In this paper, we consider only finite undirected graphs without loops or multiple edges. Notation and terminology not defined here follow those in [1] . Let $G=\left(V\left(G\right),E\left(G\right)\right)$ be a graph. The number of vertices and the number of edges in $G$ are denoted by $n$ and $e$, respectively. The degree of a vertex $v$ is denoted by $d_G\left(v\right)$. The minimum and maximum degrees of a graph $G$ are denoted by $\delta \left(G\right)$ and $\text{Δ}\left(G\right)$, respectively. A subset of $V\left(G\right)$ in a graph $G$ is called 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 maximum. The independence number of a graph $G$ is defined as the size of a maximum independent set in $G$ and is denoted by $\beta \left(G\right)$. For two disjoint vertex subsets $S$ and $T$ of $V\left(G\right)$, we define $E\left(S,T\right)$ as $\left\{e:e=ab\in E\left(G\right),a\in S,b\in T\right\}$. Namely, $E\left(S,T\right)$ is the set of all the edges in $E\left(G\right)$ such that one end vertex of each edge is in $S$ and another end vertex of the edge is in $T$. We use $K_{a,b}$ to denote a complete bipartite graph with two partition sets $X$ and $Y$ such that $\left|X\right|=a$ and $\left|Y\right|=b$. A cycle $C$ in a graph $G$ is called a Hamilton cycle of $G$ if $C$ contains all the vertices of $G$. A graph $G$ is called Hamiltonian if $G$ has a Hamilton cycle. A path $P$ in a graph $G$ is called a Hamilton path of $G$ if $P$ contains all the vertices of $G$. A graph $G$ is called traceable if $G$ has a Hamilton path.

Gutman and Trinajstić [2] introduced the concept of the first Zagreb index of a graph in 1972. Also see [3] . Let $G=\left(V\left(G\right),E\left(G\right)\right)$ be a graph. Its first Zagreb index is defined as $Z_1\left(G\right):={\displaystyle {\sum }_{u\in V\left(G\right)}{d}_G^2\left(u\right)}$. As one of the most important topological indices of a graph, the first Zagreb index of a graph has been intensively investigated. A lot of results on the first Zagreb index of a graph have been obtained. The readers are referred to the survey paper [4] and the references therein. Finding the bounds for the first Zagreb index of a graph is one of the important topics. In this paper, using two established inequalities, we obtain two lower bounds involving the independence number for the first Zagreb index of a graph. We also characterize the graphs achieving the bounds. It is noticed that in recent years, using the first Zagreb index of a graph and its variants, researchers have presented sufficient conditions for the Hamiltonian properties of graphs. Some of the conditions can be found in [5] - [15] . In this paper, we present new sufficient conditions based on the first Zagreb index for Hamiltonian graphs and traceable graphs. The main results of this paper are as follows.

Theorem 1. Let $G$ be a graph with $n$ vertices, $e$ edges, minimum degree $\delta \ge 1$, and maximum degree $\text{Δ}$. Then

(1)

$Z_1\left(G\right)\ge \beta {\delta }^2+\frac{{\delta }^2\left(2e+n-\beta \right)}{2\delta +1}.$

with equality if and only if $G$ is a regular bipartite graph.

(2)

$Z_1\left(G\right)\ge \beta {\delta }^2+\frac{e^2\left({\text{Δ}}^2+1\right)}{\left(n-\beta \right){\text{Δ}}^2}-\left(n-\beta \right).$

with equality if and only if $G$ is a bipartite graph with partition sets of $I$ and $V-I$ such that $\left|I\right|=\beta$, $d\left(u\right)=\delta$ for each $u\in I$, and $d\left(v\right)=\text{Δ}$ for each $v\in V-I$.

Theorem 2. Let $G$ be a $k$-connected ( $k\ge 2$) graph with $n\ge 3$ vertices, $e$ edges, minimum degree $\delta$, and maximum degree $\text{Δ}$.

(1) If

$Z_1\left(G\right)\le \left(k+1\right){\delta }^2+\frac{{\delta }^2\left(2e+n-k-1\right)}{2\delta +1},$

then $G$ is Hamiltonian.

(2) If

$Z_1\left(G\right)\le \left(k+1\right){\delta }^2+\frac{e^2\left({\text{Δ}}^2+1\right)}{\left(n-k-1\right){\text{Δ}}^2}-\left(n-k-1\right),$

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

Theorem 3. Let $G$ be a $k$-connected ( $k\ge 1$) with $n\ge 9$ vertices, $e$ edges, minimum degree $\delta$, and maximum degree $\text{Δ}$.

(1) If

$Z_1\left(G\right)\le \left(k+2\right){\delta }^2+\frac{{\delta }^2\left(2e+n-k-2\right)}{2\delta +1}$

then $G$ is traceable.

(2) If

$Z_1\left(G\right)\le \left(k+2\right){\delta }^2+\frac{e^2\left({\text{Δ}}^2+1\right)}{\left(n-k-2\right){\text{Δ}}^2}-\left(n-k-2\right),$

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

We will use the following results as our lemmas.

Lemma 1 [16] . Let $G$ be a $k$-connected graph of order $n\ge 3$. If $\beta \le k$, then $G$ is Hamiltonian.

Lemma 2 [16] . Let $G$ be a $k$-connected graph of order n. If $\beta \le k+1$, then $G$ is traceable.

Lemma 3 ( [17] , Theorem 6 on Page 9). Let $p\ge 1$ and $a_i$, $b_i\ge 0$ ( $1\le i\le s$). Then

$\left(\underset{i=1}{\overset{s}{{\displaystyle \sum }}}{a}_i^p\right)\left(\underset{i=1}{\overset{s}{{\displaystyle \sum }}}{b}_i^p\right)\ge \underset{i=1}{\overset{s}{{\displaystyle \sum }}}{\left(a_i+b_i\right)}^p\underset{i=1}{\overset{s}{{\displaystyle \sum }}}{\left(\frac{a_i{b}_i}{a_i+b_i}\right)}^p$

with the convention $0\cdot 0/\left(0+0\right)=0$. The equality is attained if and only if there exist constants $\lambda$, $\mu$ with $\lambda +\mu >0$ and $\lambda a_i=\mu b_i$ for every $1\le i\le s$.

Lemma 4 ( [18] , Theorem 3.20 on Page 37). Suppose $p_k$ and $q_k$ $\left(k=1,2,\cdots ,s\right)$ are real numbers with $\left|p_k|+|q_k\right|\ne 0$ $\left(k=1,2,\cdots ,s\right)$. One has the inequality

${\left(\underset{k=1}{\overset{s}{{\displaystyle \sum }}}{p}_k{q}_k\right)}^2\le \underset{k=1}{\overset{s}{{\displaystyle \sum }}}\left(p_k^2+q_k^2\right)\underset{k=1}{\overset{s}{{\displaystyle \sum }}}\frac{p_k^2{q}_k^2}{p_k^2+q_k^2}.$

Lemma 5 [19] . Let $G$ be a balanced bipartite graph of order $2n$ with bipartition ( $A$, $B$). If $d\left(x\right)+d\left(y\right)\ge n+1$ for any $x\in A$ and any $y\in B$ with $xy\notin E$, then $G$ is Hamiltonian.

Lemma 6 [20] . Let $G$ be a 2-connected bipartite graph with bipartition ( $A$, $B$), where $\left|A\right|\ge \left|B\right|$. If each vertex in $A$ has degree at least $s$ and each vertex in $B$ has degree at least $t$, then $G$ contains a cycle of length at least $2\text{min}\left(\left|B\right|,s+t-1,2s-2\right)$.

Proof of Theorem 1. Let $G$ be a graph with $n$ vertices, $e$ edges, and $\delta \ge 1$. Clearly, $\beta <n$. Let $I:=\left\{u_1,u_2,\cdots ,u_{\beta }\right\}$ be a maximum independent set in $G$ and $V-I:=\left\{v_1,v_2,\cdots ,v_{n-\beta }\right\}$. Then

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

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

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

(1) Applying Lemma 3 with $p=2$, $s=n-\beta$, $a_i=d\left(v_i\right)$, and $b_i=1$, where $i=1,2,\cdots ,n-\beta$, we have

$\left(\underset{i=1}{\overset{n-\beta }{{\displaystyle \sum }}}{d}^2\left(v_i\right)\right)\left(\underset{i=1}{\overset{n-\beta }{{\displaystyle \sum }}}{1}^2\right)\ge \left(\underset{i=1}{\overset{n-\beta }{{\displaystyle \sum }}}{\left(d\left(v_i\right)+1\right)}^2\right)\underset{i=1}{\overset{n-\beta }{{\displaystyle \sum }}}{\left(\frac{d\left(v_i\right)\cdot 1}{d\left(v_i\right)+1}\right)}^2.$

Thus

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

Therefore

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

Hence

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

So

$Z_1\left(G\right)=\underset{w\in V\left(G\right)}{{\displaystyle \sum }}{d}^2\left(w\right)=\underset{i=1}{\overset{\beta }{{\displaystyle \sum }}}{d}^2\left(u_i\right)+\underset{i=1}{\overset{n-\beta }{{\displaystyle \sum }}}{d}^2\left(v_i\right)\ge \beta {\delta }^2+\frac{{\delta }^2\left(2e+n-\beta \right)}{2\delta +1}.$

Suppose that

$Z_1\left(G\right)=\beta {\delta }^2+\frac{{\delta }^2\left(2e+n-\beta \right)}{2\delta +1}.$

In review of all the proofs above, we have ${\displaystyle {\sum }_{i=1}^{n-\beta }d\left(v_i\right)}=e$ which implies that ${\displaystyle {\sum }_{i=1}^{\beta }d\left(u_i\right)}=e$ and $G$ is a bipartite graph with partition sets of $I$ and $V-I$. In addition, $d\left(u\right)=\delta$ for each $u\in I$ and $d\left(v\right)=\delta$ for each $v\in V-I$. Therefore $G$ is a regular bipartite graph.

If $G$ is a regular bipartite graph, then a simple computation yields that

$Z_1\left(G\right)=\beta {\delta }^2+\frac{{\delta }^2\left(2e+n-\beta \right)}{2\delta +1}.$

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

(2) Applying Lemma 4 with $s=n-\beta$, $p_i=d\left(v_i\right)$ and $q_i=1$, where $i=1,2,\cdots ,n-\beta$, we have

${\left(\underset{i=1}{\overset{n-\beta }{{\displaystyle \sum }}}d\left(v_i\right)\cdot 1\right)}^2\le \underset{i=1}{\overset{n-\beta }{{\displaystyle \sum }}}\left(d^2\left(v_i\right)+1^2\right)\underset{i=1}{\overset{n-\beta }{{\displaystyle \sum }}}\frac{d^2\left(v_i\right)\cdot 1^2}{d^2\left(v_i\right)+1^2}.$

Thus

$e^2\le \left(\underset{i=1}{\overset{n-\beta }{{\displaystyle \sum }}}{d}^2\left(v_i\right)+n-\beta \right)\underset{i=1}{\overset{n-\beta }{{\displaystyle \sum }}}\frac{{\text{Δ}}^2}{{\text{Δ}}^2+1}.$

Therefore

$e^2\le \left(\underset{i=1}{\overset{n-\beta }{{\displaystyle \sum }}}{d}^2\left(v_i\right)+n-\beta \right)\left(n-\beta \right)\frac{{\text{Δ}}^2}{{\text{Δ}}^2+1}.$

Hence

$\underset{i=1}{\overset{n-\beta }{{\displaystyle \sum }}}{d}^2\left(v_i\right)\ge \frac{e^2\left({\text{Δ}}^2+1\right)}{\left(n-\beta \right){\text{Δ}}^2}-\left(n-\beta \right).$

So

$Z_1\left(G\right)=\underset{w\in V\left(G\right)}{{\displaystyle \sum }}{d}^2\left(w\right)=\underset{i=1}{\overset{\beta }{{\displaystyle \sum }}}{d}^2\left(u_i\right)+\underset{i=1}{\overset{n-\beta }{{\displaystyle \sum }}}{d}^2\left(v_i\right)\ge \beta {\delta }^2+\frac{e^2\left({\text{Δ}}^2+1\right)}{\left(n-\beta \right){\text{Δ}}^2}-\left(n-\beta \right).$

Suppose that

$Z_1\left(G\right)=\beta {\delta }^2+\frac{e^2\left({\text{Δ}}^2+1\right)}{\left(n-\beta \right){\text{Δ}}^2}-\left(n-\beta \right).$

In review of all the proofs above, we have ${\displaystyle {\sum }_{i=1}^{n-\beta }d\left(v_i\right)}=e$ which implies that ${\displaystyle {\sum }_{i=1}^{\beta }d\left(u_i\right)}=e$ and $G$ is a bipartite graph with partition sets of $I$ and $V-I$. In addition, $\left|I\right|=\beta$, $d\left(u\right)=\delta$ for each $u\in I$, and $d\left(v\right)=\text{Δ}$ for each $v\in V-I$.

If $G$ is a bipartite graph with partition sets of $I$ and $V-I$ such that $\left|I\right|=\beta$, $d\left(u\right)=\delta$ for each $u\in I$, and $d\left(v\right)=\text{Δ}$ for each $v\in V-I$, then $\delta \beta =e=\text{Δ}\left(n-\beta \right)$. A simple computation yields that

$Z_1\left(G\right)=\beta {\delta }^2+\frac{e^2\left({\text{Δ}}^2+1\right)}{\left(n-\beta \right){\text{Δ}}^2}-\left(n-\beta \right).$

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

Proof of Theorem 2. Let $G$ be a $k$-connected ( $k\ge 2$) graph with $n\ge 3$ vertices and $e$ edges satisfying exactly one of two conditions in Theorem 2. Suppose $G$ is not Hamiltonian. Then Lemma 1 implies that $\beta \ge k+1$. Let $I_1:=\left\{u_1,u_2,\cdots ,u_{\beta }\right\}$ be a maximum independent set in $G$. Then $I:=\left\{u_1,u_2,\cdots ,u_{k+1}\right\}$ is an independent set in $G$. Set $V-I=\left\{v_1,v_2,\cdots ,v_{n-k-1}\right\}$. Thus

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

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

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

(1) Applying Lemma 3 with $p=2$, $s=n-k-1$, $a_i=d\left(v_i\right)$, and $b_i=1$, where $i=1,2,\cdots ,n-k-1$, the ideas in the proof of (1) in Theorem 1, and the conditions in (1) in Theorem 2, we have

$\left(k+1\right){\delta }^2+\frac{{\delta }^2\left(2e+n-k-1\right)}{2\delta +1}\le Z_1\left(G\right)\le \left(k+1\right){\delta }^2+\frac{{\delta }^2\left(2e+n-k-1\right)}{2\delta +1}.$

Thus

$Z_1\left(G\right)=\left(k+1\right){\delta }^2+\frac{{\delta }^2\left(2e+n-k-1\right)}{2\delta +1}.$

Therefore $G$ is a regular bipartite graph with partition sets of $I$ and $V-I$ which implies that $\left|I\right|=\left|V-I\right|=\left(k+1\right)$. Lemma 5 implies that $G$ is Hamiltonian, a contradiction.

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

(2) Applying Lemma 4 with $s=n-k-1$, $p_i=d\left(v_i\right)$ and $q_i=1$, where $i=1,2,\cdots ,n-k-1$, the ideas in the proof of (2) in Theorem 1, and the conditions in (2) in Theorem 2, we have

$\begin{array}{l}\left(k+1\right){\delta }^2+\frac{e^2\left({\text{Δ}}^2+1\right)}{\left(n-k-1\right){\text{Δ}}^2}-\left(n-k-1\right)\\ \le Z_1\left(G\right)\le \left(k+1\right){\delta }^2+\frac{e^2\left({\text{Δ}}^2+1\right)}{\left(n-k-1\right){\text{Δ}}^2}-\left(n-k-1\right).\end{array}$

Thus $G$ is a bipartite graph with partition sets of $I$ and $V-I$ such that $\left|I\right|=\left(k+1\right)$, $d\left(u\right)=\delta$ for each $u\in I$, and $d\left(v\right)=\text{Δ}$ for each $v\in V-I$. $\delta \left|I\right|=e=\text{Δ}\left|V-I\right|$ and $\delta \le \text{Δ}$, we have that $\left|V-I\right|\le \left|I\right|=\left(k+1\right)$. Notice that $n\ge 2\delta +1\ge 2k+1$ otherwise $\delta \ge k\ge n/2$ and $G$ is Hamiltonian. Thus $\left|V-I\right|=\left|V\right|-\left|I\right|\ge k$. Therefore $\left|V-I\right|=k$ or $\left|V-I\right|=\left(k+1\right)$. If $\left|V-I\right|=k$, then $G$ is $K_{k,k+1}$. If $\left|V-I\right|=\left(k+1\right)$, Lemma 5 implies that $G$ is Hamiltonian, a contradiction.

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

The proof of Theorem 3 is similar to the proof of Theorem 2. For the sake of completeness, we still present a full proof of Theorem 3 below.

Proof of Theorem 3. Let $G$ be a $k$-connected ( $k\ge 1$) graph with $n\ge 9$ vertices and $e$ edges satisfying exactly one of two conditions in Theorem 3. Suppose $G$ is not traceable. Then Lemma 2 implies that $\beta \ge k+2$. Let $I_1:=\left\{u_1,u_2,\cdots ,u_{\beta }\right\}$ be a maximum independent set in $G$. Then $I:=\left\{u_1,u_2,\cdots ,u_{k+2}\right\}$ is an independent set in $G$. Set $V-I=\left\{v_1,v_2,\cdots ,v_{n-k-2}\right\}$. Thus

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

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

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

(1) Applying Lemma 3 with $p=2$, $s=n-k-2$, $a_i=d\left(v_i\right)$, and $b_i=1$, where $i=1,2,\cdots ,n-k-2$, the ideas in the proof of (1) in Theorem 1, and the conditions in (1) in Theorem 3, we have

$\left(k+2\right){\delta }^2+\frac{{\delta }^2\left(2e+n-k-2\right)}{2\delta +1}\le Z_1\left(G\right)\le \left(k+2\right){\delta }^2+\frac{{\delta }^2\left(2e+n-k-2\right)}{2\delta +1}.$

Thus

$Z_1\left(G\right)=\left(k+2\right){\delta }^2+\frac{{\delta }^2\left(2e+n-k-2\right)}{2\delta +1}.$

Therefore $G$ is a regular bipartite graph with partition sets of $I$ and $V-I$ which implies that $\left|I\right|=\left|V-I\right|=\left(k+2\right)$. Since $n=2k+4\ge 9$, we have that $k\ge 3$. Thus Lemma 5 implies that $G$ is Hamiltonian and thereby $G$ is traceable, a contradiction.

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

(2) Applying Lemma 4 with $s=n-k-2$, $p_i=d\left(v_i\right)$ and $q_i=1$, where $i=1,2,\cdots ,n-k-2$, the ideas in the proof of (2) in Theorem 1, and the conditions in (2) in Theorem 3, we have

$\begin{array}{l}\left(k+2\right){\delta }^2+\frac{e^2\left({\text{Δ}}^2+1\right)}{\left(n-k-2\right){\text{Δ}}^2}-\left(n-k-2\right)\\ \le Z_1\left(G\right)\le \left(k+2\right){\delta }^2+\frac{e^2\left({\text{Δ}}^2+1\right)}{\left(n-k-2\right){\text{Δ}}^2}-\left(n-k-2\right).\end{array}$

Thus $G$ is a bipartite graph with partition sets of $I$ and $V-I$ such that $\left|I\right|=\left(k+2\right)$, $d\left(u\right)=\delta$ for each $u\in I$, and $d\left(v\right)=\text{Δ}$ for each $v\in V-I$. Since $\delta \left|I\right|=e=\text{Δ}\left|V-I\right|$ and $\delta \le \text{Δ}$, we have that $\left|V-I\right|\le \left|I\right|=\left(k+2\right)$. Notice that $n\ge 2\delta +2\ge 2k+2$ otherwise $\delta \ge k\ge \left(n-1\right)/2$ and $G$ is traceable. Thus $\left|V-I\right|=\left|V\right|-\left|I\right|\ge k$. Therefore $\left|V-I\right|=k$ or $\left|V-I\right|=\left(k+1\right)$ or $\left|V-I\right|=\left(k+2\right)$. If $\left|V-I\right|=k$, then $G$ is $K_{k,k+2}$. If $\left|V-I\right|=\left(k+1\right)$, Lemma 6 implies that $G$ has a cycle of length at least $\left(n-1\right)$ and thereby $G$ is traceable, a contradiction. If $\left|V-I\right|=\left(k+2\right)$, since $n=2k+4\ge 9$, we have that $k\ge 3$. Thus Lemma 5 implies that $G$ is Hamiltonian and thereby $G$ is traceable, a contradiction.

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

The author would like to thank the referees for their suggestions or comments which improve the initial version of the paper.