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]:
(1)
1 =n
Pn x
CC < i i (0 x 1=2);
i
A
Pn (1 x )
(1 xi )
i
0 Qn
xi
B
i
B
@ Qn
1
=1
=1
i=1
i=1
unless x1 = x2 = : : : = xm :
With the notation
1 Pn 1=p
Qn 1=n
p
(xi > 0) and M0 (x) = plim
;
Mp (x) = n xi
!0 Mp (x) = i=1xi
i=1
(1) becomes
M0(x) < M1 (x) :
(2)
M0 (1 x) M1(1 x)
D. Segaiman [2] conjectured that
Mp (x) Mq (x)
(3)
(p q ):
Mp (1 x) Mq (1 x)
F. Chan, D. Goldberg and S. Gonek [2] gave some counter examples
when 0 < 2p =p < 2q =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 = 0 was proved to be true by Wan{Lan
Wang and Peng{Fei Wang [3]. And the case 1 p 0 1 was proved
0
1991 Mathematics Subject Classication: 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) = Mp (x)=Mp (1 x) is strictly increasing on [ 1; 1]; unless
x1 = : : : = xn ; and if (3) holds for (p; q); then also for ( q; p):
In this paper, we determine all exponents p and q such that (3) is true.
Theorem. For arbitrary n; p < q; the inequality
(4)
0 Pn
p
BB i xi
@ Pn
1
=1
1 =q
CC (0 < x 1=2)
i
A
1
=1
(1 xi)p
i=1
1 =p 0 Pn
q
CC BB i xi
A
@ Pn
(1 xi)q
i=1
holds if and only if jp + qj 3; 2p =p 2q =q when p > 0; p2p q2q when q < 0:
The proof of the suciency is contained in Sections 3, 4 and 5. In the proof
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
(4) holds, then 2p =p 2q =q for p > 0:
When q < 0; take x1 = x2 = = xn 1 = " (0 < " < 1=2) and xn = 1=2:
Then (4) becomes
(5)
or
(6)
(n 1)"p + (1=2)p =p (n 1)"q + (1=2)q =q
(n 1)(1 ")q + (1=2)q
;
(n 1)(1 ")p + (1=2)p
1
p
" +
q
1
=p (1 ")p +
=q q
1
n
1)
" + 2q (n1
1)
2p (
1
1
(1 ") + 2q (n1
1)
(7)
hence
1
1
=p
=q ;
1
n
1)
1 + 2q (n1
1)
2p (
=p
=q :
1
1)
Letting " ! 0; (6) yields
1+
1 1
n
2p (
1
1=p 1=q
1
1
(8)
1 + 2p (n 1)
1 + 2q (n 1) :
By using the Maclaurin expansion in 1=n; we obtain
(9)
1 + (p2p n) 1 + o(1=n2) 1 + (q2q n) 1 + o(1=n2):
A generalization of the Ky Fan inequality
11
So if p 2p > q 2q ; (4) would be false for suciently large n:
In the equivalent inequality of (4):
0 Pn
1 =p 0 Pn
1 =q
p
q
(1
u
)
(1
u
)
i
i
BB i
C
Bi
C
@ Pn (1 + u )p CA B@ Pn (1 + u )q CA (0 ui < 1);
i
i
1
(10)
1
=1
=1
i=1
i=1
let u1 = u2 = = un 1 = 0 and un = u (0 < u < 1); then (10) becomes
(n 1) + (1 u)p 1=p (n 1) + (1 u)q 1=q
(n 1) + (1 + u)q
:
(11)
(n 1) + (1 + u)p
Take the Maclaurin expansion of (11) in u:
(n 1) (n 2)p2 3np + 2(n2 + 2) 3
2
2
2
(12)
1 n u + n2 u
u + o(u4)
3n3
2
2
1 n2 u + n22 u2 (n 1) (n 2)q 3n33nq + 2(n + 2) u3 + o(u4 ):
Thus for u suciently small, (10) holds only if
(13)
(n 2)p2 3np (n 2)q2 3nq;
or
(14)
(p q) (n 2)(p + q) 3n 0:
So for n 3; we have
(15)
p + q n3n 2 :
Letting n ! +1; (15) yields p + q 3:
Similarly, the expansion of (10) with u1 = u2 = = un 1 = u (0 < u < 1);
un = 0 gives
(16)
p + q n 3n2 :
So we obtain p + q 3:
3. AN EQUIVALENCE PROPOSITION
In this section, we are to establish an equivalence proposition as follows:
Proposition. For p < q; the following inequalities are equivalent:
(i)
0 Pn
p
BB i ixi
@ Pn
=1
i (1 xi )p
i=1
1 =p 0 Pn
q
CC < BB i ixi
A @ Pn
1
=1
i (1 xi )q
i=1
1 =q
CC ;
A
1
12
Zhen Wang, Ji Chen, Guang{Xing Li
where i > 0; 0 < xi 1=2; i = 1; 2; : : :; n and x1; x2; : : :; xn are not all equal;
xp + yp
(ii)
(1 x)p + (1 y)p
where ; > 0; 0 < x 6= y 1=2;
=p 1
1=q
q + yq
x
< (1 x)q + (1 y)q
;
+ (1 u)p =p + (1 u)q =q
<
;
1
(iii)
1
+ (1 + u)p
+ (1 + u)q
where > 0; 0 < u < 1:
Proof. (i) obviously implies (iii).
Now suppose (iii) is true, let x > y and y=x = 1 u; x=(1 x) = k; then
0 < u < 1; 0 < k 1 and (1 y)=(1 x) = 1 + ku: So (ii) is equivalent to the
following:
u)q 1 ln + (1 u)p > 0:
(17)
f (k) = q1 ln ++(1(1+ ku
)q p + (1 + ku)p
(18)
Dierentiating f (k); one can obtain
)q 1u + (1 + ku)p 1u
f 0 (k) = +(1+(1ku
+ ku)q + (1 + ku)p
Hence
(1 + ku)p
u
= 1 + ku + (1 + ku)p
(1 + ku)q < 0:
+ (1 + ku)q
(1 u)q 1 ln + (1 u)p > 0:
f (k) f (1) = 1q ln +
+ (1 + u)q p + (1 + u)p
(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 x1 x2 xn+1; and xi are not all equal, then there exists
> 0 and = 1 n+1 = > 0 such that
(19)
nP
+1
(20)
i xip
x1p + n+1 xnp+1
=
p
p
nP
+1
i (1 xi )p (1 x1 ) + n+1 (1 xn+1 )
i=1
i=1
It is clear that (1 )(n+1
x p + x p
= (1 x1 )1p + (1n+1x )p = Rp :
1
1
n+1
) 0: Without loss of generality, we may assume
A generalization of the Ky Fan inequality
that 1 : So
Rp =
(21)
(1 )x1p +
(1 )(1 x1)p +
By the assumption, we have
(22)
0
B
R=B
@
(1 )x1 p +
(1 )(1
x1 )p +
Pn x p
i=2
Pn
i (1 xi)p
:
i=2
1 =p
CC
A
Pn x p
1
i i
i=2
Pn
i i
13
i (1 xi )p
Pn
(1 )x1q + i xi q
0i
B
B
@
=2
(1 )(1
and
i=2
n
q
x1) + P i (1
i=2
1 =q
CC ;
A
1
xi)q
1=p
p
p
R = (1 xx)1p ++n+1(1xn+1x )p
1
n+1
n+1
1=q
q
q
:
< (1 xx)1q ++n+1(1xn+1x )q
1
n+1
n+1
So we have
(23)
0 nP q 1 =q
B i ixi CC :
R<B
@ nP
A
(1 x )q
1
+1
(24)
=1
+1
i=1
i
i
Therefore, we get that (i) is true for arbitrary n and the proposition is established.
4. THREE LEMMAS
Lemma 1. If 0; < 1 ; 0 u < 1; then
(25)
(1 + u) + (1 u) (1 + u) + (1 u) :
The equality is attained if and only if u = 0 or (; ) = (0; 1):
Proof. Let '(x) = (1 + u)x + (1 u)x (0 < u < 1); then
(26)
'00 (x) = (1 + u)x ln (1 + u) 2(1 u)x ln(1 u) 2 > 0:
So we have to establish (25) only for = 1 ; i.e.
(27)
(u) = (1 + u) + (1 u) (1 + u)1 + (1 u)1 > 0;
14
Zhen Wang, Ji Chen, Guang{Xing Li
where < 0; 0 < u < 1:
(28)
(u) = 2
X1 1 n
u
2n
2n
+
2
n=0
nQ
X1 u n nQ
( k )
( k + 1)
= 2( 1) (2n)!
k
k
n
X1 n n
n
2( 1) (2un)! Q ( k) Q j k + 1j
+
2
2
1
2
=2
=2
+
2
2
=2
1
2
k=2
n=2
1
1
k=2
> 0:
This proves the lemma.
Lemma 2. If 0 < < < 1 and 0 < u 1: Let
(29)
G(u) = (1 + u) + (1 u) (1 + u) (1 u) ;
then there exists an unique u0 ; such that
(i ) G(u) > 0 for 0 < u < u0 ;
(ii) G(u) < 0 for u0 < u 1:
Proof. We have ( 1) < ( 1) < 0; hence
(30)
0 < (( 1)1 < 1:
Dene
(1 + u) 2 + (1 u) 2
(31)
g(u) = (1
+ u) 2 + (1 u) 2 (0 < u < 1):
We have
3 (1 u) 3 (
2)
(1
+
u
)
0
(32) g (u) =
(1 + u) 2 + (1 u) 2
( 2) (1 + u)
(1 u) 3 (1 + u) 2 + (1 u) 2
(1 + u) 2 + (1 u) 2 2
2
<
(1 + u) 3 (1 u) 3 2
2
2
(1 + u) + (1 u)
(1 + u) 2 + (1 u) 2
(1 + u) 3 (1 u) 3 (1 + u) 2 + (1 u) 2
1 u 3 1 u 3!
+ 6
2(
2)(1
+
u
)
=
< 0:
1+u
(1 + u) 2 + (1 u) 2 2 1 + u
3
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 u1 2 (0; 1) such that
1) :
g(u1) = (( 1)
(33)
Note that
(34)
G0(u) = (1 + u)
1
(35)
G00(u) = (
(1 + u)
1)
(1 u)
2
1
(1 + u)
+ (1
u)
2
1
(1 u)
1
;
g(u) ( 1) ;
( 1)
and from above we know that G00(u) > 0 for u 2 (0; u1); G00(u) < 0 for u 2 (u1; 1):
Because G(0) = 0 and G(1) = 2 2 < 0; we can nd a unique u0 2 (u1 ; 1)
such that G(u) > 0 in (0; u0); G(u) < 0 in (u0 ; 1):
Lemma 3. If p < q; p + q 3 and 2p =p 2q =q for p > 0; then
(1 + u)p (1 u)p (1 + u)q (1 u)q (0 u < 1);
p
q
(36)
equality occurs if and only if u = 0 or (p; q) = (1; 2):
Proof. Let
(37)
p
p
H (u) = (1 + u) p (1 u)
(1 + u)q (1 u)q (0 u < 1):
q
Then
(38)
H 0 (u) = (1 + u)p 1 + (1 u)p
1
(1 + u)q + (1 u)q :
1
1
When p 1; p 1 < q 1 1 (p 1); by Lemma 1 we obtain H 0 (u) 0:
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) = 2r =r strictly increases in (0; ln2] and
strictly decreases [1= ln2; +1); and
(39)
h(1) = h(2); h(p) h(q)
(p > 1);
we have p + q < 1 + 2 = 3; i.e. 0 < p 1 < q 1 < 1 (p 1): By Lemma 2, H 0(u)
has it's unique zero point u0 in (0; 1); such that the following is true:
H 0 (u) > 0 for 0 < u < u0; hence H (u) > H (0) = 0;
H 0 (u) < 0 for u0 < u < 1; hence H (u) > H (1) = 2p =p 2q =q 0;
and H (u0) > 0:
These establish the lemma.
16
Zhen Wang, Ji Chen, Guang{Xing Li
5. PROOF OF THE SUFFICIENCY
From the equivalence proposition in Section 3, we only need to prove the
following inequality:
(40)
+ (1 u)p =p + (1 u)q =q
<
;
1
1
+ (1 + u)p
+ (1 + u)q
where > 0; 0 < u < 1; p < q; jp + qj 3; 2p =p 2q =q when p > 0; p 2p q 2q
when q < 0:
The above inequality is equivalent to
(1 u)q 1 ln + (1 u)p > 0:
(41)
F () = 1q ln +
+ (1 + u)q p + (1 + u)p
But
1 1
1
1
(42)
F 0() = 1q +(11 u)q +(1+
q
p
p
u)
p +(1 u)
+(1+u)
= (A2 + B + C )=Q();
where
(43) Q() = + (1 u)q + (1 + u)q + (1 u)p + (1 + u)p ;
(44)
(45)
q
q
A = (1 + u) q (1 u)
(1 + u)p (1 u)p ;
p
q
q
B = (1 + u)p + (1 u)p (1 + u) q (1 u)
C = (1 u2 )p+q (1 + u)
p
p
(1 + u)q + (1 u)q (1 + u) p (1 u) ;
(1 + u) p (1 u) p :
q
p
By Lemma 3, when (p; q) 6= (1; 2) and (p; q) 6= ( 2; 1); we have A < 0 and
C > 0: If (p; q) = (1; 2); then A = 0; B = 4u3 < 0; C = 2u3(1 u2) > 0: If
(p; q) = ( 2; 1); then A = 2u3 =(1 u2)2 < 0; B = 4u3=(1 u2)3 > 0; C = 0:
Thus for all these cases, F 0() has an unique positive root 0 such that F 0() > 0
for 0 < < 0 ; F 0( < 0 for > 0 : So
(47)
F () > F (0) = F (+1) = 0 for > 0:
(46)
Now the theorem is proved.
q
(1 u) q
A generalization of the Ky Fan inequality
17
6. A CONJECTURE
Inequality (4) holds for all natural numbers n; and it is not the best result
for each xed n: We propose a conjecture for this condition as follows:
Conjecture. If p < q; jp + qj 3n=(n 2); n 3;
(48)
and
(49)
then
(50)
1 + 2p =(n 1) 1=p 1 + 2q =(n 1) 1=q when p > 0;
1 + 1=(2p(n 1)) 1=p 1 + 1=(2q (n 1)) 1=q when q < 0;
0 Pn
xi p
B
i
B
@ Pn
=1
(1 xi )p
i=1
1 =p 0 Pn q
CC < BB i xi
A
@ Pn
1
unless x1 = x2 = = xn:
=1
(1 xi )q
1 =q
CC (0 < x 1=2);
i
A
1
i=1
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.
Proc. Amer. Math. Soc., 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 generalization of Ky Fan inequality. J. Ningbo Univ., 3 (1990),
No. 1, 23{26.
Institute of Mathematical Science,
Academia Sinica,
Wuhan. Hubei, 430071,
China
Department of Mathematics,
Ningbo University,
Ningbo, Zhejiang, 315211,
China
Department of Mathematics,
University of Science and Technology of China,
Hefei, Anhui, 230026,
China
(Received December 5, 1994)