Refinement of an integral inequality
© Zhu; licensee Springer. 2012
Received: 19 January 2012
Accepted: 4 May 2012
Published: 4 May 2012
In this study, we generalize and sharpen an integral inequality raised in theory for convex and star-shaped sets and relax the conditions on the integrand.
Mathematics Subject Classification (2000): 26D15
Keywordslower and upper bounds refinement of an integral inequality a monotone form of l'Hospital's rule Mitrinović-Pečarić inequality
In the study , which investigated convex and star-shaped sets, the following interesting result was obtained.
In this note, we shall show that the convexity of the function p(t) may be replaced by the condition that is increasing, sharpen inequality (1), and obtain the following a general result using a monotone form of l'Hospital's rule, a elementary method, and Mitrinović-Pečarić inequality, respectively.
holds so that
(i) whenis increasing, we have, ;
(ii) whenis decreasing, we have, .
Furthermore, these paired numbers α and β defined in (i) and (ii) are the best constants in (2).
2 Two lemmas
Lemma 1. ([[2–5], A Monotone form of L'Hospital's rule]) Let f, g : [a, b] → ℝ be two continuous functions which are differentiable on (a, b). Further, let on(a, b). Ifis increasing (or decreasing) on (a, b), then the functions and are also increasing (or decreasing) on (a, b).
If f is decreasing the inequalities (3) are reversed.
3 A concise proof of Theorem 2
- (a)When is increasing, we have is also increasing, and is increasing by Lemma 1. At the same time, , and . So we obtain(4)
are the best constants in (4).
When is decreasing, we obtain corresponding result by the same way.
4 New elementary proof of Theorem 2
holds. Then the double inequality (5) holds, α and β are the best constants in (6) or (2).
The decreasing case can be proved similarly.
5 Other proof of Theorem 2
In what follows, we also assume that is increasing.
- (i)The left-side inequality of (7) deduces
- (ii)Let b → 0+ in the right-side inequality of (7), we can obtain
then the left-side inequality of (2) holds.
Let . Since and G(b) = β, we obtain that is α and β are the best constants in (2).
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