Int Journl of Mth Anlysis, Vol 1, 2007, no 25, 1243-1247 A Note on Feng Qi Type Integrl Inequlities Hong Yong Deprtment of Mthemtics Gungdong Business College Gungzhou City, Gungdong 510320, P R Chin hongyong59@sohucom Abstrct In this short note, by introducing prmeters α,, some sufficient conditions such tht Qi type integrl inequlity ( ) [] α dx dx hold re given, some new results re deduced Mthemtics Subject Clssifiction: 26D15 Keywords: Qi type integrl inequlity, derivtive, incresing function In [1], the following Qi type inequlity ws proved: If f C 1 [, b], f() 0, f (x) (t 2)(x ) t 3 for x [, b], nd t 3Then ( t 1 [] t dx dx) (1) If t = n + 2 Then the following Qi integrl inequlity in [2] is obtined: ( n+1 [] n+2 dx dx) (2) In recent yers, mny vluble results (see [3-7]) hve been obtined In the short note, by introducing prmeters α,, some sufficient conditions such tht Qi type integrl inequlity ( (3)
1244 Hong Yong hold re given The min result is: Theorem 01 Suppose α> 2, m =[], C 1 [, b], f (x) 0 nd [f α (x)] ( 1) ( m+1) (α ) (x (α 1)(α 2) (α m+1) ) m Then ( (4) Where [] denote the integer prt of Proof Sine f (x) 0, thus is incresing function on [, b] Let F (x) = x [f(u)] α du f(u)du) x [, b] Then F () = 0, nd we hve 1 F (x) =[] α f(u)du) ) } 1 = [] α 1 f(u)du F 1 (x), where 1 F 1 (x) =[] α 1 f(u)du) Obviously, F 1 () =[f()] α 1 0 nd 2 F 1 (x) =(α 1)[]α 2 f (x) ( 1) f(u)du) 2 (α 1)[] α 1 ( 1) f(u)du) ) } 2 = (α 1)[] α 2 ( 1) f(u)du F 2 (x), where 2 F 2 (x) =(α 1)[] α 2 ( 1) f(u)du) Obviously, F 2 () =(α 1)[f()] α 2 0 nd 3 F 2(x) =(α 1)(α 2)[] α 3 f (x) ( 1)( 2) f(u)du)
Integrl inequlities of Feng Qi type 1245 (α 1)(α 2)[] α 2 ( 1)( 2) = (α 1)(α 2)[] α 3 ( 1)( 2) F 3 (x) Finlly, we cn obtin f(u)du) 3 ) } 3 f(u)du F m 1 (x) = (α 1)(α 2) (α m + 2)[] α m+1 m+1 ( 1)( 2) ( m +2) f(u)du) Obviously, F m 1 () =(α 1) (α m + 2)[f()] α m+1 0 nd F m 1 (x) =(α 1)(α 2) (α m + 1)[]α m f (x) m ( 1)( 2) ( m +1) f(u)du) (α 1)(α 2) (α m + 1)[] α m f (x) ( 1)( 2) ( m + 1)[(x )] m = [] m+1 (α 1)(α 2) (α m + 1)[] α 1 f (x) } [] m+1 ( 1)( 2) ( m + 1)(x ) m} } 1 = [] m+1 α (α 1)(α 2) (α m + 1)[f α (x)] [] m+1 ( 1)( 2) ( m + 1)(x ) m} Since [f α (x)] ( 1) ( m +1) (α ) (α 1)(α 2) (α m +1) (x ) m, thus F m 1 (x) 0 for x [, b], it follows tht F m 1(x) is incresing on [, b], hence F m 1 (x) F m 1 () 0, it follows tht F m 2 (x) 0, thus F m 2(x) is incresing on [, + ), F m 2 (x) F m 2 () 0 we cn obtin F 1 (x) 0, it follows tht F (x) 0, thus F (x) is incresing on [, b], hence F (b) F () 0, ie ( Corollry 02 Suppose t 3, m =[t 1], C 1 [, b] on the intervl [, b], f (x) 0 nd f (x) (t m)(x ) t m 1 Then ( t 1 [] t dx dx) (5)
1246 Hong Yong Proof Setting α = t, = t 1 in (4) Corollry 03 Suppose n, k Z +, k 2, C 1 [, b], f (x) 0 nd [f k (x)] k!n! k Then (k+n 1)! ( n [] n+k dx dx) (6) Proof Setting α = n + k, = k in (4) Corollry 04 Suppose n Z +, C 1 [, b] on the intervl [, b], f (x) 0 nd f (x) 1 Then ( n+1 [] n+2 dx dx) Proof Setting α = n +2, = n + 1 in (4) Exmple: Suppose n Z +, =e x Then f (x) =e x = > 0 nd f (x) =e x 1 for x [0, 1] By Corollry 02, we obtin 1 ( 1 n+1 (e x ) n+2 dx e dx) x, it follows tht 0 e n+2 1 (n + 2)(e 1) n+1, n Z + Remrk: If n 3, t = n +2, =e x (x [0, 1]), then doesn t stisfy the condition in [1]: f (x) (t 2)(x ) t 3 ie e x nx n 1 ; doesn t lso stisfy the condition in [2]: f (n) (x) n! ie e x n! Thus (8) not be obtined by the theorems in [1] nd [2] 0 References [1] J PEČARIĆ AND T PEJKOVIĆ, Note on Feng Qi s intergl inequlity, J Inequl Pure nd Appl Mth, 5(3)(2004), Art 51 [2] FENG QI, Severl intergl inequlities, J Inequl Pure nd Appl Mth, 1(2)(2000) Art 19 [3] YIN CHEN AND JOHE KIMBALL, Note on n open problem of Feng Qi, J Inequl Pure nd Appl Mth, 7(1)(2006) Art 4
Integrl inequlities of Feng Qi type 1247 [4] LAZHAR BOUGOFFA, Note on Qi type intergl inequlities, J Inequl Pure nd Appl Mth, 4(4)(2003) Art 77 [5] TIBOR K POGANY, On n open problem of F Qi, J Inequl Pure nd Appl Mth, 3(4)(2002) Art 54 [6] S MAZOUZI AND FENG QI, On n open problem regrding n intergl inequlity, J Inequl Pure nd Appl Mth, 4(2)(2003) Art 31 [7] K W YU AND F QI, A short note on intergl inequlity, RGMIN Res Rep Coll, 4(1)(2001, Art 4 Received: December 30, 2006