By Mader W.

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31) f (a) + f (b) 1 − 2 b−a b f (x) dx ≥ max {|A| , |B| , |C|} ≤ 0 a where A := 1 b−a b x− a a+b 1 |f (x)| dx − 2 4 f (b) − f (a) 1 B := + 4 b−a a+b 2 b |f (x)| dx; a b f (x) dx − f (x) dx a+b 2 a and C := 1 b−a b x− a a+b 2 |f (x)| dx. 4. FURTHER INEQUALITIES FOR DIFFERENTIABLE CONVEX FUNCTIONS Proof. As f is convex on I, the mappings f and x − on [a, b] and we can apply Lemma 1. 32) ≥ max b x− x− a a+b 2 31 are synchronous b a+b 2 dx f (x) dx a where b A¯ := (b − a) x− a b ¯ := (b − a) B b a+b |f (x)| dx − 2 x− a a b a+b f (x) dx − 2 b a+b dx 2 x− |f (x)| dx, a a+b dx 2 x− a b f (x) dx a and b C¯ := (b − a) x− a a+b 2 b |f (x)| dx − x− a b a+b 2 |f (x)| dx.

A Proof. 29) is proved. Remark 17. The above theorem contains a sufficient condition for the differentiable mapping f such that the second inequality in the H· − H· result remains true. In what follows, we need a lemma which is interesting in itself as it provides a refinement of the celebrated Chebychev’s integral inequality (see also [66]). Lemma 1. , (f (x) − f (y)) (g (x) − g (y)) ≥ 0 for all x, y ∈ [a, b]. 30) C (f, g) ≥ max {|C (|f | , |g|)| , |C (|f | , g)| , |C (f, |g|)|} ≥ 0, where b C (f, g) := (b − a) b f (x) g (x) dx − a b f (x) dx a g (x) dx.

25) we have that C ≥ D ≥ 0 and the proposition is proved. 4. FURTHER INEQUALITIES FOR DIFFERENTIABLE CONVEX FUNCTIONS 29 4. 1. Integral Inequalities. Let us assume that I ⊆ R → R is a differentiable mapping on ˚ I, and let a, b ∈˚ I with a < b. 28) f (x) dx = a 1 b−a b a+b 2 x− a f (x) dx. 28) is proved. The following theorem holds [34]: Theorem 24. 29) b f (x) dx ≥ 0. a Proof. 29) is proved. Remark 17. The above theorem contains a sufficient condition for the differentiable mapping f such that the second inequality in the H· − H· result remains true.

### 1 - Faktoren von Graphen by Mader W.

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