Bounds for the Hyperbolic Sombor Index in Terms of the Variable Euler–Sombor Index and Relations to Other Degree-Based Invariants

1. Introduction

Topological indices play a significant role in chemistry, pharmacology, and particularly in QSPR/QSAR research, where numerous such indices have been developed to predict the physicochemical properties of molecular structures. Over the past few decades, a series of vertex-degree-based topological indices have been introduced, and they have proven highly effective in predicting molecular properties. Let $G=\left(V\right(G),E(G\left)\right)$ be a simple graph with n vertices and m edges. For a vertex $u\in V\left(G\right)$, denote its degree by $d_G\left(u\right)$. Let $\delta \left(G\right)$ and $\Delta \left(G\right)$ be the minimum and maximum degrees of G, respectively. When the graph is clear from the context, we simplify the notation to $d_u$, $\delta$ and $\Delta$. Degree-based topological indices of a graph G, denoted by $TI\left(G\right)$, are generally defined as

$$TI\left(G\right)=\sum _{uv\in E\left(G\right)}\psi (d_u,d_v),$$

where $\psi (x,y)\ge 0$ is a real-valued function satisfying the symmetric property $\psi (x,y)=\psi (y,x)$. A large number of vertex-degree-based topological indices have been introduced and studied extensively [16,28,48].

In this paper, we mainly focus on a very recently proposed index, named the Hyperbolic Sombor index $HSO\left(G\right)$, for a graph G. It was introduced by Barman and Das [7] and defined as

$$HSO\left(G\right)=\sum _{uv\in E\left(G\right)}{\displaystyle \frac{\sqrt{d_u^2+d_v^2}}{min\{d_u,d_v\}}}.$$

Among the many topological molecular descriptors that have been proposed, we only recall those that will be needed in the remainder of this text.

The first Zagreb index $M_1\left(G\right)$ and second Zagreb index $M_2\left(G\right)$ are among the oldest degree-based topological descriptors [15], defined respectively as

$$M_1\left(G\right)=\sum _{uv\in E\left(G\right)}(d_u+d_v)=\sum _{v\in V\left(G\right)}{d}_v^2$$

and

$$M_2\left(G\right)=\sum _{uv\in E\left(G\right)}{d}_u·d_v.$$

In the study of vertex-degree-based topological indices, Gutman [18] proposed a novel framework motivated by the geometric interpretation of the Euclidean distance from the origin to the point $(d_u,d_v)$. Based on this concept, the Sombor index $SO\left(G\right)$ was defined and quickly became one of the most widely studied topological indices in recent years. Its chemical applications can be found in [31,39], while mathematical properties and extremal graph results are surveyed in [17,32]. Alongside the Sombor index, the Elliptic Sombor index $ESO\left(G\right)$ [21], the Euler–Sombor index $EU\left(G\right)$ [20,47], and a variable version of the Euler-Sombor index $EU(\lambda ,G)$ [22] for a real number $\lambda$ were proposed. These indices are defined as follows:

$$SO\left(G\right)=\sum _{uv\in E\left(G\right)}\sqrt{d_u^2+d_v^2};$$

$$ESO\left(G\right)=\sum _{uv\in E\left(G\right)}\left(d_u+d_v\right)\sqrt{d_u^2+d_v^2};$$

$$EU\left(G\right)=\sum _{uv\in E\left(G\right)}\sqrt{d_u^2+d_v^2+d_u{d}_v};$$

$$EU(\lambda ,G)=\sum _{uv\in E}\sqrt{d_u^2+d_v^2+\lambda d_u{d}_v}.$$

Note that $SO\left(G\right)=EU(0,G)$, $EU\left(G\right)=EU(1,G)$ and $M_1\left(G\right)=EU(2,G)$. When $\lambda =-2$, it gives the oldest graph-theoretical irregularity measure, the Albertson index [4], denoted $Alb\left(G\right)$, which is defined as

$$Alb\left(G\right)=\sum _{uv\in E\left(G\right)}|d_u-d_v|.$$

The Forgotten Sombor index $FSO$ [29], which arises from the Euclidean distance from the origin to the squared degree pair $(d_u^2,d_v^2)$, is also considered in this paper:

$$FSO\left(G\right)=\sum _{uv\in E\left(G\right)}\sqrt{d_u^4+d_v^4}.$$

The chemical relevance of these vertex-degree-based topological indices, their comparative inequalities with other indices, and the characterization of extremal graphs for these indices have been widely investigated in recent years [1,2,3,8,9,10,11,12,13,14,19,23,25,26,27,30,33,34,35,37,38,40,41,42,43,44,45,46,49]. For a comprehensive understanding of Sombor-type degree-based topological indices, please refer to the survey [24] by Gutman.

In this paper, we first establish upper and lower bounds for the Hyperbolic Sombor index $HSO\left(G\right)$ of a graph G in terms of the variable Euler-Sombor index $EU(\lambda ,G)$. Furthermore, we provide upper and lower bounds for $HSO\left(G\right)$ in relation to several other degree-based topological invariants, namely the second Zagreb index $M_2\left(G\right)$, the Elliptic Sombor index $ESO\left(G\right)$, and the Forgotten Sombor index $FSO\left(G\right)$.

2. Bounds for $\mathit{HSO}\left(\mathit{G}\right)$ via the Variable Euler–Sombor Index $\mathit{EU}(\mathit{\lambda },\mathit{G})$

In the study of the variable Euler-Sombor index, Gutman et al. [22] pointed out that, in the general case, $EU\left(\lambda \right)$ is well-defined (i.e., real-valued) only for $\lambda \ge -2$.

Moreover‌, a second-degree polynomial approximation was obtained:

$$EU\left(\lambda \right)\approx {\displaystyle \frac{{\lambda }^2}{8}}(M_1+Alb-2SO)+{\displaystyle \frac{\lambda }{4}}(M_1-Alb)+SO.$$

The approximation is best applicable for $\lambda \in [-2,2]$.

In this section, we establish upper and lower bounds for the Hyperbolic Sombor index in terms of the variable Euler-Sombor index when $\lambda \ge -2$.

2.1. Preliminaries

We first introduce Ferrari’s method [5] for solving the quartic equation $a{x}^4+b{x}^3+c{x}^2+dx+e=0$. Define

$$P={\displaystyle \frac{c^2+12ae-3bd}{9}},Q={\displaystyle \frac{27a{d}^2+2c^3+27b^{2}e-72ace-9bcd}{54}},$$

$$D=\sqrt{Q^2-P^3},$$

$$u=\sqrt[3]{Q+D}\mathrm{or}u=\sqrt[3]{Q-D}\left(\mathrm{choose}\mathrm{the}\mathrm{one}\mathrm{with}\mathrm{larger}\mathrm{modulus}\right),$$

$$v={\displaystyle \frac{P}{u}}(\mathrm{set}v=0\mathrm{if}u=0),\omega =-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{\sqrt{3}}{2}}i,$$

$$m=\sqrt{b^2-{\displaystyle \frac{8}{3}}ac+4a\left({\omega }^{k-1}u+{\omega }^{4-k}v\right)},$$

$$S=2b^2-{\displaystyle \frac{16}{3}}ac-4a\left({\omega }^{k-1}u+{\omega }^{4-k}v\right),$$

$$T={\displaystyle \frac{8abc-16a^{2}d-2b^3}{m}}.$$

In the above three formulas for m, S and T, k can take values $1,2,3$. We should choose the triple $(m,S,T)$ such that $\left|m\right|$ is maximized. If all three $\left|m\right|$ are zero, then instead take $(m,S,T)=\left(0,b^2-{\displaystyle \frac{8}{3}}ac,0\right)$. Then the roots of the quartic equations are given as follows by Ferrari’s method.

Lemma 1. 

[5] The four roots of the quartic equation $a{x}^4+b{x}^3+c{x}^2+dx+e=0$ are given by

$$x_n={\displaystyle \frac{-b+{(-1)}^{\lceil n/2\rceil }m+{(-1)}^{n+1}\sqrt{S+{(-1)}^{\lceil n/2\rceil }T}}{4a}},$$

where $n=1,2,3,4$, and $(m,S,T)$ is the triple chosen above.

Next we will give the roots of the quartic equation $2u^4+\lambda u^3+4u^2+3\lambda u+2=0$ under certain conditions. According to Ferrari’s method, we can define the following expressions:

$$P={\displaystyle \frac{64}{9}}-{\lambda }^2,Q=8{\lambda }^2-{\displaystyle \frac{512}{27}},D=-\lambda \sqrt{{\lambda }^4+{\displaystyle \frac{128}{3}}{\lambda }^2-{\displaystyle \frac{4096}{27}}},$$

$$u=\sqrt[3]{Q+D},v={\displaystyle \frac{P}{u}},w=u+v,$$

$$m=\sqrt{{\lambda }^2-{\displaystyle \frac{64}{3}}+8w},S=2{\lambda }^2-{\displaystyle \frac{128}{3}}-8w,T=-{\displaystyle \frac{2\lambda ({\lambda }^2+64)}{m}}.$$

Lemma 2. 

Let the parameter $\lambda \in (-2,{\lambda }_0)$, where ${\lambda }_0=-{\displaystyle \frac{8}{3}}\sqrt{2\sqrt{3}-3}\approx -1.8167$. Then the quartic equation

$$2u^4+\lambda u^3+4u^2+3\lambda u+2=0$$

has exactly two real roots $u_a$ and $u_b$ in $(0,1)$, which can be expressed as

$$u_a=-{\displaystyle \frac{\lambda +m-\sqrt{S+T}}{8}},u_b=-{\displaystyle \frac{\lambda +m+\sqrt{S+T}}{8}}.$$

where $u_a<u_b$.

Proof. 

The equation is of the form $a{u}^4+b{u}^3+c{u}^2+du+e=0$ with

$$a=2,b=\lambda ,c=4,d=3\lambda ,e=2.$$

Introduce the intermediate quantities

$$P={\displaystyle \frac{c^2+12ae-3bd}{9}}={\displaystyle \frac{64}{9}}-{\lambda }^2,$$

$$Q={\displaystyle \frac{27a{d}^2+2c^3+27b^{2}e-72ace-9bcd}{54}}=8{\lambda }^2-{\displaystyle \frac{512}{27}}.$$

Then

$$Q^2=64{\lambda }^4-{\displaystyle \frac{8192}{27}}{\lambda }^2+{\displaystyle \frac{262144}{729}},$$

$$P^3={\displaystyle \frac{262144}{729}}-{\displaystyle \frac{4096}{27}}{\lambda }^2+{\displaystyle \frac{64}{3}}{\lambda }^4-{\lambda }^6.$$

Subtracting, we have

$$Q^2-P^3={\lambda }^6+{\displaystyle \frac{128}{3}}{\lambda }^4-{\displaystyle \frac{4096}{27}}{\lambda }^2={\lambda }^2\left({\lambda }^4+{\displaystyle \frac{128}{3}}{\lambda }^2-{\displaystyle \frac{4096}{27}}\right).$$

For $\lambda \in (-2,{\lambda }_0)$, set $t={\lambda }^2$. The expression in parentheses becomes the quadratic

$$f\left(t\right)=t^2+{\displaystyle \frac{128}{3}}t-{\displaystyle \frac{4096}{27}},$$

which has two real roots, the positive one is

$$t_0={\displaystyle \frac{64}{9}}(2\sqrt{3}-3)={\lambda }_0^2.$$

Since ${\lambda }^2>{\lambda }_0^2=t_0$, we have $f\left({\lambda }^2\right)>0$, hence $Q^2-P^3>0$. Moreover, $\lambda <0$, so

$$D=\sqrt{Q^2-P^3}=-\lambda \sqrt{{\lambda }^4+{\displaystyle \frac{128}{3}}{\lambda }^2-{\displaystyle \frac{4096}{27}}}>0.$$

Because $D=\sqrt{Q^2-P^3}<Q$, we get $Q-D>0$ and

$$Q+D>Q-D>0$$

Hence we set

$$u=\sqrt[3]{Q+D},v={\displaystyle \frac{P}{u}}={\displaystyle \frac{\frac{64}{9}-{\lambda }^2}{u}}.$$

Introduce the primitive cube root of unity $\omega =-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{\sqrt{3}}{2}}i$. To obtain real coefficients in the subsequent expressions we choose $k=1$, then

$${\omega }^{k-1}u+{\omega }^{4-k}v=u+v.$$

Denote ${\omega }_1=u+v$. Consequently,

$$m=\sqrt{b^2-{\displaystyle \frac{8}{3}}ac+4a({\omega }^{k-1}u+{\omega }^{4-k}v)}=\sqrt{{\lambda }^2-{\displaystyle \frac{64}{3}}+8{\omega }_1},$$

$$S=2b^2-{\displaystyle \frac{16}{3}}ac-4a({\omega }^{k-1}u+{\omega }^{4-k}v)=2{\lambda }^2-{\displaystyle \frac{128}{3}}-8{\omega }_1,$$

$$T={\displaystyle \frac{8abc-16a^{2}d-2b^3}{m}}={\displaystyle \frac{-2\lambda ({\lambda }^2+64)}{m}}.$$

From $Q^2-P^3>0$ we have $Q>P^{3/2}$. Also $P,Q,D>0$, so

$$Q+D>Q>P^{3/2},$$

which implies

$$u=\sqrt[3]{Q+D}>\sqrt{P}.$$

By the AM–GM inequality,

$${\omega }_1=u+v=u+{\displaystyle \frac{P}{u}}>2\sqrt{P}.$$

Set $t={\lambda }^2\in ({\lambda }_0^2,4)$. Then $P={\displaystyle \frac{64}{9}}-t$ and

$$t-{\displaystyle \frac{64}{3}}+8{\omega }_1>t-{\displaystyle \frac{64}{3}}+16\sqrt{{\displaystyle \frac{64}{9}}-t}=f\left(t\right).$$

Since

$$f^{\prime }\left(t\right)=1-{\displaystyle \frac{8}{\sqrt{\frac{64}{9}-t}}}<0,$$

for $t\in ({\lambda }_0^2,4)$ we have $f\left(t\right)>f\left(4\right)=4-{\displaystyle \frac{64}{3}}+16\sqrt{{\displaystyle \frac{28}{9}}}>0$. So $m^2>0$, and consequently $m>0$. Then $T>0$.

Claim 1. The equation $2u^4+2u^3+4u^2+3u+2=0$ has exactly two real roots $u_a$ and $u_b$ in $(0,1)$, with $u_a<u_b$.

Proof. In fact, let $p\left(u\right)=2u^4+\lambda u^3+4u^2+3\lambda u+2$, $u\in (0,1]$. Then

$$p\left(0\right)=2>0,p\left(1\right)=8+4\lambda >0,$$

$$p^{\prime }\left(u\right)=8u^3+3\lambda u^2+8u+3\lambda ,$$

$$p^{\prime \prime }\left(u\right)=24u^2+6\lambda u+8.$$

The discriminant of this quadratic is $36{\lambda }^2-768<0$. Hence $p^{\prime }\left(u\right)$ is strictly increasing on $(0,1]$. Since $p^{\prime }\left(0\right)=3\lambda <0$ and $p^{\prime }\left(1\right)=16+6\lambda >0$, there exists $u_0\in (0,1)$ such that $p^{\prime }\left(u_0\right)=0$. Therefore $p\left(u\right)$ first decreases and then increases on $(0,1]$. Setting $p^{\prime }\left(u_0\right)=0$ gives $\lambda =-{\displaystyle \frac{8}{3}}{u}_0$. Substituting into $p\left(u_0\right)$ yields the minimum value $p_{min}$:

$$p_{min}=-{\displaystyle \frac{2}{3}}{u}_0^4-4u_0^2+2.$$

Let $t=u_0^2\in (0,1)$. Then

$$p_{min}=-{\displaystyle \frac{2}{3}}{t}^2-4t+2.$$

Consider the equation

$$-{\displaystyle \frac{2}{3}}{t}^2-4t+2=0.$$

The positive root is $t=2\sqrt{3}-3$. Hence

$$u_0=\sqrt{2\sqrt{3}-3},{\lambda }_0=-{\displaystyle \frac{8}{3}}\sqrt{2\sqrt{3}-3}\approx -1.8167.$$

Therefore,we have

• If $\lambda \in [{\lambda }_0,0)$, then $p_{min}\ge 0$, so $p\left(u\right)\ge 0$ for all $u\in (0,1]$.
• If $\lambda \in (-2,{\lambda }_0)$, then $p_{min}<0$.

Together with $p\left(0\right)>0$ and $p\left(1\right)>0$, when $\lambda \in (-2,{\lambda }_0)$ the equation $p\left(u\right)=0$ has exactly two real roots in $(0,1)$, denoted $0<u_a<u_b<1$. □

Claim 2. The equation $2u^4+\lambda u^3+4u^2+3\lambda u+2=0$ has no real roots for $u\le 0$ and $u\ge 1$.

Proof. In fact, consider $p^{\prime }\left(u\right)$, we have

$$p^{\prime }\left(u\right)=8u^3+3\lambda u^2+8u+3\lambda =(u^2+1)(8u+3\lambda ).$$

If $u\le 0$, set $u=-t(t\ge 0)$, then $p(-t)=2t^4-\lambda t^3+4t^2-3\lambda t+2$. Because $\lambda <0$, all coefficients are positive, so $p(-t)>0$ identically. Thus no real root for $u\le 0$. If $u\ge 1$, then $\lambda \in (-2,{\lambda }_0)$ implies $3\lambda >-6$, hence $8u+3\lambda >2$. Thus $p^{\prime }\left(u\right)>0$ and $p\left(u\right)\ge p\left(1\right)>0$, so no real root for $u\ge 1$. □

Therefore the quartic possesses exactly two real roots (and two complex roots). By Ferrari’s method, the four roots are

$$x_{1,2}={\displaystyle \frac{-\lambda -m\pm \sqrt{S-T}}{8}},x_{3,4}={\displaystyle \frac{-\lambda +m\pm \sqrt{S+T}}{8}}.$$

Since $T>0$, clearly $S+T>S-T$. Because there are exactly two real roots, we must have

$$S-T<0<S+T.$$

Hence we take $x_{3,4}$ as the real ones, yielding

$$u_a={\displaystyle \frac{-\lambda +m-\sqrt{S+T}}{8}},u_b={\displaystyle \frac{-\lambda +m+\sqrt{S+T}}{8}},$$

with $u_a<u_b$.

2.2. Bounds for $HSO\left(G\right)$ via $EU(\lambda ,G)$

Lemma 3. 

Let $\delta \le x\le y\le \Delta$, ${\lambda }_0=-{\displaystyle \frac{8}{3}}\sqrt{2\sqrt{3}-3}$ and $\lambda >-2$. Define

$$f(x,y)={\displaystyle \frac{\sqrt{x^2+y^2}}{x\sqrt{x^2+y^2+\lambda xy}}}.$$

Then for $\lambda \ge 0$,

$${\displaystyle \frac{\sqrt{2}}{\Delta \sqrt{2+\lambda }}}\le f(x,y)\le {\displaystyle \frac{\sqrt{{\delta }^2+{\Delta }^2}}{\delta \sqrt{{\delta }^2+{\Delta }^2+\lambda \delta \Delta }}}.$$

For $\lambda \in [{\lambda }_0,0)$,

$${\displaystyle \frac{\sqrt{2}}{\Delta \sqrt{2+\lambda }}}\le f(x,y)\le {\displaystyle \frac{\sqrt{2}}{\delta \sqrt{2+\lambda }}}.$$

For $\lambda \in (-2,{\lambda }_0),$ when $x\le \Delta ·u_a$,

$${\displaystyle \frac{1}{\Delta }}·min\left\{{\displaystyle \frac{\sqrt{1+u_a^2}}{u_a\sqrt{1+u_a^2+\lambda u_a}}},{\displaystyle \frac{\sqrt{2}}{\sqrt{2+\lambda }}}\right\}\le f(x,y)\le {\displaystyle \frac{\sqrt{2}}{\delta \sqrt{2+\lambda }}};$$

When $\Delta ·u_a<x<\Delta ·u_b$,

$${\displaystyle \frac{1}{\Delta }}·min\left\{{\displaystyle \frac{\sqrt{{\delta }^2+{\Delta }^2}}{\delta \sqrt{{\delta }^2+{\Delta }^2+\lambda \delta \Delta }}},{\displaystyle \frac{\sqrt{2}}{\sqrt{2+\lambda }}}\right\}\le f(x,y)\le {\displaystyle \frac{\sqrt{2}}{\delta \sqrt{2+\lambda }}};$$

When $x\ge \Delta ·u_b$,

$${\displaystyle \frac{\sqrt{2}}{\Delta \sqrt{2+\lambda }}}\le f(x,y)\le {\displaystyle \frac{\sqrt{2}}{\delta \sqrt{2+\lambda }}},$$

where $u_a$ and $u_b$ are the two real roots in $(0,1)$ from Lemma 2 of the equation $2u^4+\lambda u^3+4u^2+3\lambda u+2=0$ for $\lambda \in (-2,{\lambda }_0)$.

Proof. 

Set $t={\displaystyle \frac{y}{x}}\ge 1$. Then

$$f(x,y)={\displaystyle \frac{1}{x}}·{\displaystyle \frac{\sqrt{1+t^2}}{\sqrt{1+\lambda t+t^2}}}.$$

Denote $h\left(t\right)={\displaystyle \frac{1+t^2}{1+\lambda t+t^2}}$. Then $h^{\prime }\left(t\right)={\displaystyle \frac{\lambda (t^2-1)}{{(1+t^2+\lambda t)}^2}}$.

Therefore, if $t\ge 1$, we have the following three cases:

• If $\lambda >0$, then $h^{\prime }\left(t\right)>0$, so $h\left(t\right)$ is strictly increasing.
• If $\lambda =0$, then $h\left(t\right)\equiv 1$.
• If $\lambda <0$, then $h^{\prime }\left(t\right)<0$, so $h\left(t\right)$ is strictly decreasing.

Hence, for fixed x, the function $f(x,y)={\displaystyle \frac{1}{x}}\sqrt{h\left({\displaystyle \frac{y}{x}}\right)}$ has the same monotonicity in y:

• If $\lambda \ge 0$, then $f(x,y)$ is increasing in y.
• If $\lambda \le 0$, then $f(x,y)$ is decreasing in y.

Case 1: $\lambda \ge 0$.

For fixed x, the function $f(x,y)$ is increasing in y. Hence

$${\displaystyle \frac{\sqrt{2}}{x\sqrt{2+\lambda }}}=f(x,x)\le f(x,y)\le f(x,\Delta )={\displaystyle \frac{\sqrt{x^2+{\Delta }^2}}{x\sqrt{x^2+{\Delta }^2+\lambda x\Delta }}}.$$

Note that the function $h\left(x\right)={\displaystyle \frac{\sqrt{2}}{x\sqrt{2+\lambda }}}$ decreases with x. Therefore on $x\in [\delta ,\Delta ]$,

$$maxh\left(x\right)=h\left(\delta \right)={\displaystyle \frac{\sqrt{2}}{\delta \sqrt{2+\lambda }}},minh\left(x\right)=h(\Delta )={\displaystyle \frac{\sqrt{2}}{\Delta \sqrt{2+\lambda }}}.$$

Now consider $\phi \left(x\right)=f(x,\Delta )$. Taking logariTheorems,

$$ln\phi \left(x\right)={\displaystyle \frac{1}{2}}ln(x^2+{\Delta }^2)-lnx-{\displaystyle \frac{1}{2}}ln(x^2+{\Delta }^2+\lambda x\Delta ).$$

Differentiating,

$${\displaystyle \frac{{\phi }^{\prime }\left(x\right)}{\phi \left(x\right)}}=-{\displaystyle \frac{{\Delta }^2}{x(x^2+{\Delta }^2)}}-{\displaystyle \frac{2x+\lambda \Delta }{2(x^2+{\Delta }^2+\lambda x\Delta )}}<0,$$

so $\phi$ is also decreasing. Hence for $x\in [\delta ,\Delta ]$,

$$max\phi \left(x\right)=\phi \left(\delta \right)={\displaystyle \frac{\sqrt{{\delta }^2+{\Delta }^2}}{\delta \sqrt{{\delta }^2+{\Delta }^2+\lambda \delta \Delta }}},min\phi \left(x\right)=\phi (\Delta )={\displaystyle \frac{\sqrt{2}}{\Delta \sqrt{2+\lambda }}}.$$

Since for $\lambda \ge 0$ we have

$${\displaystyle \frac{\sqrt{{\delta }^2+{\Delta }^2}}{\delta \sqrt{{\delta }^2+{\Delta }^2+\lambda \delta \Delta }}}\ge {\displaystyle \frac{\sqrt{2}}{\delta \sqrt{2+\lambda }}},$$

it follows that

$$max\left\{h\left(x\right),\phi \left(x\right)\right\}=\phi \left(\delta \right),min\left\{h\left(x\right),\phi \left(x\right)\right\}=h(\Delta ).$$

Consequently, for all $\delta \le x\le y\le \Delta$,

$$maxf(x,y)={\displaystyle \frac{\sqrt{{\delta }^2+{\Delta }^2}}{\delta \sqrt{{\delta }^2+{\Delta }^2+\lambda \delta \Delta }}},minf(x,y)={\displaystyle \frac{\sqrt{2}}{\Delta \sqrt{2+\lambda }}}.$$

Case 2: $\lambda <0$.

For fixed x, the function $f(x,y)$ is decreasing in y. Hence

$${\displaystyle \frac{\sqrt{x^2+{\Delta }^2}}{x\sqrt{x^2+{\Delta }^2+\lambda x\Delta }}}=f(x,\Delta )\le f(x,y)\le f(x,x)={\displaystyle \frac{\sqrt{2}}{x\sqrt{2+\lambda }}}.$$

Clearly $f(x,x)$ decreases with x. Thus

$$f(x,y)\le f(x,x)\le f(\delta ,\delta )={\displaystyle \frac{\sqrt{2}}{\delta \sqrt{2+\lambda }}}.$$

For the lower bound, consider $f(x,\Delta )$. Set $u={\displaystyle \frac{x}{\Delta }}\in (0,1]$. Then

$$f(x,\Delta )={\displaystyle \frac{1}{\Delta }}·{\displaystyle \frac{\sqrt{1+u^2}}{u\sqrt{1+u^2+\lambda u}}}={\displaystyle \frac{1}{\Delta }}g\left(u\right).$$

Define $h\left(u\right)=g^2\left(u\right)={\displaystyle \frac{1+u^2}{u^2(1+u^2+\lambda u)}}$. Differentiating,

$$h^{\prime }\left(u\right)=-{\displaystyle \frac{2u^4+\lambda u^3+4u^2+3\lambda u+2}{u^3{(1+u^2+\lambda u)}^2}},$$

where the denominator is always positive.

Let $p\left(u\right)=2u^4+\lambda u^3+4u^2+3\lambda u+2$, $u\in (0,1]$. By Lemma 2, we obtain that $p\left(u\right)$ has exactly two real roots $u_a$ and $u_b$ in $(0,1)$, denoted $0<u_a<u_b<1$. Moreover,

$$p\left(u\right)>0\mathrm{on}(0,u_a)\cup (u_b,1],p\left(u\right)<0\mathrm{on}(u_a,u_b).$$

Hence the monotonicity of $g\left(u\right)$ is as follows.

• If $\lambda \in [{\lambda }_0,0)$, then $p\left(u\right)\ge 0$ implies $g^{\prime }\left(u\right)\le 0$. Thus $g\left(u\right)$ is decreasing on $(0,1]$. Consequently, the minimum value is ${\displaystyle \frac{1}{\Delta }}g\left(1\right)$.
• If $\lambda \in (-2,{\lambda }_0)$, then $g\left(u\right)$ is decreasing on $(0,u_a)\cup (u_b,1]$ and increasing on $(u_a,u_b)$. Since $u\in [{\displaystyle \frac{\delta }{\Delta }},1]$, we have:

  (1)

  If ${\displaystyle \frac{\delta }{\Delta }}\le u_a$, the minimum is ${\displaystyle \frac{1}{\Delta }}min\{g\left(u_a\right),g\left(1\right)\}$;

  (2)

  If $u_a<{\displaystyle \frac{\delta }{\Delta }}<u_b$, the minimum is ${\displaystyle \frac{1}{\Delta }}min\{g\left({\displaystyle \frac{\delta }{\Delta }}\right),g\left(1\right)\}$;

  (3)

  If ${\displaystyle \frac{\delta }{\Delta }}\ge u_b$, the minimum is ${\displaystyle \frac{1}{\Delta }}g\left(1\right)$.

Here

$$g\left(1\right)={\displaystyle \frac{\sqrt{2}}{\sqrt{2+\lambda }}},g\left({\displaystyle \frac{\delta }{\Delta }}\right)={\displaystyle \frac{\sqrt{{\delta }^2+{\Delta }^2}}{\delta \sqrt{{\delta }^2+{\Delta }^2+\lambda \delta \Delta }}}.$$

□

For any edge $uv$ of G with $d_u\le d_v$. We have

$${\displaystyle \frac{\sqrt{d_u^2+d_v^2}/d_u}{\sqrt{d_u^2+d_v^2+\lambda d_u{d}_v}}}=f(d_u,d_v).$$

Then by Lemma 3 and summing over all edges of G, we obtain the bounds for $HSO\left(G\right)$ via $EU(\lambda ,G)$ as follows, where $u_a$ and $u_b$ are given by Lemma 2.

Theorem 1. 

Let G be a connected graph with m edges, maximum degree Δ and minimum degree δ. Let the parameters $\lambda >-2$ and ${\lambda }_0=-{\displaystyle \frac{8}{3}}\sqrt{2\sqrt{3}-3}$. If $\lambda \ge 0$, then

$${\displaystyle \frac{\sqrt{2}}{\Delta \sqrt{2+\lambda }}}EU(\lambda ,G)\le HSO\left(G\right)\le {\displaystyle \frac{\sqrt{{\delta }^2+{\Delta }^2}}{\delta \sqrt{{\delta }^2+{\Delta }^2+\lambda \delta \Delta }}}EU(\lambda ,G).$$

If $\lambda \in [{\lambda }_0,0)$, then

$${\displaystyle \frac{\sqrt{2}}{\Delta \sqrt{2+\lambda }}}EU(\lambda ,G)\le HSO\left(G\right)\le {\displaystyle \frac{\sqrt{2}}{\delta \sqrt{2+\lambda }}}EU(\lambda ,G).$$

If $\lambda \in (-2,{\lambda }_0)$, then when ${\displaystyle \frac{\delta }{\Delta }\le u_a}$,

$${\displaystyle \frac{1}{\Delta }}min\left\{{\displaystyle \frac{\sqrt{1+u_a^2}}{u_a\sqrt{1+u_a^2+\lambda u_a}}},{\displaystyle \frac{\sqrt{2}}{\sqrt{2+\lambda }}}\right\}\le {\displaystyle \frac{HSO\left(G\right)}{EU(\lambda ,G)}}\le {\displaystyle \frac{\sqrt{2}}{\delta \sqrt{2+\lambda }}};$$

When ${\displaystyle u_a<\frac{\delta }{\Delta }<u_b}$,

$${\displaystyle \frac{1}{\Delta }}min\left\{{\displaystyle \frac{\sqrt{{\delta }^2+{\Delta }^2}}{\delta \sqrt{{\delta }^2+{\Delta }^2+\lambda \delta \Delta }}},{\displaystyle \frac{\sqrt{2}}{\sqrt{2+\lambda }}}\right\}\le {\displaystyle \frac{HSO\left(G\right)}{EU(\lambda ,G)}}\le {\displaystyle \frac{\sqrt{2}}{\delta \sqrt{2+\lambda }}};$$

When ${\displaystyle \frac{\delta }{\Delta }\ge u_b}$,

$${\displaystyle \frac{\sqrt{2}}{\Delta \sqrt{2+\lambda }}}EU(\lambda ,G)\le HSO\left(G\right)\le {\displaystyle \frac{\sqrt{2}}{\delta \sqrt{2+\lambda }}}EU(\lambda ,G).$$

Both bounds are attained when G is regular.

Taking $\lambda =0,1\mathrm{and}2$, we respectively obtain the following three corollaries.

Corollary 1. 

[7] Let G be a connected graph with m edges, maximum degree Δ and minimum degree δ. Then

$${\displaystyle \frac{1}{\Delta }}SO\left(G\right)\le HSO\left(G\right)\le {\displaystyle \frac{1}{\delta }}SO\left(G\right).$$

Both bounds are attained when G is regular.

Corollary 2. 

Let G be a connected graph with m edges, maximum degree Δ and minimum degree δ. Then

$${\displaystyle \frac{\sqrt{6}}{3\Delta }}EU\left(G\right)\le HSO\left(G\right)\le {\displaystyle \frac{\sqrt{{\delta }^2+{\Delta }^2}}{\delta \sqrt{{\delta }^2+{\Delta }^2+\delta \Delta }}}EU\left(G\right).$$

Both bounds are attained when G is regular.

Corollary 3. 

[36] Let G be a connected graph with m edges, maximum degree Δ and minimum degree δ. Then

$${\displaystyle \frac{1}{\sqrt{2}\Delta }}{M}_1\left(G\right)\le HSO\left(G\right)\le {\displaystyle \frac{\sqrt{{\delta }^2+{\Delta }^2}}{\delta (\delta +\Delta )}}{M}_1\left(G\right).$$

Both bounds are attained when G is regular.

If $\lambda =-2$, then $Eu(\lambda ,G)=Alb\left(G\right)$. We also obtain the bounds for $HSO\left(G\right)$ in terms of $Alb\left(G\right)$.

Theorem 2. 

Let G be a connected graph with m edges, maximum degree Δ, and minimum degree δ. Then

$$\sqrt{2}m+{\displaystyle \frac{1}{\sqrt{2}\Delta }}Alb\left(G\right)\le HSO\left(G\right)\le \sqrt{2}m+{\displaystyle \frac{1}{\delta }}Alb\left(G\right).$$

Both bounds are attained when G is regular.

Proof. 

Set $d_u\le d_v$ and let $t={\displaystyle \frac{d_v}{d_u}}\ge 1$. Since

$${[\sqrt{2}+(t-1)]}^2-{\left(\sqrt{1+t^2}\right)}^2=2(\sqrt{2}-1)(t-1)\ge 0,$$

we obtain

$$\sqrt{1+t^2}\le \sqrt{2}+(t-1).$$

Consequently,

$${\displaystyle \frac{\sqrt{d_u^2+d_v^2}}{d_u}}=\sqrt{1+{\left({\displaystyle \frac{d_v}{d_u}}\right)}^2}\le \sqrt{2}+\left({\displaystyle \frac{d_v}{d_u}}-1\right)\le \sqrt{2}+{\displaystyle \frac{d_v-d_u}{\delta }}.$$

Summing over all edges yields the upper bound

$$HSO\left(G\right)\le \sqrt{2}m+{\displaystyle \frac{1}{\delta }}Alb\left(G\right).$$

Since

$${\left(\sqrt{1+t^2}\right)}^2-{\left[\sqrt{2}+{\displaystyle \frac{1}{\sqrt{2}}}(t-1)\right]}^2={\displaystyle \frac{1}{2}}{(t-1)}^2\ge 0,$$

we obtain

$$\sqrt{1+t^2}\ge \sqrt{2}+{\displaystyle \frac{1}{\sqrt{2}}}(t-1).$$

Hence

$${\displaystyle \frac{\sqrt{d_u^2+d_v^2}}{d_u}}=\sqrt{1+{\left({\displaystyle \frac{d_v}{d_u}}\right)}^2}\ge \sqrt{2}+{\displaystyle \frac{1}{\sqrt{2}}}\left({\displaystyle \frac{d_v}{d_u}}-1\right)\ge \sqrt{2}+{\displaystyle \frac{d_v-d_u}{\sqrt{2}\Delta }}.$$

Summing over all edges gives the lower bound

$$HSO\left(G\right)\ge \sqrt{2}m+{\displaystyle \frac{1}{\sqrt{2}\Delta }}Alb\left(G\right).$$

□

3. Bounds for $\mathit{HSO}\left(\mathit{G}\right)$ in Terms of ${\mathit{M}}_{\mathit{2}}\left(\mathit{G}\right)$, $\mathit{ESO}\left(\mathit{G}\right)$ and $\mathit{FSO}\left(\mathit{G}\right)$

In this section, we establish upper and lower bounds for the Hyperbolic Sombor index in terms of several other degree-based topological invariants.

Lemma 4. 

For $\delta \le x\le y\le \Delta$, let $R(x,y)={\displaystyle \frac{\sqrt{x^2+y^2}}{x^{2}y}}$. Then

$${\displaystyle \frac{\sqrt{2}}{{\Delta }^2}}\le R(x,y)\le {\displaystyle \frac{\sqrt{2}}{{\delta }^2}}.$$

Proof. 

Fix x and regard $R(x,y)={\displaystyle \frac{\sqrt{x^2+y^2}}{x^{2}y}}$ as a function of y:

$$f\left(y\right)={\displaystyle \frac{\sqrt{x^2+y^2}}{x^{2}y}}={\displaystyle \frac{1}{x^2}}·{\displaystyle \frac{\sqrt{x^2+y^2}}{y}}={\displaystyle \frac{1}{x^2}}·\sqrt{1+{\left({\displaystyle \frac{x}{y}}\right)}^2}.$$

Clearly, $f\left(y\right)$ is monotonically decreasing in y for $\delta \le x\le y\le \Delta$. Hence $R(x,y)$ is monotonically decreasing in y. Therefore, for a fixed x, $R(x,y)$ attains its maximum at $y=x$ and its minimum at $y=\Delta$; that is,

$$R(x,x)={\displaystyle \frac{\sqrt{2}}{x^2}},R(x,\Delta )={\displaystyle \frac{\sqrt{x^2+{\Delta }^2}}{x^2\Delta }}.$$

Now consider $R(x,x)$ as a function of x. It is clearly monotonically decreasing in x. So it attains its maximum ${\displaystyle \frac{\sqrt{2}}{{\delta }^2}}$ at $x=\delta$ and its minimum ${\displaystyle \frac{\sqrt{2}}{{\Delta }^2}}$ at $x=\Delta$.

Next, consider $R(x,\Delta )$. Let $h\left(x\right)=R(x,\Delta )={\displaystyle \frac{\sqrt{x^2+{\Delta }^2}}{x^2\Delta }}$. Then

$$lnh\left(x\right)={\displaystyle \frac{1}{2}}ln(x^2+{\Delta }^2)-2lnx-ln\Delta .$$

Differentiating, we obtain

$${\displaystyle \frac{h^{\prime }\left(x\right)}{h\left(x\right)}}={\displaystyle \frac{x}{x^2+{\Delta }^2}}-{\displaystyle \frac{2}{x}}={\displaystyle \frac{-x^2-2{\Delta }^2}{x(x^2+{\Delta }^2)}}<0.$$

Thus $h\left(x\right)$ is monotonically decreasing, and so is $R(x,\Delta )$ in x. Consequently, $R(x,\Delta )$ attains its maximum ${\displaystyle \frac{\sqrt{{\delta }^2+{\Delta }^2}}{{\delta }^2\Delta }}$ at $x=\delta$ and its minimum ${\displaystyle \frac{\sqrt{2}}{{\Delta }^2}}$ at $x=\Delta$.

Combining the above, the global maximum of $R(x,y)$ over $\delta \le x\le y\le \Delta$ is ${\displaystyle \frac{\sqrt{2}}{{\delta }^2}}$, and the global minimum is ${\displaystyle \frac{\sqrt{2}}{{\Delta }^2}}$. □

Theorem 3. 

Let G be a connected graph with m edges, maximum degree Δ and minimum degree δ. Then

$${\displaystyle \frac{\sqrt{2}}{{\Delta }^2}}{M}_2\left(G\right)\le HSO\left(G\right)\le {\displaystyle \frac{\sqrt{2}}{{\delta }^2}}{M}_2\left(G\right).$$

Both bounds are attained when G is regular.

Proof. 

For any edge $uv\in E\left(G\right)$, Lemma 4 gives

$${\displaystyle \frac{\sqrt{2}}{{\Delta }^2}}\le {\displaystyle \frac{\sqrt{d_u^2+d_v^2}}{d_u^2{d}_v}}\le {\displaystyle \frac{\sqrt{2}}{{\delta }^2}}.$$

Multiplying both sides by the positive factor $d_u{d}_v$ yields

$${\displaystyle \frac{\sqrt{2}}{{\Delta }^2}}{d}_u{d}_v\le {\displaystyle \frac{\sqrt{d_u^2+d_v^2}}{d_u}}\le {\displaystyle \frac{\sqrt{2}}{{\delta }^2}}{d}_u{d}_v.$$

Summing over all edges and noting that $d_u=min\{d_u,d_v\}$, we obtain

$${\displaystyle \frac{\sqrt{2}}{{\Delta }^2}}{M}_2\left(G\right)\le HSO\left(G\right)\le {\displaystyle \frac{\sqrt{2}}{{\delta }^2}}{M}_2\left(G\right),$$

where equality holds if and only if G is regular. □

Lemma 5. 

For $\delta \le x\le y\le \Delta$, let $f(x,y)={\displaystyle \frac{1}{x(x+y)}}$. Then

$${\displaystyle \frac{1}{2{\Delta }^2}}\le f(x,y)\le {\displaystyle \frac{1}{2{\delta }^2}}.$$

Proof. 

Since $x\le y\le \Delta$ and $x\ge \delta$, we have

$$x+y\ge 2x\ge 2\delta \mathrm{and}x+y\le 2y\le 2\Delta .$$

Hence

$$f(x,y)={\displaystyle \frac{1}{x(x+y)}}\le {\displaystyle \frac{1}{x·2x}}={\displaystyle \frac{1}{2x^2}}\le {\displaystyle \frac{1}{2{\delta }^2}},$$

and

$$f(x,y)={\displaystyle \frac{1}{x(x+y)}}\ge {\displaystyle \frac{1}{x·2\Delta }}\ge {\displaystyle \frac{1}{2{\Delta }^2}},$$

Both bounds are sharp: the upper bound is attained when $x=y=\delta$, and the lower bound when $x=y=\Delta$. □

Theorem 4. 

Let G be a connected graph with m edges, maximum degree Δ, and minimum degree δ. Then

$${\displaystyle \frac{1}{2{\Delta }^2}}ESO\left(G\right)\le HSO\left(G\right)\le {\displaystyle \frac{1}{2{\delta }^2}}ESO\left(G\right).$$

Both bounds are attained when G is regular.

Proof. 

For any edge $uv\in E\left(G\right)$, assume without loss of generality that $d_u\le d_v$. By Lemma 5, we have

$${\displaystyle \frac{1}{2{\Delta }^2}}\le {\displaystyle \frac{1}{d_u(d_u+d_v)}}\le {\displaystyle \frac{1}{2{\delta }^2}}.$$

Multiplying both sides by the positive factor $(d_u+d_v)\sqrt{d_u^2+d_v^2}$, we have

$${\displaystyle \frac{1}{2{\Delta }^2}}(d_u+d_v)\sqrt{d_u^2+d_v^2}\le {\displaystyle \frac{\sqrt{d_u^2+d_v^2}}{d_u}}\le {\displaystyle \frac{1}{2{\delta }^2}}(d_u+d_v)\sqrt{d_u^2+d_v^2}.$$

Summing this inequality over all edges of G and noting that $d_u=min\{d_u,d_v\}$, we obtain

$${\displaystyle \frac{1}{2{\Delta }^2}}ESO\left(G\right)\le HSO\left(G\right)\le {\displaystyle \frac{1}{2{\delta }^2}}ESO\left(G\right),$$

All the inequalities above hold with equality when G is regular. □

Lemma 6. 

Let $\delta \le x\le y\le \Delta$ and define $f(x,y)={\displaystyle \frac{\sqrt{x^2+y^2}}{x\sqrt{x^4+y^4}}}.$ Then

$${\displaystyle \frac{1}{{\Delta }^2}}\le f(x,y)\le {\displaystyle \frac{1}{{\delta }^2}}.$$

Proof. 

Let $t=y/x\ge 1$. Then

$$f(x,y)={\displaystyle \frac{1}{x^2}}\sqrt{{\displaystyle \frac{1+t^2}{1+t^4}}}.$$

Set $g\left(t\right)={\displaystyle \frac{1+t^2}{1+t^4}}$. For $t\ge 1$,

$$g^{\prime }\left(t\right)={\displaystyle \frac{2t-4t^3-2t^5}{{(1+t^4)}^2}}<0,$$

so $g\left(t\right)$ (and hence $\sqrt{g\left(t\right)}$) is strictly decreasing. Therefore, for fixed x, $f(x,y)$ decreases as y increases. Consequently,

$$\underset{y\ge x}{max}f(x,y)=f(x,x)={\displaystyle \frac{1}{x^2}},\underset{y\ge x}{min}f(x,y)=f(x,\Delta )={\displaystyle \frac{\sqrt{x^2+{\Delta }^2}}{x\sqrt{x^4+{\Delta }^4}}}.$$

Now consider $h\left(x\right)=f(x,x)=1/x^2$, which clearly decreases with x. Hence on $x\in [\delta ,\Delta ]$,

$$maxh\left(x\right)=h\left(\delta \right)={\displaystyle \frac{1}{{\delta }^2}},minh\left(x\right)=h(\Delta )={\displaystyle \frac{1}{{\Delta }^2}}.$$

For $\phi \left(x\right)=f(x,\Delta )$, take logariTheorems:

$$ln\phi \left(x\right)={\displaystyle \frac{1}{2}}ln(x^2+{\Delta }^2)-lnx-{\displaystyle \frac{1}{2}}ln(x^4+{\Delta }^4).$$

Differentiating,

$${\displaystyle \frac{{\phi }^{\prime }\left(x\right)}{\phi \left(x\right)}}=-{\displaystyle \frac{{\Delta }^2}{x(x^2+{\Delta }^2)}}-{\displaystyle \frac{2x^3}{x^4+{\Delta }^4}}<0,$$

so $\phi$ is decreasing as well. Thus for $x\in [\delta ,\Delta ]$,

$$max\phi \left(x\right)=\phi \left(\delta \right)={\displaystyle \frac{\sqrt{{\delta }^2+{\Delta }^2}}{\delta \sqrt{{\delta }^4+{\Delta }^4}}},min\phi \left(x\right)=\phi (\Delta )={\displaystyle \frac{1}{{\Delta }^2}}.$$

Since $max\left\{{\displaystyle \frac{1}{{\delta }^2}},{\displaystyle \frac{\sqrt{{\delta }^2+{\Delta }^2}}{\delta \sqrt{{\delta }^4+{\Delta }^4}}}\right\}={\displaystyle \frac{1}{{\delta }^2}},$ we have, for $\delta \le x\le y\le \Delta$,

$$maxf(x,y)={\displaystyle \frac{1}{{\delta }^2}},minf(x,y)={\displaystyle \frac{1}{{\Delta }^2}}.$$

□

Theorem 5. 

Let G be a connected graph with m edges, maximum degree Δ and minimum degree δ. Then

$${\displaystyle \frac{1}{{\Delta }^2}}FSO\left(G\right)\le HSO\left(G\right)\le {\displaystyle \frac{1}{{\delta }^2}}FSO\left(G\right).$$

Both bounds are attained when G is regular.

Proof. 

For any edge $uv$ with $d_u\le d_v$. By Lemma 6, we have

$${\displaystyle \frac{1}{{\Delta }^2}}\le {\displaystyle \frac{\sqrt{d_u^2+d_v^2}}{d_u\sqrt{d_u^4+d_v^4}}}\le {\displaystyle \frac{1}{{\delta }^2}}.$$

Multiplying by the positive factor $\sqrt{d_u^4+d_v^4}$ gives

$${\displaystyle \frac{1}{{\Delta }^2}}\sqrt{d_u^4+d_v^4}\le {\displaystyle \frac{\sqrt{d_u^2+d_v^2}}{d_u}}\le {\displaystyle \frac{1}{{\delta }^2}}\sqrt{d_u^4+d_v^4}.$$

Summing over all edges, and noting that $d_u=min\{d_u,d_v\}$, we obtain

$${\displaystyle \frac{1}{{\Delta }^2}}FSO\left(G\right)\le HSO\left(G\right)\le {\displaystyle \frac{1}{{\delta }^2}}FSO\left(G\right).$$

□

4. Discussion

The main contributions of this study are to provide several sharp upper and lower bounds for the Hyperbolic Sombor index $HSO\left(G\right)$. In Theorem 1, we establish a relationship between $HSO\left(G\right)$ and the variable Euler-Sombor index $EU(\lambda ,G)$, and give precise bounds for different ranges of the parameter $\lambda$. In particular, by taking $\lambda =0,1,2$, we obtain bounds relating $HSO\left(G\right)$ to $SO\left(G\right)$, $EU\left(G\right)$, and $M_1\left(G\right)$, respectively. Theorem 2 provides upper and lower bounds for $HSO\left(G\right)$ in terms of the Albertson index $Alb\left(G\right)$.

Moreover, Theorems 3–5 establish concise inequalities between $HSO\left(G\right)$ and $M_2\left(G\right)$, $ESO\left(G\right)$, as well as $FSO\left(G\right)$. These results enrich the mathematical theory of the Hyperbolic Sombor index and also provide tools for future research.

This paper establishes, for the first time, a connection between $HSO\left(G\right)$ and the variable Euler-Sombor index $EU(\lambda ,G)$. The index $EU(\lambda ,G)$ is a unified framework proposed by Gutman et al. [22], which yields, for different values of $\lambda$, the known indices $SO\left(G\right)$, $EU\left(G\right)$, $M_1\left(G\right)$, and $Alb\left(G\right)$. Through this unified framework, our paper not only systematically covers several existing index cases, but also reveals deeper structural relationships between $HSO\left(G\right)$ and these indices. The novelty of this methodological approach lies in the fact that, by means of Ferrari’s method for solving quartic equations, we precisely characterize the variation of the relevant functions at the critical parameter ${\lambda }_0=-{\displaystyle \frac{8}{3}}\sqrt{2\sqrt{3}-3}\approx -1.8167$, thereby obtaining piecewise sharp bounds.

Recently, Rada et al. [36] introduced the general form of the Hyperbolic Sombor index

$$HS{O}_a\left(G\right)=\sum _{uv\in E\left(G\right)}{\left({\displaystyle \frac{\sqrt{d_u^2+d_v^2}}{min\{d_u,d_v\}}}\right)}^a,$$

where a is a real number. Whether the methodology of the present paper can be extended to this general form is a direction worth exploring.

Although preliminary studies have shown that $HSO\left(G\right)$ possesses good predictive capability[6], integrating these predictive capabilities with the formulas obtained in this paper to establish bound-based prediction models remains an open problem.

Moreover, the family of Sombor-type indices is growing increasingly large, including the Sombor index, the Elliptic Sombor index, the Euler-Sombor index, the Forgotten Sombor index, and the Hyperbolic Sombor index studied in this paper. A systematic comparison among these indices and the establishment of a unified theoretical framework would contribute to a deeper understanding of the nature of degree-based topological indices.