The First Zagreb Index, the Laplacian Spectral Radius, and Some Hamiltonian Properties of Graphs

Abstract

The first Zagreb index of a graph G is defined as the sum of the squares of the degrees of all the vertices in G. The Laplacian spectral radius of a graph G is defined as the largest eigenvalue of the Laplacian matrix of the graph G. In this paper, we first establish inequalities on the first Zagreb index and the Laplacian spectral radius of a graph. Using the ideas of proving the inequalities, we present sufficient conditions involving the first Zagreb index and the Laplacian spectral radius for some Hamiltonian properties of graphs.

1. Introduction

In this section, we first introduce the definitions, terminologies, and notations to be used and the background of the research in this paper. Most of them have appeared in the author’s previous papers such as [1]. We will repeat them here. We consider only finite undirected graphs without loops or multiple edges. Notation and terminology not defined in this paper follow those in [2]. Let $G=\left(V\right(G),E(G\left)\right)$ be a graph. We use n and e to denote the number of vertices and the number of edges in G, respectively. The degree of a vertex u in G is represented by $d_G\left(u\right)$. We use $\delta \left(G\right)=d_1\left(G\right)\le d_2\left(G\right)\le \cdots \le d_n\left(G\right)=\Delta \left(G\right)$ to denote the degree sequence of a graph G. For a vertex subset S in G, we use $G\left[S\right]$ to denote the subgraph of G, which is induced by S. A subset of $V\left(G\right)$ in a graph G is an independent set if any two vertices in the subset are not adjacent. An independent set in a graph G is called a maximum independent set if its size is as large as possible. The independence number, denoted as $\beta \left(G\right)$, of a graph G is defined as the size of a maximum independent set in G. For two graphs $H_1$ and $H_2$, we define $H_1\le H_2$ as $V\left(H_1\right)=V\left(H_2\right)$ and $E\left(H_1\right)\subseteq E\left(H_2\right)$. The join of two disjoint graphs $H_1$ and $H_2$ is denoted by $H_1\vee H_2$. For two disjoint vertex subsets A and B of $V\left(G\right)$, we define $E(A,B)$ as the set of all the edges in $E\left(G\right)$ such that one end vertex of each edge is in A and another end vertex of the edge is in B. More specifically, $E(A,B):=\{e:e=ab\in E,a\in A,b\in B\}$. We use $K_p$ to denote a complete graph of order p. We also use $K_{p,q}$ to denote a complete bipartite graph with two partition sets X and Y such that $\left|X\right|=p$ and $\left|Y\right|=q$. The Laplacian matrix of a graph G, denoted as $L\left(G\right)$, is defined as $D\left(G\right)-A\left(G\right)$, where $D\left(G\right)$ is the diagonal matrix $diag(d_1,d_2,\dots ,d_n)$ of G and $A\left(G\right)$ is the adjacency matrix of G. We use $0={\lambda }_1\left(G\right)\le {\lambda }_2\left(G\right)\le \cdots \le {\lambda }_n\left(G\right)$ to denote the eigenvalues of $L\left(G\right)$. The largest eigenvalue ${\lambda }_n\left(G\right)$ of $L\left(G\right)$ is called the Lapalcian spectral radius of G. A cycle C in a graph G is a Hamilton cycle of G if all the vertices of G are on C. A graph G is called Hamiltonian if G contains a Hamilton cycle. A path P in a graph G is a Hamilton path of G if all the vertices of G are on P. A graph G is called traceable if G contains a Hamilton path.

The first Zagreb index of a graph was introduced by Gutman and Trinajstić in [3]. See also [4]. For a graph G, its first Zagreb index, denoted as $Z_1\left(G\right)$, is defined as ${\sum }_{u\in V\left(G\right)}{d}_G^2\left(u\right)$. As one of the most important topological indices of a graph, the first Zagreb index has been intensively investigated since its introduction. Many results on the first Zagreb index of a graph have been obtained. The survey paper [5] and the references therein are good resources on such results. The Laplacian spectral radius is one of the most important spectral parameters of a graph. A lot of results on the Laplacian spectral radius of a graph have been obtained. Readers interested in the results are encouraged to consult the survey papers [6,7,8] and the references therein. In this paper, we first investigate the relation between the first Zagreb index and the Laplacian spectral radius of a graph. In particular, using some classical inequalities, we obtain four inequalities on the first Zagreb index and the Laplacian spectral radius of a graph. Below are the results.

Theorem 1.

Let G be a connected graph with n vertices and e edges.

• (1) If $n\ge 2$, then

  $$Z_1\left(G\right)\le {\displaystyle \frac{{\lambda }_n^2\beta (n-\beta ){(\delta +n-\beta )}^2}{4n^2\delta }}+(n-\beta ){\Delta }^2$$

  with equality if and only if G is $K_{\beta }^c\vee H$, where $H=G[V-I]$; I is a maximum independent set in G with $\left|I\right|=\beta$ such that $d\left(u\right)=\delta =(n-\beta )$ for each $u\in I$ and $d\left(v\right)=\Delta$ for each $v\in V-I$.
• (2) Then

  $$Z_1\left(G\right)\le {\displaystyle \frac{{\lambda }_n\beta (n-\beta )(\delta +n-\beta )}{n}}-\delta \beta (n-\beta )+(n-\beta ){\Delta }^2$$

  with equality if and only if G is $K_{\beta }^c\vee H$, where $H=G[V-I]$; I is a maximum independent set in G with $\left|I\right|=\beta$ such that $d\left(u\right)=\delta =(n-\beta )$ for each $u\in I$ and $d\left(v\right)=\Delta$ for each $v\in V-I$.
• (3) If $n\ge 2$, then

  $$Z_1\left(G\right)\le {\displaystyle \frac{\beta {(n-\beta )}^2}{n^2}}{\left({\lambda }_n+{\displaystyle \frac{n{(n-\beta -\delta )}^2}{4(n-\beta )(\delta +n-\beta )}}\right)}^2+(n-\beta ){\Delta }^2$$

  with equality if and only if G is $K_{\beta }^c\vee H$, where $H=G[V-I]$; I is a maximum independent set in G with $\left|I\right|=\beta$ such that $d\left(u\right)=\delta =(n-\beta )$ for each $u\in I$ and $d\left(v\right)=\Delta$ for each $v\in V-I$.
• (4) Then

  $$Z_1\left(G\right)\le {\displaystyle \frac{{\lambda }_n(n-\beta )\left(e+\beta {\left(\sqrt{n-\beta }-\sqrt{\delta }\right)}^2\right)}{n}}+(n-\beta ){\Delta }^2$$

  with equality if and only if G is $K_{\beta ,n-\beta }$, where β is the independence number of G.

In addition, using the ideas of proving the inequalities above, in this paper, we further obtain new sufficient conditions involving the first Zagreb index and the Laplacian spectral radius for the Hamiltonian and traceable graphs. Below are the results.

Theorem 2.

Let G be a k-connected ($k\ge 2$) graph with $n\ge 3$ vertices and e edges.

• (1) If

  $$Z_1\left(G\right)\ge {\displaystyle \frac{{\lambda }_n^2(k+1)(n-k-1){(\delta +n-k-1)}^2}{4n^2\delta }}+(n-k-1){\Delta }^2,$$

  then G is Hamiltonian or $G\in \{H:K_{k,k+1}\le H\le K_{k+1}^c\vee K_k$, $\delta \left(H\right)=k$, and $d_H\left(v\right)=\Delta \ge k+1$ if v is not a vertex with the minimum degree of $H\}$.
• (2) If

  $$Z_1\left(G\right)\ge {\displaystyle \frac{{\lambda }_n(k+1)(n-k-1)(\delta +n-k-1)}{n}}-\delta (k+1)(n-k-1)+(n-k-1){\Delta }^2,$$

  then G is Hamiltonian or $G\in \{H:K_{k,k+1}\le H\le K_{k+1}^c\vee K_k$, $\delta \left(H\right)=k$, and $d_H\left(v\right)=\Delta \ge k+1$ if v is not a vertex with the minimum degree of $H\}$.
• (3) If

  $$Z_1\left(G\right)\ge {\displaystyle \frac{(k+1){(n-k-1)}^2}{n^2}}{\left({\lambda }_n+{\displaystyle \frac{n{(n-k-1-\delta )}^2}{4(n-k-1)(\delta +n-k-1)}}\right)}^2+(n-k-1){\Delta }^2,$$

  then G is Hamiltonian or $G\in \{H:K_{k,k+1}\le H\le K_{k+1}^c\vee K_k$, $\delta \left(H\right)=k$, and $d_H\left(v\right)=\Delta \ge k+1$ if v is not a vertex with the minimum degree of $H\}$.
• (4) If

  $$Z_1\left(G\right)\ge {\displaystyle \frac{{\lambda }_n(n-k-1)\left(e+(k+1){\left(\sqrt{n-k-1}-\sqrt{\delta }\right)}^2\right)}{n}}+(n-k-1){\Delta }^2,$$

  then G is Hamiltonian or G is $K_{k,k+1}$.

Theorem 3.

Let G be a k-connected ($k\ge 1$) with $n\ge 2$ vertices and e edges.

• (1) If

  $$Z_1\left(G\right)\ge {\displaystyle \frac{{\lambda }_n^2(k+2)(n-k-2){(\delta +n-k-2)}^2}{4n^2\delta }}+(n-k-2){\Delta }^2,$$

  then G is traceable or $G\in \{H:K_{k,k+2}\le H\le K_{k+2}^c\vee K_k$, $\delta \left(H\right)=k$, and $d_H\left(v\right)=\Delta \ge k+2$ if v is not a vertex with the minimum degree of $H\}$.
• (2) If

  $$Z_1\left(G\right)\ge {\displaystyle \frac{{\lambda }_n(k+2)(n-k-2)(\delta +n-k-2)}{n}}-\delta (k+2)(n-k-2)+(n-k-2){\Delta }^2,$$

  then G is traceable or $G\in \{H:K_{k,k+2}\le H\le K_{k+2}^c\vee K_k$, $\delta \left(H\right)=k$, and $d_H\left(v\right)=\Delta \ge k+2$ if v is not a vertex with the minimum degree of $H\}$.
• (3) If

  $$Z_1\left(G\right)\ge {\displaystyle \frac{(k+2){(n-k-2)}^2}{n^2}}{\left({\lambda }_n+{\displaystyle \frac{n{(n-k-2-\delta )}^2}{4(n-k-2)(\delta +n-k-2)}}\right)}^2+(n-k-2){\Delta }^2,$$

  then G is traceable or $G\in \{H:K_{k,k+2}\le H\le K_{k+2}^c\vee K_k$, $\delta \left(H\right)=k$, and $d_H\left(v\right)=\Delta \ge k+2$ if v is not a vertex with the minimum degree of $H\}$.
• (4) If

  $$Z_1\left(G\right)\ge {\displaystyle \frac{{\lambda }_n(n-k-2)\left(e+(k+2){\left(\sqrt{n-k-2}-\sqrt{\delta }\right)}^2\right)}{n}}+(n-k-2){\Delta }^2,$$

  then G is traceable or G is $K_{k,k+2}$.

2. Lemmas

We will use some existing results as our lemmas. Lemma 1 below is the Pólya-Szegő inequality in [9]. The following one is Corollary 3 on page 66 in [10].

Lemma 1

([10]). Let the real numbers $a_k$ and $b_k$ ($k=1,2,\cdots ,s$) satisfy $0<m_1\le a_k\le M_1$ and $0<m_2\le b_k\le M_2$. Then

$$\sum _{k=1}^s{a}_k^2\sum _{k=1}^s{b}_k^2\le {\displaystyle \frac{{(M_1{M}_2+m_1{m}_2)}^2}{4m_1{m}_2{M}_1{M}_2}}{\left(\sum _{k=1}^s{a}_k{b}_k\right)}^2.$$

If $M_1{M}_2>m_1{m}_2$, then the equality sign holds in the above inequality if and only if

$$\nu ={\displaystyle \frac{M_1{m}_2}{M_1{m}_2+m_1{M}_2}}s$$

is an integer; at the same time, for ν values of k, one has $(a_k,b_k)=(m_1,M_2)$ and for the remaining $s-\nu$ values of k, one has $(a_k,b_k)=(M_1,m_2)$. If $M_1{M}_2=m_1{m}_2$, the equality always holds.

Lemma 2 follows from Theorem 1 on page 61 and page 62 in [10].

Lemma 2

([10]). Let the real numbers $a_k$ and $b_k$ ($k=1,2,\cdots ,s$) satisfy $0\le m_1\le a_k\le M_1$ and $0\le m_2\le b_k\le M_2$. Then

$$m_1{M}_1\sum _{k=1}^s{b}_k^2+m_2{M}_2\sum _{k=1}^s{a}_k^2\le (M_1{M}_2+m_1{m}_2)\sum _{k=1}^s{a}_k{b}_k.$$

If $0<m_1$ and $0<m_2$, then the equality holds if and only if for each k, one has either $(a_k,b_k)=(m_1,M_2)$ or $(a_k,b_k)=(M_1,m_2)$.

Lemma 3 is Theorem $5.21$ on page 82 and page 83 in [11].

Lemma 3

([11]). Suppose $a_k$ and $b_k$$(k=1,2,\cdots ,s)$ are positive real numbers satisfying $0<\gamma \le {\displaystyle \frac{a_k}{b_k}}\le \Gamma <\infty$ for any $k\in \{1,2,\dots ,s\}$. Then

$$0\le {\left(\sum _{k=1}^s{a}_k^2\sum _{k=1}^s{b}_k^2\right)}^{{\displaystyle \frac{1}{2}}}-\sum _{k=1}^s{a}_k{b}_k\le {\displaystyle \frac{{(\Gamma -\gamma )}^2}{4(\gamma +\Gamma )}}\sum _{k=1}^s{b}_k^2.$$

The equality holds in the above inequality if and only if there exists a subsequence $(k_1,k_2,\dots ,k_t)$ of $(1,2,\dots ,s)$ such that

$$\sum _{m=1}^t{b}_{k_m}^2={\displaystyle \frac{\Gamma +3\gamma }{4(\Gamma +\gamma )}}\sum _{j=1}^s{b}_j^2,$$

${\displaystyle \frac{a_{k_m}}{b_{k_m}}}=\Gamma$$(m=1,2,\dots ,t)$, and ${\displaystyle \frac{a_k}{b_k}}=\gamma$ for every k distinct from all $k_m$.

Lemma 4 is an inequality on page 303 in [12].

Lemma 4

([12]). Let the real numbers $a_k$ and $b_k$ ($k=1,2,\cdots ,s$) satisfy $0<m_1\le a_k\le M_1$ and $0<m_2\le b_k\le M_2$. Then

$$\sum _{k=1}^s{a}_k^2\sum _{k=1}^s{b}_k^2-{\left(\sum _{k=1}^s{a}_k{b}_k\right)}^2\le {\left(\sqrt{{\displaystyle \frac{M_1}{m_2}}}-\sqrt{{\displaystyle \frac{m_1}{M_2}}}\right)}^2\sum _{k=1}^s{a}_k{b}_k\sum _{k=1}^s{b}_k^2$$

The equality holds if and only if there exists a subsequence $(k_1,k_2,\dots ,k_t)$ of $(1,2,\dots ,s)$ such that

$${\displaystyle \frac{n}{t}}=1+{\left({\displaystyle \frac{M_1}{m_1}}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\displaystyle \frac{m_2}{M_2}}\right)}^{{\displaystyle \frac{3}{2}}},$$

$a_{k_{\mu }}=M_1$, $b_{k_{\mu }}=m_2$$(\mu =1,2,\dots ,t)$, and $a_k=m_1$, $b_k=M_2$ for every k distinct from all $k_{\mu }$.

Lemma 5 below is Lemma 3 on page 392 in [13].

Lemma 5

([13]). Let $G=(V,E)$ be a connected graph of order n and $G_1$ be an induced subgraph of G with $n_1$ ($n_1<n$) vertices and average degree $r_1$ (i.e., $r_1={\sum }_{v\in V\left(G_1\right)}{d}_{G_1}\left(v\right)/n_1)$. Set $d_1={\sum }_{v\in V\left(G_1\right)}d\left(v\right)/n_1$. Then

$${\lambda }_1\left(G\right)\ge {\displaystyle \frac{n(d_1-r_1)}{n-n_1}}.$$

Moreover, if the equality holds, then $d_{G_2}\left(u\right)=s$ for all vertices $u\in V\left(G_1\right)$ and $d_{G_1}\left(v\right)=t$ for all vertices $v\in V\left(G_2\right)$, where $G_2=G[V-V\left(G_1\right)]$.

Lemma 6 below is Lemma $13.1.3$ on page 280 in [14].

Lemma 6

([14]). If G is a graph on n vertices and $2\le i\le n$, then ${\lambda }_i\left(G^c\right)=n-{\lambda }_{n-i+2}\left(G\right)$.

The next two results are from [15].

Lemma 7

([15]). Let G be a k-connected graph of order $n\ge 3$. If $\beta \le k$, then G is Hamiltonian.

Lemma 8

([15]). Let G be a k-connected graph of order n. If $\beta \le k+1$, then G is traceable.

3. Proofs

3.1. Proof of Theorem 1

We will establish the four upper bounds of the Zagreb index that involve the Laplacian spectral radius of a graph. Some ideas exhibited in the author’s previous papers such as the ones in the proofs of Theorem 3 in [1] will be used in the proofs below. Let G be a connected graph with n vertices and e edges. Let $I:=\{u_1,u_2,\dots ,u_{\beta }\}$ be a maximum independent set in G.

Proof (1) of Theorem 1.

Next, we will use Lemma 1 and Lemma 5 to prove the first inequality in Theorem 1. Since $n\ge 2$ and G is connected, we have that $\delta \ge 1$ and $\beta <n$. Applying Lemma 1 with $s=\beta$, $a_i=1$ and $b_i=d\left(u_i\right)$ (where $i=1,2,\dots ,\beta$), $m_1=1>0$, $M_1=1$, $m_2=\delta >0$, and $M_2=n-\beta$, we have that

$$\sum _{i=1}^{\beta }{1}^2\sum _{i=1}^{\beta }{d}^2\left(u_i\right)\le {\displaystyle \frac{{(\delta +n-\beta )}^2}{4\delta (n-\beta )}}{\left(\sum _{i=1}^{\beta }d\left(u_i\right)\right)}^2.$$

Thus

$${\left(\sum _{i=1}^{\beta }d\left(u_i\right)\right)}^2\ge {\displaystyle \frac{4\delta \beta (n-\beta ){\sum }_{i=1}^{\beta }{d}^2\left(u_i\right)}{{(\delta +n-\beta )}^2}}.$$

Applying Lemma 5 with $G_1=G\left[I\right]$ and $G_2=G[V-I]$, we have

$${\lambda }_n^2\ge {\displaystyle \frac{n^2{\left({\sum }_{i=1}^{\beta }d\left(u_i\right)\right)}^2}{{\left(\beta (n-\beta )\right)}^2}}\ge {\displaystyle \frac{4n^2\delta \beta (n-\beta ){\sum }_{i=1}^{\beta }{d}^2\left(u_i\right)}{{\left(\beta (n-\beta )\right)}^2{(\delta +n-\beta )}^2}}.$$

Therefore

$$\sum _{i=1}^{\beta }{d}^2\left(u_i\right)\le {\displaystyle \frac{{\lambda }_n^2\beta (n-\beta ){(\delta +n-\beta )}^2}{4n^2\delta }}.$$

Hence

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)\le {\displaystyle \frac{{\lambda }_n^2\beta (n-\beta ){(\delta +n-\beta )}^2}{4n^2\delta }}+(n-\beta ){\Delta }^2.$$

If

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)={\displaystyle \frac{{\lambda }_n^2\beta (n-\beta ){(\delta +n-\beta )}^2}{4n^2\delta }}+(n-\beta ){\Delta }^2,$$

from Lemma 1, Lemma 5, and the above proofs, we obtain $d\left(u\right)=\delta =(n-\beta )$ for each $u\in I$ and $d\left(v\right)=\Delta$ for each $v\in V-I$. Hence, G is $K_{\beta }^c\vee H$, where $H=G[V-I]$; I is a maximum independent set in G with $\left|I\right|=\beta$ such that $d\left(u\right)=\delta =(n-\beta )$ for each $u\in I$ and $d\left(v\right)=\Delta$ for each $v\in V-I$.

Suppose G is $K_{\beta }^c\vee H$, where $H=G[V-I]$; I is a maximum independent set in G with $\left|I\right|=\beta$ such that $d\left(u\right)=\delta =(n-\beta )$ for each $u\in I$ and $d\left(v\right)=\Delta$ for each $v\in V-I$. Then $G^c=K_{\beta }\cup {\left(G[V-I]\right)}^c$. Notice that ${\lambda }_1\left(K_{\beta }\right)={\lambda }_1\left({\left(G[V-I]\right)}^c\right)=0$. Thus, ${\lambda }_1\left(G^c\right)={\lambda }_2\left(G^c\right)=0$. From Lemma 6, we have ${\lambda }_n\left(G\right)=n$. A simple computation verifies that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)={\displaystyle \frac{{\lambda }_n^2\beta (n-\beta ){(\delta +n-\beta )}^2}{4n^2\delta }}+(n-\beta ){\Delta }^2.$$

This completes the proof of (1) in Theorem 1. □

Proof (2) of Theorem 1.

If $n=1$, then the inequality in (2) of Theorem 1 is true. From now on, we assume that $n\ge 2$. Thus, $\beta <n$. Next, we will use Lemma 2 and Lemma 5 to prove the second inequality in Theorem 1. Applying Lemma 2 with $s=\beta$, $a_i=1$ and $b_i=d\left(u_i\right)$ (where $i=1,2,\dots ,\beta$), $m_1=1>0$, $M_1=1$, $m_2=\delta >0$, and $M_2=n-\beta$, we have that

$$\sum _{i=1}^{\beta }{d}^2\left(u_i\right)+\delta (n-\beta )\sum _{i=1}^{\beta }{1}^2\le (\delta +n-\beta )\sum _{i=1}^{\beta }d\left(u_i\right)$$

Thus

$$\sum _{i=1}^{\beta }d\left(u_i\right)\ge {\displaystyle \frac{{\sum }_{i=1}^{\beta }{d}^2\left(u_i\right)+\delta \beta (n-\beta )}{(\delta +n-\beta )}}.$$

Applying Lemma 5 with $G_1=G\left[I\right]$ and $G_2=G[V-I]$, we have that

$${\lambda }_n\ge {\displaystyle \frac{n\left({\sum }_{i=1}^{\beta }d\left(u_i\right)\right)}{\beta (n-\beta )}}\ge {\displaystyle \frac{n\left({\sum }_{i=1}^{\beta }{d}^2\left(u_i\right)+\delta \beta (n-\beta )\right)}{\beta (n-\beta )(\delta +n-\beta )}}.$$

Therefore

$$\sum _{i=1}^{\beta }{d}^2\left(u_i\right)\le {\displaystyle \frac{{\lambda }_n\beta (n-\beta )(\delta +n-\beta )}{n}}-\delta \beta (n-\beta ).$$

Hence

$$Z_1\left(G\right)\le \sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$\le {\displaystyle \frac{{\lambda }_n\beta (n-\beta )(\delta +n-\beta )}{n}}-\delta \beta (n-\beta )+(n-\beta ){\Delta }^2.$$

If

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$={\displaystyle \frac{{\lambda }_n\beta (n-\beta )(\delta +n-\beta )}{n}}-\delta \beta (n-\beta )+(n-\beta ){\Delta }^2,$$

using the proofs similar to the ones in the third last paragraph in Proof (1) in Theorem 1, we can obtain the desired result.

Suppose G is $K_{\beta }^c\vee H$, where $H=G[V-I]$; I is a maximum independent set in G with $\left|I\right|=\beta$ such that $d\left(u\right)=\delta =(n-\beta )$ for each $u\in I$ and $d\left(v\right)=\Delta$ for each $v\in V-I$. The proofs similar to the ones in the second last paragraph in Proof (1) in Theorem 1 yield the desired result.

This completes the proof of (2) in Theorem 1. □

Proof (3) of Theorem 1.

Next, we will use Lemma 3 and Lemma 5 to prove the third inequality in Theorem 1. Since $n\ge 2$ and G is connected, we have $\beta <n$. Applying Lemma 3 with $s=\beta$, $a_i=d\left(u_i\right)$, and $b_i=1$ (where $i=1,2,\dots ,\beta$), $\gamma =\delta >0$, $\Gamma =n-\beta$, we have that

$${\left(\sum _{i=1}^{\beta }{d}^2\left(u_i\right)\sum _{i=1}^{\beta }{1}^2\right)}^{{\displaystyle \frac{1}{2}}}-\sum _{k=1}^{\beta }d\left(u_i\right)\le {\displaystyle \frac{{(n-\beta -\delta )}^2}{4(\delta +n-\beta )}}\sum _{i=1}^{\beta }{1}^2.$$

Thus

$$\sum _{k=1}^{\beta }d\left(u_i\right)\ge {\left(\beta \sum _{i=1}^{\beta }{d}^2\left(u_i\right)\right)}^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{\beta {(n-\beta -\delta )}^2}{4(\delta +n-\beta )}}.$$

Applying Lemma 5 with $G_1=G\left[I\right]$ and $G_2=G[V-I]$, we have that

$${\lambda }_n\ge {\displaystyle \frac{n\left({\sum }_{i=1}^{\beta }d\left(u_i\right)\right)}{\beta (n-\beta )}}\ge {\displaystyle \frac{n}{\beta (n-\beta )}}{\left(\beta \sum _{i=1}^{\beta }{d}^2\left(u_i\right)\right)}^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{n{(n-\beta -\delta )}^2}{4(n-\beta )(\delta +n-\beta )}}.$$

Therefore

$$\sum _{i=1}^{\beta }{d}^2\left(u_i\right)\le {\displaystyle \frac{\beta {(n-\beta )}^2}{n^2}}{\left({\lambda }_n+{\displaystyle \frac{n{(n-\beta -\delta )}^2}{4(n-\beta )(\delta +n-\beta )}}\right)}^2.$$

Hence

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$\le {\displaystyle \frac{\beta {(n-\beta )}^2}{n^2}}{\left({\lambda }_n+{\displaystyle \frac{n{(n-\beta -\delta )}^2}{4(n-\beta )(\delta +n-\beta )}}\right)}^2+(n-\beta ){\Delta }^2.$$

If

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$={\displaystyle \frac{\beta {(n-\beta )}^2}{n^2}}{\left({\lambda }_n+{\displaystyle \frac{n{(n-\beta -\delta )}^2}{4(n-\beta )(\delta +n-\beta )}}\right)}^2+(n-\beta ){\Delta }^2,$$

using the proofs similar to the ones in the third last paragraph in Proof (1) in Theorem 1, we can obtain the desired result.

Suppose G is $K_{\beta }^c\vee H$, where $H=G[V-I]$; I is a maximum independent set in G with $\left|I\right|=\beta$ such that $d\left(u\right)=\delta =(n-\beta )$ for each $u\in I$ and $d\left(v\right)=\Delta$ for each $v\in V-I$. The proofs similar to the ones in the second last paragraph in Proof (1) in Theorem 1 yield the desired result.

This completes the proof of (3) in Theorem 1. □

Proof (4) of Theorem 1.

If $n=1$, then the inequality in (2) of Theorem 1 is true. From now on, we assume that $n\ge 2$. Thus, $\beta <n$. Next, we will use Lemma 4 and Lemma 5 to prove the fourth inequality in Theorem 1. We notice that

$$\sum _{u\in I}d\left(u\right)=\left|E(I,V-I)\right|\le \sum _{v\in V-I}d\left(v\right).$$

Since ${\sum }_{u\in I}d\left(u\right)+{\sum }_{v\in V-I}d\left(v\right)=2e$, we have that

$$\sum _{u\in I}d\left(u\right)\le e\le \sum _{v\in V-I}d\left(v\right).$$

Applying Lemma 4 with $s=\beta$, $a_i=d\left(u_i\right)$ and $b_i=1$ (where $i=1,2,\dots ,\beta$), $m_1=\delta >0$, $M_1=n-\beta$, $m_2=1>0$, and $M_2=1$, we have

$$\sum _{i=1}^{\beta }{d}^2\left(u_i\right)\sum _{i=1}^{\beta }{1}^2-{\left(\sum _{k=1}^{\beta }d\left(u_i\right)\right)}^2\le {\left(\sqrt{n-\beta }-\sqrt{\delta }\right)}^2\sum _{k=1}^{\beta }d\left(u_i\right)\sum _{i=1}^{\beta }{1}^2.$$

Thus

$$\beta \sum _{i=1}^{\beta }{d}^2\left(u_i\right)\le {\left(\sum _{k=1}^{\beta }d\left(u_i\right)\right)}^2+\beta {\left(\sqrt{n-\beta }-\sqrt{\delta }\right)}^2\sum _{k=1}^{\beta }d\left(u_i\right)$$

$$\le e\sum _{k=1}^{\beta }d\left(u_i\right)+\beta {\left(\sqrt{n-\beta }-\sqrt{\delta }\right)}^2\sum _{k=1}^{\beta }d\left(u_i\right)$$

$$=\left(e+\beta {\left(\sqrt{n-\beta }-\sqrt{\delta }\right)}^2\right)\sum _{k=1}^{\beta }d\left(u_i\right).$$

Therefore

$$\sum _{k=1}^{\beta }d\left(u_i\right)\ge {\displaystyle \frac{\beta {\sum }_{i=1}^{\beta }{d}^2\left(u_i\right)}{e+\beta {\left(\sqrt{n-\beta }-\sqrt{\delta }\right)}^2}}.$$

Applying Lemma 5 with $G_1=G\left[I\right]$ and $G_2=G[V-I]$, we have that

$${\lambda }_n\ge {\displaystyle \frac{n\left({\sum }_{i=1}^{\beta }d\left(u_i\right)\right)}{\beta (n-\beta )}}\ge {\displaystyle \frac{n{\sum }_{i=1}^{\beta }{d}^2\left(u_i\right)}{(n-\beta )\left(e+\beta {\left(\sqrt{n-\beta }-\sqrt{\delta }\right)}^2\right)}}$$

Hence

$$\sum _{i=1}^{\beta }{d}^2\left(u_i\right)\le {\displaystyle \frac{{\lambda }_n(n-\beta )\left(e+\beta {\left(\sqrt{n-\beta }-\sqrt{\delta }\right)}^2\right)}{n}}.$$

Consequently,

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$\le {\displaystyle \frac{{\lambda }_n(n-\beta )\left(e+\beta {\left(\sqrt{n-\beta }-\sqrt{\delta }\right)}^2\right)}{n}}+(n-\beta ){\Delta }^2.$$

If

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$={\displaystyle \frac{{\lambda }_n(n-\beta )\left(e+\beta {\left(\sqrt{n-\beta }-\sqrt{\delta }\right)}^2\right)}{n}}+(n-\beta ){\Delta }^2,$$

from Lemma 4, Lemma 5, and the above proofs, we obtain $d\left(u\right)=\delta =(n-\beta )$ for each $u\in I$, $d\left(v\right)=\Delta$ for each $v\in V-I$, and ${\sum }_{k=1}^{\beta }d\left(u_i\right)=e$. Notice that ${\sum }_{k=1}^{\beta }d\left(u_i\right)=e$ implies that ${\sum }_{v\in V-I}d\left(v\right)=e$ and G is a bipartite graph with partition sets of I and $V-I$. Hence, G is $K_{\beta ,n-\beta }$, where $\beta$ is the independence number of G.

Suppose G is $K_{\beta ,n-\beta }$, where $\beta$ is the independence number of G. Then ${\lambda }_n=n$. A simple computation verifies that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$={\displaystyle \frac{{\lambda }_n(n-\beta )\left(e+\beta {\left(\sqrt{n-\beta }-\sqrt{\delta }\right)}^2\right)}{n}}+(n-\beta ){\Delta }^2.$$

This completes the proof of (4) in Theorem 1. □

Next, we will prove Theorem 2.

3.2. Proof of Theorem 2

We will prove the four sufficient conditions that involve the Zagreb index and the Laplacian spectral radius for Hamiltonian graphs. Some ideas exhibited in the author’s previous papers such as the ones in the proofs of Theorem 1 in [1] will be used in the proofs below. Let G be a k-connected ($k\ge 2$) graph with $n\ge 3$ vertices and e edges satisfying exactly one of the four conditions in Theorem 2. Suppose G is not Hamiltonian. Then, Lemma 7 implies that $\beta \ge k+1$. Also, we have that $n\ge 2\delta +1\ge 2k+1$; otherwise, $\delta \ge k\ge n/2$ and G is Hamiltonian. Let $I_1:=\{u_1,u_2,\dots ,u_{\beta }\}$ be a maximum independent set in G. Then $I:=\{u_1,u_2,\dots ,u_{k+1}\}$ is an independent set in G.

Proof (1) of Theorem 2.

Now, we use the ideas of proving (1) in Theorem 1 to complete this proof. Applying Lemma 1 with $s=(k+1)$, $a_i=1$, and $b_i=d\left(u_i\right)$ (where $i=1,2,\dots ,(k+1)$), $m_1=1>0$, $M_1=1$, $m_2=\delta >0$, and $M_2=n-k-1$, we have that

$$\sum _{i=1}^{k+1}{1}^2\sum _{i=1}^{k+1}{d}^2\left(u_i\right)\le {\displaystyle \frac{{(\delta +n-k-1)}^2}{4\delta (n-k-1)}}{\left(\sum _{i=1}^{k+1}d\left(u_i\right)\right)}^2.$$

Following the proof of (1) in Theorem 1, we have that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$\le {\displaystyle \frac{{\lambda }_n^2(k+1)(n-k-1){(\delta +n-k-1)}^2}{4n^2\delta }}+(n-k-1){\Delta }^2.$$

From the given conditions in this case, we have that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$={\displaystyle \frac{{\lambda }_n^2(k+1)(n-k-1){(\delta +n-k-1)}^2}{4n^2\delta }}+(n-k-1){\Delta }^2.$$

Following the proof of (1) in Theorem 1 again, we have that $d\left(u\right)=\delta =(n-k-1)$ for each $u\in I$ and $d\left(v\right)=\Delta$ for each $v\in V-I$. Hence, G is $K_{k+1}^c\vee H$, where $H=G[V-I]$; I is a maximum independent set in G with $\left|I\right|=k+1$ such that $d\left(u\right)=\delta =(n-k-1)$ for each $u\in I$ and $d\left(v\right)=\Delta$ for each $v\in V-I$.

If $n-k-1\ge k+1$, then $\delta =n-k-1\ge {\displaystyle \frac{n-k-1+k+1}{2}}={\displaystyle \frac{n}{2}}$ and G is Hamiltonian, which is a contradiction. If $n-k-1\le k$, then $n\le 2k+1$. Thus, $n=2k+1$ and $G\in \{H:K_{k,k+1}\le H\le K_{k+1}^c\vee K_k$, $\delta \left(H\right)=k$ and $d_H\left(v\right)=\Delta \ge k+1$ if v is not a vertex with the minimum degree of $H\}$.

This completes the proof of (1) in Theorem 2. □

Proof (2) of Theorem 2.

Now we use the ideas of proving (2) in Theorem 1 to complete this proof. Applying Lemma 2 with $s=(k+1)$, $a_i=1$, and $b_i=d\left(u_i\right)$ (where $i=1,2,\dots ,(k+1)$), $m_1=1>0$, $M_1=1$, $m_2=\delta >0$, and $M_2=n-k-1$, we have that

$$\sum _{i=1}^{k+1}{d}^2\left(u_i\right)+\delta (n-k-1)\sum _{i=1}^{k+1}{1}^2\le (\delta +n-k-1)\sum _{i=1}^{k+1}d\left(u_i\right).$$

Following the proof of (2) in Theorem 1, we have that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$\le {\displaystyle \frac{{\lambda }_n(k+1)(n-k-1)(\delta +n-k-1)}{n}}-\delta (k+1)(n-k-1)+(n-k-1){\Delta }^2.$$

From the given conditions in this case, we have that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$={\displaystyle \frac{{\lambda }_n(k+1)(n-k-1)(\delta +n-k-1)}{n}}-\delta (k+1)(n-k-1)+(n-k-1){\Delta }^2.$$

Using proofs similar to the ones in the third and second last paragraphs in Proof (1) in Theorem 2, we can complete the remainder of this proof.

This completes the proof of (2) in Theorem 2. □

Proof (3) of Theorem 2.

Now, we use the ideas of proving (3) in Theorem 1 to complete this proof. Applying Lemma 3 with $s=(k+1)$, $a_i=d\left(u_i\right)$, and $b_i=1$ (where $i=1,2,\dots ,(k+1)$), $\gamma =\delta >0$, $\Gamma =n-k-1$, we have that

$${\left(\sum _{i=1}^{k+1}{d}^2\left(u_i\right)\sum _{i=1}^{k+1}{1}^2\right)}^{{\displaystyle \frac{1}{2}}}-\sum _{k=1}^{k+1}d\left(u_i\right)\le {\displaystyle \frac{{(n-k-1-\delta )}^2}{4(\delta +n-k-1)}}\sum _{i=1}^{k+1}{1}^2.$$

Following the proof of (3) in Theorem 1, we have that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$\le {\displaystyle \frac{(k+1){(n-k-1)}^2}{n^2}}{\left({\lambda }_n+{\displaystyle \frac{n{(n-k-1-\delta )}^2}{4(n-k-1)(\delta +n-k-1)}}\right)}^2+(n-k-1){\Delta }^2.$$

From the given conditions in this case, we have that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$={\displaystyle \frac{(k+1){(n-k-1)}^2}{n^2}}{\left({\lambda }_n+{\displaystyle \frac{n{(n-k-1-\delta )}^2}{4(n-k-1)(\delta +n-k-1)}}\right)}^2+(n-k-1){\Delta }^2.$$

Using proofs similar to the ones in the third and second last paragraphs in Proof (1) in Theorem 2, we can complete the remainder of this proof.

This completes the proof of (3) in Theorem 2. □

Proof (4) of Theorem 2.

Now, we use the ideas of proving (4) in Theorem 1 to complete this proof. We notice that

$$\sum _{u\in I}d\left(u\right)=\left|E(I,V-I)\right|\le \sum _{v\in V-I}d\left(v\right).$$

Since ${\sum }_{u\in I}d\left(u\right)+{\sum }_{v\in V-I}d\left(v\right)=2e$, we have that

$$\sum _{u\in I}d\left(u\right)\le e\le \sum _{v\in V-I}d\left(v\right).$$

Applying Lemma 4 with $s=(k+1)$, $a_i=d\left(u_i\right)$, and $b_i=1$ (where $i=1,2,\dots ,(k+1)$), $m_1=\delta >0$, $M_1=n-k-1$, $m_2=1>0$, and $M_2=1$, we have

$$\sum _{i=1}^{k+1}{d}^2\left(u_i\right)\sum _{i=1}^{k+1}{1}^2-{\left(\sum _{i=1}^{k+1}d\left(u_i\right)\right)}^2\le {\left(\sqrt{n-k-1}-\sqrt{\delta }\right)}^2\sum _{k=1}^{k+1}d\left(u_i\right)\sum _{i=1}^{k+1}{1}^2.$$

Following the proof of (4) in Theorem 1, we have that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$\le {\displaystyle \frac{{\lambda }_n(n-k-1)\left(e+(k+1){\left(\sqrt{n-k-1}-\sqrt{\delta }\right)}^2\right)}{n}}+(n-k-1){\Delta }^2.$$

From the given conditions in this case, we have that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$={\displaystyle \frac{{\lambda }_n(n-k-1)\left(e+(k+1){\left(\sqrt{n-k-1}-\sqrt{\delta }\right)}^2\right)}{n}}+(n-k-1){\Delta }^2.$$

Following the proof of (4) in Theorem 1 again, we have that $d\left(u\right)=\delta =(n-k-1)$ for each $u\in I$ and $d\left(v\right)=\Delta$ for each $v\in V-I$, and G is a bipartite graph with partition sets of I and $V-I$.

If $n-k-1\ge k+1$, then $\delta =n-k-1\ge {\displaystyle \frac{n-k-1+k+1}{2}}={\displaystyle \frac{n}{2}}$ and G is Hamiltonian, which is a contradiction. If $n-k-1\le k$, then $n\le 2k+1$. Thus, $n=2k+1$ and G is $K_{k,k+1}$.

This completes the proof of (4) in Theorem 2. □

The proofs of Theorem 3 are similar to the proofs of Theorem 2. For the sake of completeness, we still present full proofs of Theorem 3 below.

3.3. Proof of Theorem 3

We will prove the four sufficient conditions that involve the Zagreb index and the Laplacian spectral radius for traceable graphs. Some ideas exhibited in the author’s previous papers such as the ones in the proofs of Theorem 2 in [1] will be used in the proofs below. Let G be a k-connected ($k\ge 1$) graph with $n\ge 2$ vertices and e edges satisfying exactly one of the four conditions in Theorem 3. Suppose G is not traceable. Then, Lemma 8 implies that $\beta \ge k+2$. Also, we have that $n\ge 2\delta +2\ge 2k+2$; otherwise, $\delta \ge k\ge (n-1)/2$ and G is traceable. Let $I_1:=\{u_1,u_2,\dots ,u_{\beta }\}$ be a maximum independent set in G. Then $I:=\{u_1,u_2,\dots ,u_{k+2}\}$ is an independent set in G.

Proof (1) of Theorem 3.

Now, we use the ideas of proving (1) in Theorem 1 to complete this proof. Applying Lemma 1 with $s=(k+2)$, $a_i=1$, and $b_i=d\left(u_i\right)$ (where $i=1,2,\dots ,(k+2)$), $m_1=1>0$, $M_1=1$, $m_2=\delta >0$, and $M_2=n-k-2$, we have that

$$\sum _{i=1}^{k+2}{1}^2\sum _{i=1}^{k+2}{d}^2\left(u_i\right)\le {\displaystyle \frac{{(\delta +n-k-2)}^2}{4\delta (n-k-2)}}{\left(\sum _{i=1}^{k+2}d\left(u_i\right)\right)}^2.$$

Following the proof of (1) in Theorem 1, we have that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$\le {\displaystyle \frac{{\lambda }_n^2(k+2)(n-k-2){(\delta +n-k-2)}^2}{4n^2\delta }}+(n-k-2){\Delta }^2.$$

From the given conditions in this case, we have that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$={\displaystyle \frac{{\lambda }_n^2(k+2)(n-k-2){(\delta +n-k-2)}^2}{4n^2\delta }}+(n-k-2){\Delta }^2.$$

Following the proof of (1) in Theorem 1 again, we have that $d\left(u\right)=\delta =(n-k-2)$ for each $u\in I$ and $d\left(v\right)=\Delta$ for each $v\in V-I$. Hence, G is $K_{k+2}^c\vee H$, where $H=G[V-I]$; I is a maximum independent set in G with $\left|I\right|=k+2$ such that $d\left(u\right)=\delta =(n-k-2)$ for each $u\in I$ and $d\left(v\right)=\Delta$ for each $v\in V-I$.

If $n-k-2\ge k+1$, then $\delta =n-k-2\ge {\displaystyle \frac{n-k-2+k+1}{2}}={\displaystyle \frac{n-1}{2}}$ and G is traceable, which is a contradiction. If $n-k-2\le k$, then $n\le 2k+2$. Thus, $n=2k+2$ and $G\in \{H:K_{k,k+2}\le H\le K_{k+2}^c\vee K_k$, $\delta \left(H\right)=k$ and $d_H\left(v\right)=\Delta \ge k+1$ if v is not a vertex with the minimum degree of $H\}$.

This completes the proof of (1) in Theorem 3. □

Proof (2) of Theorem 3.

Now, we use the ideas of proving (2) in Theorem 1 to complete this proof. Applying Lemma 2 with $s=(k+2)$, $a_i=1$, and $b_i=d\left(u_i\right)$ (where $i=1,2,\dots ,\beta$), $m_1=1>0$, $M_1=1$, $m_2=\delta >0$, and $M_2=n-k-2$, we have that

$$\sum _{i=1}^{k+2}{d}^2\left(u_i\right)+\delta (n-k-2)\sum _{i=1}^{k+2}{1}^2\le (\delta +n-k-2)\sum _{i=1}^{k+2}d\left(u_i\right).$$

Following the proof of (2) in Theorem 1, we have that

$$Z_1\left(G\right)\le \sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$\le {\displaystyle \frac{{\lambda }_n(k+2)(n-k-2)(\delta +n-k-2)}{n}}-\delta (k+2)(n-k-2)+(n-k-2){\Delta }^2.$$

From the given conditions in this case, we have that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$={\displaystyle \frac{{\lambda }_n(k+2)(n-k-2)(\delta +n-k-2)}{n}}-\delta (k+2)(n-k-2)+(n-k-2){\Delta }^2.$$

Using proofs similar to the ones in the third and second last paragraphs in Proof (1) in Theorem 3, we can complete the remainder of this proof.

This completes the proof of (2) in Theorem 3. □

Proof (3) of Theorem 3.

Now, we use the ideas of proving (3) in Theorem 1 to complete this proof. Applying Lemma 3 with $s=(k+2)$, $a_i=d\left(u_i\right)$, and $b_i=1$ (where $i=1,2,\dots ,(k+2)$), $\gamma =\delta >0$, $\Gamma =n-k-2$, we have that

$${\left(\sum _{i=1}^{k+2}{d}^2\left(u_i\right)\sum _{i=1}^{k+2}{1}^2\right)}^{{\displaystyle \frac{1}{2}}}-\sum _{k=1}^{k+2}d\left(u_i\right)\le {\displaystyle \frac{{(n-k-2-\delta )}^2}{4(\delta +n-k-2)}}\sum _{i=1}^{k+2}{1}^2.$$

Following the proof of (3) in Theorem 1, we have that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$\le {\displaystyle \frac{(k+2){(n-k-2)}^2}{n^2}}{\left({\lambda }_n+{\displaystyle \frac{n{(n-k-2-\delta )}^2}{4(n-k-2)(\delta +n-k-2)}}\right)}^2+(n-k-2){\Delta }^2.$$

From the given conditions in this case, we have that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$={\displaystyle \frac{(k+2){(n-k-2)}^2}{n^2}}{\left({\lambda }_n+{\displaystyle \frac{n{(n-k-2-\delta )}^2}{4(n-k-2)(\delta +n-k-2)}}\right)}^2+(n-k-2){\Delta }^2.$$

Using proofs which are similar to the ones in the third and second last paragraphs in Proof (1) in Theorem 3, we can complete the remainder of this proof.

This completes the proof of (3) in Theorem 3. □

Proof (4) of Theorem 3.

Now, we use the ideas of proving (4) in Theorem 1 to complete this proof. We notice that

$$\sum _{u\in I}d\left(u\right)=\left|E(I,V-I)\right|\le \sum _{v\in V-I}d\left(v\right).$$

Since ${\sum }_{u\in I}d\left(u\right)+{\sum }_{v\in V-I}d\left(v\right)=2e$, we have that

$$\sum _{u\in I}d\left(u\right)\le e\le \sum _{v\in V-I}d\left(v\right).$$

Applying Lemma 4 with $s=(k+2)$, $a_i=d\left(u_i\right)$, and $b_i=1$ (where $i=1,2,\dots ,(k+2)$), $m_1=\delta >0$, $M_1=n-k-2$, $m_2=1>0$, and $M_2=1$, we have

$$\sum _{i=1}^{k+2}{d}^2\left(u_i\right)\sum _{i=1}^{k+2}{1}^2-{\left(\sum _{i=1}^{k+2}d\left(u_i\right)\right)}^2\le {\left(\sqrt{n-k-2}-\sqrt{\delta }\right)}^2\sum _{i=1}^{k+2}d\left(u_i\right)\sum _{i=1}^{k+2}{1}^2.$$

Following the proof of (4) in Theorem 1, we have that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$\le {\displaystyle \frac{{\lambda }_n(n-k-2)\left(e+(k+2){\left(\sqrt{n-k-2}-\sqrt{\delta }\right)}^2\right)}{n}}+(n-k-2){\Delta }^2.$$

From the given conditions in this case, we have that

$$Z_1\left(G\right)=\sum _{u\in I}{d}^2\left(u\right)+\sum _{v\in V-I}{d}^2\left(v\right)$$

$$={\displaystyle \frac{{\lambda }_n(n-k-2)\left(e+(k+2){\left(\sqrt{n-k-2}-\sqrt{\delta }\right)}^2\right)}{n}}+(n-k-2){\Delta }^2.$$

Following the proof of (4) in Theorem 1 again, we have that $d\left(u\right)=\delta =(n-k-2)$ for each $u\in I$ and $d\left(v\right)=\Delta$ for each $v\in V-I$, and G is a bipartite graph with partition sets of I and $V-I$.

If $n-k-2\ge k+1$, then $\delta =n-k-2\ge {\displaystyle \frac{n-k-2+k+1}{2}}={\displaystyle \frac{n-1}{2}}$ and G is traceable, which is a contradiction. If $n-k-2\le k$, then $n\le 2k+2$. Thus, $n=2k+2$ and G is $K_{k,k+2}$.

This completes the proof of (4) in Theorem 3. □

4. Conclusions

In this paper, utilizing the established inequalities, we obtain new inequalities on the first Zagreb index and the Laplacian spectral radius of a graph. Based on the ideas of proving the inequalities and Chvátal and Erdös’ conditions on Hamiltonian and traceable graphs, we present new sufficient conditions involving the first Zagreb index and the Laplacian spectral radius for Hamiltonian and traceable graphs. In the future, we will adopt the methodologies exhibited in this paper to investigate relations on other graphical invariants and discover new sufficient conditions for Hamiltonian and traceable graphs.