Applied kthematics Letters
Applied Mathematics Letters 16 (2003) 557-560
www.elsevier.nl/locate/aml
New Strengthened Jordan’s Inequality and Its Applications L. DEBNATH Department of Mathematics University of Texas-Pan American Edinburg, TX 78539, U.S.A. CHANG- JIAN ZHAO Department of Mathematics, Binzhou Teachers College Shangdong, 256604, P.R. China (Received October 2001; accepted March 2002)
Abstract-The main purpose of the present article is to strengthen Jordan’s inequality. As applications, we improve some new results proved by Zhao [l]. @ 2003 Elsevier Science Ltd. All rights reserved. Keywords-Jordan
inequality, Monotone increasing,Convex functions.
1. INTRODUCTION
AND
MAIN
RESULTS
The well-known Jordan’s inequality [2] can be stated as follows: sin 2
->2 x -ii’ with equality holds if and only if x = r/2. In this paper, we will first strengthen inequality main results are given in the following result. RESULT
A XE(0’5 1
(l), and then give some applications.
Our
1. If 0 < x 5 7r/2, then
sin 2 f + &
(2 - 4x:“)
(2)
2
with equality holds if and only if x = x/2. PROOF.
Consider the following function:
f(x) = $
+ &x2,
( 1
XE 0,; .
(3)
Clearly,
f’(x) = -$ (xcosx - sinx+ $x3) 0893-9659/03/$ - see front matter @ 2003 Elsevier Science Ltd. All rights reserved. PII: SO893-9659(03)00036-3
5pe=t
by 4&W
L.DEBNATH AND C.-J. ZHAO
558
and let 23
xcosx-sinx+-x
g(x) =
. >
Consequently, g/(x)=:x2-xsinx=x
[zxYsmxJ*
_< 0
with equality holds if and only if x = 1r/2. Therefore, g(x) is a monotone nonincreasing function, and it follows that f(x) is a monotone nonincreasing function. Hence,
(4) Inequality (2) follows from (3) and (4): Furthermore, RESULT
we have the following result.
2. If 0 < x 5 r/2, then sin 2 1; X
+ $ (7T”- 4x2)
(5)
with equality holds if and only if x = ~12. PROOF.
Consider the following function: f(x)=sinx--x+-x.
3
43 73
7T
We have 3 12 f’(x) = cosx - - + -x2, A ?r3 24 f”(x) = - sinx + px, f”‘(X) = -cosx
+ $,
ft4) (x) = sin x 2 0. Then, f”‘(x) is a monotone increasing function defined on (0,7r/2], and f”(x) defined on (0,7r/2]. Notice that f”’ (;) = 48 ,,““” > 0.
is a convex function
Hence, we have
Consequently, f’( x ) is a convex function defined on [7r/4, n/2], and
, Therefore, f’(x) f(7r/2) =o. Thus,
and thus, f(x)
5 0, x E [n/4,n/2],
f(4
2 0,
The proof for the case x E [7r/4, n/2] is complete.
is a monotone nonincreasing function and
559
Jordan’s Inequality
We consider the case x E (0,7r/4] below. Since f”(x)
f”(0) = 0,
f” (i) =
is a convex function, and 12 - 7r2fi
< o
2?r2
I
5 0, x E (O,rr/4], i.e., f(x) is a concave function, and
then f”(x)
8Jz-11
f (0)= 0,
f(i)=
>.
16
*
Consequently, f(x) 2 0 for (0, X/4]. The proof is complete. 1. From the proofs of Results 1 and 2, we easily get the following inequalities. If 0 5 x 5 7r/2, then
REMARK
sinx 2 Gx 3 sinx>-x--x. ?r
- &x3, 4 ?r3
(6)
3
(7)
Equality holds in (6) and (7) if and only if x = 0 or x = n/2.
2. APPLICATIONS 1. If Ai > 0 (i = 1,2, . . . , n), ‘&
THEOREM
Ai 5 ?r, and 0 5 X 5 1. Let n 2 2 be a natural
number. Then CzA’ (3 - X2)2 cos2 $r 5 (n - 1) 2
COST
XAI, - 2 cos XT
k=l
PROOF.
C
COS
XAi
COS
XAj 5 CiX2n2s (8)
l
By using (7), we have sin $r > + (3 - X2)
(9)
x x sin --A 5 -a. 2 2
(10)
i (3 - X2) 5 sin +r 5 $r.
(11)
and
It follows from (9) and (10) that
On the other hand, by using the following inequality (see [3]): x
sin2 X?r 5 Hij 5 4sin2 -7~, 2
(12)
where Hij =
COS2
XAi +
COS2
iIAj - 2 COS XAi
CDS
XAj
COS
XT,
and noticing x x sin2 XT = 4 sin2 --A cos2 -n, 2 2 hence, X2 (3 - X2)2 COS2 +T 5 Hij 5 X2X2.
(13)
L. DEBNATH AND C.-J. ZHAO
560
If 1 5 i < j I n, taking the sum for all inequalities of (13), then
C
X2(3 -
X2)2COS2
g7T
5
l
C
Hij
5
lli
C
(14)
X2n2,
l
where c
x2 (3 - x2)2 cos2 $
= c;x2 (3 - x2)2 cos2 ;n,
(15)
I
c
x21r2 = c$Ar2,
(16)
l
and
C l
Hij =
C
(COS2
XAi +
COS2
XAj - 2 COS XAg COS XAj
COS
XX)
l
=(n-1)50s2XAk-2cosXr
c
k=l
J
COS
XAi
COS
XAj.
(17)
Putting (15), (16), and (17) in (14), we get (8). The’ proof is complete. REMARK
2. Since c;x2 (3 - xy2 co2 in - 4c;x2cos2 ;r
= c$12 (1 - x2) (5 - x2) cos2 ;r
2 0,
inequality (8) is just a strengthened version of the following new inequality proved in [l]: 4CiX2
COS2
$7T 5 (n - 1) 2
COS2
cos XAi
2
XAk - 2 COS XT
k=l
COS
XAj < CiX2r2.
(18)
l
Similarly, we can also get the following theorem by using (6). THEOREM
2. Under the hypotheses of Theorem 1, we have
~c~~(24+(l-~2)n2)~~~2~~~(n-&OS2~~-2COSh
2
COS
XAi
COS
XAj
l
k=l
09)
-cC2X2n2 -7% . REMARK
+A
3. Clearly, inequality (19) again is a strengthened version of inequality (18) because of
(24 + (1 - X2) n “) cos2 $r - 4C;4X2cos2 ;r
= fC;X
(24 + (1 + X).rr2) (1 - A) cos2 +r 2 0.
REFERENCES 1. C.-J. Zhao, On several new inequalities, Chinese Quarterly Journal of Mathematics 16 (2), 4246, (2001). 2. D.S. Mitrinovic, Analytic Inequalities, Springer-Verlag, (1970). 3. C.-J. Zhao, Generalization and strengthen of Yang Le inequality, Mathematics in Practice and Theory 30 (4), 493497,
(2000).
C.-J. Zhao, Generalization on two new Hilbert type inequalities, J. Math. (PRC) 5. E.F. Beckenbach and R. Bellman, Inequalities, Springer-Verlag, Berlin, (1961). 4.
20 (4),
413416,
(2000).