Open problem on the maximum exponential augmented Zagreb index of unicyclic graphs

Abstract

A topological index is a numerical property of a molecular graph that explains structural features of molecules. The potential of topological indices to discriminate between distinct structures is a significant topic to investigate. In this context, the exponential degree-based indices were put forward in the literature. The present work focuses on the exponential augmented Zagreb index (EAZ), which is defined for a graph G as

$$\begin{aligned} EAZ(G)=\sum \limits _{v_{i}v_{j} \in E(G)}\,e^{\displaystyle {\left( \frac{{d_i\,d_j}}{d_i+d_j-2}\right) ^{3}}}, \end{aligned}$$

where $d_i$ represents the degree of the vertex $v_i$and E(G) denotes the edge set of G. This work characterizes the maximal unicyclic graph for EAZ in terms of graph order, which was posed as an open problem in the recent article Cruz et al. (MATCH Commun Math Comput Chem 88:481-503, 2022).

1 Introduction

Throughout this article, we consider G as a simple connected graph of order n and size m having vertex set $V(G)=\lbrace v_{1},\,v_{2},\,\ldots ,v_{n}\rbrace$ and edge set E(G). A graph G is said to be unicyclic if $m=n$. The degree of vertex $v_{i} \in V(G)$ is defined as the number of vertices adjacent to $v_i$. Let $N_G(v_i)$ be the neighbor set of the vertex $v_i$, that is, $N_G(v_i)=\{v_k\in V(G):\,v_iv_k\in E(G)\}$. The double star $DS_{p,q}$ is a graph that is obtained after joining two central vertices of two stars $S_p$ and $S_q$.

Topological indices are numerical descriptors used in the field of mathematical chemistry and chem-informatics to characterize the topology of molecular structures. These indices provide a quantitative measure of the structural features of a molecule, focusing on the spatial arrangement of atoms and bonds rather than their chemical nature. The origin of topological indices lies in graph theory, where a molecule is depicted as a graph featuring atoms as vertices and bonds as edges. From a mathematical perspective, it is conceptualized as a function mapping all molecular graphs to real numbers, ensuring its invariance under graph isomorphism. These indices are employed to explain different physical and chemical properties of molecules, which supports researchers working on drug design, material science, and other branches of chemistry (Basak and Vracko 2020). After Wiener’s seminal work in 1947 Wiener (1947), a multitude of indices have been introduced in literature, utilizing various parameters such as degree, distance, eccentricity, and spectrum (Liu 2023; Hayat et al. 2023; Liu 2022; Maitreyi et al. 2023; Liu and Huang 2023; Du and Dimitrov 2020; Ghanbari 2022; Ali et al. 2021; Das et al. 2018; Gutman and Das 2004; Das and Vetrík 2023; Hosseini et al. 2022). While each category has its own set of applications and benefits, the domain of chemical graph theory is notably influenced by degree-based indices. The augmented Zagreb index (AZ) (Furtula et al. 2010) is one of the well-known degree based indices, which is formulated as

$$\begin{aligned} AZ(G)=\sum \limits _{v_iv_j \in E(G)}\,\left( \frac{{d_i\,d_j}}{d_i+d_j-2}\right) ^{3}. \end{aligned}$$

Widespread investigation of the AZ index is apparent in Chen et al. (2022); Sun et al. (2018); Cheng et al. (2021); Ali et al. (2021); Li et al. (2019, 2021); Ali (2021). To investigate the discrimination capability of topological indices, Rada (2019) proposed the exponential degree-based indices. A comprehensive identification of extremal trees for such invariants was reported in Cruz and Rada (2019). The path graph was established to be the maximal tree for the exponential Randić index (Cruz et al. 2020). Eliasi provided characterization of the maximal unicyclic and bicyclic graphs with respect to the exponential second Zagreb index in Eliasi (2022). For further insight on this direction, readers are referred to Xu et al. (2023); Cruz et al. (2021); Das et al. (2021); Wang and Wu (2022); Das and Mondal (2023); Carballosa et al. (2023). The present work focuses on the exponential augmented Zagreb index (Rada 2019), which is defined as

$$\begin{aligned} EAZ(G)=\sum \limits _{v_iv_j \in E(G)}\,e^{\displaystyle {\left( \frac{{d_i\,d_j}}{d_i+d_j-2}\right) ^{3}}}. \end{aligned}$$

Cruz and Rada (2019) determined that the star graph serves as the smallest tree for EAZ. They left the problem of identifying the maximal tree open, and this was resolved in Das et al. (2021). The distinct extremal graphs for EAZ within the set of graphs with n non-isolated vertices were identified in Cruz and Rada (2022). For more works on the EAZ index, readers are referred to Das and Mondal (2024); Das et al. (2024). Let $C_n$ and $S_{n}^{\prime }$ be cycle and unicyclic graphs of order n, respectively, where $S_n^{\prime }$ is generated from the star graph $S_n$ of order n by attaching two pendent vertices. Most of the degree based indices yield $C_n$ or $S_{n}^{\prime }$ as maximal unicyclic graph in terms of graph order n. A unified approach reported by Cruz et al. (2022) to characterize extremal unicyclic graphs for degree-based indices also confirmed this fact. However, in case of EAZ, this approach is inadequate to generate maximal unicyclic graph. As a consequence, the problem of characterizing maximal unicyclic graph for EAZ in terms of graph order n is posed as an open problem in Cruz et al. (2022). The ultimate aim of this work is to solve this problem. We are surprised to state an amazing property of this extremal graph (see, Fig. 1) that the contribution of EAZ corresponding to one edge $v_1v_2$ of this structure greater than the EAZ value of all other unicyclic graphs. Such scenario is rare to observe in extremal graph theory literature.

2 Main result

In this section, our aim is to explore the maximal unicyclic graph for the EAZ index.Let $C^{3}_{\lfloor \frac{n-3}{2} \rfloor ,\,\lceil \frac{n-3}{2} \rceil }$ be a unicyclic graph (see, Fig. 1) of order n containing a cycle $C_{3}:v_{1}v_{2}v_{3}v_{1}$ such that $\lceil \frac{n-3}{2} \rceil$ pendent edges are incident on $v_1$ and $\lfloor \frac{n-3}{2} \rfloor$ pendent edges are incident on $v_2$. We first obtain the following lemma which will be employed to generate the main outcome.

Fig. 1

Unicyclic graph $C^{3}_{\lfloor \frac{n-3}{2}\rfloor , \, \lceil \frac{n-3}{2}\rceil }$

Lemma 1

Let $C^{3}_{\lfloor \frac{n-3}{2} \rfloor ,\,\lceil \frac{n-3}{2} \rceil }$ be a unicyclic graph of order n as displayed in Fig. 1. Then there exists an edge $v_1v_2 \in E\left( C^{3}_{\lfloor \frac{n-3}{2} \rfloor ,\,\lceil \frac{n-3}{2} \rceil }\right)$ such that

$$\begin{aligned} e^{\displaystyle {\left( \frac{d_1\,d_2}{d_1+d_2-2}\right) ^3}}>\left\{ \begin{array}{ll} n\,e^{\displaystyle {\frac{(n+3)^3}{64}}} &{} \hbox { if { n} is odd,} \\ n\,e^{\displaystyle {\frac{1}{64}\left( \frac{(n+4)\,(n-2)}{n-1}\right) ^{3}}} &{} \hbox { if } n\,\, \hbox {is even.} \end{array} \right. \end{aligned}$$

Proof

Using Sage (2015), one can easily verify that the assertion holds for $n \le 9$. We now investigate the case where $n \ge 10$.

First we assume that $n=2p+1$. Then $d_1=p+1=d_2$ and

$$\begin{aligned} \displaystyle {\left( \frac{d_1\,d_2}{d_1+d_2-2}\right) ^3}=\frac{(p+1)^6}{8p^3}=\frac{1}{8}\,\left( p+2+\frac{1}{p}\right) ^3. \end{aligned}$$

(1)

Now,

$$\begin{aligned} \frac{1}{2}\,\left( p+2+\frac{1}{p}\right) -\frac{1}{2}\,(p+2)=\frac{1}{2p}. \end{aligned}$$

Then

$$\begin{aligned} \frac{1}{8}\,\left( p+2+\frac{1}{p}\right) ^3&= \frac{(p+2)^3}{8}+\frac{3\,(p+2)^2}{8p}+\frac{3\,(p+2)}{8\,p^2}+\frac{1}{8p^3}\nonumber \\&>\frac{(p+2)^3}{8}+\frac{3\,(p+2)^2}{8p}\nonumber \\&>\frac{(n+3)^3}{64}+\frac{3\,(n+7)}{16}. \end{aligned}$$

(2)

Now,

$$\begin{aligned} e^{\displaystyle {\frac{3\,(n+7)}{16}}}>1+\frac{3\,(n+7)}{16}+\frac{9\,(n+7)^{2}}{512}+\frac{27\,(n+7)^{3}}{24,576}+\frac{81\,(n+7)^{4}}{1572864}. \end{aligned}$$

For $n\ge 11$, we obtain

$$\begin{aligned} 1+\frac{3\,(n+7)}{16}+\frac{9\,(n+7)^{2}}{512}+\frac{27\,(n+7)^{3}}{24,576}+\frac{81\,(n+7)^{4}}{1572864}>n \end{aligned}$$

and hence

$$\begin{aligned} e^{\displaystyle {\frac{3\,(n+7)}{16}}}>n. \end{aligned}$$

Using the above result with (2), from (1), we obtain

$$\begin{aligned} e^{\displaystyle {\left( \frac{d_1\,d_2}{d_1+d_2-2}\right) ^3}}> e^{\displaystyle {\frac{(n+3)^3}{64}}}\times e^{\displaystyle {\frac{3\,(n+7)}{16}}}>n\,e^{\displaystyle {\frac{(n+3)^3}{64}}}. \end{aligned}$$

Next we assume that $n=2p$. Then $d_1=p+1$ and $d_2=p$. Now,

$$\begin{aligned} \displaystyle {\left( \frac{d_1\,d_2}{d_1+d_2-2}\right) ^{3}} =&\left( \frac{p\,(p+1)}{2p-1}\right) ^{3}\nonumber \\ =&\left( \frac{(p+2)\,(p-1)}{2p-1}+\frac{2}{2p-1}\right) ^{3}\nonumber \\ =&\left( \frac{(p+2)\,(p-1)}{2p-1}\right) ^{3}+\frac{8}{(2p-1)^{3}}+\frac{6\,(p+2)^{2}(p-1)^{2}}{(2p-1)^3}\nonumber \\&\quad +\frac{12\,(p+2)\,(p-1)}{(2p-1)^{3}}\nonumber \\ >&\left( \frac{(p+2)\,(p-1)}{2p-1}\right) ^{3}+\frac{6\,(p+2)^{2}(p-1)^{2}}{(2p-1)^3}\nonumber \\ =&\frac{1}{64}\left( \frac{(n+4)\,(n-2)}{n-1}\right) ^{3}+\frac{3\,(n+4)^{2}\,(n-2)^{2}}{8\,(n-1)^{3}}. \end{aligned}$$

(3)

Now one can write

$$\begin{aligned} e^{\displaystyle {\frac{3\,(n+4)^{2}\,(n-2)^{2}}{8\,(n-1)^{3}}}}>1+\frac{3\,(n+4)^{2}\,(n-2)^{2}}{8\,(n-1)^{3}}+\frac{9\,(n+4)^{4}\,(n-2)^{4}}{128\,(n-1)^{6}}. \end{aligned}$$

For $n\ge 10$, it is easy to check that

$$\begin{aligned} 1+\frac{3\,(n+4)^{2}\,(n-2)^{2}}{8\,(n-1)^{3}}+\frac{9\,(n+4)^{4}\,(n-2)^{4}}{128\,(n-1)^{6}}>n, \end{aligned}$$

which implies

$$\begin{aligned} e^{\displaystyle {\frac{3\,(n+4)^{2}\,(n-2)^{2}}{8\,(n-1)^{3}}}}>n. \end{aligned}$$

In view of above result, we obtain from (3) that

$$\begin{aligned} e^{\displaystyle {\left( \frac{d_1\,d_2}{d_1+d_2-2}\right) ^3}}> & {} e^{\displaystyle {\frac{3\,(n+4)^{2}\,(n-2)^{2}}{8\,(n-1)^{3}}}}\times e^{\displaystyle {\frac{1}{64}\left( \frac{(n+4)\,(n-2)}{n-1}\right) ^{3}}}\\> & {} n\,e^{\displaystyle {\frac{1}{64}\left( \frac{(n+4)\,(n-2)}{n-1}\right) ^{3}}}. \end{aligned}$$

Hence, the proof is done. $\square$

Lemma 2

Let G be a unicyclic graph of even order n. Then there exists $v_{\alpha },\,v_{\beta } \in V(G)$ such that $d_{\alpha }=d_{\beta }=\frac{n}{2}$ if and only if $G \cong H_k$, $k=1,\,2,\,3,\,4,\,5$, where $H_k$’s are reported in Figs. 2, 3, 4 and 5.

Proof

Suppose $n=2p$. First assume that there exists $v_{\alpha },\,v_{\beta } \in V(G)$ such that $d_{\alpha }=d_{\beta }=\frac{n}{2}=p$. We have

$$\begin{aligned} d_\alpha +d_\beta =|N_G(v_\alpha )\cup N_G(v_\beta )|+|N_G(v_\alpha )\cap N_G(v_\beta )|. \end{aligned}$$

(4)

Let $d_G(v_\alpha ,\,v_\beta )$ be the shortest distance between $v_\alpha$ and $v_\beta$ in G. Then $d_G(v_\alpha ,\,v_\beta )\ge 1$. Now we build the proof in the three cases listed below.

Fig. 2

Two unicyclic graphs $H_1$ and $H_2$

Fig. 3

Unicyclic graph $H_3$

Fig. 4

Unicyclic graph $H_4$

$\mathbf{Case\,1.}$ $d_G(v_\alpha ,\,v_\beta )=1$. Since G is a unicyclic graph, $0\le |N_G(v_\alpha )\cap N_G(v_\beta )|\le 1$. First we assume that $|N_G(v_\alpha )\cap N_G(v_\beta )|=0$. By (4), we have $|N_G(v_\alpha )\cup N_G(v_\beta )|=d_{\alpha }+d_{\beta }=n$. All the vertices in S are adjacent to either $v_\alpha$ or $v_\beta$, where $S=V(G)\backslash \{v_\alpha ,\,v_\beta \}$. Since $d_{\alpha }=d_{\beta }=\frac{n}{2}=p$ and $v_\alpha v_\beta \in E(G)$, $DS_{p-1,p-1}$ is a subgraph of G. Since G is a unicyclic graph with $d_{\alpha }=d_{\beta }=\frac{n}{2}=p$, we have $v_iv_j\in E(G)$, where $v_i\in N_G(v_\alpha )$, $v_j\in N_G(v_\beta )$; or $v_i,\,v_j\in N_G(v_\alpha )\,(\text{ or } ~v_i,\,v_j\in N_G(v_\beta ))$, that is, $G\cong H_1$ (see, Fig. 2) or $G\cong H_3$ (see, Fig. 3).

Fig. 5

Unicyclic graph $H_5$

Fig. 6

Unicyclic graph $H_0$ of order $2p-1$

Next we assume that $|N_G(v_\alpha )\cap N_G(v_\beta )|=1$. Let $v_\gamma \in N_G(v_\alpha )\cap N_G(v_\beta )$. By (4), we have $|N_G(v_\alpha )\cup N_G(v_\beta )|=d_{\alpha }+d_{\beta }-1=n-1$. Using this result with $d_{\alpha }=d_{\beta }=\frac{n}{2}=p$ and $v_\alpha v_\beta \in E(G)$, we conclude that $H_0$ (see, Fig. 6) is a subgraph of G with $|V(H_0)|=n-1$. Let $v_n\in V(G)\backslash V(H_0)$. Since G is unicyclic of order n, vertex $v_n$ is adjacent to $v_\gamma$, or $v_nv_k\in E(G)$, where $v_k\in N_G(v_\alpha )$ or $v_k\in N_G(v_\beta )$, that is, $G\cong H_4$ (see, Fig. 4) or $G\cong H_5$ (see, Fig. 5).

$\mathbf{Case\,2.}$ $d_G(v_\alpha ,\,v_\beta )=2$. In this case $|N_G(v_\alpha )\cup N_G(v_\beta )|\le n-2$. Since G is a unicyclic graph, $1\le |N_G(v_\alpha )\cap N_G(v_\beta )|\le 2$. If $|N_G(v_\alpha )\cap N_G(v_\beta )|=1$, then by (4), we obtain $d_\alpha +d_\beta =|N_G(v_\alpha )\cup N_G(v_\beta )|+|N_G(v_\alpha )\cap N_G(v_\beta )|\le n-1<n=d_\alpha +d_\beta$, a contradiction. Otherwise, $|N_G(v_\alpha )\cap N_G(v_\beta )|=2$. Again by (4), we have $|N_G(v_\alpha )\cup N_G(v_\beta )|=n-2$. Since $v_\alpha ,\,v_\beta \notin N_G(v_\alpha )\cup N_G(v_\beta )$, all the vertices in S are adjacent to $v_\alpha$ or $v_\beta$ or both, where $S=V(G)\backslash \{v_\alpha ,\,v_\beta \}$. Since $d_{\alpha }=d_{\beta }=\frac{n}{2}=p$ with $|N_G(v_\alpha )\cap N_G(v_\beta )|=2$, we must have $G\cong H_2$ (see, Fig. 2).

$\mathbf{Case\,3.}$ $d_G(v_\alpha ,\,v_\beta )\ge 3$. In this case $|N_G(v_\alpha )\cap N_G(v_\beta )|=0$ and $v_\alpha ,\,v_\beta \notin N_G(v_\alpha )\cup N_G(v_\beta )$. This with (4), we obtain $|N_G(v_\alpha )\cup N_G(v_\beta )|\le n-2<n=d_\alpha +d_\beta =|N_G(v_\alpha )\cup N_G(v_\beta )|$, a contradiction.

Conversely, we assume that $G \cong H_k,$ $k=1,\,2,\,3,\,4,\,5$. Then from Figs. 2, 3, 4 and 5, it is evident that there exists $v_{\alpha },\,v_{\beta } \in V(G)$ such that $d_{\alpha }=d_{\beta }=\frac{n}{2}$. Hence the proof is done. $\square$

Lemma 3

Let G be a unicyclic graph of even order n. If G contains an edge $v_{\alpha }v_{\beta }$ with $d_{\alpha }=d_{\beta }=\frac{n}{2},$ then

$$\begin{aligned} EAZ(G)<EAZ(C^{3}_{\lfloor \frac{n-3}{2} \rfloor ,\,\lceil \frac{n-3}{2} \rceil }). \end{aligned}$$

Proof

Employing Lemma 2, we can write $G \cong H_k,$ $k=1,\,2,\,3,\,4,\,5$. From Figs. 2, 3, 4 and 5, it is clear that

$$\begin{aligned} EAZ(H_k)=\left\{ \begin{array}{ll} e^{\frac{n^{6}}{64\,(n-2)^{3}}}+(n-4)\,e^{\left( \frac{n}{n-2}\right) ^{3}}+3e^{8} &{} \hbox { if} \ k=1,\,3, \\ (n-4)\,e^{\left( \frac{n}{n-2}\right) ^{3}}+4e^{8} &{} \hbox { if } k=2, \\ e^{\frac{n^{6}}{64\,(n-2)^{3}}}+(n-4)\,e^{\left( \frac{n}{n-2}\right) ^{3}}+2e^{\frac{27n^{3}}{(n+2)^{3}}}+e^{\frac{27}{8}} &{} \hbox { if}~ k=4, \\ e^{\frac{n^{6}}{64\,(n-2)^{3}}}+(n-5)\,e^{\left( \frac{n}{n-2}\right) ^{3}}+4e^{8} &{} \hbox { if } k=5. \\ \end{array} \right. \end{aligned}$$

One can easily find that $EAZ(H_k)<EAZ\left( C^{3}_{\lfloor \frac{n-3}{2} \rfloor ,\,\lceil \frac{n-3}{2} \rceil }\right)$ for $n=4,\,6,\,8,\,10,$ and $k=1,\,2,\,3,\,4,\,5.$ Now we consider $n \ge 12.$ Note that $d_{i} \le p$ for any $v_i \in V(H_k)$ ($k=1,\,2,\,3,\,4,\,5.$). For any $v_{i}v_{j} \in E(H_k)$ satisfying $d_i\ge d_j\ge 2$, we obtain

$$\begin{aligned} \frac{1}{d_i}+\frac{1}{d_j}-\frac{2}{d_i\,d_j}=\frac{1}{d_i}+\frac{1}{d_j}\,\left( 1-\frac{2}{d_i}\right) \ge \frac{1}{p}+\frac{1}{d_i}\,\left( 1-\frac{2}{p}\right) \ge \frac{1}{p}+\frac{1}{p}\,\left( 1-\frac{2}{p}\right) , \end{aligned}$$

which exerts

$$\begin{aligned} \displaystyle {\frac{d_i\,d_j}{d_i+d_j-2}}\le \frac{p^{2}}{2p-2}=\frac{n^{2}}{4\,(n-2)}<\frac{(n+4)\,(n-2)}{4\,(n-1)}, ~ \text{ as } ~n\ge 12. \end{aligned}$$

For any $v_{i}v_{j} \in E(H_k)$ satisfying $d_i\ge d_j=1$, we obtain

$$\begin{aligned} \frac{1}{d_i}+\frac{1}{d_j}-\frac{2}{d_i\,d_j}=1-\frac{1}{d_i}\ge \frac{1}{2}, \end{aligned}$$

that is,

$$\begin{aligned} \displaystyle {\frac{d_i\,d_j}{d_i+d_j-2}}\le 2<\frac{(n+4)\,(n-2)}{4\,(n-1)} ~ \text{ as } ~n\ge 12. \end{aligned}$$

Thus by Lemma 1, we obtain

$$\begin{aligned} EAZ(H_k)&< n\,e^{\displaystyle {\frac{1}{64}\left( \frac{(n+4)\,(n-2)}{n-1}\right) ^{3}}}<e^{\displaystyle {\left( \frac{d_1\,d_2}{d_1+d_2-2}\right) ^3}}\\&<EAZ\left( C^{3}_{\lfloor \frac{n-3}{2} \rfloor ,\,\lceil \frac{n-3}{2} \rceil }\right) \end{aligned}$$

for $k=1,\,2,\,3,\,4,\,5$. Hence the proof is completed. $\square$

We are now ready to prove our main result of this paper.

Theorem 1

Let G be a unicyclic graph of order n. Then

$$\begin{aligned} EAZ(G)\le \left\{ \begin{array}{ll} e^{\left( \frac{n\,(n+2)}{4\,(n-1)}\right) ^{3}}+\frac{n-2}{2}\,e^{\left( \frac{n+2}{n}\right) ^{3}}+\frac{n-4}{2}\,e^{\left( \frac{n}{n-2}\right) ^{3}}+2e^{8} &{}\quad \hbox {if}~n~\hbox {is even,} \\ e^{\left( \frac{(n+1)^{6}}{64\,(n-1)^{3}}\right) }+(n-3)\,e^{\left( \frac{n+1}{n-1}\right) ^{3}}+2e^{8} &{} \quad \hbox {if}~n~\hbox {is odd,} \end{array} \right. \end{aligned}$$

(5)

with equality if and only if $G\cong C^{3}_{\lfloor \frac{n-3}{2} \rfloor ,\,\lceil \frac{n-3}{2} \rceil }$.

Proof

Using Sage (2015), it is straightforward to examine the result to be true for $n\le 9$. Therefore, our task is to establish the result for $n\ge 10$. Suppose $v_iv_j \in E(G)$ with $d_i\ge d_j$. As G is a unicyclic graph, we must have $d_i+d_j\le n+1$. We consider the subsequent two cases:

$\mathbf{Case\,1.}$ $n=2p+1$. In this case $p\ge 5$. Thus we have $d_j\le p+1$ (Otherwise, $d_i\ge d_j\ge p+2$, that is, $d_i+d_j\ge 2p+4=n+3$, a contradiction). If $d_j=p+1$, then $d_i=p+1$ (Otherwise, $d_i\ge p+2$, that is, $d_i+d_j\ge 2p+3=n+2$, a contradiction) and hence $G\cong C^{3}_{\lfloor \frac{n-3}{2} \rfloor ,\,\lceil \frac{n-3}{2} \rceil }$ (see, Fig. 1). Thus we have

$$\begin{aligned} EAZ(G)=e^{\displaystyle {\left( \frac{(n+1)^{6}}{64\,(n-1)^{3}}\right) }}+(n-3)\,e^{\displaystyle {\left( \frac{n+1}{n-1}\right) ^{3}}}+2e^{8} \end{aligned}$$

and hence equality appears in (5). Otherwise, $d_j\le p$. We take into account the following two cases:

$\mathbf{Case\,1.1.}$ $d_i\le p+1$. Now,

$$\begin{aligned} \frac{1}{d_i}+\frac{1}{d_j}-\frac{2}{d_i\,d_j}&=\frac{1}{d_i}+\frac{1}{d_j}\,\left( 1-\frac{2}{d_i}\right) \ge \frac{1}{d_i}+\frac{1}{p}\,\left( 1-\frac{2}{d_i}\right) =\frac{1}{p}+\frac{1}{d_i}\,\left( 1-\frac{2}{p}\right) \\ {}&\ge \frac{1}{p}+\frac{1}{p+1}\,\left( 1-\frac{2}{p}\right) , \end{aligned}$$

that is,

$$\begin{aligned} \displaystyle {\frac{d_i\,d_j}{d_i+d_j-2}}\le \frac{p\,(p+1)}{2p-1}=\displaystyle {\frac{1}{2}\,\left( p+\frac{3}{2}+\frac{3}{2\,(2p-1)}\right) }<\frac{p+2}{2}=\frac{n+3}{4},\end{aligned}$$

as $n\ge 11$. Since G is a unicyclic graph, employing the above finding with Lemma 1, it is clear that

$$\begin{aligned} EAZ(G)&=\sum \limits _{v_iv_j\in E(G)}\,e^{\displaystyle {\left( \frac{d_i\,d_j}{d_i+d_j-2}\right) ^3}}<n\,e^{\displaystyle {\frac{(n+3)^3}{64}}}<e^{\displaystyle {\left( \frac{d_1\,d_2}{d_1+d_2-2}\right) ^3}}\\&<EAZ\left( C^{3}_{\lfloor \frac{n-3}{2} \rfloor ,\,\lceil \frac{n-3}{2} \rceil }\right) . \end{aligned}$$

Again the result (5) strictly holds.

$\mathbf{Case\,1.2.}$ $d_i\ge p+2$. Since $d_i+d_j\le n+1$, we address the two subcases listed below:

$\mathbf{Case\,1.2.1.}$ $d_i+d_j\le n$. First we have to prove that

$$\begin{aligned} \displaystyle {\frac{d_i\,d_j}{d_i+d_j-2}}<\frac{n+3}{4}. \end{aligned}$$

(6)

For $d_i\ge d_j\ge 2$, we obtain

$$\begin{aligned} \frac{1}{d_i}+\frac{1}{d_j}-\frac{2}{d_i\,d_j}\ge \frac{1}{d_j}+\frac{1}{2p+1-d_j}\,\left( 1-\frac{2}{d_j}\right) . \end{aligned}$$

Consider

$$\begin{aligned} f(x)=\frac{1}{x}+\frac{1}{2p+1-x}\,\left( 1-\frac{2}{x}\right) ,~~2\le x\le p. \end{aligned}$$

Then

$$\begin{aligned} f^{\prime }(x)&=-\frac{1}{x^2}+\frac{1}{(2p+1-x)^2}\,\left( 1-\frac{2}{x}\right) +\frac{2}{(2p+1-x)\,x^2}\\&=-\frac{(2\,p-1)\,(2p+1-2x)}{(2p+1-x)^2\,x^2}<0~~ \text{ as } x\le p. \end{aligned}$$

Thus f(x) is strictly decreasing on $2\le x\le p$, and hence

$$\begin{aligned} f(x)\ge f(p)=\frac{1}{p}+\frac{1}{p+1}\,\left( 1-\frac{2}{p}\right) =\frac{2p-1}{p(p+1)}, \end{aligned}$$

that is,

$$\begin{aligned} \displaystyle {\frac{d_i\,d_j}{d_i+d_j-2}}\le \frac{p(p+1)}{2p-1}=\displaystyle {\frac{1}{2}\,\left( p+\frac{3}{2}+\frac{3}{2\,(2p-1)}\right) }<\frac{p+2}{2}=\frac{n+3}{4}. \end{aligned}$$

The result (6) holds. For $d_i\ge d_j=1$, we obtain

$$\begin{aligned} \displaystyle {\frac{d_i\,d_j}{d_i+d_j-2}}=\frac{d_i}{d_i-1}<\frac{p+2}{2}=\frac{n+3}{4} \end{aligned}$$

as $p\ge 5$. Again the result (6) holds. Similarly, as before, we obtain

$$\begin{aligned} EAZ(G)&< n\,e^{\displaystyle {\frac{(n+3)^3}{64}}}<e^{\displaystyle {\left( \frac{d_1\,d_2}{d_1+d_2-2}\right) ^3}}<EAZ\left( C^{3}_{\lfloor \frac{n-3}{2} \rfloor ,\,\lceil \frac{n-3}{2} \rceil }\right) . \end{aligned}$$

Again the result (5) strictly holds.

$\mathbf{Case\,1.2.2.}$ $d_i+d_j=n+1$. In this case

$$\begin{aligned} \frac{1}{d_i}+\frac{1}{d_j}-\frac{2}{d_i\,d_j}=\frac{1}{d_j}+\frac{1}{2p+2-d_j}\,\left( 1-\frac{2}{d_j}\right) . \end{aligned}$$

Similarly, as $\mathbf{Case\,1.2.1}$, the function $\displaystyle {\frac{1}{x}+\frac{1}{2p+2-x}\,\left( 1-\frac{2}{x}\right) }$ is a strictly decreasing function on $x\le p$ and hence

$$\begin{aligned} \frac{1}{d_i}+\frac{1}{d_j}-\frac{2}{d_i\,d_j}=\frac{1}{d_j}+\frac{1}{2p+2-d_j}\,\left( 1-\frac{2}{d_j}\right) \ge \frac{2}{p+2}, \end{aligned}$$

that is,

$$\begin{aligned} \displaystyle {\frac{d_i\,d_j}{d_i+d_j-2}}\le \frac{p+2}{2}=\frac{n+3}{4}. \end{aligned}$$

Now,

$$\begin{aligned} EAZ(G)&\le n\,e^{\displaystyle {\frac{(n+3)^3}{64}}}<e^{\displaystyle {\left( \frac{d_1\,d_2}{d_1+d_2-2}\right) ^3}}<EAZ\left( C^{3}_{\lfloor \frac{n-3}{2} \rfloor ,\,\lceil \frac{n-3}{2} \rceil }\right) . \end{aligned}$$

Again the result (5) strictly holds.

$\mathbf{Case\,2.}$ $n=2p$. In this case $p\ge 5$. Since $d_i+d_j\le n+1=2p+1$, we must have $d_j\le p$. If $d_j=p$, then either $d_i=p$ or $d_i=p+1$. For $d_i=p+1$, we have $G\cong C^{3}_{\lfloor \frac{n-3}{2} \rfloor ,\,\lceil \frac{n-3}{2} \rceil }$ (see, Fig. 1) as G is unicyclic. Thus

$$\begin{aligned} EAZ(G)=e^{\displaystyle {\left( \frac{n\,(n+2)}{4\,(n-1)}\right) ^{3}}}+\frac{n-2}{2}\,e^{\displaystyle {\left( \frac{n+2}{n}\right) ^{3}}}+\frac{n-4}{2}\,e^{\displaystyle {\left( \frac{n}{n-2}\right) ^{3}}}+2e^{8}, \end{aligned}$$

and hence the equality holds in (5). For $d_i=p$, by Lemma 3, the result (5) strictly holds. Otherwise, $d_j\le p-1$. The remaining portion of the proof can be developed by addressing the subsequent two cases:

$\mathbf{Case\,2.1.}$ $d_i\le p+2$. Now, we obtain

$$\begin{aligned} \frac{1}{d_i}+\frac{1}{d_j}-\frac{2}{d_i\,d_j}= & {} \frac{1}{d_i}+\frac{1}{d_j}\,\left( 1-\frac{2}{d_i}\right) \\\ge & {} \frac{1}{d_i}+\frac{1}{p-1}\,\left( 1-\frac{2}{d_i}\right) =\frac{1}{p-1}+\frac{1}{d_i}\,\left( 1-\frac{2}{p-1}\right) . \end{aligned}$$

Combining the above fact with $d_i\le p+2$ and $p\ge 5$, we have

$$\begin{aligned} \frac{1}{d_i}+\frac{1}{d_j}-\frac{2}{d_i\,d_j}\ge \frac{1}{p-1}+\frac{1}{p+2}\,\left( 1-\frac{2}{p-1}\right) , \end{aligned}$$

which exerts

$$\begin{aligned} \displaystyle {\frac{d_i\,d_j}{d_i+d_j-2}}\le \frac{(p+2)\,(p-1)}{2p-1}=\frac{(n+4)\,(n-2)}{4\,(n-1)}. \end{aligned}$$

It is evident from Lemma 1 that

$$\begin{aligned} EAZ(G)\le n\,e^{\displaystyle {\frac{1}{64}\left( \frac{(n+4)\,(n-2)}{n-1}\right) ^{3}}}&<e^{\displaystyle {\left( \frac{d_1\,d_2}{d_1+d_2-2}\right) ^3}}<EAZ\left( C^{3}_{\lfloor \frac{n-3}{2} \rfloor ,\,\lceil \frac{n-3}{2} \rceil }\right) . \end{aligned}$$

Again the result (5) strictly holds.

$\mathbf{Case\,2.2.}$ $d_i\ge p+3$. Since $d_i+d_j\le n+1$, we consider the following two subcases:

$\mathbf{Case\,2.2.1.}$ $d_i+d_j\le n$. First we have to prove that

$$\begin{aligned} \displaystyle {\frac{d_i\,d_j}{d_i+d_j-2}}<\frac{(n+4)\,(n-2)}{4\,(n-1)}. \end{aligned}$$

(7)

For $d_i\ge d_j\ge 2$, we obtain

$$\begin{aligned} \frac{1}{d_i}+\frac{1}{d_j}-\frac{2}{d_i\,d_j}\ge \frac{1}{d_j}+\frac{1}{2p-d_j}\,\left( 1-\frac{2}{d_j}\right) . \end{aligned}$$

Let us consider a function

$$\begin{aligned} g(x)=\frac{1}{x}+\frac{1}{2p-x}\,\left( 1-\frac{2}{x}\right) ,~~x\le p-1. \end{aligned}$$

Similarly, as $\mathbf{Case\,1.2.1}$, the function g(x) is a strictly decreasing function on $x\le p-1$, and hence

$$\begin{aligned} g(x)\ge g(p-1)=\frac{1}{p-1}+\frac{1}{p+1}\,\left( 1-\frac{2}{p-1}\right) =\frac{2}{p+1}, \end{aligned}$$

that is,

$$\begin{aligned} \displaystyle {\frac{d_i\,d_j}{d_i+d_j-2}}\le \frac{p+1}{2}=\frac{n+2}{4}<\frac{(n+4)\,(n-2)}{4\,(n-1)}, ~ as ~n\ge 10. \end{aligned}$$

Thus (7) holds. For $d_i\ge d_j=1$, we obtain

$$\begin{aligned} \displaystyle {\frac{d_i\,d_j}{d_i+d_j-2}}=\frac{d_i}{d_i-1}<\frac{n+2}{4}<\frac{(n+4)\,(n-2)}{4\,(n-1)}, ~ as ~n\ge 10. \end{aligned}$$

Again (7) holds. Similarly, as before, we obtain

$$\begin{aligned} EAZ(G)< n\,e^{\displaystyle {\frac{1}{64}\left( \frac{(n+4)\,(n-2)}{n-1}\right) ^{3}}}&<e^{\displaystyle {\left( \frac{d_1\,d_2}{d_1+d_2-2}\right) ^3}}<EAZ\left( C^{3}_{\lfloor \frac{n-3}{2} \rfloor ,\,\lceil \frac{n-3}{2} \rceil }\right) .\\ \end{aligned}$$

$\mathbf{Case\,2.2.2.}$ $d_i+d_j=n+1$. Let us construct a function

$$\begin{aligned} h(x)=x\,(n+1-x),~~~x\le \frac{n-2}{2}. \end{aligned}$$

Subsequently, it becomes evident that the function h(x) is increasing for $x \le \frac{n-2}{2}$, and hence

$$\begin{aligned} h(x)\le h(n/2-1)=\frac{(n-2)\,(n+4)}{4}. \end{aligned}$$

Employing the aforementioned result, it is apparent that

$$\begin{aligned} \displaystyle {\frac{d_i\,d_j}{d_i+d_j-2}}=\frac{(n+1-d_j)\,d_j}{n-1}\le \frac{(n+4)\,(n-2)}{4\,(n-1)}. \end{aligned}$$

Similarly, as before, we obtain

$$\begin{aligned} EAZ(G)\le n\,e^{\displaystyle {\frac{1}{64}\left( \frac{(n+4)\,(n-2)}{n-1}\right) ^{3}}}&<e^{\displaystyle {\left( \frac{d_1\,d_2}{d_1+d_2-2}\right) ^3}}<EAZ\left( C^{3}_{\lfloor \frac{n-3}{2} \rfloor ,\,\lceil \frac{n-3}{2} \rceil }\right) . \end{aligned}$$

Again the result (5) strictly holds. This completes the proof of the theorem. $\square$

3 Concluding remarks

In this work, we have solved an open problem that was posed in Cruz et al. (2022). The maximal unicyclic graph has been characterized for the EAZ index in terms of graph order n. We are delighted to report a remarkable characteristic of this extremal graph (see, Fig. 1) that the contribution of EAZ associated with a single edge $v_1v_2$ of this structure exceeds the EAZ values of all other unicyclic graphs individually. Observing such a scenario is uncommon in the literature of extremal graph theory.