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Sharpness of Wilker and Huygens type inequalities

Abstract

We present an elementary proof of Wilker's inequality involving trigonometric functions, and establish sharp Wilker and Huygens type inequalities.

Mathematics Subject Classification 2010: 26D05.

1. Introduction

Wilker in [1] proposed two open problems:

  1. (a)

    Prove that if 0 < x < π/2, then

    sin x x 2 + tan x x > 2 . 1
    (1)
  2. (b)

    Find the largest constant c such that

    sin x x 2 + tan x x > 2 + c x 3 tan x

for 0 < x < π/2.

In [2], inequality (1) was proved, and the following inequality

2 + 2 π 4 x 3 tan x < sin x x 2 + tan x x < 2 + 8 45 x 3 tan x for 0 < x < π 2 ,
(2)

where the constants 2 π 4 and 8 45 are best possible, was also established.

Wilker type inequalities (1) and (2) have attracted much interest of many mathematicians and have motivated a large number of research papers involving different proofs and various generalizations and improvements (cf. [213] and the references cited therein). A brief survey of some old and new inequalities associated with trigonometric functions can be found in [14]. These include (among other results) Wilker's inequality.

Another inequality which is of interest to us is Huygens [15] inequality, which asserts that

2 sin x x + tan x x > 3 for all 0 < x < π 2 .
(3)

Neuman and Sándor [16] have pointed out that (3) implies (1). In [17], Zhu established some new inequalities of the Huygens type for trigonometric and hyperbolic functions. Baricz and Sándor [18] pointed out that inequalities (1) and (3) are simple consequences of the arithmetic-geometric mean inequality together with the well-known Lazarević-type inequality [[19], p. 238]

( cos x ) 1 / 3 < sin x x for all 0 < x < π 2 ,
(4)

or equivalently,

sin x x 2 tan x x > 1 for all 0 < x < π 2 .

Wu and Srivastava [[7], Lemma 3] established another inequality

x sin 2 + x tan x > 2 for all 0 < x < π 2 .
(5)

In [20], Chen and Cheung showed that Wilker inequality (1), Huygens inequality (3), Lazarević-type inequality (4) and Wu-Srivastava inequality (5) can be grouped into the following inequality chain:

( sin x / x ) 2 + tan x / x 2 > 2 ( sin x / x ) + tan x / x 3 > sin x x 2 tan x x 3 > 1 > 2 1 / ( sin x / x ) 2 + 1 / ( tan x / x ) , 0 < x < π 2 ,
(6)

in terms of the arithmetic, geometric and harmonic means.

In this article, we present an elementary proof of Wilker's inequality (2), and we establish sharp Wilker and Huygens type inequalities.

The following elementary power series expansions are useful in our investigation.

sin x = n = 0 ( - 1 ) n x 2 n + 1 ( 2 n + 1 ) ! , x < ,
(7)
cos x = n = 0 ( - 1 ) n x 2 n ( 2 n ) ! , x < ,
(8)
tan x = n = 1 2 2 n ( 2 2 n - 1 ) ( - 1 ) n - 1 B 2 n ( 2 n ) ! x 2 n - 1 , x < π 2 ,
(9)
x cot x = n = 0 ( - 1 ) n B 2 n ( 2 x ) 2 n ( 2 n ) ! , 0 < x < π .
(10)

where B n (n = 0, 1,2,...) are Bernoulli numbers, defined by

t e t - 1 = n = 0 B n t n n ! .

The following lemma is also needed in the sequel.

Lemma 1. [21] Let a n and b n > 0, n = 0,1, 2,... be real numbers with a n b n n = 1 being strictly increasing (respectively, decreasing). If the power series A ( x ) := n = 0 a n x n and B ( x ) := n = 0 b n x n are convergent for |x| < R, then the function A(x)/B(x) is strictly increasing (respectively, decreasing) on (0, R).

2. An elementary proof of Wilker's inequality (2)

Proof of (2). Consider the function

f ( x ) : = sin x x 2 + tan x x - 2 x 3 tan x , 0 < x < π 2 .

By using power series expansions (7) and (8), we obtain

- x 6 sin 2 x f ( x ) = - 3 x 2 sin ( 2 x ) + 5 sin 3 x cos x + 5 x - 2 x 3 + 2 x cos 4 x - 7 x cos 2 x = - 3 x 2 + 5 4 sin ( 2 x ) - 5 8 sin ( 4 x ) - 2 x 3 + 9 4 x + 1 4 x cos ( 4 x ) - 5 2 x cos ( 2 x ) = n = 4 ( - 1 ) n u n ( x ) = 16 945 x 9 - 16 1575 x 11 + 16 7425 x 13 - 11072 42567525 x 15 + n = 8 ( - 1 ) n u n ( x ) ,

where

u n ( x ) : = ( ( 2 n - 9 ) 4 n - 1 + 6 n 2 - 2 n ) 4 n ( 2 n + 1 ) ! x 2 n + 1 .

Elementary calculations reveal that, for 0 < x < π/2 and n ≥ 8,

u n + 1 ( x ) u n ( x ) = 8 x 2 6 n 2 + 10 n + 4 + ( 2 n - 7 ) 4 n ( n + 1 ) ( 2 n + 3 ) ( 24 n 2 - 8 n + ( 2 n - 9 ) 4 n ) < 8 π 2 2 6 n 2 + 10 n + 4 + ( 2 n - 7 ) 4 n ( n + 1 ) ( 2 n + 3 ) ( 24 n 2 - 8 n + ( 2 n - 9 ) 4 n ) = π 2 2 n + 3 12 n 2 + 20 n + 8 + 2 ( 2 n - 7 ) 4 n ( n + 1 ) ( 24 n 2 - 8 n + ( 2 n - 9 ) 4 n ) .

Write

α n : = π 2 2 n + 3 12 n 2 + 20 n + 8 + 2 ( 2 n - 7 ) 4 n ( n + 1 ) ( 24 n 2 - 8 n + ( 2 n - 9 ) 4 n ) .

It is easy to see that for n ≥ 8,

12 n 2 + 20 n + 8 + 2 ( 2 n - 7 ) 4 n ( n + 1 ) ( 24 n 2 - 8 n + ( 2 n - 9 ) 4 n ) < 1 .

Hence for all 0 < x < π/2 and n ≥ 8,

u n + 1 ( x ) u n ( x ) < α n < π 2 2 n + 3 < 1 .

Therefore, for fixed x (0, π/2), the sequence n u n (x) is strictly decreasing with regard to n ≥ 8. Hence, for 0 < x < π/2,

- x 6 sin 2 x f ( x ) > 16 945 x 9 - 16 1575 x 11 + 16 7425 x 13 - 11072 42567525 x 15 = x 9 16 945 - 16 1575 x 2 + 16 7425 x 4 - 11072 42567525 x 6 > 0 .

Hence f(x) is strictly decreasing on (0, π/2). Noting that lim t 0 + f ( t ) = 8 45 , we have

16 π 4 = lim t ( π / 2 ) - f ( t ) < f ( x ) = sin x x 2 + tan x x - 2 x 3 tan x < lim t 0 + f ( t ) = 8 45

for all x 0 , π 2 , with the constants 16 π 4 and 8 45 being best possible. This completes the proof of (2).

3. Sharp Wilker's inequality

By using power series expansions (8) and (9), we have, for 0 < x < π/2,

sin x x 2 + tan x x = 1 2 x 2 = - 1 2 x 2 cos ( 2 x ) + tan x x = 2 + k = 3 ( 2 ( 2 2 k - 1 ) B 2 k - ( - 1 ) k ) 2 2 k - 1 ( 2 k ) ! x 2 k - 2 .

It is well known [[22], p. 805] that

2 ( 2 k ) ! ( 2 π ) 2 k < B 2 k < 2 ( 2 k ) ! ( 2 π ) 2 k ( 1 - 2 1 - 2 k ) , k 1 .
(11)

By (11), we find that

2 ( 2 2 k - 1 ) B 2 k > 2 ( 2 2 k - 1 ) 2 ( 2 k ) ! ( 2 π ) 2 k > 1 , k 3 .

Hence, we have

sin x x 2 + tan x x > 2 + k = 3 n ( 2 ( 2 2 k - 1 ) B 2 k - ( - 1 ) k ) 2 2 k - 1 ( 2 k ) ! x 2 k - 2 = 2 + 8 45 x 4 + 16 315 x 6 + 104 4725 x 8 + 592 66825 x 10 + + ( 2 ( 2 2 n - 1 ) B 2 n - ( - 1 ) n ) 2 2 n - 1 ( 2 n ) ! x 2 n - 2 .
(12)

Motivated by (12), we are now in a position to establish our first main result.

Theorem 1. (i) For 0 < x < π/2, we have

2 + 8 45 x 4 + 16 315 x 5 tan x < sin x x 2 + tan x x < 2 + 8 45 x 4 + 2 π 6 x 5 tan x .
(13)

The constants 16 315 and 2 π 6 are best possible.

  1. (ii)

    For 0 < x < π/2, we have

    2 + 8 45 x 4 + 16 315 x 6 + 104 4725 x 7 tan x < sin x x 2 + tan x x < 2 + 8 45 x 4 + 16 315 x 6 + 2 π 8 x 7 tan x .
    (14)

The constants 104 4725 and 2 π 8 are best possible.

Proof. We only prove inequality (13). The proof of (14) is analogous.

Consider the function

g ( x ) : = sin x x 2 + tan x x - 2 - 8 45 x 4 x 5 tan x = sin ( 2 x ) 2 x 7 + 1 x 6 - 2 cot x x 5 - 8 cot x 45 x , 0 < x < π 2 .

By using power series expansions (7) and (10), we find that

g ( x ) = n = 3 β n 2 2 n - 1 x 2 n - 6 ,

where

β n : = 4 B 2 n ( 2 n ) ! + B 2 ( n - 2 ) 45 ( 2 n - 4 ) ! + ( - 1 ) n 2 ( 2 n + 1 ) ! , n 3 .

By (11), we obtain

4 B 2 n ( 2 n ) ! + B 2 ( n - 2 ) 45 ( 2 n - 4 ) ! > 4 ( 2 n ) ! 2 ( 2 n ) ! ( 2 π ) 2 n + 1 45 ( 2 n - 4 ) ! 2 ( 2 n - 4 ) ! ( 2 π ) 2 n - 4 = 180 + 2 ( 2 π ) 4 45 ( 2 π ) 2 n .

By induction on n, it is easy to see that

180 + 2 ( 2 π ) 4 45 ( 2 π ) 2 n > 2 ( 2 n + 1 ) ! for all n 3 .

Hence β n > 0 for n ≥ 3, and we have

g ( x ) = n = 4 b n 2 2 n - 1 ( 2 n - 6 ) x 2 n - 7 > 0 , 0 < x < π 2 .

Therefore, g(x) is strictly increasing on (0, π/2). Noting that lim t ( π / 2 ) - g ( t ) = 2 π 6 , we have

16 315 = lim t 0 + g ( t ) < g ( x ) = sin x x 2 + tan x x - 2 - 8 45 x 4 x 5 tan x < lim t ( π / 2 ) - g ( t ) = 2 π 6

for all x 0 , π 2 , with the constants 16 315 and 2 π 6 being best possible. This completes the proof of (13).

Remark 1. Inequality (14) is sharper than (13). On the other hand, there is no strict comparison between inequalities (2) and (13). There is no strict comparison between inequalities (2) and (14) either.

In view of inequalities (13) and (14), we propose the following conjecture.

Conjecture 1. For 0 < x < π/2 and n ≥ 3, we have

2 + k = 3 n ( 2 ( 2 2 k - 1 ) B 2 k - ( - 1 ) k ) 2 2 k - 1 ( 2 k ) ! x 2 k - 2 + ( 2 ( 2 2 n + 2 - 1 ) B 2 n + 2 - ( - 1 ) n + 1 ) 2 2 n + 1 ( 2 n + 2 ) ! x 2 n - 1 tan x < sin x x 2 + tan x x < 2 + k = 3 n ( 2 ( 2 2 k - 1 ) B 2 k - ( - 1 ) k ) 2 2 k - 1 ( 2 k ) ! x 2 k - 2 + 2 π 2 n x 2 n - 1 tan x .

4. Sharp the Wu-Srivastava inequality

By using power series expansion (10), we obtain for 0 < x < π/2,

csc 2 x = - ( cot x ) = 1 x 2 + k = 1 2 2 k ( 2 k - 1 B 2 k ) ( 2 k ) ! x 2 k - 2 .
(15)

Hence for 0 < x < π/2,

x sin x 2 + x tan x = x 2 csc 2 x + x cot x = 2 + k = 2 2 2 k + 1 ( k - 1 ) B 2 k ( 2 k ) ! x 2 k = 2 + 2 45 x 4 + 8 945 x 6 + 2 1575 x 8 + 16 93555 x 10 + .
(16)

Motivated by (16), we establish our second main result:

Theorem 2. (i) For 0 < x < π/2, we have

x sin x 2 + x tan x < 2 + 2 45 x 3 tan x .
(17)

The constant 2 45 is best possible.

  1. (ii)

    For 0 < x < π/2, we have

    x sin x 2 + x tan x < 2 + 2 45 x 4 + 8 945 x 5 tan x .
    (18)

The constant 8 945 is best possible.

Proof. We only prove inequality (18). The proof of (17) is analogous.

Consider the function

G ( x ) : = x sin x 2 + x tan x - 2 - 2 45 x 4 x 5 tan x = A ( x ) B ( x ) ,

where

A ( x ) : = x sin x 2 + x tan x - 2 - 2 45 x 4 = n = 1 2 2 n + 5 ( n + 1 ) B 2 ( n + 2 ) ( 2 n + 4 ) ! x 2 n + 4 = n = 1 a n x 2 n + 4

with

a n : = 2 2 n + 5 ( n + 1 ) B 2 ( n + 2 ) ( 2 n + 4 ) ! ,

and

B ( x ) : = x 5 tan x = n = 1 2 2 n ( 2 2 n - 1 ) B 2 n ( 2 n ) ! x 2 n + 4 = n = 1 b n x 2 n + 4

with

b n : = 2 2 n ( 2 2 n - 1 ) B 2 n ( 2 n ) ! .

We claim that the function G(x) is strictly decreasing on (0, π/2). By Lemma 1, it suffices to show that

a n b n > a n + 1 b n + 1 , n 1 .
(19)

It is known [[23], p. 96] that

2 ( 2 n ) ! ( 2 π ) 2 n < B 2 n < π 2 ( 2 n ) ! 3 ( 2 π ) 2 n , n 1 .
(20)

By using (20), we obtain

a n b n = 2 5 ( n + 1 ) B 2 ( n + 2 ) ( 2 n + 4 ) ! ( 2 n ) ! ( 2 2 n - 1 ) B 2 n > 192 ( n + 1 ) π 2 ( 2 π ) 4 ( 4 n - 1 )

and

a n + 1 b n + 1 = 2 5 ( n + 2 ) B 2 ( n + 3 ) ( 2 n + 6 ) ! ( 2 n + 2 ) ! ( 2 2 n + 2 - 1 ) B 2 ( n + 1 ) < 16 π 2 ( n + 2 ) 3 ( 2 π ) 4 ( 4 n + 1 - 1 ) .

So (19) is a consequence of the elementary inequality

192 ( n + 1 ) π 2 ( 2 π ) 4 ( 4 n - 1 ) > 16 π 2 ( n + 2 ) 3 ( 2 π ) 4 ( 4 n + 1 - 1 ) ,

which is equivalent to

36 ( n + 1 ) 4 n - 1 > π 4 ( n + 2 ) 4 n + 1 - 1 , n 1 .
(21)

The proof of the inequality (21) is not difficult, and is left with the readers. This proves the claim.

Noting that lim t 0 + G ( t ) = 8 945 , we have

G ( x ) < lim t 0 + G ( t ) = 8 945 for all x 0 , π 2

with the constant 8 945 being best possible. This completes the proof of (18).

In view of inequalities (17) and (18), we propose the following conjecture.

Conjecture 2. For 0 < x < π/2 and n ≥ 1,

x sin x 2 + x tan x < 2 + k = 2 n ( k - 1 ) 2 2 k + 1 B 2 k ( 2 k ) ! x 2 k + n 2 2 n + 3 B 2 ( n + 1 ) ( 2 n + 2 ) ! x 2 n + 1 tan x .
(22)

5. Skarp Huygens inequality

By using power series expansions (7) and (9), for 0 < x < π/2, we have

2 sin x x + tan x x = 3 + k = 3 2 2 k ( 2 2 k - 1 ) B 2 k 4 k - ( - 1 ) k 2 x 2 k - 2 ( 2 k - 1 ) ! .

By (11), we find that

2 2 k ( 2 2 k - 1 ) B 2 k 4 k > 2 2 k ( 2 2 k - 1 ) 4 k 2 ( 2 k ) ! ( 2 π ) 2 k = ( 2 2 k - 1 ) ( 2 k - 1 ) ! π 2 k > 1 , k 3 .

Hence we have

2 sin x x + tan x x > 3 + k = 3 n 2 2 k ( 2 2 k - 1 ) B 2 k 4 k - ( - 1 ) k 2 x 2 k - 2 ( 2 k - 1 ) ! = 3 + 3 20 x 4 + 3 56 x 6 + 7 320 x 8 + 3931 443520 x 10 + + ( 2 2 n ( 2 2 n - 1 ) B 2 n ) 4 n - ( - 1 ) n 2 x 2 n - 2 ( 2 n - 1 ) ! .
(23)

Motivated by (23), we establish our third main result:

Theorem 3. (i) For 0 < x < π/2, we have

3 + 3 20 x 3 tan x < 2 sin x x + tan x x < 3 + 2 π 4 x 3 tan x .
(24)

The constants 3 20 and 2 π 4 are best possible.

  1. (ii)

    For 0 < x < π/2, we have

    3 + 3 20 x 4 + 3 56 x 5 tan x < 2 sin x x + tan x x < 3 + 3 20 x 4 + 2 π 6 x 5 tan x .
    (25)

The constants 3 56 and 2 π 6 are best possible.

Proof. We only prove inequality (25). The proof of (24) is analogous.

Consider the function

h ( x ) : = 2 sin x x + tan x x - 3 - 3 20 x 4 x 5 tan x = 2 cos x x 6 + 1 x 6 - 3 cot x x 5 - 3 cot x 20 x , 0 < x < π 2 .

By using power series expansions (8) and (10), we find that

h ( x ) = n = 3 c n x 2 n - 6 ,

where

c n : = 3 2 2 n B 2 n ( 2 n ) ! + 3 2 2 n - 4 B 2 ( n - 2 ) 20 ( 2 n - 4 ) ! + ( - 1 ) n 2 ( 2 n ) ! , n 3 .

By (11), we obtain

3 2 2 n B 2 n ( 2 n ) ! + 3 2 2 n - 4 B 2 ( n - 2 ) 20 ( 2 n - 4 ) ! > 3 2 2 n ( 2 n ) ! 2 ( 2 n ) ! ( 2 π ) 2 n + 3 2 2 n - 4 20 ( 2 n - 4 ) ! 2 ( 2 n - 4 ) ! ( 2 π ) 2 n - 4 = 60 + 3 π 4 10 π 2 n .

By induction on n, it is easy to show that

60 + 3 π 4 10 π 2 n > 2 ( 2 n ) ! for all n 3 .

Hence c n > 0 for n ≥ 3, and we have

h ( x ) = n = 4 c n ( 2 n - 6 ) x 2 n - 7 > 0 for all 0 < x < π 2 .

Therefore, h(x) is strictly increasing on (0, π/2). Noting that lim t 0 + h ( t ) = 3 56 and lim t ( π / 2 ) - h ( t ) = 2 π 6 , we have

3 56 = lim t 0 + h ( t ) < h ( x ) = 2 sin x x + tan x x - 3 - 3 20 x 4 x 5 tan x < lim t ( π / 2 ) - h ( t ) = 2 π 6

for all x 0 , π 2 with the constants 3 56 and 2 π 6 being possible. This completes the proof of (25).

Remark 2. There is no strict comparison between inequalities (24) and (25).

In view of inequalities (24) and (25), we propose the following conjecture.

Conjecture 3. For 0 < x < π/2 and n ≥ 2, we have

3 + k = 3 n 2 2 k ( 2 2 k - 1 ) B 2 k 4 k - ( - 1 ) k 2 ( 2 k - 1 ) ! x 2 k - 2 + 2 2 n + 2 ( 2 2 n + 2 - 1 ) B 2 n + 2 4 ( n + 1 ) - ( - 1 ) n + 1 2 ( 2 n + 1 ) ! x 2 n - 1 tan x < 2 sin x x + tan x x < 3 + k = 3 n 2 2 k ( 2 2 k - 1 ) B 2 k 4 k - ( - 1 ) k 2 ( 2 k - 1 ) ! x 2 k - 2 + 2 π 2 n x 2 n - 1 tan x .
(26)

References

  1. Wilker JB: Problem E 3306. Am Math Mon 1989, 96: 55. 10.2307/2323260

    Article  MathSciNet  Google Scholar 

  2. Sumner JS, Jagers AA, Vowe M, Anglesio J: Inequalities involving trigonometric functions. Am Math Mon 1991, 98: 264–267. 10.2307/2325035

    Article  MathSciNet  Google Scholar 

  3. Guo BN, Qiao BM, Qi F, Li W: On new proofs of Wilker inequalities involving trigonometric functions. Math Inequal Appl 2003, 6: 19–22.

    MathSciNet  MATH  Google Scholar 

  4. Mortitc C: The natural approach of Wilker-Cusa-Huygens inequalities. Math Inequal Appl 2011, 14: 535–541.

    MathSciNet  MATH  Google Scholar 

  5. Neuman E: On Wilker and Huygens type inequalities. Math Inequal Appl, in press.

  6. Pinelis I: L'Hospital rules of monotonicity and Wilker-Anglesio inequality. Am Math Mon 2004, 111: 905–909. 10.2307/4145099

    Article  MathSciNet  MATH  Google Scholar 

  7. Wu SH, Srivastava HM: A weighted and exponential generalization of Wilker's inequality and its applications. Integr Trans Spec Funct 2007, 18: 529–535. 10.1080/10652460701284164

    Article  MathSciNet  MATH  Google Scholar 

  8. Wu SH, Srivastava HM: A further refinement of Wilker's inequality. Integr Trans Spec Funct 2008, 19: 757–765. 10.1080/10652460802340931

    Article  MathSciNet  MATH  Google Scholar 

  9. Zhang L, Zhu L: A new elementary proof of Wilker's inequalities. Math Inequal Appl 2008, 11: 149–151.

    MathSciNet  MATH  Google Scholar 

  10. Zhu L: A new simple proof of Wilker's inequality. Math Inequal Appl 2005, 8: 749–750.

    MathSciNet  MATH  Google Scholar 

  11. Zhu L: On Wilker-type inequalities. Math Inequal Appl 2007, 10: 727–731.

    MathSciNet  MATH  Google Scholar 

  12. Zhu L: Some new Wilker-type inequalities for circular and hyperbolic functions. Abstr Appl Anal 2009, 2009: 9. (Article ID 485842)

    MathSciNet  MATH  Google Scholar 

  13. Zhu L: A source of inequalities for circular functions. Comput Math Appl 2009, 58: 1998–2004. 10.1016/j.camwa.2009.07.076

    Article  MathSciNet  MATH  Google Scholar 

  14. Srivastava R: Some families of integral, trigonometric and other related inequalities. Appl Math Inf Sci 2011, 5: 342–360.

    MathSciNet  Google Scholar 

  15. Huygens C: Oeuvres Completes 1888–1940. Société Hollondaise des Science, Haga;

  16. Neuman E, Sándor J: On some inequalities involving trigonometric and hyperbolic functions with emphasis on the Cusa-Huygens, Wilker, and Huygens inequalities. Math Inequal Appl 2010, 13: 715–723.

    MathSciNet  MATH  Google Scholar 

  17. Zhu L: Some new inequalities of the Huygens type. Comput Math Appl 2009, 58: 1180–1182. 10.1016/j.camwa.2009.07.045

    Article  MathSciNet  MATH  Google Scholar 

  18. Baricz A, Sándor J: Extensions of generalized Wilker inequality to Bessel functions. J Math Inequal 2008, 2: 397–406.

    Article  MathSciNet  MATH  Google Scholar 

  19. Mitrinović DS: Analytic Inequalities. Springer-Verlag, Berlin; 1970.

    Chapter  Google Scholar 

  20. Chen CP, Cheung WS: Wilker- and Huygens-type inequalities and solution to Oppenheim's problem. Integr Trans Spec Funct, in press.

  21. Ponnusamy S, Vuorinen M: Asymptotic expansions and inequalities for hypergeometric functions. Mathematika 1997, 44: 278–301. 10.1112/S0025579300012602

    Article  MathSciNet  MATH  Google Scholar 

  22. Abramowitz M, Stegun IA: Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. In Appl Math Ser National Bureau of Standards. Volume 55. 9th edition. Washington, D.C; 1972.

    Google Scholar 

  23. Kuang J-Ch: Applied Inequalities. 3rd edition. Shandong Science and Technology Press, Jinan City, Shandong Province, China (Chinese); 2004.

    Google Scholar 

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Acknowledgements

The research is supported in part by the Research Grants Council of the Hong Kong SAR, Project No. HKU7016/07P

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Chen, CP., Cheung, WS. Sharpness of Wilker and Huygens type inequalities. J Inequal Appl 2012, 72 (2012). https://doi.org/10.1186/1029-242X-2012-72

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Keywords

  • inequalities
  • trigonometric functions