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Article

On New Generalized Mitrinović-Adamović-Type Inequalities

1
Department of Mathematics, K. K. M. College, Manwath 431505, India
2
Department of Mathematics, National Kaohsiung Normal University, Kaohsiung 824004, Taiwan
*
Author to whom correspondence should be addressed.
Mathematics 2025, 13(7), 1174; https://doi.org/10.3390/math13071174
Submission received: 2 March 2025 / Revised: 30 March 2025 / Accepted: 31 March 2025 / Published: 2 April 2025
(This article belongs to the Special Issue New Aspects of Differentiable and Not Differentiable Function Theory)

Abstract

:
In this paper, we establish new generalized Mitrinović-Adamović-type inequalities in a wider range ( 0 , π ) by using the monotonicity of certain functions. These inequalities contain sharp and tractable bounds for the function sin x x 3 . All the main results are also true in ( π , 0 ) due to the symmetry of the curves involved.

1. Introduction

An obvious relation cos x < sin x x , x ( 0 , π / 2 ) was refined in [1,2] as
cos x < sin x x 3 , x 0 , π 2 .
The inequality (1) is known in the literature as Mitrinović–Adamović inequality. In recent years, some mathematicians tried to refine and extend the inequality (1). For instance, the inequality
cos 4 x 2 < sin x x 3 , x 0 , π 2
appeared in the references [3,4,5]. Mortici [6] and Chouikha [7] independently obtained, respectively, the following double inequalities:
cos x + x 4 15 23 x 6 1890 < sin x x 3 < cos x + x 4 15 , x 0 , π 2
and
cos x + x 3 1 x 2 63 sin x 15 < sin x x 3 < cos x + x 3 sin x 15 , x 0 , π 2 .
We observe that it is not difficult to show the validity of (1) in an interval 0 , π . In this direction, Zhu [8] achieved the inequalities
cos x < sin x x 3 < 8 π 3 + 1 8 π 3 cos x , x 0 , π 2 ,
8 π 3 + 1 8 π 3 cos x < sin x x 3 < 1 + cos x 2 , x π 2 , π
and
1 32 π 5 x 2 cos x + 32 π 5 x 2 < sin x x 3 < 1 2 15 x 2 cos x + 2 15 x 2 , x 0 , π 2 ,
1 1 2 π 2 x 2 cos x + 1 2 π 2 x 2 < sin x x 3 < 1 32 π 5 x 2 cos x + 32 π 5 x 2 , x π 2 , π .
On the other hand, W.-D. Jiang [9] very recently maintained the uniformity and sharpness of the bounds for sin x x 3 in a wider range ( 0 , π ) and established a better double inequality
cos x + 21 5 x 3 sin x 63 + x 2 < sin x x 3 < cos x + π 2 15 x 3 sin x π 2 x 2 , x ( 0 , π )
along with
cos x + x 3 sin x 15 + 5 21 x 2 < sin x x 3 < cos x + x 3 sin x 15 + π 6 960 16 π 2 x 2 , x ( 0 , π / 2 ) .
We also refer the reader to [10,11,12,13,14,15,16,17,18,19,20,21,22,23,24] and the references therein for more information on this topic.
In this work, by using the monotonicity of certain functions, we establish new generalized Mitrinović–Adamović-type inequalities in a wider range ( 0 , π ) . We obtain sharp bounds for sin x x 3 in a wider range ( 0 , π ) and refine double inequalities (2)–(3).

2. Lemmas for Bernoulli Numbers

Recall that the Bernoulli numbers B n can be generated by
z e z 1 = n = 0 B n z n n ! = 1 z 2 + k = 1 B 2 k z 2 k ( 2 k ) ! , z < 2 π ,
and all of the Bernoulli numbers B 2 k + 1 for k N equal 0. For more details, we refer the interested readers to the research monograph [25].
To prove our main results, we need the following important lemmas.
Lemma 1
(see [8,9,25]). Let B 2 k ( k N ) be the even indexed Bernoulli numbers. Then, for | x | < π , the following identities hold:
x sin x = 1 + k = 1 2 ( 2 2 k 1 1 ) | B 2 k | ( 2 k ) ! x 2 k ,
cot x = 1 x k = 1 2 2 k | B 2 k | ( 2 k ) ! x 2 k 1 ,
1 sin 2 x = 1 x 2 + k = 1 2 2 k ( 2 k 1 ) | B 2 k | ( 2 k ) ! x 2 k 2 ,
cos x sin 2 x = 1 x 2 k = 1 2 ( 2 k 1 ) ( 2 2 k 1 1 ) | B 2 k | ( 2 k ) ! x 2 k 2 ,
cos x sin 3 x = 1 x 3 k = 2 2 2 k ( 2 k 1 ) ( k 1 ) | B 2 k | ( 2 k ) ! x 2 k 3 ,
1 sin 3 x = 1 x 3 + 1 2 k = 2 ( 2 2 k 2 ) ( 2 k 1 ) ( 2 k 2 ) | B 2 k | ( 2 k ) ! x 2 k 3 + 1 2 x + 1 2 k = 1 ( 2 2 k 2 ) | B 2 k | ( 2 k ) ! x 2 k 1 ,
cos x sin 4 x = 1 3 x 3 1 6 x k = 1 ( 2 2 k 2 ) ( 2 k 1 ) 6 · ( 2 k ) ! | B 2 k | x 2 k 2 k = 2 ( 2 2 k 2 ) ( 2 k 1 ) ( 2 k 2 ) ( 2 k 3 ) 6 · ( 2 k ) ! | B 2 k | x 2 k 4 ,
and
1 sin 4 x = 1 x 4 + k = 1 2 2 k ( 2 k 1 ) ( 2 k 2 ) ( 2 k 3 ) 6 · ( 2 k ) ! | B 2 k | x 2 k 4 + 2 3 x 2 + k = 1 2 2 k + 1 ( 2 k 1 ) 3 · ( 2 k ) ! | B 2 k | x 2 k 2 .
Lemma 2
(see [26]). For k N , we have
| B 2 k | > 2 ( 2 k ) ! π 2 k ( 2 2 k 1 ) ,
where B 2 k are the even indexed Bernoulli numbers.
The following well-known Qi’s inequality for Bernoulli numbers is crucial in this paper.
Lemma 3
(see [27]). For k N , the even indexed Bernoulli numbers satisfy
| B 2 k + 2 | | B 2 k | > 2 2 k 1 1 2 2 k + 1 1 ( 2 k + 1 ) ( 2 k + 2 ) π 2 .
We now establish some new auxiliary results for even indexed Bernoulli numbers.
Lemma 4.
For k = 4 , 5 , 6 , , we have
| B 2 k | > 2 k 3 ( 2 2 k 1 ) .
Proof. 
It is obvious that
3 ( 2 k ) ! > π 2 k · k , for k = 4 , 5 , 6 , .
From this, we write
2 ( 2 k ) ! π 2 k ( 2 2 k 1 ) > 2 k 3 ( 2 2 k 1 ) , for k = 4 , 5 , 6 , .
Combining this with the inequality (4), we obtain the inequality (6). □
Lemma 5.
For k = 4 , 5 , 6 , , it is true that
| B 2 k + 2 | | B 2 k | > 2 ( 2 k 1 ) ( 2 k + 1 ) ( 2 k + 2 ) 3 ( 2 2 k + 2 1 ) .
Proof. 
Since,
2 2 k + 2 1 > 2 2 k + 1 1
and
3 ( 2 2 k 1 1 ) > 2 π 2 ( 2 k 1 ) ,
for k = 4 , 5 , 6 , , we obtain
3 ( 2 2 k 1 1 ) ( 2 2 k + 2 1 ) > 2 π 2 ( 2 k 1 ) ( 2 2 k + 1 1 ) ,
for k = 4 , 5 , 6 , . Equivalently, we write
( 2 2 k 1 1 ) ( 2 k + 1 ) ( 2 k + 2 ) π 2 ( 2 2 k + 1 1 ) > 2 ( 2 k 1 ) ( 2 k + 1 ) ( 2 k + 2 ) 3 ( 2 2 k + 2 1 ) , k = 4 , 5 , 6 , .
Combining this with (5), we obtain the desired inequality (7). □
Lemma 6.
Let k = 3 , 4 , 5 , . Then, it holds that
| B 2 k + 2 | | B 2 k | > ( 2 k + 1 ) ( 2 k + 2 ) P ( k ) 3 16 k ( 2 2 k + 2 2 ) ( 2 k + 1 ) 14 k ( 2 k + 1 ) · 2 2 k + 2 15 · 2 2 k + 2 ,
where P ( k ) = 16 ( 2 k 1 ) ( 2 2 k 1 1 ) + 48 ( 2 2 k 1 1 ) 24 ( 2 2 k 2 ) 7 · 2 2 k ( 2 k 1 ) .
Proof. 
In view of Lemma 3, it suffices to prove
( 2 2 k 1 1 ) π 2 ( 2 2 k + 1 1 ) > 16 ( 2 k 1 ) ( 2 2 k 1 1 ) + 48 ( 2 2 k 1 1 ) 24 ( 2 2 k 2 ) 7 · 2 2 k ( 2 k 1 ) 3 16 k ( 2 k + 1 ) ( 2 2 k + 2 2 ) 14 k ( 2 k + 1 ) · 2 2 k + 2 15 · 2 2 k + 2
for k = 3 , 4 , 5 , . Equivalently,
π 2 ( 2 2 k + 1 1 ) 16 ( 2 k 1 ) ( 2 2 k 1 1 ) + 48 ( 2 2 k 1 1 ) 24 ( 2 2 k 2 ) 7 · 2 2 k ( 2 k 1 ) < 3 ( 2 2 k 1 1 ) 16 k ( 2 k + 1 ) ( 2 2 k + 2 2 ) 14 k ( 2 k + 1 ) · 2 2 k + 2 15 · 2 2 k + 2 .
A direct computation gives
( 4 π 2 · 2 2 k 2 π 2 ) ( 2 k · 2 2 k 2 2 k 32 k + 16 ) < ( 3 · 2 2 k 6 ) ( 16 k 2 · 2 2 k 64 k 2 + 8 k · 2 2 k 60 · 2 2 k 32 k ) ,
which can be put as
( 48 k 2 + 24 k + 4 π 2 ) · 2 2 k + ( 32 π 2 k + 360 ) · 2 2 k + ( 384 k 2 + 192 k + 32 π 2 ) > ( 8 π 2 k + 180 ) · 2 2 k + 288 k 2 + ( 66 π 2 + 144 k ) · 2 2 k + 64 π 2 k .
Since
384 k 2 + 192 k + 32 π 2 > 64 π 2 k ,
for k = 3 , 4 , 5 , , we only need to prove that
Q ( k ) > 0 , for k = 3 , 4 , 5 , ,
where
Q ( k ) = ( 48 k 2 + 24 k + 4 π 2 8 π 2 k 180 ) · 2 2 k + ( 32 π 2 k + 360 288 k 2 66 π 2 144 k ) .
Now, k 3 implies
Q ( k ) ( 48 k 2 + 24 k + 4 π 2 8 π 2 k 180 ) · 64 + ( 32 π 2 k + 360 288 k 2 66 π 2 144 k ) = 2784 k 2 + 1392 k + 190 π 2 480 π 2 k 11160 2784 · 3 k + 1392 k + 190 π 2 480 π 2 k 11160 = ( 9744 480 π 2 ) k + ( 190 π 2 11160 ) ( 9744 480 π 2 ) · 3 + ( 190 π 2 11160 ) = ( 29232 + 190 π 2 ) 12600 > 0 .

3. Some New Generalized Mitrinović–Adamović-Type Inequalities

We are now in a position to state and prove our main results.
Theorem 1.
The function
f ( x ) = sin x x 3 cos x ( 1 cos x ) 2
is strictly decreasing on ( 0 , π ) .
Proof. 
The differentiation of f ( x ) gives
( 1 cos x ) 4 f ( x ) = ( 1 cos x ) ( 1 cos x ) 3 sin 2 x ( x cos x sin x ) x 4 + sin x 2 sin x sin 3 x x 3 cos x x 3 ,
and then deduces
x 4 ( 1 cos x ) 3 f ( x ) = ( 1 cos x ) ( 3 x sin 2 x cos x 3 sin 3 x + x 4 sin x ) 2 sin x ( x sin 3 x x 4 cos x ) = 3 x sin 2 x cos x 3 sin 3 x + x 4 sin x 3 x sin 2 x + x sin 4 x + 3 sin 3 x cos x + x 4 sin x cos x = sin 3 x · ϕ ( x ) ,
where
ϕ ( x ) = 3 x cot x 3 + x 4 1 sin 2 x 3 x sin x + x sin x + 3 cos x + x 4 cos x sin 2 x .
Using power series expansions in Lemma 1, and known series expansions of sin x and cos x , we have
ϕ ( x ) = 2 x 2 k = 1 3 · 2 2 k ( 2 k ) ! | B 2 k | x 2 k + k = 1 2 2 k ( 2 k 1 ) ( 2 k ) ! | B 2 k | x 2 k + 2 k = 1 6 ( 2 2 k 1 1 ) ( 2 k ) ! | B 2 k | x 2 k + k = 0 ( 1 ) k ( 2 k + 1 ) ! x 2 k + 2 + k = 1 3 ( 1 ) k ( 2 k ) ! x 2 k k = 1 2 ( 2 2 k 1 1 ) ( 2 k 1 ) ( 2 k ) ! | B 2 k | x 2 k + 2 = k = 1 3 · 2 2 k + 2 ( 2 k + 2 ) ! | B 2 k + 2 | x 2 k + 2 + k = 1 2 2 k ( 2 k 1 ) ( 2 k ) ! | B 2 k | x 2 k + 2 k = 1 6 ( 2 2 k + 1 1 ) ( 2 k + 2 ) ! | B 2 k + 2 | x 2 k + 2 + k = 1 ( 1 ) k ( 2 k + 1 ) ! x 2 k + 2 + k = 1 3 ( 1 ) k + 1 ( 2 k + 2 ) ! x 2 k + 2 k = 1 2 ( 2 2 k 1 1 ) ( 2 k 1 ) ( 2 k ) ! | B 2 k | x 2 k + 2 : = k = 1 a k ( 2 k + 2 ) ! x 2 k + 2 ,
where
a k = 2 2 k ( 2 k + 1 ) ( 2 k + 2 ) ( 2 k 1 ) | B 2 k | 2 ( 2 k + 1 ) ( 2 k + 2 ) ( 2 2 k 1 1 ) ( 2 k 1 ) | B 2 k | 3 · 2 2 k + 2 | B 2 k + 2 | 6 ( 2 2 k + 1 1 ) | B 2 k + 2 | + ( 1 ) k ( 2 k + 2 ) + 3 ( 1 ) k + 1 = 2 ( 2 k 1 ) ( 2 k + 1 ) ( 2 k + 2 ) | B 2 k | 6 ( 2 2 k + 2 1 ) | B 2 k + 2 | + ( 2 k + 2 ) ( 1 ) k + 3 ( 1 ) k + 1 .
We calculate a 1 = a 2 = 0 and a 3 = 128 3 .
Next, we claim that a k < 0 for k = 4 , 5 , 6 , 7 , 8 , . First, we need to prove a k < 0 for k = 4 , 6 , 8 , , that is,
( 2 k + 2 ) + 2 ( 2 k 1 ) ( 2 k + 1 ) ( 2 k + 2 ) | B 2 k | < 3 + 6 ( 2 2 k + 2 1 ) | B 2 k + 2 |
for k = 4 , 6 , 8 , , and then a k < 0 for k = 5 , 7 , 9 , , that is,
( 1 2 k ) + 2 ( 2 k 1 ) ( 2 k + 1 ) ( 2 k + 2 ) | B 2 k | < 6 ( 2 2 k + 2 1 ) | B 2 k + 2 |
for k = 5 , 7 , 9 , . Now, by Lemma 4 and Lemma 5, we, respectively, have
( 2 k + 2 ) < 3 ( 2 2 k + 2 1 ) | B 2 k + 2 | , k = 3 , 4 , 5 ,
and
2 ( 2 k 1 ) ( 2 k + 1 ) ( 2 k + 2 ) | B 2 k | < 3 ( 2 2 k + 2 1 ) | B 2 k + 2 | , k = 4 , 5 , 6 , .
Combining (10) with (11), we obtain
( 2 k + 2 ) + 2 ( 2 k 1 ) ( 2 k + 1 ) ( 2 k + 2 ) | B 2 k | < 6 ( 2 2 k + 2 1 ) | B 2 k + 2 | , k = 4 , 5 , 6 , ,
which implies (8). Also, the inequality (11) implies (9). Thus, a k 0 for k = 1 , 2 , 3 , 4 , 5 , . Therefore, we conclude that ϕ ( x ) < 0 , x ( 0 , π ) and consequently f ( x ) < 0 , x ( 0 , π ) . This shows that f ( x ) is strictly decreasing on ( 0 , π ) . The proof is completed. □
The following new generalized Mitrinović–Adamović-type inequality can be established by applying Theorem 1.
Theorem 2.
The following double inequalities are valid:
(i)
The double inequality
cos x + 8 π 3 ( 1 cos x ) 2 < sin x x 3 < cos x + 4 15 ( 1 cos x ) 2 , x 0 , π 2
holds with the best possible constants 8 π 3 and 4 15 .
(ii)
The double inequality
cos x + 1 4 ( 1 cos x ) 2 < sin x x 3 < cos x + 4 15 ( 1 cos x ) 2 , x 0 , π
holds with the best possible constants 1 4 and 4 15 .
Proof. 
Applying Theorem 1, we have
f ( 0 + ) > f ( x ) > f ( π / 2 ) ,
and
f ( 0 + ) > f ( x ) > f ( π ) .
From the limits
lim x 0 + f ( x ) = 4 15 , lim x π / 2 f ( x ) = 8 π 3 , lim x π f ( x ) = 1 4 ,
we can show the desired inequalities (12) and (13). □
Remark 1.
The left inequality of (13) can be written as
1 + cos x 2 2 < sin x x 3 , x ( 0 , π ) .
This inequality was first proved by Neuman and Sándor [4] and reappeared in [15]. However, it was shown to be true on 0 , π 2 only.
Remark 2.
It can be observed that the right inequality of (13) is in fact true for x > 0 . This fact can be verified graphically at https://www.desmos.com/calculator (accessed on 1 March 2025). Let us write
L ( x ) = cos x + 21 5 x 3 sin x 63 + x 2 , U ( x ) = cos x + π 2 15 x 3 sin x π 2 x 2
and
L 1 ( x ) = cos x + 1 4 ( 1 cos x ) 2 , U 1 ( x ) = cos x + 4 15 ( 1 cos x ) 2
for x ( 0 , π ) . Then, we compare graphically the bounds in (13) with those in (2).
The numerical calculations and graphs in Figure 1 show that the lower bound L 1 ( x ) of sin x x 3 in (13) is sharper than the corresponding lower bound L ( x ) in (2) for x > λ 1.5301 , whereas the upper bound U 1 ( x ) of sin x x 3 in (13) is uniformly sharper than the corresponding upper bound U ( x ) in (2).
In what follows, we further refine the bounds of sin x x 3 .
Theorem 3.
The function
g ( x ) = 1 x 6 sin x x 3 cos x 4 15 ( 1 cos x ) 2
is strictly increasing on ( 0 , π ) .
Proof. 
Differentiating g ( x ) with respect to x yields
x 10 g ( x ) = x 3 sin 2 x cos x + x 3 sin x 3 x 2 cos x 8 15 x 3 ( 1 cos x ) sin x 12 15 x 2 ( 1 cos x ) 2 9 sin 3 x x 3 cos x 4 15 x 3 ( 1 cos x ) 2 = 3 x sin 2 x cos x + 7 15 x 4 sin x 7 5 x 3 cos x + 8 15 x 4 sin x cos x 12 15 x 2 12 15 x 2 cos 2 x 9 sin 3 x + 9 x 3 cos x + 12 5 x 3 24 5 x 3 cos x + 12 5 x 3 cos 2 x = 3 x sin 2 x cos x + 7 15 x 4 sin x + 14 5 x 3 cos x + 8 15 x 4 sin x cos x + 16 5 x 3 8 5 x 3 sin 2 x 9 sin 3 x .
This can be written as x 10 sin 3 x = ψ ( x ) , where
ψ ( x ) = 3 x cot x + 7 15 x 4 1 sin 2 x + 14 5 x 3 cos x sin 3 x + 8 15 x 4 cos x sin 2 x + 16 5 x 3 1 sin 3 x 8 5 x 3 1 sin x 9 .
Utilizing Lemma 1, we obtain
ψ ( x ) = 3 k = 1 3 · 2 2 k ( 2 k ) ! | B 2 k | x 2 k + 7 15 x 2 + k = 1 7 15 2 ( 2 k 1 ) ( 2 k ) ! | B 2 k | x 2 k + 2 + 14 5 k = 2 14 5 2 2 k ( 2 k 1 ) ( k 1 ) ( 2 k ) ! | B 2 k | x 2 k + 8 15 x 2 k = 1 8 15 2 ( 2 k 1 ) ( 2 2 k 1 1 ) ( 2 k ) ! | B 2 k | x 2 k + 2 + 16 5 + k = 2 8 5 ( 2 2 k 2 ) ( 2 k 1 ) ( 2 k 2 ) ( 2 k ) ! | B 2 k | x 2 k + 8 5 x 2 + k = 1 8 5 ( 2 2 k 2 ) ( 2 k ) ! | B 2 k | x 2 k + 2 8 5 x 2 k = 1 16 5 ( 2 2 k 1 1 ) ( 2 k ) ! | B 2 k | x 2 k + 2 9 = x 2 + k = 1 8 5 ( 2 2 k 2 ) ( 2 k 1 ) ( 2 k 2 ) ( 2 k ) ! 14 5 2 2 k ( 2 k 1 ) ( k 1 ) ( 2 k ) ! 3 · 2 2 k ( 2 k ) ! | B 2 k | x 2 k + k = 1 7 15 2 2 k ( 2 k 1 ) ( 2 k ) ! + 8 5 ( 2 2 k 2 ) ( 2 k ) ! 16 5 ( 2 2 k 1 1 ) ( 2 k ) ! 16 15 ( 2 k 1 ) ( 2 2 k 1 1 ) ( 2 k ) ! | B 2 k | x 2 k + 2 = x 2 + k = 1 8 ( 2 2 k 2 ) ( 2 k 1 ) ( 2 k 2 ) 14 · 2 2 k ( 2 k 1 ) ( k 1 ) 15 · 2 2 k 5 · ( 2 k ) ! | B 2 k | x 2 k + k = 1 7 · 2 2 k ( 2 k 1 ) + 24 ( 2 2 k 2 ) 48 ( 2 2 k 1 1 ) 16 ( 2 k 1 ) ( 2 2 k 1 1 ) 15 · ( 2 k ) ! | B 2 k | x 2 k + 2 = x 2 + k = 0 16 k ( 2 k + 1 ) ( 2 2 k + 2 2 ) 14 k ( 2 k + 1 ) · 2 2 k + 2 15 · 2 2 k + 2 5 · ( 2 k + 2 ) ! | B 2 k + 2 | x 2 k + 2 + k = 1 7 · 2 2 k ( 2 k 1 ) + 24 ( 2 2 k 2 ) 48 ( 2 2 k 1 1 ) 16 ( 2 k 1 ) ( 2 2 k 1 1 ) 15 · ( 2 k ) ! | B 2 k | x 2 k + 2 = k = 3 1 5 · ( 2 k ) ! b k x 2 k + 2
where
b k = 16 k ( 2 2 k + 2 2 ) ( 2 k + 1 ) 14 k ( 2 k + 1 ) · 2 2 k + 2 15 · 2 2 k + 2 ( 2 k + 1 ) ( 2 k + 2 ) | B 2 k + 2 | 16 ( 2 k 1 ) ( 2 2 k 1 1 ) + 48 ( 2 2 k 1 1 ) 24 ( 2 2 k 2 ) 7 · 2 2 k ( 2 k 1 ) 3 | B 2 k | .
By Lemma 6, b k > 0 . This proves that ψ ( x ) and hence g ( x ) is positive, implying that g ( x ) is strictly increasing on ( 0 , π ) . The proof is completed. □
Theorem 4.
The following double inequalities are valid:
(i)
The double inequality
cos x + 4 15 ( 1 cos x ) 2 1 945 x 6 < sin x x 3 < cos x + 4 15 ( 1 cos x ) 2 256 π 6 1 15 2 π 3 x 6 , x 0 , π 2
holds with the best possible constants 1 945 and 256 π 6 1 15 2 π 3 .
(ii)
The double inequality
cos x + 4 15 ( 1 cos x ) 2 1 945 x 6 < sin x x 3 < cos x + 4 15 ( 1 cos x ) 2 1 15 π 6 x 6 , x ( 0 , π )
holds with the best possible constants 1 945 and 1 15 π 6 .
Proof. 
Applying Theorem 3, we obtain
g ( 0 + ) < g ( x ) < g ( π / 2 ) ,
and
g ( 0 + ) < g ( x ) < g ( π ) .
The double inequalities (14) and (15) follow due to the limits
lim x 0 g ( x ) = 1 945 , lim x π / 2 g ( x ) = 64 π 6 4 15 8 π 3 , and lim x π g ( x ) = 1 15 π 6 .
Remark 3.
If we suppose
L 2 ( x ) = cos x + 4 15 ( 1 cos x ) 2 1 945 x 6
and
U 2 ( x ) = cos x + 4 15 ( 1 cos x ) 2 1 15 π 6 x 6 ,
then it is obvious that U 2 ( x ) < U 1 ( x ) for all x ( 0 , π ) . Based on Figure 2 and some numerical calculations, it is found that L 1 ( x ) < L 2 ( x ) for x ( 0 , λ * ) where λ * 1.5969 .
The curves E 1 ( x ) = L 2 ( x ) L ( x ) , and E 2 ( x ) = U ( x ) U 2 ( x ) depicted in the following Figure 3 show that the double inequality (2) is completely refined to (15).
Lemma 7.
Let
T ( k ) = ( 2 2 k 1 1 ) · [ 16 k 3 ( 2 2 k + 2 1 ) ( 2176 k + 1980 ) · 2 2 k + 544 k + 720 ] ( 2 2 k + 1 1 ) ( 2 k 1 ) π 2 · ( 2 2 k + 90 k 136 ) .
Then, T ( k ) > 0 for k = 7 , 8 , 9 , .
Proof. 
We rewrite T ( k ) as
T ( k ) = ( 2 2 k + 1 4 ) · [ 4 k 3 ( 2 2 k + 2 1 ) ( 544 k + 495 ) · 2 2 k + 136 + 180 ] ( 2 2 k + 1 1 ) ( 2 k 1 ) π 2 · ( 2 2 k + 90 k 136 ) .
Clearly,
2 2 k + 2 2 k > 2 2 k + ( 90 k 132 ) , for k = 7 , 8 , 9 , .
Therefore
( 2 2 k + 1 4 ) > ( 2 2 k + 90 k 136 ) , k = 7 , 8 , 9 , .
Similarly,
16 k 3 + 2 π 2 > ( 544 + 4 π 2 ) k + 495 , for k = 7 , 8 , 9 ,
implies
( 16 k 3 + 2 π 2 ) · 2 2 k + ( 136 + 2 π 2 ) k + ( 180 π 2 ) > [ ( 544 + 4 π 2 ) k + 495 ] · 2 2 k .
Then, we have
4 k 3 ( 2 2 k + 2 1 ) ( 544 k + 495 ) · 2 2 k + 136 k + 180 > ( 2 2 k + 1 1 ) ( 2 k 1 ) π 2
for k = 7 , 8 , 9 , . From (16) and (17), we obtain T ( k ) > 0 for k = 7 , 8 , 9 , .
It is still possible to refine the inequalities of Theorem 3. We present the following more refined results.
Theorem 5.
The function
h ( x ) = sin x x 3 cos x 4 15 ( 1 cos x ) 2 x 2 ( 1 cos x ) 2
is strictly increasing on ( 0 , π ) .
Proof. 
By the differentiation of h ( x ) , we have
x ( 1 cos x ) 4 · h ( x ) = x ( 1 cos x ) 3 sin 2 x cos x x 3 3 sin 3 x x 4 + 7 15 sin x + 8 15 sin x cos x ( 2 2 cos x + 2 x sin x ) sin 3 x x 3 7 15 cos x 4 15 4 15 cos 2 x = 3 sin 2 x cos x x 2 5 sin 3 x x 3 + x sin x + x sin x cos x 3 sin 2 x x 2 + sin 4 x x 2 + 5 sin 3 x cos x x 3 2 15 cos x + 2 15 + 6 15 sin 2 x + 8 15 sin 2 x cos x : = sin 4 x · λ ( x ) ,
where
λ ( x ) = 3 x 2 cos x sin 2 x 5 x 3 1 sin x + x 1 sin 3 x + x cos x sin 3 x 3 x 2 1 sin 2 x + 1 x 2 + 5 x 3 cot x 2 15 cos x sin 4 x + 2 15 1 sin 4 x + 6 15 1 sin 2 x + 8 15 cos x sin 2 x .
We plan to use the series expansions given in Lemma 1 to obtain
λ ( x ) = 3 x 4 k = 1 3 ( 2 2 k 2 ) ( 2 k 1 ) ( 2 k ) ! | B 2 k | x 2 k 4 5 x 4 k = 1 5 ( 2 2 k 2 ) ( 2 k ) ! | B 2 k | x 2 k 4 + 1 x 2 + k = 2 ( 2 2 k 2 ) ( 2 k 1 ) ( 2 k 2 ) 2 · ( 2 k ) ! | B 2 k | x 2 k 2 + 1 2 + k = 1 ( 2 2 k 2 ) 2 · ( 2 k ) ! | B 2 k | x 2 k + 1 x 2 k = 2 2 2 k ( 2 k 1 ) ( k 1 ) ( 2 k ) ! | B 2 k | x 2 k 2 3 x 2 k = 1 3 · 2 2 k ( 2 k 1 ) ( 2 k ) ! | B 2 k | x 2 k 4 + 1 x 2 + 5 x 4 k = 1 5 · 2 2 k ( 2 k ) ! | B 2 k | x 2 k 4 + 2 45 x 3 + 1 45 x + k = 1 ( 2 2 k 2 ) ( 2 k 1 ) 45 · ( 2 k ) ! | B 2 k | x 2 k 2 + k = 2 ( 2 2 k 2 ) ( 2 k 1 ) ( 2 k 2 ) ( 2 k 3 ) 45 · ( 2 k ) ! | B 2 k | x 2 k 4 + 2 15 x 4 + k = 1 2 2 k ( 2 k 1 ) ( 2 k 2 ) ( 2 k 3 ) 45 · ( 2 k ) ! | B 2 k | x 2 k 4 + 4 45 x 2 + k = 1 4 45 2 2 k ( 2 k 1 ) ( 2 k ) ! | B 2 k | x 2 k 2 + 6 15 x 2 + k = 1 6 15 2 2 k ( 2 k 1 ) ( 2 k ) ! | B 2 k | x 2 k 2 + 8 15 x 2 k = 1 8 15 ( 2 2 k 2 ) ( 2 k 1 ) ( 2 k ) ! | B 2 k | x 2 k 2 .
After simplifying, λ ( x ) can be written as
λ ( x ) = 1 2 + 1 45 x + 181 45 x 2 + 2 45 x 3 + 2 15 x 4 + k = 1 ( 2 2 k 2 ) ( 2 k ) ! | B 2 k | x 2 k + k = 1 [ 4 · 2 2 k + 2 2 k 2 ] ( 2 k 1 ) 45 · ( 2 k ) ! | B 2 k | x 2 k 2 + k = 2 [ 2 2 k 2 2 2 k ] ( 2 k 1 ) ( k 1 ) ( 2 k ) ! | B 2 k | x 2 k 2 + k = 1 ( 2 k 2 ) ( 2 k 3 ) 45 3 · 2 2 k ( 2 k 1 ) ( 2 k ) ! | B 2 k | x 2 k 4 k = 1 [ 3 ( 2 k 1 ) + 5 ] ( 2 2 k 2 ) ( 2 k ) ! | B 2 k | x 2 k 4 + k = 2 ( 2 2 k 2 ) ( 2 k 1 ) ( 2 k 2 ) ( 2 k 3 ) 45 · ( 2 k ) ! | B 2 k | x 2 k 4 + k = 1 [ 3 · 2 2 k 4 ( 2 2 k 2 ) ] 2 ( 2 k 1 ) 15 · ( 2 k ) ! | B 2 k | x 2 k 2 .
Then, we have
λ ( x ) = 1 2 + 1 45 x + 181 45 x 2 + 2 45 x 3 + 2 15 x 4 + k = 2 ( 2 2 k 2 2 ) ( 2 k 2 ) ! | B 2 k 2 | x 2 k 2 + k = 1 ( 5 · 2 2 k 2 ) ( 2 k 1 ) 45 · ( 2 k ) ! | B 2 k | x 2 k 2 k = 2 2 ( 2 k 1 ) ( k 1 ) ( 2 k ) ! | B 2 k | x 2 k 2 + k = 0 2 k ( 2 k 1 ) 45 3 · 2 2 k + 2 ( 2 k + 1 ) ( 2 k + 2 ) ! | B 2 k + 2 | x 2 k 2 k = 0 ( 6 k + 8 ) ( 2 2 k + 2 2 ) ( 2 k + 2 ) ! | B 2 k + 2 | x 2 k 2 + k = 1 ( 2 2 k + 2 2 ) ( 2 k + 1 ) ( 2 k ) ( 2 k 1 ) 45 · ( 2 k + 2 ) ! | B 2 k + 2 | x 2 k 2 k = 1 ( 2 2 k 8 ) · 2 ( 2 k 1 ) 15 · ( 2 k ) ! | B 2 k | x 2 k 2 ,
which is equivalent to
λ ( x ) = 1 2 + 1 45 x + 181 45 x 2 + 2 45 x 3 + 2 15 x 4 + 1 1 x 2 7 3 x 2 + k = 1 ( 2 2 k 2 2 ) ( 2 k 2 ) ! | B 2 k 2 | x 2 k 2 + k = 1 ( 5 · 2 2 k 2 ) 45 2 ( k 1 ) ( 2 k ) ! ( 2 k 1 ) | B 2 k | x 2 k 2 + k = 1 2 k ( 2 k 1 ) 45 3 · 2 2 k + 2 ( 2 k + 1 ) ( 2 k + 2 ) ! | B 2 k + 2 | x 2 k 2 k = 1 ( 6 k + 8 ) ( 2 2 k + 2 2 ) ( 2 k + 2 ) ! | B 2 k + 2 | x 2 k 2 + k = 1 ( 2 2 k + 2 2 ) ( 2 k + 1 ) ( 2 k ) ( 2 k 1 ) 45 · ( 2 k + 2 ) ! | B 2 k + 2 | x 2 k 2 k = 1 2 ( 2 2 k 8 ) ( 2 k 1 ) 15 · ( 2 k ) ! | B 2 k | x 2 k 2 ,
that is,
λ ( x ) = 3 2 + 1 45 x + 31 45 x 2 + 2 45 x 3 + 2 15 x 4 + k = 1 c k ( 2 k 2 ) ! x 2 k 2 .
Here,
c k = ( 2 2 k 2 2 ) | B 2 k 2 | + [ 5 · 2 2 k 2 90 ( k 1 ) ] 90 k | B 2 k | + [ 2 k ( 2 k 1 ) 135 ] · 2 2 k + 2 45 ( 2 k + 2 ) ( 2 k ) ( 2 k 1 ) | B 2 k + 2 | ( 6 k + 8 ) ( 2 2 k + 2 2 ) ( 2 k + 2 ) ( 2 k + 1 ) ( 2 k ) ( 2 k 1 ) | B 2 k + 2 | + ( 2 2 k + 2 2 ) 45 ( 2 k + 2 ) | B 2 k + 2 | ( 2 2 k 8 ) 15 k | B 2 k | .
We write c k conveniently as
c k = ( 2 2 k 2 2 ) | B 2 k 2 | + ( 5 · 2 2 k 90 k + 88 ) 90 k | B 2 k | + ( 4 k 2 2 k 135 ) · 2 2 k 45 k ( k + 1 ) ( 2 k 1 ) | B 2 k + 2 | + ( 2 2 k + 2 2 ) ( 8 k 3 272 k 360 ) 45 ( 2 k + 2 ) ( 2 k + 1 ) ( 2 k ) ( 2 k 1 ) | B 2 k + 2 | ( 2 2 k 8 ) 15 k | B 2 k | = ( 2 2 k 2 2 ) | B 2 k 2 | + [ 16 k 3 ( 2 2 k + 2 1 ) ( 2176 k + 1980 ) · 2 2 k + 544 k + 720 ] 45 ( 2 k + 2 ) ( 2 k + 1 ) ( 2 k ) ( 2 k 1 ) | B 2 k + 2 | ( 2 2 k + 90 k 136 ) 90 k | B 2 k | .
From this, we obtain c 1 = 85 54 . So,
λ ( x ) = 2 27 + 1 45 x + 31 45 x 2 + 2 45 x 3 + 2 15 x 4 + k = 2 c k ( 2 k 2 ) ! x 2 k 2
where the function in the parentheses is strictly decreasing on ( 0 , π ) . Hence,
λ ( x ) > 2 27 + 1 45 π + 31 45 π 2 + 2 45 π 3 + 2 15 π 4 + k = 2 c k ( 2 k 2 ) ! x 2 k 2 = ( π 3 + 31 π 2 + 2 π + 6 ) 45 π 4 2 27 + 37703 38178 × ( 2 ! ) x 2 + 449 1890 × ( 4 ! ) x 4 + 79073 135135 × ( 6 ! ) x 6 + 56513281 8918910 × ( 8 ! ) x 8 + 13078211909763 186943326570 × ( 10 ! ) x 10 + k = 7 c k ( 2 k 2 ) ! x 2 k 2 .
The values of c k ’s in the above expression for k = 2 , 3 , , 6 are calculated using the Bernoulli numbers ([25], Chapter 23). For proving λ ( x ) > 0 on ( 0 , π ) , it suffices to prove c k > 0 for k = 7 , 8 , 9 , . Now, we rewrite c k as follows:
c k = ( 2 2 k 2 2 ) | B 2 k 2 | + | B 2 k | × [ 16 k 3 ( 2 2 k + 2 1 ) ( 2176 k + 1980 ) · 2 2 k + 544 k + 720 ] 45 ( 2 k + 2 ) ( 2 k + 1 ) ( 2 k ) ( 2 k 1 ) | B 2 k + 2 | | B 2 k | ( 2 2 k + 90 k 136 ) 90 k .
Since
16 k 3 ( 2 2 k + 2 1 ) ( 2176 k + 1980 ) · 2 2 k + 544 k + 720 > 0 for k = 7 , 8 , 9 , ,
by using Qi’s inequality (5) and applying Lemma 7. we have
c k > ( 2 2 k 2 2 ) | B 2 k 2 | + | B 2 k | × [ 16 k 3 ( 2 2 k + 2 1 ) ( 2176 k + 1980 ) · 2 2 k + 544 k + 720 ] ( 2 2 k 1 1 ) 45 ( 2 k ) ( 2 k 1 ) ( 2 2 k + 1 1 ) π 2 ( 2 2 k + 90 k 136 ) 90 k = ( 2 2 k 2 2 ) | B 2 k 2 | + | B 2 k | · T ( k ) 90 k ( 2 k 1 ) ( 2 2 k + 1 1 ) π 2 > 0 .
Thus, λ ( x ) > 0 , implying that h ( x ) > 0 , x ( 0 , π ) . Therefore, h ( x ) is strictly increasing on ( 0 , π ) . The proof is completed. □
Theorem 6.
The following double inequalities are valid:
(i)
The double inequality
cos x + 4 15 4 945 x 2 ( 1 cos x ) 2 < sin x x 3 < cos x + 4 15 4 π 2 4 15 8 π 3 x 2 ( 1 cos x ) 2 , x 0 , π 2
holds with the best possible constants 4 945 and 4 π 2 4 15 8 π 3 .
(ii)
The double inequality
cos x + 4 15 4 945 x 2 ( 1 cos x ) 2 < sin x x 3 < cos x + 4 15 1 60 π 2 x 2 ( 1 cos x ) 2 , x ( 0 , π )
holds with the best possible constants 4 945 and 1 60 π 2 .
Proof. 
Applying Theorem 5, we obtain
h ( 0 + ) < h ( x ) < h ( π / 2 ) ,
and
h ( 0 + ) < h ( x ) < h ( π ) .
Then, using the limits
lim x 0 + h ( x ) = 4 945 , lim x π / 2 h ( x ) = 4 π 2 4 15 8 π 3 , and lim x π h ( x ) = 1 60 π 2 ,
we can prove our conclusion. □
Remark 4.
It is worth noting that the double inequality (21) is a refinement of (15). Because
4 15 ( 1 cos x ) 2 1 945 x 6 < 4 15 4 945 x 2 ( 1 cos x ) 2 1 x 2 2 < cos x
and
4 15 1 60 π 2 x 2 ( 1 cos x ) 2 < 4 15 ( 1 cos x ) 2 1 15 π 6 x 6 cos x < 1 2 x 2 π 2
where the right-side inequalities are true due to the double inequality 1 x 2 2 < cos x < 1 4 x 2 π 2 as proved in [28].
Remark 5.
In our new generalized Mitrinović–Adamović-type inequalities, we have compared our bounds of sin x x 3 in a larger interval ( 0 , π ) with the old ones, i.e., with the corresponding bounds of (2). A similar numerical and/or graphical comparison can be made to show that the double inequalities (14) and (20) are superior to (3). In conclusion, we obtained stronger and superior bounds for the function sin x x 3 than those in (2)–(3). Moreover, all the main results of the paper hold in ( π , 0 ) because of the the symmetry of curves involved. Thus, our bounds provide the better alternatives.
Remark 6.
Our results can be used to bound the so-called sinc function, i.e., sin x x , which is extensively used in mathematics, physics and engineering. For instance, the double inequality (13) can be written as
cos x + 1 4 ( 1 cos x ) 2 1 / 3 < sin x x < cos x + 4 15 ( 1 cos x ) 2 1 / 3 , x 0 , π .
Remark 7.
An interesting application of our new generalized Mitrinović–Adamović-type inequalities is to find the value 3 π 8 of the integral 0 sin x x 3 d x . It is known that there are some complex standard methods; however, it is difficult to evaluate 0 p sin x x 3 d x , for any p > 0 . In such a case, we need to rely on an approximate value of the integral. Here, we can better approximate the integral 0 p sin x x 3 d x , 0 < p π by using one of our main results. In particular, we approximate 0 π sin x x 3 d x by selecting the inequality (21) as the best candidate whose bounds are tractable. Thus, integrating (21) from 0 to π, we have
I 1 < 0 π sin x x 3 d x < I 2 ,
where
I 1 = 0 π cos d x + 0 π 4 15 4 945 x 2 ( 1 cos x ) 2 d x ,
and
I 2 = 0 π cos d x + 0 π 4 15 1 60 π 2 x 2 ( 1 cos x ) 2 d x .
After expanding ( 1 cos x ) 2 , using simple formulae of integration and integrating by parts, we obtain
I 1 = ( 361 2 π 2 ) π 945 and I 2 = 47 π 120 17 240 π .
Then,
0 π sin x x 3 d x I 1 + I 2 2 = 1 2 ( 361 2 π 2 ) π 945 + 47 π 120 17 240 π 1.1712
which is very close to the exact value 1.1896 .

4. Conclusions

In this paper, we established various new generalized Mitrinović–Adamović-type inequalities as follows:
  • (See Theorem 2):
    (i)
    The double inequality
    cos x + 8 π 3 ( 1 cos x ) 2 < sin x x 3 < cos x + 4 15 ( 1 cos x ) 2 , x 0 , π 2
    holds with the best possible constants 8 π 3 and 4 15 .
    (ii)
    The double inequality
    cos x + 1 4 ( 1 cos x ) 2 < sin x x 3 < cos x + 4 15 ( 1 cos x ) 2 , x 0 , π
    holds with the best possible constants 1 4 and 4 15 .
  • (See Theorem 4):
    (i)
    The double inequality
    cos x + 4 15 ( 1 cos x ) 2 1 945 x 6 < sin x x 3 < cos x + 4 15 ( 1 cos x ) 2 256 π 6 1 15 2 π 3 x 6 , x 0 , π 2
    holds with the best possible constants 1 945 and 256 π 6 1 15 2 π 3 .
    (ii)
    The double inequality
    cos x + 4 15 ( 1 cos x ) 2 1 945 x 6 < sin x x 3 < cos x + 4 15 ( 1 cos x ) 2 1 15 π 6 x 6 , x ( 0 , π )
    holds with the best possible constants 1 945 and 1 15 π 6 .
  • (See Theorem 6):
    (i)
    The double inequality
    cos x + 4 15 4 945 x 2 ( 1 cos x ) 2 < sin x x 3 < cos x + 4 15 4 π 2 4 15 8 π 3 x 2 ( 1 cos x ) 2 , x 0 , π 2
    holds with the best possible constants 4 945 and 4 π 2 4 15 8 π 3 .
    (ii)
    The double inequality
    cos x + 4 15 4 945 x 2 ( 1 cos x ) 2 < sin x x 3 < cos x + 4 15 1 60 π 2 x 2 ( 1 cos x ) 2 , x ( 0 , π )
    holds with the best possible constants 4 945 and 1 60 π 2 .
In these new generalized Mitrinović–Adamović-type inequalities, we have compared our bounds of sin x x 3 in a larger interval ( 0 , π ) with the old ones. We believe that our results will assist us in obtaining novel expression results related to other generalized Mitrinović–Adamović-type inequalities in future studies.

Author Contributions

Writing—original draft, Y.J.B. and W.-S.D.; writing—review and editing, Y.J.B. and W.-S.D. All authors contributed equally to the manuscript and read and approved the final version of the manuscript.

Funding

Wei-Shih Du is partially supported by Grant No. NSTC 113-2115-M-017-004 of the National Science and Technology Council of the Republic of China.

Data Availability Statement

The original contributions presented in this study are included in the article. Further inquiries can be directed to the corresponding author.

Acknowledgments

The authors wish to express their sincere thanks to the anonymous referees for their valuable suggestions and comments.

Conflicts of Interest

The authors declare no conflicts of interest.

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Figure 1. Graphs of the lower and upper bounds of sin x x 3 in (2) and (13) for x ( 0 , π ) .
Figure 1. Graphs of the lower and upper bounds of sin x x 3 in (2) and (13) for x ( 0 , π ) .
Mathematics 13 01174 g001
Figure 2. Graphs of the lower bounds of sin x x 3 in (2), (13) and (15) for x ( 0 , π ) .
Figure 2. Graphs of the lower bounds of sin x x 3 in (2), (13) and (15) for x ( 0 , π ) .
Mathematics 13 01174 g002
Figure 3. Graphs of the functions E 1 ( x ) and E 2 ( x ) for x ( 0 , π ) .
Figure 3. Graphs of the functions E 1 ( x ) and E 2 ( x ) for x ( 0 , π ) .
Mathematics 13 01174 g003
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Bagul, Y.J.; Du, W.-S. On New Generalized Mitrinović-Adamović-Type Inequalities. Mathematics 2025, 13, 1174. https://doi.org/10.3390/math13071174

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Bagul YJ, Du W-S. On New Generalized Mitrinović-Adamović-Type Inequalities. Mathematics. 2025; 13(7):1174. https://doi.org/10.3390/math13071174

Chicago/Turabian Style

Bagul, Yogesh J., and Wei-Shih Du. 2025. "On New Generalized Mitrinović-Adamović-Type Inequalities" Mathematics 13, no. 7: 1174. https://doi.org/10.3390/math13071174

APA Style

Bagul, Y. J., & Du, W.-S. (2025). On New Generalized Mitrinović-Adamović-Type Inequalities. Mathematics, 13(7), 1174. https://doi.org/10.3390/math13071174

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