Unicyclic and bicyclic graphs with minimal augmented Zagreb index

Abstract

The augmented Zagreb index of a graph G is defined as

$$ \operatorname{AZI}(G)=\sum_{uv\in E(G)} \biggl( \frac{d_{u}d_{v}}{d_{u}+d_{v}-2} \biggr)^{3}, $$

where $E(G)$ is the edge set, and $d_{u}$, $d_{v}$ are the degrees of vertices u and v in G, respectively. This new molecular structure descriptor, introduced by Furtula et al. (J. Math. Chem. 48:370-380, 2010), has proven to be a valuable predictive index in the study of the heat of formation in heptanes and octanes. In this paper, the n-vertex unicyclic graphs with the minimal and the second minimal AZI indices and the n-vertex bicyclic graphs with the minimal AZI index are determined.

1 Introduction

Let $G=(V,E)$ be a simple, finite and undirected graph of order $n=|V|$ and size $m=|E|$. For $v\in V(G)$, the degree of v, denoted by $d_{v}$, is the number of edges incident to v. A vertex of degree one is said to be a pendent vertex. The maximum vertex degree is denoted by Δ, the minimum vertex degree is denoted by δ, and the minimum non-pendent vertex degree is denoted by $\delta_{1}$. If u and v are two adjacent vertices of G, then the edge connecting them will be denoted by uv [1]. Other notations used in this work are standard and mainly taken from [1–3].

A description of the structure or shape of molecules is very helpful in predicting the activity and properties of molecules in complex experiments. Molecular descriptors play a significant role in mathematical chemistry, especially in QSPR/QSAR investigations. Among them, topological indices [4] play an important role. Today, many topological indices exist that have various applications in chemistry [2, 5, 6]. Here, a relatively new topological index is considered. In 2010, Furtula et al. [7] proposed a new, vertex-degree-based graph topological index called the augmented Zagreb index (AZI), defined as

$$ \operatorname{AZI}(G)=\sum_{uv\in E(G)} \biggl( \frac{d_{u}d_{v}}{d_{u}+d_{v}-2} \biggr)^{3}, $$

and showed that it is a valuable predictive index in the study of the heat of formation in heptanes and octanes. Moreover, Gutman and Tošovič [8] recently tested the correlation abilities of 20 vertex-degree-based topological indices for the case of standard heats of formation and normal boiling points of octane isomers. They found that the AZI index yields the best results. Consequently, the AZI index should be preferred in designing quantitative, structure-property relations.

Furtula et al. [7] obtained several tight upper and lower bounds of the AZI index of a chemical tree and showed that among all trees, the star graph has the minimal AZI value. Huang et al. [9] and Wang et al. [10] provided particular bounds on the AZI indices of connected graphs and characterized the corresponding extremal graphs. Ali et al. [11] established inequalities between AZI and several other vertex-degree-based topological indices. Ali et al. [12] proposed tight upper bounds for the AZI of chemical bicyclic and unicyclic graphs and provided a Nordhaus-Gaddum-type result for the AZI index.

In this paper, the n-vertex unicyclic graphs are determined with the minimal and the second minimal AZI indices. Additionally, the n-vertex bicyclic graphs in which the AZI index attains its minimal value are obtained.

2 Preliminaries

Some of the auxiliary results provided below will be used in the main theorem proofs. For convenience, let $A(x,y)=(\frac{xy}{x+y-2})^{3}$ for $x, y\geq1$ with $x+y>2$. Obviously, $A(x,y)=A(y,x)$.

Lemma 2.1

([9])

• (i) $A(1,y)$ is decreasing for $y\geq2$.

• (ii) $A(2,y)=8$.

• (iii) If $y\geq3$ is fixed, then $A(x,y)$ is increasing for $x\geq3$.

Lemma 2.2

([9])

$A(1, \Delta)\leq A(1, i)\leq A(1, 2)=A(2,j)< A(3, 3)\leq A(k, l)\leq A(\Delta, \Delta)$, where $2\leq i, j\leq\Delta$ and $3\leq k\leq l\leq\Delta$.

Lemma 2.3

Let $f(x)=xA(1,x+2)-(x+1)A(1,x+3)$ with $x>0$. Then $f(x)$ increases in x.

Proof

Let $h(x)=xA(1,x+2)=\frac{x(x+2)^{3}}{(x+1)^{3}}$. Therefore, $f(x)=h(x)-h(x+1)$.

Then

$$ h^{\prime}(x)=\frac{(x+2)^{2}(x^{2}+2)}{(x+1)^{4}} $$

and

$$ h^{\prime\prime}(x)=-\frac{12(x+2)}{(x+1)^{5}}< 0. $$

Applying Lagrange’s mean value theorem, $f^{\prime}(x)=h^{\prime}(x)-h^{\prime}(x+1)=-h^{\prime\prime}(\xi)>0$ with $x\leq\xi\leq x+1$. Therefore, $f(x)$ increases in x, which completes the proof. □

Lemma 2.4

Let $g(x,y)=A(x,y)-A(x-1,y)$, with x and y as positive integers, and $y< x+2$. Then $g(x,y)$ strictly decreases in x for fixed $y\geq2$.

Proof

First, the partial derivative $g(x,y)$ is considered with respect to x,

$$ \frac{\partial g(x,y)}{\partial x}= 3y^{3}(y-2) \biggl(\frac {x}{(x+y-2)^{2}}+ \frac{x-1}{(x+y-3)^{2}} \biggr) \biggl(\frac{x}{(x+y-2)^{2}} -\frac{x-1}{(x+y-3)^{2}} \biggr). $$

Now, we will show that

$$ \frac{x}{(x+y-2)^{2}}-\frac{x-1}{(x+y-3)^{2}}< 0. $$

Let $m_{1}(t)=\frac{t}{(t+y-2)^{2}}$ for $y-t<2$. Then the function $m_{1}(t)$ decreases in t because

$$ m_{1}^{\prime}(t)=\frac{y-2-t}{(t+y-2)^{3}}< 0. $$

Therefore, $\frac{\partial g(x,y)}{\partial x}<0$, from which it follows that $g(x,y)$ strictly decreases in x. □

Lemma 2.5

Let $l_{1}(x)=2A(x-2,3)-A(x-1,3)$ for $x\geq5$. Then $l_{1}(x)>8$.

Proof

From the definition of $A(x,y)$, the following is obtained:

$$ l_{1}(x)=27\times \biggl(\frac{2(x-2)^{3}}{(x-1)^{3}}-\frac{(x-1)^{3}}{x^{3}} \biggr). $$

Therefore,

$$\begin{aligned} l_{1}^{\prime}(x) =&81\times\frac{(\sqrt{2}-1)x^{3}+(3-2\sqrt {2})x^{2}-3x+1}{(x-1)^{2}x^{2}} \\ &{}\times \biggl(\frac{\sqrt{2}(x-2)}{(x-1)^{2}}+\frac{x-1}{x^{2}} \biggr). \end{aligned}$$

$l_{1}^{\prime}(x)>0$ must be demonstrated. Let $m_{2}(t)=(\sqrt{2}-1)t^{3}+(3-2\sqrt{2})t^{2}-3t+1$, so $m_{2}^{\prime}(t)=3(\sqrt{2}-1)t^{2}+2(3-2\sqrt{2})t-3$, which is positive for $t\geq5$, implying that $m_{2}(t)$ is an increasing function for t. Thus, $m_{2}(t)\geq m_{2}(5)>0$. Therefore, $l^{\prime}(x)>0$ and $l_{1}(x)\geq l_{1}(5)= 8.95725>8$. □

Lemma 2.6

Let $l_{2}(x)= (\frac{3(x+2)}{x+3} )^{3}+ (\frac{3(n-x-2)}{n-x-1} )^{3}$, where $n\geq6$ and $0\leq x\leq n-4$. Then $l_{2}(x)\geq (\frac{3(n+1)}{n+4} )^{3}+ (\frac {3(2n-1)}{2n+2} )^{3}$.

Proof

Consider the derivative $l_{2}(x)$ with respect to x,

$$\begin{aligned} l_{2}^{\prime}(x) =&3^{4}\times \biggl( \frac {(x+2)(n-x-1)^{2}+(x+3)^{2}(n-x-2)}{(x+3)^{2}(n-x-1)^{2}} \biggr) \\ &{}\times \biggl(\frac{(x+2)(n-x-1)^{2}-(x+3)(n-x-2)^{2}}{(x+3)^{2}(n-x-1)^{2}} \biggr). \end{aligned}$$

Now, the expression $(x+2)(n-x-1)^{2}-(x+3)(n-x-2)^{2}$ must be discussed. Let $m_{3}(t)=(t+2)(n-t-1)^{2}$ with $0\leq t\leq n-4$. Then $m_{3}(t)-m_{3}(t+1)=(t+2)(n-t-1)^{2}-(t+3)(n-t-2)^{2}$ and $m_{3}^{\prime}(t)=(n-t-1)(n-3t-5)$. Obviously, $n-t-1>0$.

If $t>\frac{n-5}{3}$, then $m_{3}^{\prime}(t)<0$, implying that $m_{3}(t)$ is decreasing for t. Therefore, $l_{2}^{\prime}(x)>0$ and $l_{2}(x)\geq l_{2}(\frac{n-5}{3})$.

If $t<\frac{n-5}{3}$, then $m_{3}^{\prime}(t)>0$, implying $m_{3}(t)< m_{3}(t+1)$. Therefore, $l_{2}^{\prime}(x)<0$ and $l_{2}(x)\geq l_{2}(\frac{n-5}{3})$.

Therefore, $l_{2}(x)\geq (\frac{3(n+1)}{n+4} )^{3}+ (\frac {3(2n-1)}{2n+2} )^{3}$ for $0\leq x\leq n-4$. □

Lemma 2.7

Let $l_{3}(x)=2 (\frac{3(x+1)}{x+4} )^{3}- (\frac{3(x-1)}{x} )^{3}$, where $x\geq6$. Then $l_{3}(x)\geq l_{3}(6)$.

Proof

We have

$$ l_{3}^{\prime}(x)=3^{4}\times \biggl( \frac{\sqrt{6}(x+1)}{(x+4)^{2}}+\frac {x-1}{x^{2}} \biggr) \biggl(\frac{\sqrt{6}(x+1)x^{2}-(x-1)(x+4)^{2}}{(x+4)^{2}x^{2}} \biggr). $$

Let $m_{4}(t)=\sqrt{6}(t+1)t^{2}-(t-1)(t+4)^{2}$ for $t\geq6$. It is easily seen that $m_{4}^{\prime}(t)=3(\sqrt{6}-1)t^{2}+2(\sqrt{6}-7)t-8$. By simple calculation, $m_{4}^{\prime}(t)>0$ for $t\geq6$. Then $m_{4}(t)\geq m_{4}(6)>0$ and $l_{3}^{\prime}(x)>0$, implying that $l_{3}(x)$ is increasing in x. Therefore, $l_{3}(x)\geq l_{3}(6)$ for $x\geq6$. □

3 On the AZI indices of unicyclic graphs

In this section, the n-vertex unicyclic graphs are determined with the minimal and the second minimal AZI indices. The technique from [13] is used for these determinations.

For $r\neq s$, a graph G is said to be $(r,s)$-biregular if each vertex has degree r or s, and each vertex of degree r is adjacent to some vertices of degree s and vice versa. Let $\phi_{1}$ be the class of connected graphs whose pendent vertices are adjacent to the vertices of maximum degree and all other edges have at least one end-vertex of degree two. Let $\phi_{2}$ be the class of connected graphs with vertices that are of degree at least two. Additionally, all of the edges have at least one end-vertex of degree two.

Lemma 3.1

([10])

Let G be a connected graph of order $n\geq3$ with m edges, p pendent vertices, maximum degree Δ and minimum non-pendent vertex degree $\delta_{1}$. Then

$$ \operatorname{AZI}(G)\geq p \biggl(\frac{\Delta}{\Delta-1} \biggr)^{3}+(m-p) \biggl(\frac{\delta _{1}^{2}}{2\delta_{1}-2} \biggr)^{3} $$

with equality if and only if G is isomorphic to a $(1,\Delta )$-biregular graph or G is isomorphic to a regular graph or $G\in\phi_{1}$ or $G\in\phi_{2}$.

Let ${\mathcal{U}}_n$ be the set of n-vertex unicyclic graphs. Let ${\mathcal{U}}_{n,p}$ be the set of unicyclic graphs with n vertices and p pendent vertices, and let $C_{n,p}$ be the unicyclic graph formed from $C_{n-p}$ by attaching p pendent vertices to a vertex of the cycle $C_{n-p}$, where $0\leq p\leq n-3$. Clearly $C_{n-p,0}=C_{n}$.

Lemma 3.2

([9])

Let $G\in {\mathcal{U}}_{n,p}$, where $0\leq p\leq n-3$. Then

$$ \operatorname{AZI}(G)\geq p \biggl(\frac{p+2}{p+1} \biggr)^{3}+8(n-p) $$

with equality if and only if $G\cong C_{n,p}$.

Lemma 3.3

For every positive integer n, graphs $C_{n,p}$ with $0\leq p\leq n-3$, it holds that

$$ \operatorname{AZI}(C_{n,n-3})< \operatorname{AZI}(C_{n,n-4})< \cdots<\operatorname{AZI}(C_{n,1})<\operatorname{AZI}(C_{n,0}). $$

Proof

Consider the function

$$ f_{1}(x)=x \biggl(\frac{x+2}{x+1} \biggr)^{3}+8(n-x) \quad \mbox{for }0\leq x\leq n-3. $$

Then

$$ f_{1}^{\prime}(x)=-\frac{x(7x^{3}+28x^{2}+42x+24)}{(x+4)^{4}}< 0 \quad \mbox{as }x\geq1. $$

Thus, $f(x)$ is a decreasing function for $1\leq x\leq n-3$. Because $f_{1}(1)=8n-\frac{37}{8}$ and $f_{1}(0)=8n$, then

$$ \operatorname{AZI}(C_{n,n-3})< \operatorname{AZI}(C_{n,n-4})< \cdots<\operatorname{AZI}(C_{n,1})<\operatorname{AZI}(C_{n,0}). $$

 □

Theorem 3.4

Among all graphs in ${\mathcal{U}}_n$ with $n\geq3$, $C_{n,n-3}$ is the unique graph with the minimal AZI index, which is equal to $\frac{(n-3)(n-1)^{3}}{(n-2)^{3}}+24$.

Proof

By Lemma 3.1 and Lemma 3.2, $C_{n,p}$ is attained and is the unique graph with the minimal AZI index among all graphs in ${\mathcal{U}}_{n,p}$ with $0\leq p\leq n-3$. Applying Lemma 3.3, $C_{n,n-3}$ is the unique graph with the minimal AZI index in ${\mathcal{U}}_n$. It is clear that $\operatorname{AZI}(C_{n,n-3})=\frac{(n-3)(n-1)^{3}}{(n-2)^{3}}+24$. □

The vertices of $C_{3}$ are consecutively labeled by $v_{1}$, $v_{2}$, $v_{3}$. Let $Q_{n}(n_{1},n_{2},n_{3})$ be the unicyclic graph formed by attaching $n_{i}$ pendent vertices to $v_{i}$, where $n_{i}\geq0$ for $i=1,2,3$, $n_{1}\geq n_{2}\geq n_{3}$, and $\sum_{i=1}^{3} n_{i}=n-3$.

Lemma 3.5

Let $G\cong Q_{n}(n_{1},n_{2},n_{3})$ with $n_{1}\geq n_{2}\geq 1$ and $G^{\prime}\cong Q_{n}(n_{1}+1,n_{2}-1,n_{3})$ (see Figure  1). Then $\operatorname{AZI}(G^{\prime})<\operatorname{AZI}(G)$.

Figure 1

Transformation σ from Lemma  3.5 .

Proof

Consider the transformation σ depicted in Figure 1. By applying the transformation σ to G, a pendent edge is cut from $v_{2}$ and attached to $v_{1}$. By Lemma 2.3 and Lemma 2.4, the change of the AZI index after applying this transformation is

$$\begin{aligned}& \operatorname{AZI} (G)-\operatorname{AZI}\bigl(G^{\prime}\bigr) \\& \quad = n_{1}A(1,n_{1}+2)+n_{2}A(1,n_{2}+2)+A(n_{1}+2,n_{2}+2)+A(n_{1}+2,n_{3}+2) \\& \qquad {} +A(n_{2}+2,n_{3}+2)- \bigl[(n_{1}+1)A(1,n_{1}+3)+(n_{2}-1)A(1,n_{2}+1) \\& \qquad {} +A(n_{1}+3,n_{2}+1)+A(n_{1}+3,n_{3}+2)+A(n_{2}+1,n_{3}+2) \bigr] \\& \quad = \bigl[n_{1}A(1,n_{1}+2)-(n_{1}+1)A(1,n_{1}+3) \bigr]- \bigl[(n_{2}-1)A(1,n_{2}+1) \\& \qquad {} -n_{2}A(1,n_{2}+2) \bigr]+ \bigl[A(n_{1}+2,n_{2}+2)-A(n_{1}+3,n_{2}+1) \bigr] \\& \qquad {} + \bigl[A(n_{2}+2,n_{3}+2)-A(n_{2}+1,n_{3}+2) \bigr] \\& \qquad {} - \bigl[A(n_{1}+3,n_{3}+2)-A(n_{1}+2,n_{3}+2) \bigr] \\& \quad = f(n_{1})-f(n_{2}-1)+A(n_{1}+2,n_{2}+2)-A(n_{1}+3,n_{2}+1) \\& \qquad {} +g(n_{2}+2,n_{3}+2)-g(n_{1}+3,n_{3}+2). \end{aligned}$$

By Lemma 2.3, the expression $f(n_{1})-f(n_{2}-1)$ is positive for $n_{1}>n_{2}-1$. By Lemma 2.4, $g(n_{2}+2,n_{3}+2)>g(n_{1}+3,n_{3}+2)$ for $n_{1}\geq n_{2}\geq n_{3}$. Note that $(n_{1}+2)(n_{2}+2)>(n_{1}+3)(n_{2}+1)$, so

$$ \frac{(n_{1}+2)^{3}(n_{2}+2)^{3}}{(n_{1}+n_{2}+2)^{3}}>\frac {(n_{1}+3)^{3}(n_{2}+1)^{3}}{(n_{1}+n_{2}+2)^{3}}, $$

that is, $A(n_{1}+2,n_{2}+2)>A(n_{1}+3,n_{2}+1)$. Thus, it has been shown that after the transformation σ, the AZI index of G decreases. □

Theorem 3.6

Among all graphs in ${\mathcal{U}}_n$ with $n\geq4$.

• (i) For $n=4$ or $n\geq6$, $C_{n,n-4}$ is the unique graph with the second minimal AZI index, which is equal to $(n-4) (\frac{n-2}{n+1} )^{3}+32$.

• (ii) For $n=5$, $Q_{5}(1,1,0)$ is the unique graph with the second minimal AZI index, which is equal to $\frac{2\text{,}185}{64}$.

Proof

By Lemma 3.2 and Lemma 3.3, the second minimal AZI index of graphs in ${\mathcal{U}}_n$ with $n\geq4$ is achieved by the graphs in ${\mathcal{U}}_{n,n-3}\setminus \{C_{n,n-3}\}$ and $C_{n,n-4}$. Now, two cases are considered.

Case 1. $n=4$.

Obviously, $C_{4}$ is the unique graph with the second minimal AZI index, which is equal to 32.

Case 2. $n\geq5$.

Note that $n_{3}\geq1$, so $n_{2}\geq n_{3}\geq1$. By Lemma 3.5, it is determined that $Q_{n}(n-4,1,0)$ is the unique graph with the minimal AZI index among all of the graphs in ${\mathcal{U}}_{n,n-3}\setminus \{C_{n,n-3}\}$. Note that

$$\begin{aligned} \operatorname{AZI} \bigl(Q_{n}(n-4,1,0) \bigr) =&(n-4)A(1,n-2)+A(1,3)+A(2,n-2) \\ &{}+A(2,3)+A(3,n-2) \end{aligned}$$

and

$$ \operatorname{AZI}(C_{n,n-4})=(n-4)A(1,n-2)+2A(2,n-2)+2A(2,2). $$

Then

$$\begin{aligned} \operatorname{AZI} \bigl(Q_{n}(n-4,1,0) \bigr)-\operatorname{AZI}(C_{n,n-4}) =&A(3,n-2)+A(1,3)-16 \\ =&A(3,n-2)-\frac{101}{8}. \end{aligned}$$

By Lemma 2.1(iii), $A(3,n-2)$ is increasing for $n\geq5$. Using a simple calculation, it can be shown that $\operatorname{AZI}(Q_{5}(1,1,0))<\operatorname{AZI}(C_{5,1})$ for $n=5$ and $\operatorname{AZI} (Q_{n}(n-4,1,0) )>\operatorname{AZI}(C_{n,n-4})$ for $n\geq6$.

Thus, it follows that $Q_{5}(1,1,0)$ for $n=5$ is the unique graph with the second minimal AZI index, which is equal to $\frac{2\text{,}185}{64}$, and $C_{n,n-4}$ for $n\geq6$ is the unique graph with the second minimal AZI index, which is equal to $(n-4) (\frac{n-2}{n+1} )^{3}+32$. □

4 On the AZI indices of bicyclic graphs

In this section, the n-vertex bicyclic graphs are determined with the minimal AZI index for $n\geq5$.

Let ${\mathcal{B}}_{n,p}$ be the set of bicyclic graphs with n vertices and p pendent vertices for $0\leq p\leq n-4$. Let $D^{r,t}_{n,p}$ be the n-vertex bicyclic graph by identifying one vertex of two cycles $C_{r}$ and $C_{t}$ and attaching p pendent vertices to the common vertex, where $r\geq t\geq3$ and $0\leq p\leq n-5$.

Lemma 4.1

([9])

Let G be a bicyclic graph with $n\geq5$ vertices and p pendent vertices, where $0\leq p\leq n-5$. Then

$$ \operatorname{AZI}(G)\geq\frac{p(p+4)^{3}}{(p+3)^{3}}+8(n-p+1) $$

with equality holding if and only if $G\cong D^{r,t}_{n,p}$, where $r\geq t\geq3$.

Let ${\mathcal{D}}_{n,p}$ be the set of graphs $D^{r,t}_{n,p}$ with $3\leq t\leq r\leq n-p-2$ and $r+t=n-p+1$. Let $D_{n,p}$ be any graph in ${\mathcal{D}}_{n,p}$.

Lemma 4.2

For the graphs in ${\mathcal{D}}_{n,p}$ with $0\leq p\leq n-5$ and $n\geq5$, it holds that

$$ \operatorname{AZI}(D_{n,n-5})< \operatorname{AZI}(D_{n,n-6})< \cdots<\operatorname{AZI}(D_{n,1})<\operatorname{AZI}(D_{n,0}). $$

Proof

Let $H(x)=\frac{x(x+4)^{3}}{(x+3)^{3}}+8(n-x+1)$, where $0\leq x\leq n-5$.

Note that

$$ H^{\prime}(x)=\frac{(x+4)^{2}(x^{2}+4x+12)-8(x+3)^{4}}{(x+3)^{4}}< 0. $$

Thus, $H(x)$ is decreasing for x. The result follows. □

Let $C^{*}_{4}$ be the bicyclic graph obtained by adding an edge to the cycle $C_{4}$. Label the vertices of $C^{*}_{4}$ by $v_{1}$, $v_{2}$, $v_{3}$, $v_{4}$ with $d_{v_{1}}=d_{v_{3}}=3$, $d_{v_{2}}=d_{v_{4}}=2$, respectively. Let $S_{n}(n_{1},n_{2},n_{3},n_{4})$ be the graph formed from $C^{*}_{4}$ by attaching $n_{i}$ pendent vertices to $v_{i}$, where $n_{i}\geq0$ for $i=1,2,3,4$, $n_{1}\geq n_{3}$, $n_{2}\geq n_{4}$ and $\sum_{i=1}^{4} n_{i}=n-4$ (see Figure 2). For convenience, let $B_{n}^{1}\cong S_{n}(n-4, 0, 0, 0)$, $B_{n}^{2}\cong S_{n}(0, n-4, 0, 0)$, $B_{n}^{3}\cong S_{n}(0, k_{1}, 0, k_{2})$ with $k_{1}+k_{2}=n-4$, $B_{n}^{4}\cong S_{n}(k_{3}, 0, k_{4}, 0)$ with $k_{3}+k_{4}=n-4$ and $B_{n}^{5}\cong S_{n}(k_{5}, k_{6}, k_{7}, k_{8})$ with $k_{5}+k_{6}+k_{7}+k_{8}=n-4$ (see Figure 3).

Figure 2

Graph $\pmb{S_{n}(n_{1},n_{2},n_{3},n_{3})}$ .

Figure 3

Graphs $\pmb{B^{1}_{n}}$ , $\pmb{B^{2}_{n}}$ , $\pmb{B^{3}_{n}}$ , $\pmb{B^{4}_{n}}$ and $\pmb{B^{5}_{n}}$ .

Lemma 4.3

Let $G\in {\mathcal{B}}_{n,n-4}$ with $n\geq5$. Then

$$ \operatorname{AZI}(G)\geq(n-4) \biggl(\frac{n-1}{n-2} \biggr)^{3}+ \biggl(\frac{3(n-1)}{n} \biggr)^{3}+32 $$

with equality if and only if $G\cong S_{n}(n-4,0,0,0)$.

Proof

For $n=5$, $G\cong S_{5}(1,0,0,0)$ or $G\cong S_{5}(0,1,0,0)$.

From the definition of the AZI index, the following is obtained:

$$ \operatorname{AZI}\bigl(S_{5}(1,0,0,0)\bigr)=A(1,4)+2A(4,2)+A(4,3)+2A(3,2) $$

and

$$ \operatorname{AZI}\bigl(S_{5}(0,1,0,0)\bigr)=A(1,3)+3A(3,3)+2A(3,2). $$

By simple calculation, $\operatorname{AZI}(S_{5}(1,0,0,0))<\operatorname{AZI}(S_{5}(0,1,0,0))$. Lemma 4.3 obviously holds.

For $n\geq6$, G is isomorphic to one of the graphs $B^{1}_{n}$, $B^{2}_{n}$, $B^{3}_{n}$, $B^{4}_{n}$, $B^{5}_{n}$ shown in Figure 3. By the definition of the AZI index,

$$\begin{aligned}& \operatorname{AZI}\bigl(B^{1}_{n}\bigr)=(n-4)A(1,n-1)+2A(2,n-1)+A(3,n-1) +2A(2,3), \\& \operatorname{AZI}\bigl(B^{2}_{n}\bigr)=(n-4)A(1,n-2)+2A(3,n-2)+A(3,3)+2A(2,3), \\& \operatorname{AZI}\bigl(B^{3}_{n}\bigr)=k_{1}A(1,k_{1}+2)+k_{2}A(1,k_{2}+2)+2A(3,k_{1}+2) \\& \hphantom{\operatorname{AZI}\bigl(B^{3}_{n}\bigr)=}{}+2A(3,k_{2}+2)+A(3,3), \\& \operatorname{AZI}\bigl(B^{4}_{n}\bigr)=k_{3}A(1,k_{3}+3)+k_{4}A(1,k_{4}+3)+2A(2,k_{3}+3) \\& \hphantom{\operatorname{AZI}\bigl(B^{4}_{n}\bigr)=}{}+2A(2,k_{4}+3)+A(k_{3}+3,k_{4}+3) \end{aligned}$$

and

$$\begin{aligned} \operatorname{AZI}\bigl(B^{5}_{n}\bigr) =&k_{5}A(1,k_{5}+3)+k_{6}A(1,k_{6}+2)+k_{7}A(1,k_{7}+3) \\ &{}+k_{8}A(1,k_{8}+2)+A(k_{5}+3,k_{6}+2)+A(k_{5}+3,k_{8}+2) \\ &{}+A(k_{5}+3,k_{7}+3)+A(k_{7}+3,k_{8}+2)+A(k_{7}+3,k_{6}+2). \end{aligned}$$

Claim 1

$\operatorname{AZI}(B^{1}_{n})<\operatorname{AZI}(B^{2}_{n})$.

Using Lemma 2.2 and Lemma 2.5, we get

$$\begin{aligned} \operatorname{AZI}\bigl(B^{2}_{n}\bigr)-\operatorname{AZI} \bigl(B^{1}_{n}\bigr) =&(n-4) \bigl(A(1,n-2)-A(1,n-1) \bigr) \\ &{}+ \bigl(2A(n-2,3)-A(n-1,3)-A(3,2) \bigr) \\ &{}+ \bigl(A(3,3)-A(3,2) \bigr)>0. \end{aligned}$$

This proves Claim 1.

Claim 2

$\operatorname{AZI}(B^{1}_{n})<\operatorname{AZI}(B^{3}_{n})$.

Note that $k_{1}+k_{2}=n-4$. Using Lemma 2.1, Lemma 2.2, Lemma 2.6 and Lemma 2.7, we have

$$\begin{aligned}& \operatorname{AZI}\bigl(B^{3}_{n}\bigr)-\operatorname{AZI} \bigl(B^{1}_{n}\bigr) \\& \quad = k_{1}A(1,k_{1}+2)+k_{2}A(1,k_{2}+2)-(n-4)A(1,n-1)+2 \bigl(A(k_{1}+2,3) \\& \qquad {} +A(k_{2}+2,3) \bigr)-A(n-1,3)+A(3,3)-4A(3,2) \\& \quad \geq k_{1}A(1,n-2)+k_{2}A(1,n-2)-(n-4)A(1,n-1)+2 \bigl(A(k_{1}+2,3) \\& \qquad {} +A(k_{2}+2,3) \bigr)-A(n-1,3)+A(3,3)-4A(3,2) \\& \quad = (n-4) \bigl(A(1,n-2)-A(1,n-1) \bigr)+2 \bigl(A(k_{1}+2,3)+A(k_{2}+2,3) \bigr) \\& \qquad {} -A(n-1,3)+A(3,3)-4A(3,2) \\& \quad > 2 \bigl(A(k_{1}+2,3)+A(k_{2}+2,3) \bigr)-A(n-1,3)+A(3,3)-4A(3,2) \\& \quad > 2 \biggl(\frac{3(n+1)}{n+4} \biggr)^{3}- \biggl( \frac{3(n-1)}{n} \biggr)^{3} \\& \qquad {}+2 \biggl(\frac{3(2n-1)}{2n+2} \biggr)^{3}+ \biggl(\frac{9}{4} \biggr)^{3}-4 \times2^{3} \\& \quad \geq l_{3}(6)+2\times3^{3}\times \biggl( \frac{2n-1}{2n+2} \biggr)^{3}+ \biggl(\frac {9}{4} \biggr)^{3}-4\times2^{3}. \end{aligned}$$

Let $q(x)= (\frac{2x-1}{2x+2} )^{3}$, so $q^{\prime}(x)=\frac {9(2x-1)^{2}}{(2x+2)^{4}}>0$, implying that $q(x)$ is strictly increasing for x. Then

$$\begin{aligned} \operatorname{AZI}\bigl(B^{3}_{n}\bigr)-\operatorname{AZI} \bigl(B^{1}_{n}\bigr) >&2 \biggl(\frac{3\times7}{10} \biggr)^{3} - \biggl(\frac{3\times5}{6} \biggr)^{3}+2 \times3^{3}\times \biggl(\frac{11}{14} \biggr)^{3} \\ &{}+ \biggl(\frac{9}{4} \biggr)^{3}-4\times2^{3}>0. \end{aligned}$$

This proves Claim 2.

Claim 3

$\operatorname{AZI}(B^{1}_{n})<\operatorname{AZI}(B^{4}_{n})$.

Because $k_{3}+k_{4}=n-4$, using Lemma 2.1 and Lemma 2.2, we have

$$\begin{aligned}& \operatorname{AZI}\bigl(B^{4}_{n}\bigr)-\operatorname{AZI} \bigl(B^{1}_{n}\bigr) \\& \quad = k_{3}A(1,k_{3}+3)+k_{4}A(1,k_{4}+3)-(n-4)A(1,n-1) \\& \qquad {} +A(k_{3}+3,k_{4}+3)-A(n-1,3) \\& \quad > \biggl(\frac{(k_{3}+3)(k_{4}+3)}{k_{3}+k_{4}+4} \biggr)^{3}- \biggl( \frac{3(n-1)}{n} \biggr)^{3} \\& \quad = \biggl(\frac{k_{3}k_{4}+3(n-4)+9}{n} \biggr)^{3}- \biggl( \frac{3(n-1)}{n} \biggr)^{3}\quad \mbox{as }k_{3}+k_{4}=n-4 \\& \quad > 0. \end{aligned}$$

This proves Claim 3.

Claim 4

$\operatorname{AZI}(B^{1}_{n})<\operatorname{AZI}(B^{5}_{n})$.

The following two cases are presented.

Case 1. $k_{5}\geq2$, $k_{6}\geq2$.

Using Lemma 2.1 and Lemma 2.2, it follows that

$$\begin{aligned} \operatorname{AZI}\bigl(B^{5}_{n}\bigr) >& k_{5}A(1,n-2)+k_{6}A(1,n-2)+k_{7}A(1,n-2)+k_{8}A(1,n-2) \\ &{} +A(5,4)+A(5,3)+A(5,2)+A(3,2)+A(3,4) \\ >& (n-4)A(1,n-1)+ \biggl(\frac{20}{7} \biggr)^{3}+ \biggl( \frac{5}{2} \biggr)^{3}+ \biggl(\frac{12}{5} \biggr)^{3}+2\times2^{3} \\ >& (n-4)A(1,n-1)+2\times2^{3}+3^{3}+2\times2^{3} \\ >& (n-4)A(1,n-1)+3^{3}\times \biggl(\frac{n-1}{n} \biggr)^{3}+4\times2^{3} \\ =& (n-1)A(1,n-1)+A(3,n-1)+2A(n-1,2)+2A(3,2) \\ =& \operatorname{AZI}\bigl(B^{1}_{n}\bigr). \end{aligned}$$

Case 2. $k_{5}=2$, $k_{6}=1$ or $k_{5}=1$, $k_{6}=2$ or $k_{5}=1$, $k_{6}=1$.

If $k_{5}=2$ and $k_{6}=1$, then according to the definition of graph $S_{n}(n_{1}, n_{2}, n_{3}, n_{4})$, $B_{n}^{5}\cong S_{7}(2, 1, 0, 0)$ or $S_{8}(2, 1, 1, 0)$ or $S_{8}(2, 1, 0, 1)$ or $S_{9}(2, 1, 1, 1)$ or $S_{9}(2, 1, 2, 0)$ or $S_{10}(2, 1, 2, 1)$.

By the definition of the AZI index and some calculations, the following are obtained:

$$\begin{aligned}& \operatorname{AZI} \bigl(S_{7}(2, 1, 0, 0) \bigr)\doteq65.922> \operatorname{AZI}\bigl(B_{7}^{1}\bigr)\doteq54.187, \\& \operatorname{AZI} \bigl(S_{8}(2, 1, 1, 0) \bigr)\doteq78.424> \operatorname{AZI}\bigl(B_{8}^{1}\bigr)\doteq56.440, \\& \operatorname{AZI} \bigl(S_{8}(2, 1, 0, 1) \bigr)\doteq80.313> \operatorname{AZI}\bigl(B_{8}^{1}\bigr)\doteq56.440, \\& \operatorname{AZI} \bigl(S_{9}(2, 1, 1, 1) \bigr)\doteq95.248> \operatorname{AZI}\bigl(B_{9}^{1}\bigr)\doteq58.427, \\& \operatorname{AZI} \bigl(S_{9}(2, 1, 2, 0) \bigr)\doteq88.955> \operatorname{AZI}\bigl(B_{9}^{1}\bigr)\doteq58.427 \end{aligned}$$

and

$$ \operatorname{AZI} \bigl(S_{10}(2, 1, 2, 1) \bigr)\doteq107.580> \operatorname{AZI}\bigl(B_{10}^{1}\bigr)\doteq60.226. $$

In a similar way, we can verify the inequality $\operatorname{AZI}(B_{n}^{1})<\operatorname{AZI}(B_{n}^{5})$ for each of the cases for $k_{5}=1$, $k_{6}=2$ or $k_{5}=1$, $k_{6}=1$. The details are omitted. This proves Claim 4.

Thus, the result follows from Claims 1-4.  □

Theorem 4.4

For the graphs in ${\mathcal{B}}_n$ with $n\geq5$, it holds that $D^{3,3}_{n,n-5}$ is the unique graph with the minimal AZI index, which is equal to $(n-5) (\frac{n-1}{n-2} )^{3}+48$.

Proof

Using Lemma 4.3, among all of the graphs in ${\mathcal{B}}_{n,n-4}$, $S_{n}(n-4,0,0,0)$ is the unique graph with the minimal AZI index, which is equal to $(n-4) (\frac{n-1}{n-2} )^{3}+ (\frac {3(n-1)}{n} )^{3}+32$. Using Lemma 4.1 and Lemma 4.2, among all of the graphs in ${\mathcal{B}}_{n,p}$ with $0\leq p\leq n-5$, $D^{3,3}_{n,n-5}$ is the unique graph with the minimal AZI index, which is equal to $(n-5) (\frac{n-1}{n-2} )^{3}+48$.

Note that

$$\begin{aligned}& \operatorname{AZI} \bigl(S_{n}(n-4,0,0,0) \bigr)-\operatorname{AZI} \bigl(D^{3,3}_{n,n-5} \bigr) \\& \quad = (n-4) \biggl(\frac{n-1}{n-2} \biggr)^{3}+ \biggl( \frac{3(n-1)}{n} \biggr)^{3} \\& \qquad {}+32-(n-5) \biggl(\frac{n-1}{n-2} \biggr)^{3}-48 \\& \quad = \biggl(\frac{n-1}{n-2} \biggr)^{3}+ \biggl( \frac{3(n-1)}{n} \biggr)^{3}-16. \end{aligned}$$

Let $g(x)= (\frac{x-1}{x-2} )^{3}+ (\frac{3(x-1)}{x} )^{3}-16$, where $x\geq5$. We have

$$\begin{aligned} g^{\prime}(x) =& \frac{3^{4}(x-1)^{2}}{x^{4}}-\frac{3(x-1)^{2}}{(x-2)^{4}} \\ =& \frac{3(x-1)^{2}}{x^{4}(x-2)^{2}} \bigl(\sqrt{27}(x-2)^{2}+x^{2} \bigr) \\ &{}\times \bigl(\sqrt [4]{27}(x-2)+x \bigr) \bigl(\sqrt[4]{27}(x-2)-x \bigr) \\ >& 0 \end{aligned}$$

for $x\geq5$. Then $g(n)\geq g(5)= (\frac{4}{3} )^{3}+ (\frac {12}{5} )^{3}-16>0$. Therefore, this completes the proof of Theorem 4.4. □