A GENERALIZATION of the KY FAN INEQUALITY Mq(X)

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)

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

M (x)= x (x > 0) and M (x) = lim M (x)= x ;

p i i 0 p i

p!0

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

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

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

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)

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

2 (n1)

; (7) 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);

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

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 ;

(14) (p q ) (n 2)(p + q ) 3n 0:

So for n 3; wehave

(15) p + q :

n 2

Letting n ! +1; (15) yields p + q 3:

Similarly, the expansion of (10) with u = u = = u = u (0

1 2 n1

u = 0 gives

(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

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

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

1=p 1=q

p q

+(1 u) +(1u)

(iii) < ;

p q

+(1+ u) +(1+u)

where >0; 0

Pro of. (i) obviously implies (iii).

Now supp ose (iii) is true, let x>y and y=x =1u; x=(1 x)=k; then

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

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 i i

x + x

1 n+1

i=1

n+1

= (20)

p p

n+1

(1 x ) + (1 x )

1 n+1 n+1

(1 x )

i i

i=1

x + x

1 1

n+1

= = 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

p p

( )x + x

1 1 i i

i=2

(21) R = :

p p

( )(1 x ) + (1 x )

1 1 i i

i=2

By the assumption, wehave

0 1

1=p

p p

( )x x +

1 1 i i

B C

i=2

B C

(22) R =

@ A

p p

( )(1 x ) + (1 x )

1 1 i i

i=2

1 0

1=q

q q

( )x + x

1 1 i i

C B

i=2

C B

;

A @

q q

( )(1 x ) + (1 x )

1 1 i i

i=2

and

1=p

x + x

1 n+1

n+1

(23) R =

(1 x ) + (1 x )p

1 n+1 n+1

1=q

x + x

1 n+1

n+1

< :

(1 x ) + (1 x )q

1 n+1 n+1

So wehave

1=q

0 1

n+1

i i

B C

i=1

B C

(24) R< :

@ A

n+1

(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. Let '(x) = (1 + u) +(1u) (0

2 2

x 00 x

ln (1 u) > 0: ln (1 + u) (1 u) (26) ' (x) = (1 + u)

So wehave to establish (25) only for =1 ; i.e.

1 1

(27) (u) = (1 + u) +(1u) (1 + u) +(1 u) > 0;

14 Zhen Wang, Ji Chen, Guang{Xing Li

where < 0; 0

(28) (u)=2 u

2n 2n

n=0

2n1 2n1 X

Q Q

( k +1) ( k ) =2 ( 1)

(2n)!

k =2 k =2

n=2

X 2n1 2n1

Q Q

2 ( 1) ( k ) j k +1j

(2n)!

k =2 k =2

n=2

> 0:

This proves the lemma.

Lemma 2. If 0 < < <1 and 0

(29) G(u) = (1 + u) +(1 u) (1 + u) (1 u) ;

then there exists an unique u ; such that

(i ) G(u) > 0 for 0

(ii) G(u) < 0 for u

Pro of. Wehave ( 1) < ( 1) < 0; hence

( 1

(30) 0 < < 1:

( 1)

De ne

2 2

(1 + u) +(1 u)

(31) g (u)= (0

2 2

(1 + u) +(1 u)

Wehave

3 3

( 2) (1 + u) (1 u)

(32) g (u)=

2 2

(1 + u) +(1 u)

3 3 2 2

( 2) (1 + u) (1 u) (1 + u) +(1u)

2 2

(1 + u) +(1u)

3 3

(1 + u) (1 u) <

2 2

(1 + u) +(1u)

2 2

(1 + u) +(1 u)

2 2 3 3

(1 + u) +(1 u) (1 + u) (1 u)

3 3

+ 6

2( 2)(1 + u) 1u 1u

= < 0:

2 2

1+ u 1+u

(1 + u) +(1u)

A generalization of the Ky Fan inequality 15

So g (u) is strictly decreasing with g (0) = 1 and g (1) = 0: Hence there exists a

unique u 2 (0; 1) such that

( 1)

(33) g (u )= :

( 1)

Note that

1 1 0 1 1

; (1 + u) (1 u) (34) G (u)= (1 + u) (1 u)

( 1)

00 2 2

(35) G (u)= ( 1) (1 + u) +(1u) g(u) ;

( 1)

00 00

and from ab ovewe know that G (u) > 0 for u 2 (0;u );G (u) < 0 for u 2 (u ; 1):

1 1

Because G(0) = 0 and G(1) = 2 2 < 0; we can nd a unique u 2 (u ; 1)

0 1

such that G(u) > 0in(0;u );G(u)<0in(u ;1):

0 0

p q

Lemma 3. If p 0; then

p p q q

(1 + u) (1 u) (1 + u) (1 u)

(36) (0 u< 1);

p q

equality occurs if and only if u =0 or (p; q )= (1;2):

Pro of. Let

p p q q

(1 + u) (1 u) (1 + u) (1 u)

(37) H (u)= (0 u< 1):

p q

Then

0 p1 p1 q 1 q1

(38) H (u)= (1 + u) +(1 u) (1 + u) +(1 u) :

When p 1;p1

Thus H (u) H (0) = 0 with equality if and only if u = 0 or (p; q )= (1;2):

When p> 1; since the function h(r )=2 =r strictly increases in (0; ln 2] and

strictly decreases [1= ln 2; +1); and

(39) h(1) = h(2); h(p) h(q ) (p>1);

wehavep+q<1+ 2 = 3; i.e. 0

has it's unique zero p oint u in (0; 1); such that the following is true:

H (u) > 0 for 0 H(0) = 0;

0 p q

H (u) < 0 for u H(1) = 2 =p 2 =q 0;

and H (u ) > 0:

0 These establish the lemma.

16 Zhen Wang, Ji Chen, Guang{Xing Li

5. PROOF OF THE SUFFICIENCY

From the equivalence prop osition in Section 3, we only need to prove the

following inequality:

1=p 1=q

p q

+(1 u) +(1u)

(40) < ;

p q

+(1+ u) +(1+u)

p q p q

where > 0; 0 0;p2 q2

when q<0:

The ab ove inequality is equivalentto

q p

1 +(1u) 1 +(1 u)

(41) F ()= ln ln > 0:

q p

q +(1+u) p +(1+ u)

But

1 1 1 1 1 1

(42) F ()=

q q p p

q +(1u) +(1+u) p +(1u) +(1+u)

=(A + B + C )=Q();

where

q q p p

(43) Q()= +(1 u) +(1+u) +(1u) +(1+u) ;

p p q q

(1 + u) (1 u) (1 + u) (1 u)

; (44) A =

q p

q q

(1 + u) (1 u)

p p

(45) B = (1 + u) +(1 u)

p p

(1 + u) (1 u)

q q

(1 + u) +(1u) ;

q q p p

(1 + u) (1 u) (1 + u) (1 u)

2 p+q

(46) C =(1u ) :

q p

By Lemma 3, when (p; q ) 6=(1;2) and (p; q ) 6=(2;1); wehaveA<0 and

3 3 2

C>0:If (p; q )= (1;2); then A =0;B = 4u < 0;C =2u(1 u ) > 0: If

3 2 2 3 2 3

(p; q )=(2;1); then A = 2u =(1 u ) < 0;B =4u =(1 u ) > 0;C =0:

0 0

Thus for all these cases, F () has an unique p ositiveroot such that F () > 0

for 0 << ; F (<0 for > : So

0 0

(47) F () >F(0) = F (+1)=0 for >0:

Now the theorem is proved.

A generalization of the Ky Fan inequality 17

6. A CONJECTURE

Inequality (4) holds for all natural numb ers n; and it is not the b est result

for each xed n: We prop ose a conjecture for this condition as follows:

Conjecture. If p

1=p 1=q

p q

(48) 1+2 =(n 1) 1+2 =(n 1) when p> 0;

and

1=p 1=q

p q

(49) 1+ 1=(2 (n 1)) 1+ 1=(2 (n 1)) when q<0;

0 1 0 1

1=p 1=q

n n

then

P P

p q

x x

i i

B C B C

i=1 i=1

B C B C

(50) < (0

n n

@ A @ A

P P

p q

(1 x ) (1 x )

i i

i=1 i=1

unless x = x = = x :

1 2 n

REFERENCES

1. E. F. Beckenbach, R. Bellman: Inequalities. Springer{Verlag 1961, p.5.

2. F. Chan, D. Goldberg, S. Gonek: On extensions of an inequality among means.

Pro c. Amer. Math. So c., 42 (1974), 202{207.

3. W.{L. Wang, P. {F. Wang: A class of inequalities for the symmetric functions

(Chinese). Acta. Math. Sinica, 27 (1984), 485{497.

4. G.{X. Li, J. Chen: An extension of Ky Fan inequality (Chinese). Hunan Math.

Comm., 1989, No. 4, 37{39.

5. Z. Wang, J. Chen: A generalizationofKyFan inequality. J. Ningb o Univ., 3 (1990),

No. 1, 23{26.

Institute of Mathematical Science, (Received Decemb er 5, 1994)

Academia Sinica,

Wuhan. Hub ei, 430071,

China

Department of Mathematics,

Ningb o University,

Ningb o, Zhejiang, 315211,

China

Department of Mathematics,

University of Science and Technology of China,

Hefei, Anhui, 230026, China