Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 4, Article 56, 2002
ON GENERALIZATIONS OF L. YANG’S INEQUALITY
CHANG-JIAN ZHAO AND LOKENATH DEBNATH
D EPARTMENT OF M ATHEMATICS
S HANGHAI U NIVERSITY
S HANGHAI 200436
T HE P EOPLE ’ S R EPUBLIC OF C HINA ;
AND
D EPARTMENT OF M ATHEMATICS
B INZHOU T EACHERS C OLLEGE
B INZHOU , S HANDONG 256604
T HE P EOPLE ’ S R EPUBLIC OF C HINA .
chjzhao@163.com
D EPARTMENT OF M ATHEMATICS
U NIVERSITY OF T EXAS -PAN A MERICAN
E DINBURG , T EXAS 78539-2999, U.S.A
Debnathl@panam.edu
Received 03 December, 2001; accepted 02 June, 2002
Communicated by Feng Qi
A BSTRACT. A geometric inequality due to L. Yang involving cosine and sine functions is generalized.
Key words and phrases: L. Yang’s inequality, Theory of distribution of functional values, Jensen inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
The well-known inequality due to L. Yang [1, pp. 116–118] can be stated as follows:
If A > 0, B > 0, A + B ≤ π and 0 ≤ λ ≤ 1, then
(1.1)
cos2 λA + cos2 λB − 2 cos λA · cos λB · cos λπ ≥ sin2 λπ.
The equality in (1.1) holds if and only if λ = 0 or A + B = π.
L. Yang’s inequality plays an important role in the theory of distribution of values of functions. See [1].
In this paper we will generalize inequality (1.1).
ISSN (electronic): 1443-5756
c 2002 Victoria University. All rights reserved.
085-01
2
C HANG - JIAN Z HAO AND L OKENATH D EBNATH
2. M AIN R ESULTS
In this paper, we use the following notations and abbreviations:
n
n!
=
;
r
r!(n − r)!
λ
λ
[2]
[1]
xij = (π + Ai + Aj ), xij = (π − Ai − Aj );
2
2
λ
λ
[4]
[3]
xij = (π + Ai − Aj ), xij = (π − Ai + Aj );
2
2
2
2
Hij = cos λAi + cos λAj − 2 cos λAi · cos λAj · cos λπ.
Lemma 2.1 ([2]). If Ai > 0, Aj > 0, Ai + Aj ≤ π for 1 ≤ i, j ≤ n, and −1 ≤ λ ≤ 1, then
2
Hij − sin λπ = 4
4
Y
[k]
sin xij .
k=1
Proof. By using the following two identities
cos λ (Ai + Aj ) cos λ (Ai − Aj ) = cos2 λAi + cos2 λAj − 1,
cos λ (Ai + Aj ) + cos λ (Ai − Aj ) = 2 cos λAi cos λAj ,
it is easy to observe that
Hij − sin2 λπ = cos2 λAi + cos2 λAj − 1 − 2 cos λAi cos λAj cos λπ + cos2 λπ
= cos2 λπ + cos λ (Ai + Aj ) cos λ (Ai − Aj )
− [cos λ (Ai + Aj ) + cos λ (Ai − Aj )] cos λπ
= [cos λπ − cos λ (Ai + Aj )] [cos λπ − cos λ (Ai − Aj )]
λ
λ
= 4 sin (π + Ai + Aj ) sin (π − Ai − Aj )
2
2
λ
λ
× sin (π + Ai − Aj ) sin (π − Ai + Aj ) .
2
2
The proof is complete.
Pn
Theorem 2.2. If Ai > 0, λk > 0 (i = 1, 2, . . . , n; k = 1, 2, 3, 4), i=1 Ai ≤ π, n ≥ 2 being a
natural number, and −1 ≤ λ ≤ 1, then
" n
#2
n
X
X
n
2
2
sin λπ ≤ (n − 1 + cos λπ)
cos λAk − cos λπ
cos λAi
2
i=1
k=1
(2.1)
4
P4
X
λ
n
k
k=1
≤
sin2 λπ +
sin4 θij ,
Q4
2
64 k=1 λk 1≤i<j≤n
where
[k]
k=1 λk xij
.
P4
λ
k
k=1
P4
(2.2)
θij =
The equalities in (2.1) hold if and only if λ = 0.
J. Inequal. Pure and Appl. Math., 3(4) Art. 56, 2002
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L. YANG ’ S I NEQUALITY
3
Proof. Since y = sin x is a continuous and convex (or concave, resp.) function on [−π, 0] (or
[k]
[0, π], resp.), and xij ∈ [−π, 0] (or [0, π], resp.) for −1 ≤ λ ≤ 0 (or 0 ≤ λ ≤ 1, resp.), and
using Jensen’s inequality (see [3]), we observe that
P4
[k]
k=1 λk sin xij
sin θij ≤ (or ≥ , resp.)
.
P4
λ
k
k=1
Consequently
4
X
(2.3)
!4
4
X
(sin θij )4 ≥
λk
k=1
[k]
On the other hand, since λk sin −xij
1
4
(2.4)
[k]
λk sin xij
.
k=1
4
X
!4
[k]
≥ 0 or λk sin xij ≥ 0, resp. , then
[k]
λk sin −xij
v
u 4
uY
[k]
4
≥t
λk sin xij .
k=1
k=1
v
!
u 4
4
X
Y
u
1
[k]
[k]
4
λk sin xij ≥ t
or
λk sin xij , resp. .
4 k=1
k=1
From (2.3) and (2.4), we obtain
4
Y
(2.5)
[k]
sin xij
4
k=1 λk
Q
44 4k=1 λk
P4
≤
k=1
sin4 θij .
By using Lemma 2.1, we have
4
k=1 λk
Q
64 4k=1 λk
P4
2
sin λπ ≤ Hij ≤
(2.6)
sin4 θij + sin2 λπ.
Summing both sides of (2.6) for 1 ≤ i < j ≤ n yields
X
(2.7)
sin2 λπ ≤
1≤i<j≤n
X
Hij ≤
1≤i<j≤n
X
1≤i<j≤n
!
4
λ
k=1 k
sin4 θij + sin2 λπ .
Q4
64 k=1 λk
P4
It is not difficult to see that
n
sin λπ =
sin2 λπ.
2
1≤i<j≤n
X
(2.8)
2
Direct computing yields
(2.9)
X
1≤i<j≤n
4
k=1 λk
Q
64 4k=1 λk
!
P4
sin4 θij + sin2 λπ
4
X
k=1 λk
sin4 θij
Q4
64 k=1 λk 1≤i<j≤n
P4
=
J. Inequal. Pure and Appl. Math., 3(4) Art. 56, 2002
n
+
sin2 λπ,
2
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4
C HANG - JIAN Z HAO AND L OKENATH D EBNATH
and
(2.10)
X
Hij =
1≤i<j≤n
X
(cos2 λAi + cos2 λAj − 2 cos λAi cos λAj cos λπ)
1≤i<j≤n
=
X
(cos2 λAi + cos2 λAj ) −
1≤i<j≤n
= (n − 1)
X
2 cos λAi cos λAj cos λπ
1≤i<j≤n
n
X

cos2 λAk − cos λπ 
!2
cos λAi
i=1
k=1
= (n − 1 + cos λπ)
n
X
n
X
cos2 λAk − cos λπ
k=1
−
n
X

cos2 λAi 
i=1
n
X
!2
cos λAi
.
i=1
Putting (2.8), (2.9) and (2.10) into (2.7), inequality (2.7) reduces to inequality (2.1). The
proof is complete.
(2.11)
1
4
(i = 1, 2, 3, 4) in the right-hand of (2.7), we find that
!
4
P4
X
λ
k=1 k
sin4 θij + sin2 λπ
Q4
64 k=1 λk
1≤i<j≤n


!
4
P4
P4
[k] 4
X
k=1 λk xij
k=1 λk
 Q
=
sin P
+ sin2 λπ 
4
4
64
λ
λ
k=1 k
k=1 k
1≤i<j≤n


!!4
4
X
X
[k]
4 sin 1
=
xij
+ sin2 λπ 
4
1≤i<j≤n
k=1
X
4 λ
2
=
4 sin π + sin λπ
2
1≤i<j≤n
n
λπ
=4
sin2
.
2
2
Remark 2.3. If taking λi =
Therefore, inequality (2.1) reduces to the following inequality
(2.12)
!2
n
n
X
X
n
sin2 λπ ≤ (n − 1 + cos λπ)
cos2 λAk − cos λπ
cos λAi
2
i=1
k=1
n
λ
≤4
sin2 π.
2
2
The equalities in (2.12) hold if and only if λ = 0.
Letting n = 2 in (2.12), we have the following result: If A > 0, B > 0, A + B ≤ π, and
−1 ≤ λ ≤ 1, then
(2.13)
λ
sin2 λπ ≤ cos2 λA + cos2 λB − 2 cos λA cos λB cos λπ ≤ 4 sin2 π.
2
The equalities in (2.13) holds if and only if λ = 0.
It is obvious that inequality (2.13) is the same as L. Yang’s inequality (1.1).
J. Inequal. Pure and Appl. Math., 3(4) Art. 56, 2002
http://jipam.vu.edu.au/
L. YANG ’ S I NEQUALITY
5
P
Theorem 2.4. If Ai > 0 and λk > 0 for i = 1, 2, . . . , n and k = 1, 2, 3, 4, ni=1 Ai ≤ π, n ≥ 2,
−1 ≤ λ ≤ 1, then
n
X
X
n
2
0 ≤ (n − 1)
cos λAk −
(2.14)
sin2 λπ − cos λπ
cos λAi cos λAj
2
1≤i,j≤n
k=1
i6=j
P4
≤
4
k=1 λk
sin4 θij ,
Q4
128 k=1 λk 1≤i,j≤n
i6=j
X
where
[k]
k=1 λk xij
.
P4
λ
k
k=1
P4
θij =
The equalities in (2.14) hold if and only if λ = 0.
Proof. It follows from inequality (2.6) that
4
k=1 λk
Q
64 4k=1 λk
P4
2
0 ≤ Hij − sin λπ ≤
(2.15)
sin4 θij .
Summing both sides of (2.15) for i 6= j, first over j from 1 to n and then over i from 1 to n of
the resulting inequality, then
4
P4
X
X
λ
n
k
k=1
(2.16)
0≤
Hij − 2
sin2 λπ ≤
sin4 θij .
Q4
2
64
λ
k=1 k 1≤i,j≤n
1≤i,j≤n
i6=j
i6=j
On the other hand
n
X
X
X
(2.17)
Hij = 2(n − 1)
cos2 λAk − 2 cos λπ
cos λAi cos λAj .
1≤i,j≤n
i6=j
k=1
1≤i,j≤n
i6=j
Putting (2.17) into (2.16), we get inequality (2.14).
(2.18)
1
4
for k = 1, 2, 3, 4, then
n
X
X
n
2
0 ≤ (n − 1)
cos λAk −
sin2 λπ − cos λπ
cos λAi cos λAj
2
1≤i,j≤n
k=1
Remark 2.5. In (2.14), if λk =
i6=j
n
λ
≤
sin4 π.
2
2
The equalities in (2.18) hold if and only if λ = 0.
Letting n = 2, then (2.18) reduces to inequality (2.13).
R EFERENCES
[1] L. YANG, Distribution of Values and New Research, Science Press, Beijing, 1982. (Chinese).
[2] Ch.-J. ZHAO, Further research of L. Yang’s inequality, J. Binzhou Teachers College, 12(4) (1996),
32–34. (Chinese).
[3] D. S. MITRINOVIĆ, Analytic Inequalities, Springer-Verlag, 1970.
J. Inequal. Pure and Appl. Math., 3(4) Art. 56, 2002
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