Resolving an Open Problem on the Exponential Arithmetic–Geometric Index of Unicyclic Graphs

Abstract

Recently, the exponential arithmetic–geometric index ($EAG$) was introduced. The exponential arithmetic–geometric index ($EAG$) of a graph G is defined as $EAG\left(G\right)={\displaystyle \sum _{v_i{v}_j\in E\left(G\right)}}{e}^{{\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}}$, where $d_i$ represents the degree of the vertex $v_i$ in G. The characterization of extreme structures in relation to graph invariants from the class of unicyclic graphs is an important problem in discrete mathematics. Cruz et al., 2022 proposed a unified method for finding extremal unicyclic graphs for exponential degree-based graph invariants. However, in the case of $EAG$, this method is insufficient for generating the maximal unicyclic graph. Consequently, the same article presented an open problem for the investigation of the maximal unicyclic graph with respect to this invariant. This article completely characterizes the maximal unicyclic graph in relation to $EAG$.

1. Introduction

In recent years, graph theory has become a crucial tool in the field of chemistry, particularly in the study of molecular structures. A topological index is a numerical quantity associated with a molecular graph that describes specific physicochemical properties. Since Wiener introduced the first such index [1], many other indices have been developed depending on various graph parameters, including degree, distance, and eccentricity [2,3,4,5,6,7]. Among these, degree-based indices have received significant attention since the 1970s due to their strong correlation with molecular properties. One important example is the arithmetic–geometric index, introduced in [8]. Let G be a simple connected graph, whose vertex set $V\left(G\right)=\{v_1,v_2,\dots ,v_n\}$, with the edge set being denoted as $E\left(G\right)$. The degree of the j-th vertex $v_j$ is denoted by $d_j$. The arithmetic–geometric index ($AG$) is defined as

$$\begin{array}{c}\hfill AG\left(G\right)=\sum _{v_i{v}_J\in E\left(G\right)}{\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}.\end{array}$$

Numerous publications have extensively studied the mathematical properties of $AG$, specifically the extremal problems and bounds. Shegehalli et al. [9,10] computed $AG$ for different classes of graphs. Milovanović et al. [11] established some upper bounds on $AG$ for simple connected graphs. Hertz et al. [12] identified extremal chemical graphs for $AG$. Maximal chemical trees with respect to $AG$ were characterized in [13]. Additional research on $AG$ can be found in [14,15,16,17]. To improve the discriminative power of topological indices, exponential versions were introduced in 2019 [18]. In [19,20], researchers provided a characterization of the extremal trees for the exponential Randić index. Das et al. [21] explored extremal graphs for the exponential atom–bond connectivity index. Jahanbani et al. [22] determined the minimum value of the exponential forgotten index for trees. Das and Mondal [23] examined sharp bounds on the exponential geometric–arithmetic index of bipartite graphs. Further studies on this concept include [24,25,26,27,28,29]. In this paper, we focus on the exponential arithmetic–geometric index ($EAG$), which is defined as

$$\begin{array}{c}\hfill EAG\left(G\right)=\sum _{v_i{v}_J\in E\left(G\right)}{e}^{{\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}}.\end{array}$$

Previous studies have systematically characterized extremal unicyclic graphs for various degree-based indices. For example, Moon and Park [30] investigated extremal unicyclic graphs for the geometric–arithmetic index, while Cruz et al. [31] addressed the same problem with respect to $AG$. Liu et al. [32] presented a complete classification of extremal unicyclic graphs for the Lanzhou index. The maximal unicyclic graphs for the exponential second Zagreb and augmented Zagreb indices were described in [33,34]. For more insights into extremal unicyclic graphs, readers can refer to [35,36,37,38,39,40,41,42]. Cruz et al. [37] introduced a unified method for the identification of extremal unicyclic graphs for exponential degree-based indices, which was successfully applied to many well-known indices. However, they discovered that this method is insufficient for generating the maximal unicyclic graph in the case of $EAG$. Because of this, the following open problem was posed in [37]:

Problem 1

([37]). Characterize the maximal unicyclic graph with respect to $EAG$ in terms of graph order.

This paper aims to fully solve this problem by applying advanced combinatorial methods. Our goal is to develop new methods that will help us to identify the maximal unicyclic graph with respect to $EAG$.

2. Main Result

In this section, we address and resolve the open problem concerning the exponential arithmetic–geometric index of unicyclic graphs. To achieve this, we first establish the following essential results:

Lemma 1.

Let $x\ge 2$. Then, $3x^5-49x^4+310x^3-936x^2+1366x-775>0$.

Proof. 

Let $f\left(x\right)=3x^5-49x^4+310x^3-936x^2+1366x-775$. Now,

$$\begin{array}{c}\hfill 3x^5-49x^4+310x^3-936x^2+1366x-775\hfill \\ \hfill =x^4(3x-49)+x^2(310x-936)+1366x-775>0\hfill \end{array}$$

for $x\ge 17$. Using Mathematica [43], we can visualize the function with a plot, which clearly shows that $f\left(x\right)>0$ for all x in the interval $[2,17]$. □

Lemma 2.

Let

$$f\left(x\right)={\displaystyle \frac{x}{2\sqrt{x-1}}}-{\displaystyle \frac{x-1}{2\sqrt{3(x-4)}}},x\ge 2.$$

Then, $f\left(x\right)$ is a strictly increasing function on $x\ge 2$.

Proof. 

Since

$$f\left(x\right)={\displaystyle \frac{x}{2\sqrt{x-1}}}-{\displaystyle \frac{x-1}{2\sqrt{3(x-4)}}},$$

we have

$$f^{\prime }\left(x\right)={\displaystyle \frac{1}{4\sqrt{x-1}}}\left({\displaystyle \frac{x-2}{x-1}}\right)-{\displaystyle \frac{1}{4\sqrt{3}\sqrt{x-4}}}\left({\displaystyle \frac{x-7}{x-4}}\right).$$

We have to prove that $f^{\prime }\left(x\right)>0$, i.e.,

$${\displaystyle \frac{1}{\sqrt{x-1}}}\left({\displaystyle \frac{x-2}{x-1}}\right)>{\displaystyle \frac{1}{\sqrt{3}\sqrt{x-4}}}\left({\displaystyle \frac{x-7}{x-4}}\right),$$

that is,

$$3{(x-2)}^2{(x-4)}^3>(x-7){(x-1)}^3,$$

that is,

$$3x^5-49x^4+310x^3-936x^2+1366x-775>0,$$

which is true by Lemma 1 $(x\ge 2)$. This proves the result. □

Lemma 3.

Let

$$g\left(x\right)={\displaystyle \frac{x}{2\sqrt{x-1}}}-{\displaystyle \frac{x+3}{2\sqrt{2(x+1)}}},x\ge 4.$$

Then, $g\left(x\right)$ is a strictly increasing function on $x\ge 4$.

Proof. 

Since $x\ge 4$, one can easily see that

$$g^{\prime }\left(x\right)={\displaystyle \frac{1}{4\sqrt{x-1}}}\left({\displaystyle \frac{x-2}{x-1}}\right)-{\displaystyle \frac{1}{4\sqrt{2(x+1)}}}\left({\displaystyle \frac{x-1}{x+1}}\right)>0.$$

Therefore, $g\left(x\right)$ is a strictly increasing function on $x\ge 4$. □

Lemma 4.

Let

$$q\left(x\right)={\displaystyle \frac{x+3}{2\sqrt{2(x+1)}}}-{\displaystyle \frac{x+5}{4\sqrt{x+1}}},x>0.$$

Then, $q\left(x\right)$ is a strictly increasing function on $x>0$.

Proof. 

Since $x>0$, one can easily see that

$$q^{\prime }\left(x\right)={\displaystyle \frac{1}{4\sqrt{2}}}{\displaystyle \frac{(x-1)}{{(x+1)}^{3/2}}}-{\displaystyle \frac{1}{8}}{\displaystyle \frac{(x-3)}{{(x+1)}^{3/2}}}>0.$$

Therefore, $q\left(x\right)$ is a strictly increasing function on $x>0$. □

Lemma 5.

For $b\le d_j\le d_i\le a$,

$${\displaystyle \frac{d_i+d_j}{\sqrt{d_i{d}_j}}}\le {\displaystyle \frac{a+b}{\sqrt{ab}}}$$

with equality if and only if $d_i=a$, $d_j=b$.

Proof. 

For $b\le d_j\le d_i\le a$, we obtain

$${\displaystyle \frac{d_i}{d_j}}\le {\displaystyle \frac{a}{b}},{\displaystyle \frac{d_j}{d_i}}\ge {\displaystyle \frac{b}{a}},\mathrm{and}\mathrm{hence}{\left({\displaystyle \frac{d_i}{d_j}}\right)}^{1/4}-{\left({\displaystyle \frac{d_j}{d_i}}\right)}^{1/4}\le {\left({\displaystyle \frac{a}{b}}\right)}^{1/4}-{\left({\displaystyle \frac{b}{a}}\right)}^{1/4}.$$

Using the above result, we obtain

$${\left({\left({\displaystyle \frac{d_i}{d_j}}\right)}^{1/4}+{\left({\displaystyle \frac{d_j}{d_i}}\right)}^{1/4}\right)}^2={\left({\left({\displaystyle \frac{d_i}{d_j}}\right)}^{1/4}-{\left({\displaystyle \frac{d_j}{d_i}}\right)}^{1/4}\right)}^2+4\le {\left({\left({\displaystyle \frac{a}{b}}\right)}^{1/4}-{\left({\displaystyle \frac{b}{a}}\right)}^{1/4}\right)}^2+4,$$

$$\mathrm{that}\mathrm{is},\sqrt{{\displaystyle \frac{d_i}{d_j}}}+\sqrt{{\displaystyle \frac{d_j}{d_i}}}\le \sqrt{{\displaystyle \frac{a}{b}}}+\sqrt{{\displaystyle \frac{b}{a}}},\mathrm{that}\mathrm{is},{\displaystyle \frac{d_i+d_j}{\sqrt{d_i{d}_j}}}\le {\displaystyle \frac{a+b}{\sqrt{ab}}}.$$

Moreover, the equality holds if and only if $d_i=a$, $d_j=b$. □

Let $S_{n,3}$ be a unicyclic graph of order n obtained by adding an edge to the star graph $S_n(S_n$ is a star graph of order n).

Theorem 1.

Let G be a unicyclic graph of order n. Then,

$$\begin{array}{c}\hfill EAG\left(G\right)\le (n-3)e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}+2e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}+e\end{array}$$

(1)

with equality if and only if $G\cong S_{n,3}$.

Proof. 

Let $∆$ be the maximum degree in the unicyclic graph G. If $∆=n-1$, then $G\cong S_{n,3}$ with

$$EAG\left(G\right)=(n-3)e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}+2e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}+e,$$

and hence the equality holds. For $n\le 9$, in accordance to Sage [44], one can easily check that the result (1) holds with equality if and only if $G\cong S_{n,3}$. Otherwise, $∆\le n-2$ and $n\ge 10$. For any pendant edge $v_i{v}_j\in E\left(G\right)(1=d_j\le d_i\le n-2<n-1)$, by Lemma 5, we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}<{\displaystyle \frac{n}{2\sqrt{n-1}}}.\end{array}$$

(2)

For any non-pendant edge $v_i{v}_j\in E\left(G\right)(2\le d_j\le d_i\le n-2<n-1)$, by Lemma 5, we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}<{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}.\end{array}$$

(3)

For any edge $v_i{v}_j\in E\left(G\right)$, from (2) and (3), we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}<{\displaystyle \frac{n}{2\sqrt{n-1}}}\end{array}$$

(4)

as

$${\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}<{\displaystyle \frac{n}{2\sqrt{n-1}}}.$$

Let k be the length of the cycle in the unicyclic graph G. Then, $k\ge 3$. We consider the following two cases:

Case 1.

$k=3$. Let $v_1,v_2,v_3$ be the three vertices on the cycle in G. We assume that $d_1\ge d_2\ge d_3\ge 2$. We consider the following cases:

Case 1.1.

$d_2=2$. In this case, $d_2=d_3=2$; hence,

$${\displaystyle \frac{d_2+d_3}{2\sqrt{d_2{d}_3}}}=1.$$

Let $E_1=\{v_1{v}_2,v_2{v}_3,v_3{v}_1\}$. Using the above result with (3) and (4), we obtain

$$\begin{array}{cc}\hfill EAG\left(G\right)& =\sum _{v_i{v}_j\in E\left(G\right)}{e}^{{\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}}\hfill \\ \hfill & =e^{{\displaystyle \frac{d_1+d_2}{2\sqrt{d_1{d}_2}}}}+e^{{\displaystyle \frac{d_2+d_3}{2\sqrt{d_2{d}_3}}}}+e^{{\displaystyle \frac{d_3+d_1}{2\sqrt{d_3{d}_1}}}}+\sum _{v_i{v}_j\in E\left(G\right)\setminus E_1}{e}^{{\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}}\hfill \\ \hfill & <2e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}+e+(n-3)e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}.\hfill \end{array}$$

Inequality (1) holds strictly.

Case 1.2.

$d_2=3$. If $d_3=3$, then similarly, by Case 1.1, we obtain

$$\begin{array}{cc}\hfill EAG\left(G\right)& =e^{{\displaystyle \frac{d_1+d_2}{2\sqrt{d_1{d}_2}}}}+e^{{\displaystyle \frac{d_2+d_3}{2\sqrt{d_2{d}_3}}}}+e^{{\displaystyle \frac{d_3+d_1}{2\sqrt{d_3{d}_1}}}}+\sum _{v_i{v}_j\in E\left(G\right)\setminus E_1}{e}^{{\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}}\hfill \\ \hfill & <2e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}+e+(n-3)e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}.\hfill \end{array}$$

The result (1) strictly holds. Otherwise, $d_3=2$. Let $v_4$ be the vertex adjacent to the vertex $v_2$ other than $v_1$ and $v_3$. Since $d_2=3$ and $d_3=2$, we have

$${\displaystyle \frac{d_2+d_3}{2\sqrt{d_2{d}_3}}}=\sqrt{{\displaystyle \frac{25}{24}}}.$$

Again, since $d_2=3$ and $d_4\le n-4$, by Lemma 5, we obtain

$${\displaystyle \frac{d_2+d_4}{2\sqrt{d_2{d}_4}}}\le {\displaystyle \frac{n-1}{2\sqrt{3(n-4)}}}.$$

Since $n\ge 10$, by Lemma 2, we obtain

$$f\left(n\right)={\displaystyle \frac{n}{2\sqrt{n-1}}}-{\displaystyle \frac{n-1}{2\sqrt{3(n-4)}}}\ge f\left(10\right)={\displaystyle \frac{5}{3}}-\sqrt{{\displaystyle \frac{9}{8}}}>0.606.$$

From the above results, we obtain

$$\begin{array}{cc}\hfill e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}+e>e^{{\displaystyle \frac{n-1}{2\sqrt{3(n-4)}}}+0.606}+e>1.83e^{{\displaystyle \frac{n-1}{2\sqrt{3(n-4)}}}}+e& >e^{{\displaystyle \frac{n-1}{2\sqrt{3(n-4)}}}}+0.83e^{{\displaystyle \frac{\sqrt{n+2}}{2\sqrt{3}}}}+e\hfill \\ \hfill & \ge e^{{\displaystyle \frac{n-1}{2\sqrt{3(n-4)}}}}+1.83e\hfill \\ \hfill & >e^{{\displaystyle \frac{n-1}{2\sqrt{3(n-4)}}}}+e^{\sqrt{{\displaystyle \frac{25}{24}}}}\hfill \\ \hfill & \ge e^{{\displaystyle \frac{d_2+d_3}{2\sqrt{d_2{d}_3}}}}+e^{{\displaystyle \frac{d_2+d_4}{2\sqrt{d_2{d}_4}}}}.\hfill \end{array}$$

Let $E_2=E_1\cup \left\{v_2{v}_4\right\}$. By (3) and (4), we obtain

$$e^{{\displaystyle \frac{d_1+d_2}{2\sqrt{d_1{d}_2}}}}<e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}},e^{{\displaystyle \frac{d_3+d_1}{2\sqrt{d_3{d}_1}}}}<e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}},\mathrm{and}{e}^{{\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}}<e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}\mathrm{for}\mathrm{any}{v}_i{v}_j\in E\left(G\right)\setminus E_2.$$

Using the above results with $|E\left(G\right)\setminus E_2|=n-4$, we obtain

$$\begin{array}{cc}\hfill EAG\left(G\right)& =e^{{\displaystyle \frac{d_1+d_2}{2\sqrt{d_1{d}_2}}}}+e^{{\displaystyle \frac{d_2+d_3}{2\sqrt{d_2{d}_3}}}}+e^{{\displaystyle \frac{d_3+d_1}{2\sqrt{d_3{d}_1}}}}+e^{{\displaystyle \frac{d_2+d_4}{2\sqrt{d_2{d}_4}}}}+\sum _{v_i{v}_j\in E\left(G\right)\setminus E_2}{e}^{{\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}}\hfill \\ \hfill & <2e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}+e+(n-3)e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}.\hfill \end{array}$$

The result (1) strictly holds.

Case 1.3.

$d_2\ge 4$. Let $S_1=\{v_2{v}_j\in E\left(G\right)|\mathrm{for}\mathrm{all}{v}_j\in N_G\left(v_2\right)\}$, $S=S_1\cup \left\{v_1{v}_3\right\}$ and $E_3=E\left(G\right)\setminus S$. Since G is a unicyclic graph of order n, we have $d_1+d_2\le n+1$. Thus, we have $d_1\le n-3$, $4\le d_2\le {\displaystyle \frac{n+1}{2}}$, and $d_3\ge 2$. Since $d_1\ge d_2\ge d_3$, by Lemma 5, we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{d_1+d_2}{2\sqrt{d_1{d}_2}}}\le {\displaystyle \frac{n+1}{4\sqrt{n-3}}},{\displaystyle \frac{d_2+d_3}{2\sqrt{d_2{d}_3}}}\le {\displaystyle \frac{n+5}{4\sqrt{n+1}}},{\displaystyle \frac{d_1+d_3}{2\sqrt{d_1{d}_3}}}\le {\displaystyle \frac{n-1}{2\sqrt{2(n-3)}}}<{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}.\end{array}$$

(5)

Since $|E_3|=|E\left(G\right)\setminus S|=n-d_2-1$, by (4), we obtain

$$\begin{array}{cc}\hfill \sum _{v_i{v}_j\in E_3}{e}^{{\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}}& <(n-d_2-1)e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}.\hfill \end{array}$$

(6)

Claim 1.

$$2e^{{\displaystyle \frac{n+3}{2\sqrt{2(n+1)}}}}+e^{{\displaystyle \frac{n+1}{4\sqrt{n-3}}}}+e^{{\displaystyle \frac{n+5}{4\sqrt{n+1}}}}<2e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}+e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}.$$

Proof of Claim 1.

For $10\le n\le 11$, in accordance to Mathematica [43], one can easily check that the result holds. Otherwise, $n\ge 12$. By Lemma 3, the function $g\left(x\right)={\displaystyle \frac{x}{2\sqrt{x-1}}}-{\displaystyle \frac{x+3}{2\sqrt{2(x+1)}}}$ is increasing on $x\ge 12$; hence,

$$g\left(x\right)\ge g\left(12\right)={\displaystyle \frac{6}{\sqrt{11}}}-{\displaystyle \frac{15}{2\sqrt{26}}}>0.338.$$

Since $n\ge 12$, from the above, we have

$${\displaystyle \frac{n}{2\sqrt{n-1}}}-{\displaystyle \frac{n+3}{2\sqrt{2(n+1)}}}>0.338>ln1.4,$$

$$\begin{array}{c}\hfill \mathrm{that}\mathrm{is},e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}-{\displaystyle \frac{n+3}{2\sqrt{2(n+1)}}}}>1.4,\mathrm{that}\mathrm{is},e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}>1.4e^{{\displaystyle \frac{n+3}{2\sqrt{2(n+1)}}}}.\end{array}$$

(7)

By Lemma 4, the function $q\left(x\right)={\displaystyle \frac{x+3}{2\sqrt{2(x+1)}}}-{\displaystyle \frac{x+5}{4\sqrt{x+1}}}$ is increasing on $x\ge 12$; hence,

$$q\left(x\right)\ge q\left(12\right)={\displaystyle \frac{15}{2\sqrt{26}}}-{\displaystyle \frac{17}{4\sqrt{13}}}>0.29.$$

Since $n\ge 12$, from the above, we have

$${\displaystyle \frac{n+3}{2\sqrt{2(n+1)}}}-{\displaystyle \frac{n+5}{4\sqrt{n+1}}}>0.29>ln\left({\displaystyle \frac{5}{4}}\right),$$

$$\begin{array}{c}\hfill \mathrm{that}\mathrm{is},e^{{\displaystyle \frac{n+3}{2\sqrt{2(n+1)}}}-{\displaystyle \frac{n+5}{4\sqrt{n+1}}}}>{\displaystyle \frac{5}{4}},\mathrm{that}\mathrm{is},0.8e^{{\displaystyle \frac{n+3}{2\sqrt{2(n+1)}}}}>e^{{\displaystyle \frac{n+5}{4\sqrt{n+1}}}}.\end{array}$$

(8)

Since $n>5$, one can easily see that

$${\displaystyle \frac{n+1}{4\sqrt{n-3}}}<{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}},\mathrm{that}\mathrm{is},e^{{\displaystyle \frac{n+1}{4\sqrt{n-3}}}}<e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}.$$

Using the above result with (7) and (8), we obtain

$$\begin{array}{cc}\hfill 2e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}+e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}& >2.8e^{{\displaystyle \frac{n+3}{2\sqrt{2(n+1)}}}}+e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}\hfill \\ \hfill & >2e^{{\displaystyle \frac{n+3}{2\sqrt{2(n+1)}}}}+e^{{\displaystyle \frac{n+5}{4\sqrt{n+1}}}}+e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}\hfill \\ \hfill & >2e^{{\displaystyle \frac{n+3}{2\sqrt{2(n+1)}}}}+e^{{\displaystyle \frac{n+5}{4\sqrt{n+1}}}}+e^{{\displaystyle \frac{n+1}{4\sqrt{n-3}}}}.\hfill \end{array}$$

This completes the proof of Claim 1. □

For $v_j\in N_G\left(v_2\right)\setminus \{v_1,v_3\}$, we have $4\le d_2\le {\displaystyle \frac{n+1}{2}}$, $1\le d_j\le {\displaystyle \frac{n-3}{2}}$, and by Lemma 5, we obtain

$${\displaystyle \frac{d_2+d_j}{2\sqrt{d_2{d}_j}}}\le {\displaystyle \frac{n+3}{2\sqrt{2(n+1)}}}.$$

Using the above result with (5), we obtain

$$\begin{array}{cc}\hfill \sum _{v_j:v_j\in N_G\left(v_2\right)}{e}^{{\displaystyle \frac{d_2+d_j}{2\sqrt{d_2{d}_j}}}}& =e^{{\displaystyle \frac{d_1+d_2}{2\sqrt{d_1{d}_2}}}}+e^{{\displaystyle \frac{d_2+d_3}{2\sqrt{d_2{d}_3}}}}+\sum _{v_j:v_j\in N_G\left(v_2\right)\setminus \{v_1,v_3\}}{e}^{{\displaystyle \frac{d_2+d_j}{2\sqrt{d_2{d}_j}}}}\hfill \\ \hfill & \le e^{{\displaystyle \frac{n+1}{4\sqrt{n-3}}}}+e^{{\displaystyle \frac{n+5}{4\sqrt{n+1}}}}+(d_2-2)e^{{\displaystyle \frac{n+3}{2\sqrt{2(n+1)}}}}\hfill \\ \hfill & \le e^{{\displaystyle \frac{n+1}{4\sqrt{n-3}}}}+e^{{\displaystyle \frac{n+5}{4\sqrt{n+1}}}}+2e^{{\displaystyle \frac{n+3}{2\sqrt{2(n+1)}}}}+(d_2-4)e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}\hfill \end{array}$$

(9)

as

$${\displaystyle \frac{n+3}{2\sqrt{2(n+1)}}}<{\displaystyle \frac{n}{2\sqrt{n-1}}}.$$

Using (5), (6), (9) and Claim 1, we obtain

$$\begin{array}{cc}\hfill EAG\left(G\right)& =\sum _{v_i{v}_j\in E\left(G\right)}{e}^{{\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}}\hfill \\ \hfill & =e^{{\displaystyle \frac{d_1+d_3}{2\sqrt{d_1{d}_3}}}}+\sum _{v_j:v_j\in N_G\left(v_2\right)}{e}^{{\displaystyle \frac{d_2+d_j}{2\sqrt{d_2{d}_j}}}}+\sum _{v_i{v}_j\in E_3}{e}^{{\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}}\hfill \\ \hfill & <e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}+e^{{\displaystyle \frac{n+1}{4\sqrt{n-3}}}}+e^{{\displaystyle \frac{n+5}{4\sqrt{n+1}}}}+2e^{{\displaystyle \frac{n+3}{2\sqrt{2(n+1)}}}}+(d_2-4)e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}+(n-d_2-1)e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}\hfill \\ \hfill & <(n-3)e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}+2e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-2)}}}}<(n-3)e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}+2e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}+e.\hfill \end{array}$$

The result (1) strictly holds.

Case 2.

$k\ge 4$. Let $v_1,v_2,\dots ,v_k$ be the vertices on the cycle $C_k(k\ge 4)$. We can assume that $d_1=max\{d_i:v_i\in V\left(C_k\right)\}$. Then, $d_1\le n-k+2\le n-2$ and $d_i\ge 2(2\le i\le k)$. Then, by Lemma 5, for $v_1{v}_2\in E\left(C_k\right)$ and $v_1{v}_k\in E\left(C_k\right)$, we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{d_1+d_2}{2\sqrt{d_1{d}_2}}}\le {\displaystyle \frac{n}{2\sqrt{2(n-2)}}}\mathrm{and}{\displaystyle \frac{d_1+d_k}{2\sqrt{d_1{d}_k}}}\le {\displaystyle \frac{n}{2\sqrt{2(n-2)}}}.\end{array}$$

Since $v_2{v}_3\in E\left(C_k\right)$ and $v_3{v}_4\in E\left(C_k\right)$ with $2\le d_2,d_3,d_4\le {\displaystyle \frac{n-k+4}{2}}\le {\displaystyle \frac{n}{2}}$, by Lemma 5, we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{d_2+d_3}{2\sqrt{d_2{d}_3}}}\le {\displaystyle \frac{n+4}{4\sqrt{n}}}\mathrm{and}{\displaystyle \frac{d_3+d_4}{2\sqrt{d_3{d}_4}}}\le {\displaystyle \frac{n+4}{4\sqrt{n}}}.\end{array}$$

(10)

Claim 2.

$$2e^{{\displaystyle \frac{n+4}{4\sqrt{n}}}}<e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}+e.$$

Proof of Claim 2.

For $10\le n\le 12$, in accordance to Mathematica [43], one can easily check that the result holds. Otherwise, $n\ge 13$. Let us consider a function

$$h\left(x\right)={\displaystyle \frac{x}{\sqrt{x-1}}}-{\displaystyle \frac{x+4}{2\sqrt{x}}},x\ge 13.$$

Then, we have

$$h^{\prime }\left(x\right)={\displaystyle \frac{1}{4x^{3/2}{(x-1)}^{3/2}}}\left[2x^{3/2}(x-2)-(x-4){(x-1)}^{3/2}\right]>0.$$

Thus, $h\left(x\right)$ is an increasing function on $x\ge 13$ and, hence,

$$h\left(x\right)\ge h\left(13\right)={\displaystyle \frac{13}{\sqrt{12}}}-{\displaystyle \frac{17}{2\sqrt{13}}}>1.395>2ln2.$$

From the above, we obtain

$${\displaystyle \frac{n}{\sqrt{n-1}}}-{\displaystyle \frac{n+4}{2\sqrt{n}}}>2ln2,$$

that is,

$$e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}-{\displaystyle \frac{n+4}{4\sqrt{n}}}}>2,$$

that is,

$$2e^{{\displaystyle \frac{n+4}{4\sqrt{n}}}}<e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}<e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}+e.$$

This completes the proof of Claim 2. □

Let p be the number of pendant vertices in G. Since G has a cycle length of at least 4, we have $p\le n-4$. Since ${\displaystyle \frac{n+1}{\sqrt{2(n-1)}}}<{\displaystyle \frac{n}{\sqrt{n-1}}}$, using (3), we obtain

$$\begin{array}{c}\hfill \sum _{{\displaystyle \genfrac{}{}{0pt}{}{v_i{v}_j\in E\left(G\right)\setminus \{v_2{v}_3,v_3{v}_4\},}{d_i\ge d_j\ge 2}}}{e}^{{\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}}<(n-p-2)e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}\le 2e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}+(n-p-4)e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}.\end{array}$$

Using the above result with (2), (10) and Claim 2, we obtain

$$\begin{array}{cc}\hfill EAG\left(G\right)& =\sum _{{\displaystyle \genfrac{}{}{0pt}{}{v_i{v}_j\in E\left(G\right),}{d_i\ge d_j=1}}}{e}^{{\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}}+\sum _{{\displaystyle \genfrac{}{}{0pt}{}{v_i{v}_j\in E\left(G\right),}{d_i\ge d_j\ge 2}}}{e}^{{\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}}\hfill \\ \hfill & <p{e}^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}+e^{{\displaystyle \frac{d_2+d_3}{2\sqrt{d_2{d}_3}}}}+e^{{\displaystyle \frac{d_3+d_4}{2\sqrt{d_3{d}_4}}}}+\sum _{{\displaystyle \genfrac{}{}{0pt}{}{v_i{v}_j\in E\left(G\right)\setminus \{v_2{v}_3,v_3{v}_4\},}{d_i\ge d_j\ge 2}}}{e}^{{\displaystyle \frac{d_i+d_j}{2\sqrt{d_i{d}_j}}}}\hfill \\ \hfill & <p{e}^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}+2e^{{\displaystyle \frac{n+4}{4\sqrt{n}}}}+(n-p-4)e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}+2e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}\hfill \\ \hfill & <(n-3)e^{{\displaystyle \frac{n}{2\sqrt{n-1}}}}+2e^{{\displaystyle \frac{n+1}{2\sqrt{2(n-1)}}}}+e.\hfill \end{array}$$

The result (1) strictly holds. This completes the proof of the theorem. □

3. Concluding Remarks

The exponential arithmetic–geometric index (denoted as $EAG$) is a recently introduced topological index in the field of mathematical chemistry. This index was first brought into the literature by Rada [18]. In [37], the authors proposed an open problem: to characterize the unicyclic graph that attains the maximum $EAG$ value among all unicyclic graphs of a given order n. In this paper, we address and resolve this open problem by providing a complete characterization of an extremal graph. Our findings show that, among all unicyclic graphs with n vertices, the graph $S_{n,3}$ uniquely maximizes the $EAG$ index. This result not only settles the open problem posed by Cruz, Rada, and Sánchez but also contributes to a deeper understanding of the behavior of the $EAG$ index in relation to graph structures.