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

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 that is associated with a molecular graph, used to describe specific physicochemical properties. Since Wiener introduced the first such index [43], many other indices have been developed depending on various graph parameters, including degree, distance, and eccentricity [2,3,15,26,27,28,30]. 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 [37]. Let G be a simple connected graph, whose vertex set $V\left(G\right)=\{v_1,v_2,\dots ,v_n\}$ and the edge set $E\left(G\right)$. 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. [38,39] calculated $AG$ for different classes of graphs. Milovanović et al. [31] determined some upper bounds on $AG$ for simple connected graphs. Li and Zhang [25] provided sharp bounds on $AG$ of line graphs. Hertz et al. [23] identified the extremal chemical graphs for $AG$. Maximal chemical trees with respect to $AG$ were characterized in [41]. More studies on $AG$ can be found in references [22,36,45]. To improve the discriminative ability of topological indices, researchers introduced exponential versions of these indices in 2019 [35]. In [8,10], researchers provided a characterization of the extremal trees for the exponential Randić index. Das et al. [14] explored the extremal graphs for the exponential atom-bond connectivity index. Jahanbani et al. [24] determined the lowest value of the exponential forgotten index for trees. Das and Mondal [16] examined sharp bounds on the exponential geometric-arithmetic index of bipartite graphs. More studies on this concept can be found in references [4,5,6,9,11,17]. In this paper, we focus on the exponential arithmetic-geometric index ($EAG$), which is defined as

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

The investigation of extremal structures within different classes of graphs is a critical area of research in discrete mathematics. In particular, the identification of extremal unicyclic graphs is a major challenge. Moon and Park [32] explored the extremal values of the geometric-arithmetic index for unicyclic graphs and identified those that attained these extreme values. Extremal unicyclic graphs for $AG$ were characterized in [42]. Liu et al. [29] provided a complete classification of extremal unicyclic graphs for the Lanzhou index. In [18,20], researchers described the maximal unicyclic graph for exponential second Zagreb and augmented Zagreb indices. For more insights into extremal unicyclic graphs, readers can refer to [1,7,12,13,19,21,33,34]. Cruz et al.[12] 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 to generate the maximal unicyclic graph in the case of $EAG$. Because of this, the following open problem was posed in [12]:

Problem 1. [12] 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 result.

Lemma 1. 

Let

$$f\left(x\right)={\displaystyle \frac{x}{2\sqrt{x-1}}}-{\displaystyle \frac{x-1}{2\sqrt{3(x-4)}}},x\ge 5.$$

Then $f\left(x\right)$ is a strictly increasing function on $x\ge 5$.

Proof. 

Since

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

we have

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

For $x=5$, we have $f^{\prime }\left(x\right)>0$. Otherwise, $x\ge 6$. One can easily see that

$${\displaystyle \frac{1}{\sqrt{x-1}}}>{\displaystyle \frac{1}{\sqrt{3}\sqrt{x-4}}}\mathrm{and}{\displaystyle \frac{x-2}{x-1}}>{\displaystyle \frac{x-7}{x-4}},$$

and hence $f^{\prime }\left(x\right)>0$. This proves the result. □

Lemma 2. 

Let

$$g\left(x\right)={\displaystyle \frac{x}{2\sqrt{x-1}}}-{\displaystyle \frac{x+3}{2\sqrt{2(x+1)}}},x\ge 5.$$

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 3. 

Let

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

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

Proof. 

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. □

Lemma 4. 

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 $S_n^{\prime }$ 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 $G\cong S_n^{\prime }$.

Proof. 

Let $\Delta$ be the maximum degree in the unicyclic graph G. If $\Delta =n-1$, then $G\cong S_n^{\prime }$ 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 $n\le 9$, by Sage [40], one can easily check that the result (1) holds with equality if and only if $G\cong S_n^{\prime }$. Otherwise, $\Delta \le n-2$ 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 4, 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 4, 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 $d_2=d_3=2$ and 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

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

The result (1) strictly holds.

Case 1.2. $d_2=3$. If $d_3=3$, then similarly, by $\mathbf{Case}\mathbf{1}.\mathbf{1}$, 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 4, 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 1, 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

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

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 $S_1=\{v_2{v}_j\in E\left(G\right)|forall{v}_j\in N_G\left(v_2\right)\}$, $S=S_1\cup \left\{v_1{v}_3\right\}$ and $E_3=E\left(G\right)\setminus S$. In this case $d_1\le n-3$, $4\le d_2\le {\displaystyle \frac{n+1}{2}}$, and $d_3\ge 2$. Since $d_1\ge d_2\ge d_3$, by Lemma 4, 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 $10\le n\le 11$, by Mathematica [44], one can easily check that the result holds. Otherwise, $n\ge 12$. By Lemma 2, 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 $x\ge 12$, and 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 3, 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 $x\ge 12$, and 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 4, 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

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

The result (1) strictly holds.

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 4, for $v_1{v}_2\in E\left(C_k\right)$ and $v_1{v}_k\in E\left(C_k\right)$, we obtain

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

(10)

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 4, we obtain

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

(11)

Claim 2.

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

Proof of Claim 2. For $10\le n\le 12$, by Mathematica [44], 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 cycle length 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

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

(12)

Using the above result with (2), (11) 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

An open problem that was stated in [12] has been resolved in this paper. It has been determined that the maximal unicyclic graph with respect to $EAG$ can be described in terms of graph order n.