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Article

On Coefficient Inequalities for Functions of Symmetric Starlike Related to a Petal-Shaped Domain

1
Department of Mathematics, Abdul Wali Khan University Mardan, Mardan 23200, Pakistan
2
Department of Mathematics, College of Sciences, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia
3
Department of Mathematics, “1 Decembrie 1918” University of Alba Iulia, 510009 Alba Iulia, Romania
*
Author to whom correspondence should be addressed.
Axioms 2025, 14(3), 165; https://doi.org/10.3390/axioms14030165
Submission received: 5 February 2025 / Revised: 22 February 2025 / Accepted: 22 February 2025 / Published: 24 February 2025
(This article belongs to the Special Issue Theory of Functions and Applications, 2nd Edition)

Abstract

:
The research on coefficient inequalities in various classes of univalent holomorphic functions focuses on interpreting their coefficients through the coefficients associated with Carathéodory functions. Therefore, researchers can investigate the behavior of coefficient functionals by applying the known inequalities for Carathéodory functions. This study will explore various coefficient inequalities employing the techniques developed for the previously discussed family of functions. These coefficient inequalities include the Krushkal, Zalcman, and Fekete-Szegö inequalities, along with the second and third Hankel determinants. The class of symmetric starlike functions linked with a petal-shaped domain is the primary focus of our study.

1. Introduction and Definitions

To ensure a clear understanding of the key terminology utilized throughout our primary findings, we first present some preliminary concepts. First, let us consider D = ξ C : ξ < 1 to denote the open unit disk, and let A represent the family of holomorphic functions normalized by the conditions g ( 0 ) = 0 and g ( 0 ) 1 = 0 . This indicates that g A possesses a form
g ( ξ ) = ξ + u = 1 d u ξ u , ξ D .
Also, keep in mind that a holomorphic function is univalent within the domain D if it assigns each value in D at most once. Specifically, for g to be univalent in D , it means that g ξ 1 = g ξ 2 only when ξ 1 = ξ 2 , where ξ 1 , ξ 2 D . We use the symbol S to represent the family of univalent functions defined by the series expansion (1). In 1907, Köebe established this family.
In 1916, Bieberbach [1] put forward a notable result in function theory, which became known as the “Bieberbach conjecture”. The conjecture asserts that for g S , the inequality d u u holds for u 2 . He also demonstrated that this holds specifically for u = 2 . It is apparent that numerous prominent scholars have utilized different techniques to approach this result. Löwner [2] solved this result for u = 3 , while Spence and Schaeffer [3] addressed it by applying the variational technique and the Löwner differential equation. Jenkins [4] later established the coefficient inequality d 3 3 through the application of quadratic differentials. Schiffer and Garabedian [5] employed the variational technique to establish that d 4 4 . Meanwhile, Schiffer and Pederson [6] utilized the Schiffer–Garabedian inequality ([7], p. 108) to show that d 5 5 . Furthermore, Ozawa [8,9] and Pederson [10] independently proved that d 6 6 by utilizing Grunsky’s inequality ([7], p. 60). Numerous scholars have attempted to resolve this conjecture for years, but none have succeeded in proving it for u 7 . De Branges [11] successfully resolved the result for u 2 in 1985 by employing hypergeometric functions.
From 1916 to 1985, researchers focused on determining the estimates of the u-th coefficient for subfamilies of univalent functions such as starlike S , convex C , and close-to-convex K , trying to solve the problem. Some of the key classes are defined as:
S = g S : Re ξ g ξ g ξ > 0 , ξ D , C = g S : Re ξ g ξ g ξ > 0 , ξ D , K = g S : Re ξ g ξ h ξ > 0 with h S , ξ D .
By setting h ( ξ ) = ξ , the K family is transformed into the class of bounded turning functions, represented by the notation BT . In 1992, researchers [12] examined a univalent function ϕ in D that fulfilled the conditions ϕ 0 > 0 and Re ϕ > 0 . They also noted that the region ϕ D is star-shaped around the fixed point ϕ 0 = 1 and is symmetric across the real axis. By utilizing the concept of subordination, they established the following subclass within the family S .
S ϕ = g S : ξ g ξ g ξ ϕ ξ , ξ D .
They focused on several results, including the distortion theorem, growth properties and covering theorem. In recent years, scholars have explored various subclasses of the collection S as potential choices for the class S ϕ . A recent study identifies the following classes as particularly significant.
(i). 
S ( u 1 ) L S ( Ψ u 1 ξ ) [13] with Ψ u 1 ξ = 1 + u u + 1 ξ + 1 u + 1 ξ u for u 4 ,
(ii). 
S sin S 1 + sin ( ξ ) [14] , S cosh S cosh ( ξ ) [15],
(iii). 
S c a r S 1 + 2 3 ξ + 1 3 ξ 2 [16], S c S ( 1 + ξ + 1 2 ξ 2 ) [17].
The determinant D δ , η g , where η , δ N = 1 , 2 , , is known as the Hankel determinant. It is formed from the coefficients of the function g that resides in the class S
D δ , η g = d η d η + 1 d η + δ 1 d η + 1 d η + 2 d η + δ d η + δ 1 d η + δ d η + 2 δ 2 .
Pommerenke [18,19] made this contribution. In particular, the second- and third-order Hankel determinants are recognized as the following determinants:
D 2 , 1 g = 1 d 2 d 2 d 3 = d 3 d 2 2 , D 2 , 2 g = d 2 d 3 d 3 d 4 = d 2 d 4 d 3 2 , D 3 , 1 g = 1 d 2 d 3 d 2 d 3 d 4 d 3 d 4 d 5 = d 3 d 2 d 4 d 3 2 d 4 d 4 d 2 d 3 + d 5 d 3 d 2 2 .
The first two determinants have been the focus of detailed studies in the literature for numerous subclasses of univalent functions. The research by the authors in [20,21,22] is notable for establishing the estimates of the second determinant. For further details on the analysis of this determinant, refer to the articles [23,24,25,26].
The third-order determinant presented above is the most complex problem to analyze, particularly in terms of determining its sharp estimates. Babalola [27] became the first to analyze the estimates of the third-order determinant for S , BT and K classes in 2010. Zaprawa [28] improved the Babalola’s results in 2017 by utilizing an innovative approach. In 2019, Lecko et al. [29] established the sharp estimate of D 3 , 1 g for functions within the family S 1 / 2 . Additionally, In 2022, Arif et al. [30] presented the sharp estimate D 3 , 1 g 16 / 225 for g S 3 l in their publication. Readers may find the study of [31,32,33,34,35] noteworthy, as they determined sharp estimates for the third-order Hankel determinant in new subclasses of univalent functions.
Using the function ϕ with the constraints as given above, we now consider the following Ma–Minda type Sakaguchi starlike functions
S s ϕ = g S : 2 g ξ g ξ g ξ ϕ ξ , ξ D .
It is to be noted that class S s ϕ with ϕ z = 1 + z 1 z consists of starlike functions about symmetric points and was introduced by Sakaguchi [36] in 1959. Also, he proved that the family S s is a subclass of the set K of close-to-convex functions and also includes both odd starlike functions and convex functions, considering the origin as a reference point. Several researchers studied numerous subclasses of univalent functions that are linked with the Sakaguchi function. In [37], Zaprawa investigated the estimates of initial logarithmic coefficients for the family S s . Later on, Zaprawa [38] estimated the logarithmic coefficients, Zalcman inequalities, and Hankel determinants for symmetric starlike functions related to the exponential function. Furthermore, Faisal et al. [39] studied the initial coefficients, Zalcman and Krushkal inequalities, along with the Hankel determinants for the family S S G . The authors also studied the inverse and logarithmic properties of the same family.
In [40], Arora and Kumar study the petal-shaped domain Γ P specified by w C : sinh w 1 < 1 based on the families presented in [13,14,15,16,17]. This domain is functionally defined through P ( ξ ) = 1 + sinh 1 ξ , which plays a key role in establishing the family. P ( ξ ) is clearly recognized as a Ma–Minda function. Remember that sinh 1 ξ is a multivalued function with branch cuts along the segments ι , ι and ι , ι on the imaginary axis. As a result, it remains analytic in D . A petal-shaped domain finds applications in numerous disciplines, such as mathematics, biology, computer graphics, architecture, botany, materials science, engineering, art, and geometry. Moreover, Arora and Kumar presented a unique subfamily of starlike functions that is defined as
S P : = g A : ξ g ( ξ ) g ( ξ ) 1 + sinh 1 ξ , ξ D .
Based on the given definition, we deduce that g S P if and only if there exists a holomorphic function q ( ξ ) P ( ξ ) such that
g ξ = ξ exp 0 ξ q ξ 1 t d t .
Later, Barukab et al. [41] examined sharp estimates for coefficient inequalities and the third Hankel determinant in the BT s family. In 2022, Shi et al. [42] calculated the second Hankel determinant of logarithmic coefficients for the family BT s . Based on this definition, we present the class S sinh of symmetric starlike functions linked with a petal-shaped domain and is given as
S sinh = g S : 2 ξ g ξ g ξ g ξ 1 + 4 5 ξ + 1 5 ξ 4 , ξ D .
In this research, our aim is to investigate the sharp estimates of coefficient inequalities, Krushkal, Zalcman and Fekete-Szegö inequalities for the class S sinh of symmetric starlike function related to a petal-shaped domain. In addtion, we also study the estimates of D 2 , 2 g , D 2 , 3 g , and D 3 , 1 g for the above-mentioned class.

2. A Set of Lemmas

Let us consider P to denote the Carathéodory class, which contains all holomorphic functions p in D that satisfy Re p ( ξ ) > 0 and and follow the given normalization:
p ξ = 1 + i = 1 ς i ξ i , ξ D .
Lemma 1
([43]). Let p P be given by (4) and K 0 , 1 with K 2 K 1 T K , then
ς 3 2 K ς 1 ς 2 + T ς 1 3 2 .
Lemma 2
([44]). Let p P be given by (4). Then
ς i 2 , i 1 .
Lemma 3
([45]). If ϱ ,   ν ,  ρ and ϖ satisfy ϱ 0 , 1 and ρ 0 , 1 with
ϖ + ρ ρ + ϱ 2 + 2 ν + ϖ ρ 2 8 1 ϱ ϱ + 2 ϱ ρ + ϖ 2 1 ρ ρ 4 ρ 2 1 ρ 2 1 ϱ ϱ .
Let p P , be given by (4). Then
ν ς 1 4 + ϱ ς 2 2 + 2 ρ ς 1 ς 3 3 2 ϖ ς 1 2 ς 2 ς 4 2 .
Lemma 4
([7]). Let p P be given by (4), then
ς q + p ϑ ς q ς p 2 max 1 , 2 ϑ 1 = 2 for ϑ 0 , 1 ; 2 2 ϑ 1 o t h e r w i s e . .
Lemma 5
([46]). If p P be given by (4). Then for τ ,   σ ,   ζ D ¯ 1 .
2 ς 2 = 4 ς 1 2 τ + ς 1 2 ,
4 ς 3 = 2 4 ς 1 2 τ ς 1 τ 2 4 ς 1 2 ς 1 + 2 σ 1 τ 2 4 ς 1 2 + ς 1 3 , 8 ς 4 = ς 1 2 3 τ + τ 2 + 3 + 4 τ 4 ς 1 2 τ 4 1 τ 2 4 ς 1 2
τ 1 σ ς + σ 2 τ ¯ ζ 1 σ 2 + ς 1 4 .

3. Coefficient Inequalities

This section explores the calculation of the initial four coefficient inequalities for functions within the family S sinh .
Theorem 1.
If g S sinh has the series expansion (1). Then
d 2 1 2 , d 3 1 2 , d 4 1 4 , d 5 1 4 .
These outcomes are sharp.
Proof. 
Let g S sinh . Then
2 ξ g ξ g ξ g ξ = 1 + sinh 1 w ξ , ξ D .
If p P , we can express it using the Schwarz function w ξ as follows
p ξ = 1 + w ξ 1 w ξ = 1 + ς 1 ξ + ς 2 ξ 2 + ς 3 ξ 3 + ,
or correspondingly,
w ξ = p ξ 1 p ξ + 1 = ς 1 ξ + ς 2 ξ 2 + ς 3 ξ 3 + 2 + ς 1 ξ + ς 2 ξ 2 + ς 3 ξ 3 + .
Using (1), it follows
2 ξ g ξ g ξ g ξ = 1 + 2 d 2 ξ + 2 d 3 ξ 2 + 4 d 4 2 d 2 d 3 ξ 3 + 4 d 5 2 d 3 2 ξ 4 + .
By simplification and using the series expansion of (12), it yields
1 + sinh 1 w ξ = 1 + 1 2 ς 1 ξ + 1 2 ς 2 1 4 ς 1 2 ξ 2 + 1 2 ς 3 1 2 ς 1 ς 2 + 5 48 ς 1 3 ξ 3 + 1 32 ς 1 4 1 4 ς 2 2 1 2 ς 1 ς 3 + 5 16 ς 1 2 ς 2 + 1 2 ς 4 ξ 4 + .
Comparing (13) and (14), we obtain
d 2 = 1 4 ς 1 ,
d 3 = 1 4 ς 2 1 8 ς 1 2 ,
d 4 = 1 96 ς 1 3 3 32 ς 1 ς 2 + 1 8 ς 3 ,
d 5 = 1 32 ς 2 2 + 3 64 ς 1 2 ς 2 1 8 ς 1 ς 3 + 1 8 ς 4 .
Implementing (6) in (15), we obtain
d 2 1 2 .
Reordering (16), we obtain
d 3 = 1 4 ς 2 1 2 ς 1 2 .
Using (8), we have
d 3 1 2 .
We can write (17) as
d 4 = 1 8 ς 3 2 3 8 ς 1 ς 2 + 1 12 ς 1 3 .
From (5), let
K = 3 8 and T = 1 12 .
It is clear that 0 K 1 and K T with
K 2 K 1 = 3 32 T .
Thus all the conditions of (5) are satisfied. Hence we have
d 4 1 4 .
We can rewrite (18) as
d 5 = 1 8 1 4 ς 2 2 + 2 1 2 ς 1 ς 3 3 2 1 4 ς 1 2 ς 2 ς 4 . = 1 8 ν ς 1 4 + d ς 2 2 + 2 ρ ς 1 ς 3 3 2 ϖ ς 1 2 ς 2 ς 4 ,
where
ν = 0 , ϱ = 1 4 , ρ = 1 2 , ϖ = 1 4 ,
are such that
ϖ + ρ ρ + ϱ 2 + 2 ν + ϖ ρ 2 8 1 ϱ ϱ + 2 ϱ ρ + ϖ 2 1 ρ ρ 4 ϱ ρ 2 1 ρ 2 1 ϱ ,
ϱ 0 , 1 , ρ 0 , 1 , therefore by (7) and (19), we have
d 5 1 4 .
The inequalities are sharp, and the functions listed below attain equality
2 ξ g ξ g ξ g ξ = 1 + ξ 1 6 ξ 3 + 3 40 ξ 5 + , 2 ξ g ξ g ξ g ξ = 1 + ξ 2 1 6 ξ 6 + 3 40 ξ 10 + , 2 ξ g ξ g ξ g ξ = 1 + ξ 3 1 6 ξ 9 + 3 40 ξ 15 + , 2 ξ g ξ g ξ g ξ = 1 + ξ 4 1 6 ξ 12 + 3 40 ξ 20 + .
The required proof is accomplished. □

4. Fekete-Szegö Inequality

The determinant D 2 , 1 g corresponds to a specific Fekete-Szegö inequality d 3 ϑ d 2 2 , where ϑ is a complex parameter. The Fekete-Szegö inequality [47] is among the first problems about the coefficients of univalent holomorphic functions, established in 1933. We compute the Fekete-Szegö inequality bound for g S sinh in this second section.
Theorem 2.
If g S sinh , then
d 3 ϑ d 2 2 max 1 2 , ϑ 4 , for ϑ C .
This inequality is sharp.
Proof. 
Putting (15) and (16), we obtain
d 3 ϑ d 2 2 = 1 4 ς 2 1 8 ς 1 2 ϑ 1 16 ς 1 2 .
Application of (8), leads us to
d 3 ϑ d 2 2 1 4 max 2 , 2 2 + ϑ 2 1 .
After the simplification, we obtain
d 3 ϑ d 2 2 max 1 2 , ϑ 4 .
This outcome is best possible and is obtained by
2 ξ g ξ g ξ g ξ = 1 + ξ 2 1 6 ξ 6 + 3 40 ξ 10 + .

5. Zalcman Inequalities

In 1960, Lawrence Zalcman presented a conjecture stating that the coefficients of a function in S adhere to the sharp inequality
| d u 2 d 2 u 1 | ( u 1 ) 2 .
The Köebe function and its rotations provide equality. We can now calculate the Zalcman inequalities for g S sinh .
Theorem 3.
If g S sinh , be given by (1). Then
d 5 d 3 2 1 4 .
This is the finest possible inequality.
Proof. 
Using (16) and (18), we obtain
d 5 d 3 2 = 3 32 ς 2 2 + 7 64 ς 1 2 ς 2 1 64 ς 1 4 1 8 ς 1 ς 3 + 1 8 ς 4 .
After the simplifying we have
d 5 d 3 2 = 1 8 1 8 ς 1 4 + 3 34 ς 2 2 + 2 1 2 ς 1 ς 3 3 2 7 12 ς 1 2 ς 2 ς 4 . = 1 8 ν ς 1 4 + ϱ ς 2 2 + 2 ρ ς 1 ς 3 3 2 ϖ ς 1 2 ς 2 ς 4 ,
where
ν = 1 8 , ϱ = 3 34 , ρ = 1 2 , ϖ = 7 12 ,
are such that
ϖ + ρ ρ + ϱ 2 + 2 ν + ϖ ρ 2 8 1 ϱ ϱ + 2 ϱ ρ + ϖ 2 1 ρ ρ 4 ϱ ρ 2 1 ρ 2 1 ϱ ,
ϱ 0 , 1 , ρ 0 , 1 , therefore by (7) and (20), we have
d 5 d 3 2 1 4 .
This result is sharp and is achieved by
2 ξ g ξ g ξ g ξ = 1 + ξ 4 1 6 ξ 12 + 3 40 ξ 20 + .
Theorem 4.
If g belongs to S sinh , and given by (1). Then
d 2 d 3 d 4 1 4 .
This inequality is best possible.
Proof. 
Putting (15)–(17), we have
d 2 d 3 d 4 = 1 8 ς 3 2 5 8 ς 1 ς 2 + 1 3 ς 1 3 .
From (5), we have
0 K = 5 8 1 , K = 5 8 T = 1 3 ,
and
K 2 K 1 = 5 32 T = 1 3 .
Using (5), we obtain
d 2 d 3 d 4 1 4 .
This outcome is sharp. Equality is achieved from
2 ξ g ξ g ξ g ξ = 1 + ξ 3 1 6 ξ 9 + 3 40 ξ 15 + .

6. Krushkal Inequalities

In [48], Krushkal proved the Zalcman conjecture for u 6 by applying the holomorphic homotopy of univalent functions, which was originally presented in an unpublished article [49] for u 2 . It was also proved that, for g S :
d u t d 2 t u 1 2 t u 1 u t , with u , t 2 .
The estimation of Zalcman inequalities are now presented here in this section.
Theorem 5.
If g S sinh be given by (1). Then
d 5 d 2 4 1 4 .
The outcome is sharp.
Proof. 
From (15) and (18), we obtain
d 5 d 2 4 = 1 256 ς 1 4 + 3 64 ς 1 2 ς 2 1 32 ς 2 2 1 8 ς 1 ς 3 + 1 8 ς 4 .
After the simplifying we have
d 5 d 2 4 = 1 8 1 32 ς 1 4 + 1 4 ς 2 2 + 2 1 2 ς 1 ς 3 3 2 1 4 ς 1 2 ς 2 ς 4 .
Comparing the right side of (21) with
ν ς 1 4 + ϱ ς 2 2 + 2 ρ ς 1 ς 3 3 2 ϖ ς 1 2 ς 2 ς 4 ,
we obtain
ν = 1 32 , ϱ = 1 4 , ρ = 1 2 , ϖ = 1 4 .
It follows that
ϖ + ρ ρ + ϱ 2 + 2 ν + ϖ ρ 2 8 1 ϱ ϱ + 2 ϱ ρ + ϖ 2 1 ρ ρ = 0.0292968 ,
and
4 ϱ ρ 2 1 ρ 2 1 ϱ = 0.046875 .
From (7), we deduce that
d 5 d 2 4 1 4 .
The equality is attained from
2 ξ g ξ g ξ g ξ = 1 + ξ 4 1 6 ξ 12 + 3 40 ξ 20 + .
Theorem 6.
If g S sinh , and given by (1). Then
d 4 d 2 3 1 4 .
The result is best possible.
Proof. 
Putting (15) and (17), we have
d 4 d 2 3 = 1 8 ς 3 2 3 8 ς 1 ς 2 + 1 24 ς 1 3 .
From (5), let
K = 3 8 and T = 1 24 ,
and 0 K 1 , K T with
K 2 K 1 = 0.09375 T .
Thus, all the conditions of (5) are satisfied. Hence, we have
d 4 d 2 3 1 4 .
The equality is attained from
2 ξ g ξ g ξ g ξ = 1 + ξ 3 1 6 ξ 9 + 3 40 ξ 15 + .

7. Second and Third Hankel Determinants

Finally, we study the estimates of the Hankel determinants D 2 , 2 g , D 2 , 3 g and D 3 , 1 g for g S sinh .
Theorem 7.
If g S sinh and has the form (1), then
D 2 , 2 g = d 2 d 4 d 3 2 1 4 .
This outcome is best possible.
Proof. 
From (15)–(17), we have
D 2 , 2 g = 5 384 ς 1 4 + 5 128 ς 1 2 ς 2 + 1 32 ς 1 ς 3 1 16 ς 2 2 .
Applying (9) and (10) to write ς 2 and ς 3 in terms of ς 1 and observing that we can write ς 1 = ς , we achieve
D 2 , 2 g = 1 768 ς 1 4 + 1 256 4 ς 2 ς 2 τ 1 128 4 ς 2 ς 2 τ 2 1 64 4 ς 2 2 τ 2 + 1 64 ς 4 ς 2 1 τ 2 σ ,
Invoking τ = l , σ 1 with l 1 and applying the triangle inequality we have
D 2 , 2 g 1 768 ς 4 + 1 256 4 ς 2 ς 2 l + 1 128 4 ς 2 ς 2 l 2 + 1 64 4 ς 2 2 l 2 + 1 64 ς 4 ς 2 1 l 2 : = φ ς , l .
It is now a straightforward task to illustrates that φ ς , l 0 on 0 , 1 , and hence φ ς , l φ ς , 1 . Thus
D 2 , 2 g 1 768 ς 4 + 1 256 ς 2 4 ς 2 + 1 64 4 ς 2 2 + 1 128 4 ς 2 ς 2 : = φ ς , 1 .
By simple computation, it follows that φ ς , 1 gets its maxima at 0 . Hence
D 2 , 2 g 1 4 .
The required D 2 , 2 g is sharp. Equality is achieved from
2 ξ g ξ g ξ g ξ = 1 + ξ 2 1 6 ξ 6 + 3 40 ξ 10 + .
Theorem 8.
If g S sinh , and has the form (1). Then
D 2 , 3 g = d 3 d 5 d 4 2 840959 11215872 .
Proof. 
Plugging (16)–(18), with ς 1 = ς we obtain
D 2 , 3 g = 1 9216 ς 6 36 ς 4 ς 2 + 120 ς 3 ς 3 + 63 ς 2 ς 2 2 144 ς 2 ς 4 72 ς ς 2 ς 3 72 ς 2 3 + 288 ς 2 ς 4 144 ς 3 2 .
Let m = 4 ς 2 in (9)–(11). Now using these lemmas, we obtain
36 ς 4 ς 2 = 18 ς 6 + ς 4 s τ , 120 ς 3 ς 3 = 30 ς 4 m τ 2 + 60 ς 3 m 1 τ 2 σ + 60 ς 4 m τ + 30 ς 6 , 63 ς 2 ς 2 2 = 63 4 ς 6 + 63 2 ς 4 m τ + 63 4 ς 2 m 2 τ 2 , 144 ς 2 ς 4 = 18 ς 6 + 18 ς 4 m τ 3 54 ς 4 m τ 2 + 54 ς 4 m τ + 72 ς 2 m τ 2 72 ς 3 m τ σ 1 τ 2 72 ς 2 m τ ¯ 1 τ 2 σ 2 + 72 ς 2 m 1 τ 2 1 σ 2 ζ + 72 ς 3 m 1 τ 2 σ , 72 ς ς 2 ς 3 = 9 ς 2 m 2 τ 3 9 ς 4 m τ 2 + 18 ς m 2 τ 1 τ 2 σ + 18 ς 2 m 2 τ 2 + 27 ς 4 m τ + 18 ς 3 m 1 τ 2 σ + 9 ς 6 , 72 ς 2 3 = 9 ς 6 + 27 ς 4 m τ + 27 ς 2 m 2 τ 2 + 9 m 3 τ 3 , 288 ς 2 ς 4 = 18 ς 6 + 18 ς 4 m τ 3 54 ς 4 m τ 2 + 72 ς 4 m τ + 72 ς 2 m τ 2 72 ς 3 m τ σ 1 τ 2 72 ς 2 m τ ¯ 1 τ 2 σ 2 + 72 ς 2 m 1 τ 2 1 σ 2 ζ + 72 ς 3 m 1 τ 2 σ + 18 ς 2 m 2 τ 4 54 ς 2 m 2 τ 3 + 54 ς 2 m 2 τ 2 + 72 m 2 τ 3 72 ς m 2 τ 2 1 τ 2 σ 72 m 2 τ τ ¯ 1 τ 2 σ 2 + 72 m 2 τ 1 τ 2 1 σ 2 ζ + 72 ς m 2 τ 1 τ 2 σ , 144 ς 3 2 = 9 ς 2 m 2 τ 4 36 ς m 2 τ 2 1 τ 2 σ 36 ς 2 m 2 τ 3 18 ς 4 m τ 2 + 36 ς 4 m τ + 36 m 2 1 τ 2 2 σ 2 + 72 ς m 2 τ 1 τ 2 σ + 36 ς 2 m 2 τ 2 + 36 ς 3 m 1 τ 2 σ + 9 ς 6 .
Plugging the above expressions in (22), we obtain
D 2 , 3 g = 1 9216 6 ς 3 m 1 τ 2 σ 9 τ 3 m 3 + 72 τ 3 m 2 3 ς 4 m τ 2 45 4 ς 2 τ 2 m 2 9 ς 2 τ 3 m 2 + 9 ς 2 m 2 τ 4 36 m 2 1 τ 2 2 σ 2 + 3 2 ς 4 τ m 18 ς τ m 2 1 τ 2 σ 1 4 ς 6 36 ς τ 2 m 2 1 τ 2 σ 72 τ m 2 1 τ 2 τ ¯ σ 2 + 72 τ m 2 1 τ 2 1 σ 2 ζ .
Since m = 4 ς 2 ,
D 2 , 3 g = 1 9216 L 0 ς , τ + L 1 ς , τ σ + L 2 ς , τ σ 2 + L 3 ς , τ , σ ζ ,
where
L 0 ς , τ = 1 4 ς 6 + 4 ς 2 4 ς 2 36 τ 3 45 4 ς 2 τ 2 3 ς 4 τ 2 + 3 2 ς 4 τ , L 1 ς , τ = 4 ς 2 1 τ 2 4 ς 2 18 ς τ 36 ς τ 2 + 6 ς 3 , L 2 ς , τ = 4 ς 2 1 τ 2 4 ς 2 36 τ 2 36 , L 3 ς , τ , σ = 4 ς 2 1 τ 2 1 σ 2 72 τ 4 ς 2 .
By replacing τ by τ and σ by κ , if we apply the statement ζ 1 , it follows
D 2 , 3 g 1 9216 L 0 ς , τ + L 1 ς , τ κ + L 2 ς , τ κ 2 + L 3 ς , τ , σ . 1 9216 F ς , τ , κ ,
where
F ς , τ , κ = R 0 ς , τ + R 1 ς , τ κ + R 2 ς , τ κ 2 + R 3 ς , τ 1 κ 2 ,
with
R 0 ς , τ = 1 4 ς 6 + 4 ς 2 4 ς 2 36 τ 3 + 45 4 ς 2 τ 2 + 3 ς 4 τ 2 + 3 2 ς 4 τ , R 1 ς , τ = 4 ς 2 1 τ 2 4 ς 2 18 ς τ + 36 ς τ 2 + 6 ς 3 , R 2 ς , τ = 4 ς 2 1 τ 2 4 ς 2 36 τ 2 + 36 , R 3 ς , τ = 4 ς 2 1 τ 2 72 τ 4 ς 2 .
Now, we are to maximize F ς , τ , κ in the closed cuboid Ξ : 0 , 2 × 0 , 1 × 0 , 1 .
For this purpose, we have to find the maxima of F ς , τ , κ in the interior of Ξ , in the interior of its faces and on the edges.
1. Interior points of cuboid Ξ:
Suppose ς , τ , κ 0 , 2 × 0 , 1 × 0 , 1 . Then, differentiate F ς , τ , κ partially about the parameter κ , and it yields
F κ = 6 4 ς 2 1 τ 2 κ 12 τ 12 4 ς 2 τ 1 + ς 3 τ 4 ς 2 2 τ + 1 + ς 2 .
Taking F κ = 0 , we obtain
κ = ς 3 τ 4 ς 2 2 τ + 1 + ς 2 12 τ 12 4 ς 2 1 τ = κ 1 .
If κ 1 should belong to 0 , 1 , then it is possible only if
3 ς τ 4 ς 2 2 τ + 1 + ς 3 < 12 τ 12 4 ς 2 1 τ .
and
ς 2 > 4 .
At this stage, we will recognize the solution that fulfills both inequalities (24) and (25) as an optimal point.
Therefore we establish that ς 2 > 4 , and a simple analysis reveals that (24) is not valid for all τ 0 , 1 . As a result, no optimal point for F exists within 0 , 2 × 0 , 1 × 0 , 1 .
2. Interior of all the six faces of cuboid Ξ:
(i) if ς = 0 , we acquire
r 1 ( τ , κ ) = 576 τ 3 + 1 τ 2 κ 2 τ 1 2 + 2 τ = F ( 0 , τ , κ ) .
Differentiating r 1 ( τ , κ ) about the parameter κ , we have
r 1 κ = 1152 κ ( 1 τ 2 ) ( τ 1 ) 2 .
But, r 1 κ 0 for τ , κ 0 , 1 . As a result, no optimal point for r 1 exists within 0 , 1 × 0 , 1 .
(ii) when ς = 2 , we achieve
F ( 2 , τ , κ ) = 16 .
(iii) substituting τ = 0 , it yields
r 2 ( ς , κ ) = 1 4 ς 6 + 6 ς 3 κ ( 4 ς 2 ) + 36 κ 2 ( 4 ς 2 ) 2 = F ( ς , 0 , κ ) .
Differentiating r 2 ( ς , κ ) about the parameter κ , then about the parameter ς we have
r 2 κ = 6 ς 3 ( 4 ς 2 ) + 72 κ ( 4 ς 2 ) 2 .
and
r 2 ς = 3 2 ς 5 12 ς 4 κ + 18 ς 2 κ 4 ς 2 144 ς κ 2 ( 4 ς 2 ) .
A calculation shows that r 2 ( ς , κ ) has no optimal solution in 0 , 2 × 0 , 1 .
(iv) selecting τ = 1 , we have
r 3 ( ς , κ ) = 1 4 ς 6 + ( 4 ς 2 ) ( 4 ς 2 ) 45 4 ς 2 + 36 + 9 2 ς 4 = F ( ς , 1 , κ ) .
It is clear that
r 3 ς = 42 ς 5 144 ς 3 216 ς .
We see that r 3 ( ς ) < 0 for 0 , 2 . This implies that r 3 ( ς ) is a decreasing function of ς , so
r 3 ( ς ) 576 .
(v) If we choose κ = 0 , we find that
r 4 ( ς , τ ) = 1 4 ς 6 36 ς 4 τ 3 + 288 ς 2 τ 3 576 τ 3 + 33 4 ς 6 τ 2 78 ς 4 τ 2 + 180 ς 2 τ 2 3 2 ς 6 τ + 78 ς 4 τ 576 ς 2 τ + 1152 τ = F ( ς , τ , 0 ) .
Now, differentiating partially about the parameter ς , then about the parameter τ and simplifying we have
r 4 ς = 3 2 ς 5 144 ς 3 τ 3 + 576 ς τ 3 + 99 2 ς 5 τ 2 312 ς 3 τ 2 + 360 ς τ 2 9 ς 5 τ + 312 ς 3 τ 1152 ς τ .
and
r 4 τ = 108 ς 4 τ 2 + 864 ς 2 τ 2 1728 τ 2 + 33 2 ς 6 τ 156 ς 4 τ + 360 ς 2 τ 3 2 ς 6 + 78 ς 4 576 ς 2 + 1152 .
A numerical calculation shows that the solution does not exists for the system of equations
r 4 ς = 0 and r 4 τ = 0
in 0 , 2 × 0 , 1 .
(vi) Choosing κ = 1 , we obtain
r 5 ( ς , τ ) = 576 6 ς 5 576 τ 4 + 36 ς 4 + 1 4 ς 6 288 ς 2 + 576 τ 3 + 24 ς 3 288 ς 2 τ 3 + 36 ς 4 τ 3 + 33 4 ς 6 τ 2 3 2 ς 6 τ 36 ς 5 τ 4 18 ς 5 τ 3 + 42 ς 5 τ 2 + 288 ς 3 τ 4 + 18 ς 5 τ + 144 ς 3 τ 3 312 ς 3 τ 2 576 ς τ 4 144 ς 3 τ 288 ς τ 3 36 ς 4 τ 4 + 288 ς 2 τ 4 + 180 ς 2 τ 2 78 ς 4 τ 2 + 6 ς 4 τ + 288 ς τ + 576 ς τ 2 = F ( ς , τ , 1 ) .
Partial derivative of r 5 ( ς , τ ) about the parameter ς and then about the parameter τ , we have
r 5 ς = 432 ς 2 τ + 3 2 ς 5 576 τ 4 30 ς 4 + 72 ς 2 288 τ 3 + 576 τ 2 + 144 ς 3 + 432 ς 2 τ 3 90 ς 4 τ 3 + 99 2 ς 5 τ 2 144 ς 3 τ 4 9 ς 5 τ + 144 ς 3 τ 3 312 ς 3 τ 2 + 576 ς τ 4 + 24 ς 3 τ 576 ς τ 3 180 ς 4 τ 4 + 864 ς 2 τ 4 936 ς 2 τ 2 + 210 ς 4 τ 2 + 90 ς 4 τ + 360 ς τ 2 576 ς + 288 τ .
and
r 5 τ = 2304 τ 3 + 1728 τ 2 864 ς 2 τ 2 + 108 ς 4 τ 2 + 33 2 ς 6 τ 3 2 ς 6 144 ς 5 τ 3 54 ς 5 τ 2 + 84 ς 5 τ + 1152 ς 3 τ 3 + 18 ς 5 + 432 ς 3 τ 2 624 ς 3 τ 2304 ς τ 3 144 ς 3 864 ς τ 2 144 ς 4 τ 3 + 1152 ς 2 τ 3 + 360 ς 2 τ 156 ς 4 τ + 6 ς 4 + 288 ς + 1152 ς τ .
A simple computation illustrates that there exists a unique solution ς , τ 0.4196 , 0.7198 for the system of equations
r 5 ς = 0 and r 5 τ = 0 ,
in 0 , 2 × 0 , 1 . Thus, we have
F ( ς , τ , 1 ) 691.009 .
3. On the Edges of Cuboid Ξ:
(i) By setting τ = 0 and κ = 0 , it yields
F ( ς , 0 , 0 ) = 1 4 ς 6 16 .
(ii) By choosing τ = 0 and κ = 1 , we find that
r 6 ( ς ) = 1 4 ς 6 6 ς 5 + 24 ς 3 + 36 ς 4 288 ς 2 + 576 = F ( ς , 0 , 1 ) .
Differentiating r 6 ( ς ) about the parameter ς , we have
r 6 ( ς ) = 3 2 ς 5 30 ς 4 + 72 ς 2 + 144 ς 3 576 ς .
By simple computation, we indicates that r 6 ( ς ) achieves its maxima at 0 . Thus
r 6 ( ς ) 576 .
(iii) By selecting ς = 0 and τ = 0 , we obtain
r 7 κ = 576 κ 2 = F ( 0 , 0 , κ ) .
Since, r 7 ( κ ) > 0 for 0 , 1 . This shows that r 7 ( κ ) is a increasing function of κ , therefore
F ( 0 , 0 , κ ) = r 7 ( κ ) 576 .
(iv) We note that F ( ς , 1 , κ ) is free of κ , and we obtain
r 8 ( ς ) = F ( ς , 1 , 1 ) = F ( ς , 1 , 0 ) .
r 8 ( ς ) = 7 ς 6 36 ς 4 108 ς 2 + 576 .
By simple computation, we indicates that r 8 ( ς ) achieves its maxima at 0 . Thus
r 8 ( ς ) 576 .
(v) By taking ς = 0 and τ = 1 , it yields
F ( 0 , 1 , κ ) 576 .
(vi) By choosing ς = 2 , it becomes
F ( 2 , τ , κ ) = 16 .
We see that F ( 2 , τ , κ ) is free of κ , τ , ς . It follows
F ( 2 , 1 , κ ) = F ( 2 , τ , 1 ) = F ( 2 , τ , 0 ) = F ( 2 , 0 , κ ) = 16 .
(vii) By setting ς = 0 and κ = 1 , we achieve
F ( 0 , τ , 1 ) = 576 τ 4 + 576 τ 3 + 576 = r 9 ( τ ) .
It is clear that
r 9 ( τ ) = 2304 τ 3 + 1728 τ 2 .
Now, r 9 τ = 0 , gives τ 0.750 at which maxima is acquired for r 9 ( τ ) . Hence
F ( 0 , τ , 1 ) 2547 4 .
(viii) By substituting ς = 0 and κ = 0 , we find that
F ( 0 , τ , 0 ) = 576 τ 3 + 1152 τ = r 10 ( τ ) .
Differentiating r 10 ( τ ) about the parameter τ , we have
r 10 ( τ ) = 1728 τ 2 + 1152 .
Now r 10 τ = 0 , we obtain τ 0.8164 at which maxima is acquired for r 10 ( τ ) . Thus
F ( 0 , τ , 0 ) 256 2 3 .
Hence, from the above situations, we achieve
F ς , τ , κ 840959 1217 on 0 , 2 × 0 , 1 × 0 , 1 .
By using Equation (23), it follows
D 2 , 3 g 1 9216 F ς , τ , κ 840959 11215872 .
The required proof is completed. □
Conjecture 1.
If g S sinh , and has the form (1). Then
D 2 , 3 g 1 16 .
This inequality is sharp and equality will be attained from
2 ξ g ξ g ξ g ξ = 1 + ξ 3 1 6 ξ 9 + 3 40 ξ 15 + .
Theorem 9.
If g S sinh . Then
D 3 , 1 g 387691 4939776 .
Proof. 
The determinant D 3 , 1 g is described as follows:
D 3 , 1 g = 2 d 2 d 3 d 4 d 3 3 d 4 2 + d 3 d 5 d 2 2 d 5 .
Plugging (15)–(18), with ς 1 = ς we obtain
D 3 , 1 g = 1 9216 11 ς 6 105 ς 4 ς 2 + 120 ς 3 ς 3 + 189 ς 2 ς 2 2 216 ς 2 ς 4 + 72 ς ς 2 ς 3 216 ς 2 3 + 288 ς 2 ς 4 144 ς 3 2 .
Let m = 4 ς 2 in (9)–(11). Now, using these lemmas, we obtain
105 ς 4 ς 2 = 105 2 ς 6 + ς 4 m τ , 120 ς 3 ς 3 = 30 ς 4 m τ 2 + 60 ς 3 m 1 τ 2 σ + 60 ς 4 m τ + 30 ς 6 , 189 ς 2 ς 2 2 = 189 4 ς 6 + 189 2 ς 4 m τ + 189 4 ς 2 m 2 τ 2 , 216 ς 2 ς 4 = 108 1 τ 2 ς 2 τ ¯ σ 2 m 81 ς 4 m τ 2 + 108 1 τ 2 ς 3 σ m + 108 ς 2 m τ 2 + 27 ς 4 m τ 3 + 27 ς 6 + 108 1 σ 2 1 τ 2 ς 2 ζ m + 81 ς 4 m τ 108 1 τ 2 ς 3 m σ τ , 72 ς ς 2 ς 3 = 9 ς 2 m 2 τ 3 9 ς 4 m τ 2 + 18 ς τ m 2 1 τ 2 σ + 18 ς 2 τ 2 m 2 + 18 ς 3 m σ 1 τ 2 + 27 ς 4 m τ + 9 ς 6 , 216 ς 2 3 = 27 ς 6 + 81 ς 4 m τ + 81 ς 2 m 2 τ 2 + 27 m 3 τ 3 , 288 ς 2 ς 4 = 18 ς 6 + 18 ς 4 m τ 3 + 72 ς 4 m τ + 72 ς 2 m τ 2 72 1 τ 2 m τ ¯ σ 2 ς 2 72 1 τ 2 ς 3 m σ τ 54 ς 4 m τ 2 + 72 1 σ 2 1 τ 2 ς 2 ζ m + 18 ς 2 m 2 τ 4 54 ς 2 m 2 τ 3 + 72 1 τ 2 ς 3 σ m + 54 ς 2 m 2 τ 2 + 72 m 2 τ 3 72 ς m 2 τ 2 1 τ 2 σ 72 τ m 2 τ ¯ 1 τ 2 σ 2 + 72 1 σ 2 1 τ 2 m 2 ζ τ + 72 1 τ 2 m 2 τ σ ς , 144 ς 3 2 = 36 1 τ 2 2 m 2 σ 2 + 9 ς 2 m 2 τ 4 36 1 τ 2 m 2 τ 2 σ ς 18 ς 4 m τ 2 + 36 ς 2 m 2 τ 2 36 ς 2 m 2 τ 3 + 72 1 τ 2 m 2 τ σ ς + 36 ς 4 m τ + 9 ς 6 + 36 1 τ 2 ς 3 σ m .
Plugging the above expressions in (26), we obtain
D 3 , 1 g = 1 9216 6 ς 4 τ 2 m + 3 ς 4 τ m 36 ς 2 τ 2 m 9 ς 4 τ 3 m 27 ς 2 τ 3 m 2 + 9 ς 2 m 2 τ 4 36 m 2 1 τ 2 2 σ 2 27 τ 3 m 3 + 72 τ 3 m 2 + 36 ς 3 τ m 1 τ 2 σ + 36 1 τ 2 m τ ¯ σ 2 ς 2 36 1 σ 2 1 τ 2 ς 2 ζ m + 18 ς τ m 2 1 τ 2 σ 36 1 τ 2 τ 2 m 2 ς σ 72 1 τ 2 τ m 2 τ ¯ σ 2 1 4 ς 6 + 72 1 σ 2 1 τ 2 m 2 τ ζ + 6 1 τ 2 ς 3 m σ + 9 4 ς 2 τ 2 m 2 .
Since m = 4 ς 2 ,
D 3 , 1 g = 1 9216 N 0 ς , τ + N 1 ς , τ σ + N 2 ς , τ σ 2 + χ ς , τ , σ ζ ,
where σ , τ , ζ D ¯ , and
N 0 ς , τ = 1 4 ς 6 + 4 ς 2 4 ς 2 9 ς 2 τ 4 36 τ 3 + 9 4 ς 2 τ 2 + 6 ς 4 τ 2 + 3 ς 4 τ 36 ς 2 τ 2 9 ς 4 τ 3 , N 1 ς , τ = 4 ς 2 1 τ 2 4 ς 2 18 ς τ 36 ς τ 2 + 36 ς 3 τ + 6 ς 3 , N 2 ς , τ = 4 ς 2 1 τ 2 4 ς 2 36 τ 2 36 + 36 ς 2 τ ¯ , χ ς , τ , σ = 4 ς 2 1 τ 2 1 σ 2 36 ς 2 + 72 τ 4 ς 2 .
By replacing τ by τ and σ by κ , if we apply the statement ζ 1 , and it follows
D 3 , 1 g 1 9216 N 0 ς , τ + N 1 ς , τ κ + N 2 ς , τ κ 2 + χ ς , τ , σ . 1 9216 E ς , τ , κ ,
where
E ς , τ , κ = c 0 ς , τ + c 1 ς , τ κ + c 2 ς , τ κ 2 + c 3 ς , τ 1 κ 2 ,
with
c 0 ς , τ = 1 4 ς 6 + 4 ς 2 4 ς 2 9 ς 2 τ 4 + 36 τ 3 + 9 4 ς 2 τ 2 + 6 ς 4 τ 2 + 3 ς 4 τ + 36 ς 2 τ 2 + 9 ς 4 τ 3 , c 1 ς , τ = 4 ς 2 1 τ 2 4 ς 2 18 ς τ + 36 ς τ 2 + 36 ς 3 τ + 6 ς 3 , c 2 ς , τ = 4 ς 2 1 τ 2 4 ς 2 36 τ 2 + 36 + 36 ς 2 τ , c 3 ς , τ = 4 ς 2 1 τ 2 36 ς 2 + 72 4 ς 2 τ .
Now, we are to maximize E ς , τ , κ in the closed cuboid Ξ : 0 , 2 × 0 , 1 × 0 , 1 .
For this purpose, we have to find the maxima of E ς , τ , κ in the interior of Ξ , in the interior of its faces and on the edges.
1. Interior points of cuboid Ξ:
Suppose ς , τ , κ 0 , 2 × 0 , 1 × 0 , 1 . Then, on differentiating E ς , τ , κ partially about the parameter κ , it yields
E κ = 6 4 ς 2 ( 1 τ 2 ) κ 12 ( τ 1 ) 4 ς 2 τ 1 + ς 2 + ς 3 τ 4 ς 2 2 τ + 1 + ς 2 6 τ + 1 .
Taking E κ = 0 , gives
κ = ς 3 τ 4 ς 2 2 τ + 1 + ς 2 6 τ + 1 12 ( τ 1 ) 4 ς 2 1 τ ς 2 = κ .
If κ should belong to 0 , 1 , then it is possible only if
ς 3 6 τ + 1 + 3 ς τ 4 ς 2 2 τ + 1 + 12 1 τ 2 4 ς 2 < 12 ( 1 τ ) ς 2 .
and
ς 2 > 4 1 τ 2 τ .
Now, only that solution which meets both the inequalities (28) and (29) will be accepted as a critical point.
Suppose U ( τ ) = 4 1 τ 2 τ . Thus, U ( τ ) is decreasing over 0 , 1 . Thus ς 2 > 0 , and a straightforward task illustrates that (28) is not hold for all values of τ 0 , 1 . As a result, no optimal point for E exists within 0 , 2 × 0 , 1 × 0 , 1 .
2. Interior of all the six faces of cuboid Ξ:
(i) Choosing ς = 0 , we achieve
e 1 ( τ , κ ) = 576 τ 3 + ( 1 τ 2 ) ( κ 2 τ 1 2 + 2 τ ) = E ( 0 , τ , κ ) .
Differentiate e 1 ( τ , κ ) partially about the parameter κ , and we obtain
e 1 κ = 1152 κ ( 1 τ 2 ) ( τ 1 ) 2 .
But, e 1 κ 0 for τ , κ 0 , 1 . Hence, we have found no maxima for E ( 0 , τ , κ ) in 0 , 1 × 0 , 1 .
(ii) Setting ς = 2 , we have
E ( 2 , τ , κ ) = 16 .
(iii) Taking τ = 0 , we obtain
E ( ς , 0 , κ ) = e 2 ( ς , κ ) = 1 4 ς 6 + ( 4 ς 2 ) 6 ς 3 κ 72 ς 2 κ 2 + 36 ς 2 + 144 κ 2 .
Putting e 2 κ = 0 , we achieve
κ = ς 3 24 ς 2 2 = κ 0 .
κ 0 must lie within 0 , 1 for the provided range of κ , that is only achievable if ς > ς 0 , ς 0 1.4592733 . Also, substituting the value of κ in e 2 ς = 0 and simplifying we obtain
e 2 ς = 3 ς 9 584 ς 7 + 3456 ς 5 6912 ς 3 + 4608 ς = 0 .
The solution to (30) can be found using computation, which is ς 1.3069302 in the interval 0 , 2 . As a result, there is no optimal point for e 2 ( ς , κ ) in 0 , 2 × 0 , 1 .
(iv) Considering τ = 1 , we have
e 3 ς , κ = 1 4 ς 6 + ( 4 ς 2 ) ( 4 ς 2 ) ( 36 + 45 4 ς 2 ) + 36 ς 2 + 18 ς 4 = E ( ς , 1 , κ ) .
Then
e 3 ς = 39 ς 5 72 ς 3 + 72 ς .
By taking e 3 ς = 0 , we achieve ς 0.848263 at which e 3 ς , κ attains its maxima, that is
e 3 ( ς , κ ) 590.162 .
(v) If we choose κ = 0 , we find that
e 4 ( ς , τ ) = 9 ς 6 τ 4 72 ς 4 τ 4 + 144 ς 2 τ 4 + 288 ς 2 τ 3 576 τ 3 15 4 ς 6 τ 2 + 6 ς 4 τ 2 + 36 ς 2 τ 2 36 ς 4 + 1 4 ς 6 9 ς 6 τ 3 3 ς 6 τ + 84 ς 4 τ 576 ς 2 τ + 144 ς 2 + 1152 τ = E ( ς , τ , 0 ) .
Now, differentiating partially about the parameter ς , then about the parameter τ and simplifying, we have
e 4 ς = 3 2 ς 5 + 54 ς 5 τ 4 288 ς 3 τ 4 + 288 ς τ 4 + 576 ς τ 3 45 2 ς 5 τ 2 + 24 ς 3 τ 2 + 72 ς τ 2 54 ς 5 τ 3 18 ς 5 τ + 336 ς 3 τ 144 ς 3 1152 ς τ + 288 ς .
and
e 4 τ = 36 ς 6 τ 3 288 ς 4 τ 3 + 576 ς 2 τ 3 + 864 ς 2 τ 2 1728 τ 2 15 2 ς 6 τ + 12 ς 4 τ + 72 ς 2 τ 27 ς 6 τ 2 3 ς 6 + 84 ς 4 576 ς 2 + 1152 .
From computation, we find that the solution does not exists for the system of equations
e 4 ς = 0 and e 4 τ = 0
in 0 , 2 × 0 , 1 .
(vi) Taking κ = 1 , we obtain
e 5 ( ς , τ ) = 576 + 9 ς 6 τ 4 15 4 ς 6 τ 2 9 ς 6 τ 3 3 ς 6 τ 36 ς 5 τ 4 + 18 ς 5 τ 3 + 42 ς 5 τ 2 + 288 ς 3 τ 4 18 ς 5 τ 312 ς 3 τ 2 576 ς τ 4 288 ς τ 3 6 ς 5 108 ς 4 τ 4 432 ς 2 τ 3 576 τ 4 + 36 ς 4 + 1 4 ς 6 288 ς 2 + 576 τ 3 + 24 ς 3 + 432 ς 2 τ 4 + 180 ς 2 τ 2 30 ς 4 τ 2 + 108 ς 4 τ 3 24 ς 4 τ + 288 ς τ + 576 ς τ 2 + 144 ς 2 τ = E ( ς , τ , 1 ) .
Partial derivative of e 5 ( ς , τ ) about the parameter ς and then about the parameter τ , and we have
e 5 ς = 432 ς 3 τ 3 + 54 ς 5 τ 4 54 ς 5 τ 3 45 2 ς 5 τ 2 432 ς 3 τ 4 18 ς 5 τ 120 ς 3 τ 2 + 864 ς τ 4 864 ς τ 3 + 3 2 ς 5 180 ς 4 τ 4 576 τ 4 30 ς 4 + 72 ς 2 288 τ 3 + 576 τ 2 + 144 ς 3 + 864 ς 2 τ 4 936 ς 2 τ 2 + 210 ς 4 τ 2 + 90 ς 4 τ 3 90 ς 4 τ + 288 ς τ + 360 ς τ 2 96 ς 3 τ 576 ς + 288 τ .
and
e 5 τ = 36 ς 6 τ 3 15 2 ς 6 τ 27 ς 6 τ 2 3 ς 6 144 ς 5 τ 3 + 54 ς 5 τ 2 + 84 ς 5 τ + 1152 ς 3 τ 3 18 ς 5 624 ς 3 τ 2304 ς τ 3 864 ς τ 2 432 ς 4 τ 3 1296 ς 2 τ 2 2304 τ 3 + 1728 τ 2 + 1728 ς 2 τ 3 + 360 ς 2 τ 60 ς 4 τ + 324 ς 4 τ 2 24 ς 4 + 288 ς + 1152 ς τ + 144 ς 2 .
A simple computation illustrates that there exists a unique solution ς , τ 0.6652 , 0.7409 for the system of equations
e 5 ς = 0 and e 5 τ = 0 ,
in 0 , 2 × 0 , 1 . Hence
E ( ς , τ , 1 ) = e 5 ς , τ 723.304 .
3. On the Edges of Cuboid Ξ:
(i) By selecting τ = 0 and κ = 0 , we find that
E ( ς , 0 , 0 ) = 1 4 ς 6 36 ς 4 + 144 ς 2 = e 6 ( ς ) .
Differentiating e 6 ( ς ) about the parameter ς , we have
e 6 ( ς ) = 3 2 ς 5 144 ς 3 + 288 ς .
We note that e 6 ( ς ) = 0 for the critical point ς 1.4295 at which e 6 ( ς ) obtains its maxima. Thus
e 6 ( ς ) 146.06 .
(ii) By substituting τ = 0 and κ = 1 , it yields
E ( ς , 0 , 1 ) = 1 4 ς 6 6 ς 5 + 24 ς 3 + 36 ς 4 288 ς 2 + 576 = e 7 ( ς ) .
Differentiating e 7 ( ς ) about the parameter ς , we have
e 7 ( ς ) = 3 2 ς 5 30 ς 4 + 72 ς 2 + 144 ς 3 576 ς .
As e 7 ( ς ) < 0 for 0 , 2 . This implies that e 7 ( ς ) is a decreasing function of ς , so
E ( ς , 0 , 1 ) = e 7 ( ς ) 576 .
(iii) By choosing ς = 0 and τ = 0 , we obtain
e 8 κ = 576 κ 2 = E ( 0 , 0 , κ ) .
Since e 8 ( κ ) > 0 for 0 , 1 . This reveals that e 8 ( κ ) is increasing and maxima is acquired when κ = 1 . Then
e 8 ( κ ) 576 .
(iv) We note that E ( ς , 1 , κ ) is free of κ . It follows
e 9 ( ς ) = E ( ς , 1 , 1 ) = E ( ς , 1 , 0 ) .
e 9 ( ς ) = 13 2 ς 6 18 ς 4 + 36 ς 2 + 576 .
Differentiate e 9 ( ς ) partially about the parameter ς , and we obtain
e 9 ( ς ) = 39 ς 5 72 ς 3 + 72 ς .
By taking e 9 ( ς ) = 0 we achieve ς 0.8482 at which e 9 ( ς ) achieves its maxima. Thus
e 9 ( ς ) 590.16 .
(v) By selecting ς = 0 and τ = 1 , we achieve
E ( 0 , 1 , κ ) = 576 .
(vi) By taking ς = 2 , it yields
E ( 2 , τ , κ ) = 16 .
We see that E ( 2 , τ , κ ) is free of κ , τ , ς . It follows
E ( 2 , 1 , κ ) = E ( 2 , τ , 1 ) = E ( 2 , τ , 0 ) = E ( 2 , 0 , κ ) = 16 .
(vii) By substituting ς = 0 and κ = 1 , we find that
e 10 ( τ ) = 576 τ 4 + 576 τ 3 + 576 = E 0 , τ , 1 .
It follows that
e 10 ( τ ) = 2304 τ 3 + 1728 τ 2 .
Now, e 10 τ = 0 , we achieve τ 0.75 at which e 10 τ achieves its maxima. Hence
e 10 ( τ ) 2574 4 .
(viii) By taking ς = 0 and κ = 0 , we have
E 0 , τ , 0 = 576 τ 3 + 1152 τ = e 11 ( τ ) .
Clearly,
e 11 ( τ ) = 1728 τ 2 + 1152 .
We know that e 11 τ = 0 gives τ 0.816 at which e 11 τ obtain its maximum value, which is given by
E 0 , τ , 0 = e 11 τ 256 2 3 .
Hence, from the above situations, we achieve
E ς , τ , κ 387691 536 on 0 , 2 × 0 , 1 × 0 , 1 .
By using Equation (27), it follows
D 3 , 1 g 1 9216 E ς , τ , κ 387691 4939776 .
Thus, we have completed the proof. □
Conjecture 2.
If g S sinh , and has the form (1). Then
D 3 , 1 g 1 16 .
This inequality is sharp and equality will be attained from
2 ξ g ξ g ξ g ξ = 1 + ξ 3 1 6 ξ 9 + 3 40 ξ 15 + .

8. Conclusions

In this work, we explore symmetric starlike functions linked to a petal-shaped domain. We have established the sharp coefficient inequalities for the above-mentioned class. The discussed coefficient inequalities include the Krushkal, Zalcman, and Fekete-Szegö inequalities, as well as the second and third Hankel determinants. We present several conjectures derived from our estimated problems, which are well supported by our findings. The sharpness of the results and the proposed conjectures distinguish this work from existing studies. This research tries to increase focus on coefficient-related problems involving inverse and logarithmic inverse functions for specific classes of analytic functions. For example, researchers can showcase their skills in various directions [50,51,52], including Toeplitz determinants, majorization results, convolution properties, and partial sum inequalities.

Author Contributions

Conceptualization: R.K.A.; Formal analysis: M.A. (Muhammad Abbas) and M.A. (Muhammad Arif); Funding acquisition: D.B.; Investigation: M.A. (Muhammad Abbas) and M.A. (Muhammad Arif); Supervision: D.B. and M.A. (Muhammad Arif); Writing—original draft: M.A. (Muhammad Abbas); Writing—review and editing, M.A. (Muhammad Arif). All authors have read and agreed to the published version of the manuscript.

Funding

This research received no specific funding.

Data Availability Statement

No new data were created in this study.

Acknowledgments

The author Reem K. Alhefthi would like to extend their sincere appreciation to the Researchers Supporting Project Number (RSPD2025R802) of King Saud University, Riyadh, Saudi Arabia.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

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MDPI and ACS Style

Abbas, M.; Alhefthi, R.K.; Breaz, D.; Arif, M. On Coefficient Inequalities for Functions of Symmetric Starlike Related to a Petal-Shaped Domain. Axioms 2025, 14, 165. https://doi.org/10.3390/axioms14030165

AMA Style

Abbas M, Alhefthi RK, Breaz D, Arif M. On Coefficient Inequalities for Functions of Symmetric Starlike Related to a Petal-Shaped Domain. Axioms. 2025; 14(3):165. https://doi.org/10.3390/axioms14030165

Chicago/Turabian Style

Abbas, Muhammad, Reem K. Alhefthi, Daniel Breaz, and Muhammad Arif. 2025. "On Coefficient Inequalities for Functions of Symmetric Starlike Related to a Petal-Shaped Domain" Axioms 14, no. 3: 165. https://doi.org/10.3390/axioms14030165

APA Style

Abbas, M., Alhefthi, R. K., Breaz, D., & Arif, M. (2025). On Coefficient Inequalities for Functions of Symmetric Starlike Related to a Petal-Shaped Domain. Axioms, 14(3), 165. https://doi.org/10.3390/axioms14030165

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