# Selected Topics on Hermite-Hadamard Inequalities and by S.S.Dragomir C.E.M.Pearce By S.S.Dragomir C.E.M.Pearce By S.S.Dragomir C.E.M.Pearce

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Remark 20. 38) b a+b 2 f (x) dx − f a = 1 b−a b q (x) f (x) dx a where q (x) :=   a − x, x ∈ a, a+b 2  b − x, x ∈ a+b 2 ,b which will be more appropriate, later, for our purposes. The following theorem also holds : Theorem 28. Let f : I ⊆ R → R be a differentiable mapping on ˚ I, a, b ∈˚ I, p with a < b and p > 1. 39) f a+b 2 1 − b−a b a 1 1 (b − a) p f (x) dx ≤ 2 (p + 1) p1 b 1 q q |f (x)| dx a . 4. FURTHER INEQUALITIES FOR DIFFERENTIABLE CONVEX FUNCTIONS 35 Proof. Using H¨ older’s inequality we have that: 1 b−a 1 b−a ≤ b p (x) f (x) dx a 1 p b p |p (x)| dx × a 1 q b 1 b−a q |f (x)| dx .

62) 0 ≤ A (ap , bp ) − Lpp (a, b) 1 1 2 (b − a) p (p − 1) [B (p + 1, p + 1)] p Lp−2 ≤ p(p−2) (a, b) , 2 (p−1) where B is Euler’s Beta function. Proof. 63) 0 ≤ A (ap , bp ) − Lpp (a, b) ≤ p+1 1 1 (b − a) p [B (p + 1, p + 1)] p 2 1 q b p−2 q p (p − 1) x dx a where B is Euler’s Beta function. As b bpq−2q+1 − apq−2q+1 bp−q+1 − ap−q+1 xpq−2q dx = = , pq − 2q + 1 p−q+1 a and bp−q+1 − ap−q+1 = Lp−q p−q (a, b) (b − a) , p−q+1 , 2. 63) , that 0 ≤ A (ap , bp ) − Lpp (a, b) p−q p+1 1 1 1 q ≤ (b − a) p [B (p + 1, p + 1)] p p (p − 1) Lp−q (a, b) (b − a) q 2 1 1 2 = (b − a) p (p − 1) [B (p + 1, p + 1)] p Lp−2 p(p−2) (a, b) , 2 (p−1) as p−q =p−2 q and p−q = p (p − 2) ; p−1 and the proposition is proved.

The following inequality for the geometric and identric mean also holds: Proposition 20. Let p > 1 and 0 < a < b. Then we have the inequality:   1 p I (a, b) 1 [B (p + 1, p + 1)] 2 . 66) 1≤ ≤ exp  (b − a) G (a, b) 2 L2− 2p (a, b) p+1 5. FURTHER INEQUALITIES FOR TWICE DIFFERENTIABLE CONVEX FUNCTIONS 51 Proof. 66) . We shall now point out some applications of Theorem 33 for special means. Proposition 21. Let p ≥ 2 and 0 ≤ a < b. 67) p (p − 1) 2 (b − a) Lp−2 p−2 (a, b) 12 3 p (p − 1) (b − a) p−3 Lp−3 (a, b) .

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