₂∈(0, 1].If f is (h₁, m₁)-convex and nonnegative on [0, ∞), and if g is (h₂, m₂)-convex and nonnegative on [0, ∞) with t∈(0, 1], then the following inequality holds,

$ \begin{array}{l} \frac{1}{{{h_1}\left( {\frac{1}{2}} \right){h_2}\left( {\frac{1}{2}} \right)}}f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)\\ \le \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} + \\ {m_1}{m_2}\int_0^1 {f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right)g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right){\rm{d}}t} + \\ {m_1}{N_1}\left( {a,b} \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( t \right){\rm{d}}t} + \\ {m_1}{m_2}{N_2}\left( {a,b} \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right){\rm{d}}t} + \\ {m_1}{M_1}\left( {a,b} \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( t \right){\rm{d}}t} + \\ {m_2}{M_2}\left( {a,b} \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} , \end{array} $

(16)

when

$ {N_1}\left( {a,b} \right) = {m_1}f\left( {\frac{b}{{{m_1}}}} \right)g\left( a \right) + {m_2}f\left( a \right)g\left( {\frac{b}{{{m_2}}}} \right), $

$ {N_2}\left( {a,b} \right) = {m_2}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{a}{{m_2^2}}} \right) + {m_1}f\left( {\frac{a}{{m_1^2}}} \right)g\left( {\frac{b}{{{m_2}}}} \right), $

$ {M_1}\left( {a,b} \right) = {m_1}f\left( {\frac{a}{{m_1^2}}} \right)g\left( a \right) + {m_2}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{b}{{{m_2}}}} \right), $

$ {M_2}\left( {a,b} \right) = {m_2}f\left( a \right)g\left( {\frac{a}{{m_2^2}}} \right) + {m_1}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{b}{{{m_2}}}} \right). $

Proof We can write

$ \frac{{a + b}}{2} = \frac{{ta + \left( {1 - t} \right)b}}{2} + \frac{{\left( {1 - t} \right)a + tb}}{2}. $

Since f is (h₁, m₁)-convex and nonnegative on [0, ∞), and g is (h₂, m₂)-convex and nonnegative on [0, ∞), so we have

$ \begin{array}{l} f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)\\ = f\left( {\frac{{ta + \left( {1 - t} \right)b}}{2} + \frac{{\left( {1 - t} \right)a + tb}}{2}} \right) \times \\ g\left( {\frac{{ta + \left( {1 - t} \right)b}}{2} + \frac{{\left( {1 - t} \right)a + tb}}{2}} \right)\\ \le {h_1}\left( {\frac{1}{2}} \right){h_2}\left( {\frac{1}{2}} \right)\left[ {f\left( {ta + \left( {1 - t} \right)b} \right) + } \right.\\ \left. {{m_1}f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right)} \right] \times \\ \left[ {g\left( {ta + \left( {1 - t} \right)b} \right) + {m_2}g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right)} \right]\\ \le {h_1}\left( {\frac{1}{2}} \right){h_2}\left( {\frac{1}{2}} \right)\left\{ {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right) + } \right.\\ {m_1}{m_2}f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right)g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right) + \\ {m_2}\left[ {\left( {{h_1}\left( t \right)f\left( a \right) + {m_1}{h_1}\left( {1 - t} \right)f\left( {\frac{b}{{{m_1}}}} \right)} \right)} \right] \times \\ \left[ {g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right)} \right] + {m_1}f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right) \times \\ \left. {\left[ {{h_2}\left( t \right)g\left( a \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right]} \right\}\\ \le {h_1}\left( {\frac{1}{2}} \right){h_2}\left( {\frac{1}{2}} \right)\left\{ {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right) + } \right.\\ {m_1}{m_2}f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right)g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right) + \\ {m_2}\left[ {\left( {{h_1}\left( t \right)f\left( a \right) + {m_1}{h_1}\left( {1 - t} \right)f\left( {\frac{b}{{{m_1}}}} \right)} \right)} \right] \times \\ \left[ {{h_2}\left( t \right)g\left( {\frac{b}{{{m_2}}}} \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( {\frac{a}{{m_2^2}}} \right)} \right] + \\ {m_1}\left[ {{h_1}\left( t \right)f\left( {\frac{b}{{{m_1}}}} \right) + {m_1}{h_1}\left( {1 - t} \right)f\left( {\frac{a}{{m_1^2}}} \right)} \right] \times \\ \left. {\left[ {{h_2}\left( t \right)g\left( a \right) + {m_2}{h_2}\left( {1 - t} \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right]} \right\}\\ = {h_1}\left( {\frac{1}{2}} \right){h_2}\left( {\frac{1}{2}} \right)\left\{ {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right) + } \right.\\ {m_1}{m_2}f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right)g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right) + \\ \left[ {{m_1}f\left( {\frac{b}{{{m_1}}}} \right)g\left( a \right) + {m_2}f\left( a \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right]{h_1}\left( t \right){h_2}\left( t \right) + \\ \left[ {m_2^2f\left( a \right)g\left( {\frac{a}{{m_2^2}}} \right) + {m_1}{m_2}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right]{h_1}\left( t \right){h_2}\left( {1 - t} \right) + \\ \left[ {{m_1}{m_2}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{b}{{{m_2}}}} \right) + m_1^2f\left( {\frac{a}{{m_1^2}}} \right)g\left( a \right)} \right]{h_1}\left( {1 - t} \right){h_2}\left( t \right) + \\ \left[ {{m_1}m_2^2f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{a}{{m_2^2}}} \right)} \right] + \\ \left. {\left. {m_1^2{m_2}f\left( {\frac{a}{{m_1^2}}} \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right]{h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right)} \right\}. \end{array} $

Integrating both side of the above inequality on [0, 1], with respect to t, and by the fact that

$ \begin{array}{*{20}{c}} {\int_0^1 {f\left( {ta + \left( {1 - t} \right)b} \right)g\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} }\\ { = \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} ,} \end{array} $

we obtain

$ \begin{array}{l} f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)\\ \le {h_1}\left( {\frac{1}{2}} \right){h_2}\left( {\frac{1}{2}} \right)\left\{ {\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} + } \right.\\ {m_1}{m_2}\int_0^1 {f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right)g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right){\rm{d}}t} + \\ \left[ {{m_1}f\left( {\frac{b}{{{m_1}}}} \right)g\left( a \right) + {m_2}f\left( a \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right]\int_0^1 {{h_1}\left( t \right){h_2}\left( t \right){\rm{d}}t} + \\ {m_2}\left[ {{m_2}f\left( a \right)g\left( {\frac{a}{{m_2^2}}} \right) + {m_1}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right] \times \\ \int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} + {m_1}\left[ {{m_2}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{b}{{{m_2}}}} \right) + } \right.\\ \left. {{m_1}f\left( {\frac{a}{{m_1^2}}} \right)g\left( a \right)} \right]\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( t \right){\rm{d}}t} , \end{array} $

which completes the proof.

Remark 2.5 If we choose m₁=m₂=1, h₁(t)=t, h₂(t)=t^s in (16) for some s∈(0, 1), then we obtain the inequaltiy (10).

Remark 2.6 If we choose m₁=m₂=1 and h₁(t)=h₂(t)=t in (16), we can obtain the inequality (7).

Corollary 2.3 Let f:[0, ∞)→ℝ, be an (h, m)-convex and nonnegative on [0, ∞) with m∈(0, 1] and t∈[0, 1]. If h∈L¹([0, 1]), f∈L¹([a, b]) with a, b∈[0, ∞), a < b, then the following inequality holds,

$ \begin{array}{*{20}{c}} {\frac{2}{{h\left( {\frac{1}{2}} \right)}}f\left( {\frac{{a + b}}{2}} \right) \le }\\ {\frac{1}{{b - a}}\left[ {\int_a^b {f\left( x \right){\rm{d}}x} + m\int_a^b {f\left( {\frac{x}{m}} \right){\rm{d}}x} } \right]\left[ {f\left( a \right) + mf\left( {\frac{b}{m}} \right)} \right] \times }\\ {\int_0^1 {h\left( t \right){\rm{d}}t} + m\left[ {f\left( {\frac{b}{m}} \right) + mf\left( {\frac{a}{{{m^2}}}} \right)} \right]\int_0^1 {h\left( {1 - t} \right){\rm{d}}t} .} \end{array} $

(17)

Proof We choose g(x)=1 for all x∈[a, b], and h₂(t)=t, m₂=1 in (16), then we can obtain the following inequality

$ \begin{array}{l} \frac{2}{{{h_1}\left( {\frac{1}{2}} \right)}}f\left( {\frac{{a + b}}{2}} \right)\\ \le \frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} + {m_1}\int_0^1 {f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right){\rm{d}}t} + \\ \left[ {f\left( a \right) + {m_1}f\left( {\frac{b}{{{m_1}}}} \right)} \right]\int_0^1 {{h_1}\left( {1 - t} \right){\rm{d}}t} + \\ {m_1}\left[ {f\left( {\frac{b}{{{m_1}}}} \right) + {m_1}f\left( {\frac{a}{{m_1^2}}} \right)} \right]\int_0^1 {{h_1}\left( {1 - t} \right){\rm{d}}t} . \end{array} $

By the fact that

$ \int_0^1 {f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right){\rm{d}}t} = \frac{1}{{b - a}}\int_a^b {f\left( {\frac{x}{{{m_1}}}} \right){\rm{d}}x} , $

we can obtain the result.

Remark 2.7 If in Corollary 2.3 we choose h(t)=t^s, m=1, we can obtain the following inequality for s-convex functions,

$ {2^s}f\left( {\frac{{a + b}}{2}} \right) \le \frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} + \frac{{f\left( a \right) + f\left( b \right)}}{{s + 1}}, $

which is the Remark 4 in Ref. [11].

Theorem 2.5 Let h₁, h₂:[0, 1]→(0, ∞), f, g:[0, ∞)→ℝ, be functions such that h₁h₂∈L¹([0, 1]), fg∈L¹([a, b]) with a, b∈[0, ∞), a < b. If f is (h₁, m₁)-convex and nonnegative on [0, ∞), and if g is (h₂, m₂)-convex and nonnegative on [0, ∞) with m₁, m₂∈(0, 1] and t∈[0, 1], then the following inequality holds,

$ \begin{array}{*{20}{c}} {\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} }\\ { \le f\left( a \right)P\left( {{h_1}\left( t \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right) + }\\ {{m_1}f\left( {\frac{b}{{{m_1}}}} \right)P\left( {{h_1}\left( {1 - t} \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right),} \end{array} $

(18)

where

$ \begin{array}{l} P\left( {{h_1}\left( t \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right)\\ = g\left( a \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( t \right){\rm{d}}t} + {m_2}g\left( {\frac{b}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( t \right){h_2}\left( {1 - t} \right){\rm{d}}t} , \end{array} $

$ \begin{array}{l} P\left( {{h_1}\left( {1 - t} \right),{h_2}\left( t \right),{h_2}\left( {1 - t} \right)} \right)\\ = g\left( a \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( t \right){\rm{d}}t} + \\ \;\;\;{m_2}g\left( {\frac{b}{{{m_2}}}} \right)\int_0^1 {{h_1}\left( {1 - t} \right){h_2}\left( {1 - t} \right){\rm{d}}t} . \end{array} $

Proof From the proof of Theorem 2.2, we are easy to obtain the inequality (18).

Remark 2.8 If we choose m₁=m₂=1, h₁(t)=t and h₂(t)=t^s in (18), we can obtain the inequality (8).

Remark 2.9 If we choose m₁=m₂=1, h₁(t)=t^s₁ and h₂(t)=t^s₂ in (18), we can obtain the inequality (9).

Remark 2.10 If we choose m₁=m₂=1, h₁(t)=t and h₂(t)=t in (18), we can obtain the inequality (6).

Remark 2.11 From Remark 1.1, if we choose proper values of m_i and h_i(t) (i=1, 2) in inequalities (12), (16) and (18), we can botain the corresponding inequalities under the condition of different convexity. For example,

1) If we choose h₁(t)=h₂(t)=t in (16) and (18), we can obtain the following Hadamard-type inequalities for products of m-convex function,

$ \begin{array}{l} 4f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)\\ \le \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} + \\ {m_1}{m_2}\int_0^1 {f\left( {\frac{{\left( {1 - t} \right)a}}{{{m_1}}} + \frac{{tb}}{{{m_1}}}} \right)g\left( {\frac{{\left( {1 - t} \right)a}}{{{m_2}}} + \frac{{tb}}{{{m_2}}}} \right){\rm{d}}t} + \\ \frac{1}{3}\left[ {{N_1}\left( {a,b} \right) + {m_1}{m_2}{N_2}\left( {a,b} \right)} \right] + \\ \frac{1}{6}\left[ {{m_1}{M_1}\left( {a,b} \right) + {m_2}{M_2}\left( {a,b} \right)} \right], \end{array} $

(19)

and

$ \begin{array}{*{20}{c}} {\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} }\\ { \le \frac{1}{3}\left[ {f\left( a \right)g\left( a \right) + {m_1}{m_2}f\left( {\frac{b}{{{m_1}}}} \right)g\left( {\frac{b}{{{m_2}}}} \right)} \right] + \frac{1}{6}{N_1}\left( {a,b} \right),} \end{array} $

(20)

where N₁(a, b), N₂(a, b), M₁(a, b), M₂(a, b) are as in Theorem 2.4.

2) If we choose m₁=m, h₁(t)=t; m₂=1, h₂(t)=t^s in (16) and (18), we can obtain the Hadamard-type inequalities for products of m-convex function and s-convex function as follows,

$ \begin{array}{*{20}{c}} {{2^{s + 1}}f\left( {\frac{{a + b}}{2}} \right)g\left( {\frac{{a + b}}{2}} \right)}\\ { \le \frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} + }\\ {m\int_0^1 {f\left( {\frac{{\left( {1 - t} \right)a}}{m} + \frac{{tb}}{m}} \right)g\left( {\left( {1 - t} \right)a + tb} \right){\rm{d}}t} + }\\ {\frac{1}{{s + 2}}\left[ {{{N'}_1}\left( {a,b} \right) + m{{N'}_2}\left( {a,b} \right)} \right] + }\\ {m{{M'}_1}\left( {a,b} \right)B\left( {s + 1,2} \right) + {{M'}_2}\left( {a,b} \right)B\left( {2,s + 1} \right),} \end{array} $

(21)

and

$ \begin{array}{*{20}{c}} {\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right){\rm{d}}x} }\\ { \le \frac{1}{{s + 2}}{{N'}_1}\left( {a,b} \right) + f\left( a \right)g\left( b \right)B\left( {2,s + 1} \right) + }\\ {mf\left( {\frac{b}{m}} \right)g\left( a \right)B\left( {s + 1,2} \right),} \end{array} $

(22)

where $ B\left( {x, y} \right) = \smallint _{_0}^{^1}{t^{x-1}}{\left( {1-t} \right)^{y-1}}{\rm{d}}t $, and

$ {{N'}_1}\left( {a,b} \right) = mf\left( {\frac{b}{m}} \right)g\left( a \right) + f\left( a \right)g\left( b \right), $

$ {{N'}_2}\left( {a,b} \right) = f\left( {\frac{b}{m}} \right)g\left( a \right) + mf\left( {\frac{a}{{{m^2}}}} \right)g\left( b \right), $

$ {{M'}_1}\left( {a,b} \right) = mf\left( {\frac{b}{{{m^2}}}} \right)g\left( a \right) + f\left( {\frac{b}{m}} \right)g\left( b \right), $

$ {{M'}_2}\left( {a,b} \right) = f\left( a \right)g\left( a \right) + mf\left( {\frac{b}{m}} \right)g\left( b \right). $