New strengthened Jordan's inequality and its applications

New strengthened Jordan's inequality and its applications

Applied kthematics Letters Applied Mathematics Letters 16 (2003) 557-560 www.elsevier.nl/locate/aml New Strengthened Jordan’s Inequality and Its Ap...

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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).