Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. 7 (1996), 9{17. A GENERALIZATION OF THE KY FAN INEQUALITY Zhen Wang, Ji Chen, Guang{Xing Li Dedicated to the memory of Professor Dragoslav S. Mitrinovic A certain extension of the Ky Fan inequality is proved by means of elementary calculus. 1. INTRODUCTION The following inequality due to Ky Fan was recorded in [1]: 0 1 1=n n n P Q x x i i B C i=1 i=1 B C < (0 x 1=2); (1) i n n @ A Q P (1 x ) (1 x ) i i i=1 i=1 unless x = x = ::: = x : 1 2 m With the notation 1=p 1=n n n P Q 1 p M (x)= x (x > 0) and M (x) = lim M (x)= x ; p i i 0 p i p!0 n i=1 i=1 (1) b ecomes M (x) M (x) 1 0 < : (2) M (1 x) M (1 x) 0 1 D. Segaiman [2] conjectured that M (x) M (x) p q (3) (p q ): M (1 x) M (1 x) p q F. Chan, D. Goldberg and S. Gonek [2]gave some counter examples p q when 0 < 2 =p < 2 =q or p + q>9:In addition, they proved that (3) is true for p + q =0>p or 0 p 1 q 2: Recently the case p = 1 and q =0was proved to b e true by Wan{Lan Wang and Peng{Fei Wang [3]. And the case 1 p 0 1was proved 0 1991 Mathematics Sub ject Classi cation: 26D15 9 10 Zhen Wang, Ji Chen, Guang{Xing Li by Guang{Xing Li and Ji Chen [4]. Zhen Wang and Ji Chen [5] proved that the function R(p)= M (x)=M (1 x) is strictly increasing on [1; 1]; unless p p x = ::: = x ; and if (3) holds for (p; q ); then also for (q; p): 1 n In this pap er, we determine all exp onents p and q such that (3) is true. Theorem. For arbitrary n; p < q ; the inequality 1 1 0 0 1=p 1=q n n P P p q x x i i C C B B i=1 i=1 C C B B (4) (0 <x 1=2) i n n A A @ @ P P p q (1 x ) (1 x ) i i i=1 i=1 p q p q holds if and only if jp + q j3; 2 =p 2 =q when p> 0;p2 q2 when q<0: The pro of of the suciency is contained in Sections 3, 4 and 5. In the pro of we assume pq 6=0; otherwise by letting p or q ! 0; it is easy to see that (4) is also true. In Section 2, we will prove the necessity. 2. PROOF OF THE NECESSITY In [2], it is proved that (4) and p<q are equivalent when n =2; and that if p q (4) holds, then 2 =p 2 =q for p> 0: When q<0;take x = x = = x = " (0 <" <1=2) and x =1=2: 1 2 n1 n Then (4) b ecomes 1=p 1=q q q p p (n 1)" +(1=2) (n 1)" +(1=2) ; (5) p p q q (n 1)(1 ") +(1=2) (n 1)(1 ") +(1=2) or 1=p 1=p 1 1 p p " + (1 ") + p p 2 (n1) 2 (n1) (6) : 1=q 1=q 1 1 q q " + (1 ") + q q 2 (n1) 2 (n1) Letting " ! 0; (6) yields 1=p 1 1+ p 2 (n1) ; (7) 1 1=q 1 1+ q 2 (n1) hence 1=p 1=q 1 1 (8) 1+ 1+ : p q 2 (n 1) 2 (n1) By using the Maclaurin expansion in 1=n; we obtain p 1 2 q 1 2 (9) 1+ (p2 n) +o(1=n ) 1+(q2 n) + o(1=n ): A generalization of the Ky Fan inequality 11 p q So if p 2 >q2 ; (4) would b e false for suciently large n: In the equivalent inequality of (4): 0 1 1 0 1=p 1=q n n P P p q (1 u ) (1 u ) i i B C C B i=1 i=1 B C C B (10) (0 u < 1); i n n @ A A @ P P p q (1 + u ) (1 + u ) i i i=1 i=1 let u = u = = u = 0 and u = u (0 <u<1); then (10) b ecomes 1 2 n1 n 1=p 1=q p q (n 1) + (1 u) (n 1)+(1 u) : (11) p q (n 1)+(1+ u) (n 1)+(1+ u) Take the Maclaurin expansion of (11) in u: 2 2 (n 1) (n 2)p 3np +2(n +2) 2 2 2 3 4 (12) 1 u + u u +o(u ) 2 3 n n 3n 2 2 (n1) (n 2)q 3nq +2(n +2) 2 2 2 3 4 u+ u u +o(u ): 1 2 3 n n 3n Thus for u suciently small, (10) holds only if 2 2 (13) (n 2)p 3np (n 2)q 3nq ; or (14) (p q ) (n 2)(p + q ) 3n 0: So for n 3; wehave 3n (15) p + q : n 2 Letting n ! +1; (15) yields p + q 3: Similarly, the expansion of (10) with u = u = = u = u (0 <u<1); 1 2 n1 u = 0 gives n 3n (16) p + q : n 2 So we obtain p + q 3: 3. AN EQUIVALENCE PROPOSITION In this section, we are to establish an equivalence prop osition as follows: Prop osition. For p<q; the fol lowing inequalities areequivalent: 0 1 0 1 1=p 1=q n n P P p q x x i i i i B C B C i=1 i=1 B C B C (i) < ; n n @ A @ A P P p q (1 x ) (1 x ) i i i i i=1 i=1 12 Zhen Wang, Ji Chen, Guang{Xing Li where > 0; 0 <x 1=2;i=1;2;:::;n and x ;x ;:::;x are not al l equal; i i 1 2 n 1=p 1=q p p q q x + y x + y (ii) < ; p p q q (1 x) + (1 y ) (1 x) + (1 y ) where ; > 0; 0 <x6=y 1=2; 1=p 1=q p q +(1 u) +(1u) (iii) < ; p q +(1+ u) +(1+u) where >0; 0<u<1: Pro of. (i) obviously implies (iii). Now supp ose (iii) is true, let x>y and y=x =1u; x=(1 x)=k; then 0 <u<1; 0<k1 and (1 y )=(1 x)= 1+ku: So (ii) is equivalenttothe following: q p 1 + (1 u) + (1 u) 1 (17) f (k )= ln ln > 0: q p q + (1 + ku) p + (1 + ku) Di erentiating f (k ); one can obtain q1 p1 (1 + ku) u (1 + ku) u 0 f (k )= + q p + (1 + ku) + (1 + ku) (18) p q u (1 + ku) (1 + ku) = < 0: p q 1+ku + (1 + ku) + (1 + ku) Hence q p +(1u) +(1 u) 1 1 ln ln (19) f (k ) f (1) = > 0: q p q +(1+u) p +(1+u) (ii) is established. We will use induction to show that (i) is true if (ii) holds. At rst, (ii) is the case n = 2 of (i). Now assume that (i) holds for some n (n 2): Let 1=2 x x x ; and x are not all equal, then there exists 1 2 n+1 i > 0 and = = > 0 such that 1 n+1 n+1 P p x p i i p x + x 1 n+1 i=1 n+1 = (20) p p n+1 P (1 x ) + (1 x ) 1 n+1 n+1 p (1 x ) i i i=1 p p x + x 1 1 n+1 p = = R : p p (1 x ) + (1 x ) 1 1 n+1 It is clear that ( )( ) 0: Without loss of generality,wemay assume 1 n+1 A generalization of the Ky Fan inequality 13 that : So 1 n P p p ( )x + x 1 1 i i i=2 p (21) R = : n P p p ( )(1 x ) + (1 x ) 1 1 i i i=2 By the assumption, wehave 0 1 1=p n P p p ( )x x + 1 1 i i B C i=2 B C (22) R = n @ A P p p ( )(1 x ) + (1 x ) 1 1 i i i=2 1 0 1=q n P q q ( )x + x 1 1 i i C B i=2 C B ; n A @ P q q ( )(1 x ) + (1 x ) 1 1 i i i=2 and 1=p p p x + x 1 n+1 n+1 (23) R = p (1 x ) + (1 x )p 1 n+1 n+1 1=q q q x + x 1 n+1 n+1 < : q (1 x ) + (1 x )q 1 n+1 n+1 So wehave 1=q 0 1 n+1 P q x i i B C i=1 B C (24) R< : @ A n+1 P q (1 x ) i i i=1 Therefore, we get that (i) is true for arbitrary n and the prop osition is established. 4. THREE LEMMAS Lemma 1. If 0; < 1 ; 0 u<1; then (25) (1 + u) +(1u) (1 + u) +(1u) : The equality is attained if and only if u =0 or ( ; )= (0;1): x x Pro of.