Let \(G = (V, E)\) be a graph. The Gutman-Milovanović index of a graph \(G\) is defined as \(\sum\limits_{uv \in E} (d(u) d(v))^{\alpha}(d(u) + d(v))^{\beta}\), where \(\alpha\) and \(\beta\) are any real numbers and \(d(u)\) and \(d(v)\) are the degrees of vertices \(u\) and \(v\) in \(G\), respectively. In this note, we present sufficient conditions based on the Gutman-Milovanović index with \(\alpha > 0\) and \(\beta >0\) for some Hamiltonian properties of a graph. We also present upper bounds for the Gutman-Milovanović index of a graph for different ranges of \(\alpha\) and \(\beta\).
We consider only finite undirected graphs without loops or multiple edges. Notations and terminologies not defined here follow those in [1]. Let \(G = (V, E)\) be a graph with \(n\) vertices and \(e\) edges, the degree of a vertex \(v\) in \(G\) is denoted by \(d_G(v)\). We use \(\delta\) and \(\Delta\) to denote the minimum degree and maximum degree of \(G\), respectively. The complement of a graph \(G\) is denoted by \(G^c\). We use \(G[S]\) to denote a subgraph induced by a subset \(S\) of \(V(G)\). A set of vertices in a graph \(G\) is independent if the vertices in the set are pairwise nonadjacent. A maximum independent set in a graph \(G\) is an independent set of largest possible size. The independence number, denoted \(\gamma(G)\), of a graph \(G\) is the cardinality of a maximum independent set in \(G\). For disjoint vertex subsets \(X\) and \(Y\) of \(V\), we use \(E(X, Y)\) to denote the set of all the edges in \(E\) such that one end vertex of each edge is in \(X\) and another end vertex of the edge is in \(Y\). Namely, \(E(X, Y) := \{\, f : f = xy \in E, x \in X, y \in Y \,\}\). We use \(G_1 \vee G_2\) to denote the the join of two disjoint graphs \(G_1\) and \(G_2\). The complete graph of order \(p\) is denoted by \(K_p\). A cycle \(C\) in a graph \(G\) is called a Hamiltonian cycle of \(G\) if \(C\) contains all the vertices of \(G\). A graph \(G\) is called Hamiltonian if \(G\) has a Hamiltonian cycle. A path \(P\) in a graph \(G\) is called a Hamiltonian path of \(G\) if \(P\) contains all the vertices of \(G\). A graph \(G\) is called traceable if \(G\) has a Hamiltonian path.
A variety of topological indexes for a graph have been introduced. Some notable ones are the Wiener index [11]. the Randić index [10], and the first Zagreb index and the second Zagreb index [5] of a graph. In 2020, Gutman, Milovanović, and Milovanović [4] introduced the concept of the Gutman-Milovanović index. For a graph \(G\), its Gutman-Milovanović index, denoted \(M_{\alpha, \, \beta}(G)\), is defined as \(\sum\limits_{uv \in E} (d(u) d(v))^{\alpha}(d(u) + d(v))^{\beta},\) where \(\alpha\) and \(\beta\) are any real numbers. The Gutman-Milovanović index is a generalization of both the first Zagreb index and second Zagreb index of a graph \(G\) since \(M_{0, 1}(G)\) and \(M_{1, 0}(G)\) are respectively the same as the first Zagreb index and second Zagreb index of a graph \(G\). Some recent results on the Gutman-Milovanović index of a graph can be found in [3] and [9]. Motivated by the topological index conditions for some Hamiltonian properties of a graph in [6], [7], and [8], we in this note present sufficient conditions based on the Gutman-Milovanović index with \(\alpha > 0\) and \(\beta >0\) for some Hamiltonian properties of a graph. We also present upper bounds for the Gutman-Milovanović index of a graph for different ranges of \(\alpha\) and \(\beta\). The main results are as follows.
Theorem 1.1. Let \(G\) be a \(k\)-connected (\(k \geq 2\)) graph with \(n \geq 3\) vertices. If \(\alpha > 0\), \(\beta >0\), and \[\begin{aligned} M_{\alpha, \, \beta}(G) \geq& (k + 1)(n – k – 1)^{\alpha + 1} \Delta^{\alpha} (n – k – 1 + \Delta)^{\beta} \\ &+ \,(n – k – 1)(n – k – 2)2^{\beta – 1} \Delta^{2\alpha + \beta}, \end{aligned}\] then \(G\) is Hamiltonian or \(G\) is \(K_{k + 1}^{c} \vee K_{k}\).
Theorem 1.2. Let \(G\) be a \(k\)-connected (\(k \geq 1\)) graph with \(n\) vertices. If \(\alpha > 0\), \(\beta > 0\), and \[\begin{aligned} M_{\alpha, \, \beta}(G) \geq& (k + 2)(n – k – 2)^{\alpha + 1} \Delta^{\alpha} (n – k – 2 + \Delta)^{\beta} \\ &+ (n – k – 2)(n – k – 3)2^{\beta – 1} \Delta^{2\alpha + \beta}, \end{aligned}\] then \(G\) is traceable or \(G\) is \(K_{k + 2}^{c} \vee K_{k}\).
Theorem 1.3. Let \(G\) be a graph with \(n\) vertices and \(\delta \geq 1\). If \(\alpha > 0\) and \(\beta > 0\), then \[\begin{aligned} M_{\alpha, \, \beta}(G) \leq \gamma (n – \gamma)^{\alpha + 1} \Delta^{\alpha} (n – \gamma + \Delta)^{\beta} + (n – \gamma)(n – \gamma – 1)2^{\beta – 1} \Delta^{2\alpha + \beta}, \end{aligned}\] with equality if and only if \(G\) is \(K_{\gamma}^{c} \vee K_{n – \gamma}\).
Theorem 1.4. Let \(G\) be a graph with \(n\) vertices and \(\delta \geq 1\). If \(\alpha > 0\) and \(\beta < 0\), then \[M_{\alpha, \, \beta}(G) \leq \gamma (n – \gamma)^{\alpha + 1} (2 \delta)^{\beta} \Delta^{\alpha} + (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \delta^{\beta} \Delta^{2\alpha},\] with equality if and only if \(G\) is \(K_{n}\).
Theorem 1.5. Let \(G\) be a graph with \(n\) vertices and \(\delta \geq 1\). If \(\alpha < 0\) and \(\beta > 0\), then \[M_{\alpha, \, \beta}(G) \leq \gamma (n – \gamma) \delta^{2\alpha}(n – \gamma + \Delta)^{\beta} + (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \delta^{2 \alpha} \Delta^{\beta},\] with equality if and only if \(G\) is \(K_{n}\).
Theorem 1.6. Let \(G\) be a graph with \(n\) vertices and \(\delta \geq 1\). If \(\alpha < 0\) and \(\beta < 0\), then \[M_{\alpha, \, \beta}(G) \leq (n – \gamma)(n + \gamma – 1) 2^{\beta – 1} \delta^{2 \alpha + \beta},\] with equality if and only if \(G\) is \(K_{n}\).
We will use the following results as our lemmas.
Lemma 2.1. [2] Let \(G\) be a \(k\)-connected graph of order \(n \geq 3\). If \(\gamma \leq k\), then \(G\) is Hamiltonian.
Lemma 2.2. [2] Let \(G\) be a \(k\)-connected graph of order \(n\). If \(\gamma \leq k + 1\), then \(G\) is traceable.
Proof of Theorem 1.1. Let \(G\) be a \(k\)-connected (\(k \geq 2\)) graph with \(n \geq 3\) vertices satisfying the conditions in Theorem 1.1. Suppose \(G\) is not Hamiltonian. Then Lemma 2.1 implies that \(\gamma \geq k + 1\). Also, we have that \(n \geq 2 \delta + 1 \geq 2 k + 1\) otherwise \(\delta \geq k \geq n/2\) and \(G\) is Hamiltonian. Let \(I_1 := \{\, u_1, u_2, …, u_{\gamma} \,\}\) be a maximum independent set in \(G\). Then \(I := \{\, u_1, u_2, …, u_{k + 1} \,\}\) is an independent set in \(G\). Clearly, \(\delta \leq d(x) \leq n – (k + 1)\) for each \(x \in I\) and \(\delta \leq d(y) \leq \Delta\) for each \(y \in V – I\). Thus we have that
\[\begin{aligned} S_1 :=& \sum\limits_{uv \in E, u \in I, v \in V – I}(d(u) d(v))^{\alpha}(d(u) + d(v))^{\beta} \\ \leq& \sum\limits_{u \in I, v \in V – I} \left((n – k – 1)^{\alpha} \Delta^{\alpha} (n – k – 1 + \Delta)^{\beta} \right) \\ \leq& (n – k – 1)^{\alpha} \Delta^{\alpha} (n – k – 1 + \Delta)^{\beta} (n – k – 1) (k + 1) \\ =& (k + 1)(n – k – 1)^{\alpha + 1} \Delta^{\alpha} (n – k – 1 + \Delta)^{\beta}. \end{aligned}\]
We further have that
\[\begin{aligned} S_2 :=& \sum\limits_{uv \in E, u \in V – I, v \in V – I, u \neq v } (d(u) d(v))^{\alpha}(d(u) + d(v))^{\beta}\\ \leq& \sum\limits_{u \in V – I, v \in V – I, u \neq v }\left(\Delta^{\alpha} \Delta^{\alpha} (\Delta + \Delta)^{\beta})\right)\\ \leq& \frac{1}{2} (n – k – 1)(n – k – 2) \, 2^{\beta} \Delta^{2\alpha + \beta}\\ =& (n – k – 1)(n – k – 2) \, 2^{\beta – 1} \Delta^{2\alpha + \beta}. \end{aligned}\]
Therefore
\[\begin{aligned} M_{\alpha, \, \beta}(G) =& S_1 + S_2\\ \leq&(k + 1)(n – k – 1)^{\alpha + 1} \Delta^{\alpha} (n – k – 1 + \Delta)^{\beta} + \,(n – k – 1)(n – k – 2)2^{\beta – 1} \Delta^{2\alpha + \beta}. \end{aligned}\]
From the conditions in Theorem 1.1, we have that \[\begin{aligned} M_{\alpha, \, \beta}(G) =& (k + 1)(n – k – 1)^{\alpha + 1} \Delta^{\alpha} (n – k – 1 + \Delta)^{\beta}\\ &+ \,(n – k – 1)(n – k – 2)2^{\beta – 1} \Delta^{2\alpha + \beta}. \end{aligned}\]
Thus \[\begin{aligned} S_1 =& (k + 1)(n – k – 1)^{\alpha + 1} \Delta^{\alpha} (n – k – 1 + \Delta)^{\beta},\\ S_2 =& (n – k – 1)(n – k – 2) \, 2^{\beta – 1} \Delta^{2\alpha + \beta}. \end{aligned}\]
Therefore \(d(x) = n – k – 1\) for each \(x \in I\), \(G[V – I]\) is a complete graph of order \((n – k – 1)\), and \(d(y) = \Delta = n – 1\) for each \(y \in V – I\), Therefore \(G\) is \(K_{k + 1}^{c} \vee K_{n – (k + 1)}\). Since \(n \geq 2k + 1\), we have that \(|V – I| \geq k\). If \(|V – I| \geq (k + 1)\), then \(G\) is Hamiltonian, a contradiction. Thus \(|V – I| = k\) and \(G\) is \(K_{k + 1}^{c} \vee K_{k}\). ◻
Proof of Theorem 1.2. If \(n = 1\) or \(n = 2\), then it is trivial that \(G\) is traceable. Let \(G\) be a \(k\)-connected (\(k \geq 1\)) graph with \(n \geq 3\) vertices satisfying the conditions in Theorem 1.2. Suppose \(G\) is not traceable. Then Lemma 2.2 implies that \(\gamma \geq k + 2\). Also, we have that \(n \geq 2 \delta + 2 \geq 2 k + 2\) otherwise \(\delta \geq k \geq (n – 1)/2\) and \(G\) is traceable. Let \(I_1 := \{\, u_1, u_2, …, u_{\gamma} \,\}\) be a maximum independent set in \(G\). Then \(I := \{\, u_1, u_2, …, u_{k + 2} \,\}\) is an independent set in \(G\). Clearly, \(\delta \leq d(x) \leq n – (k + 2)\) for each \(x \in I\) and \(\delta \leq d(y) \leq \Delta\) for each \(y \in V – I\). Using the arguments similar to the ones in Proof of Theorem 1.1, we have that
\[\begin{aligned} S_1 :=& \sum\limits_{uv \in E, u \in I, v \in V – I}(d(u) d(v))^{\alpha}(d(u) + d(v))^{\beta}\\ \leq& (k + 2)(n – k – 2)^{\alpha + 1} \Delta^{\alpha} (n – k – 2 + \Delta)^{\beta},\\ S_2 :=& \sum\limits_{uv \in E, u \in V – I, v \in V – I, u \neq v } (d(u) d(v))^{\alpha}(d(u) + d(v))^{\beta}\\ \leq& (n – k – 2)(n – k – 3) \, 2^{\beta – 1} \Delta^{2\alpha + \beta}. \end{aligned}\]
Therefore
\[\begin{aligned} M_{\alpha, \,\, \beta}(G) =& S_1 + S_2 \\ \leq &(k + 2)(n – k – 2)^{\alpha + 1} \Delta^{\alpha} (n – k – 2 + \Delta)^{\beta} + \,(n – k – 2)(n – k – 3)2^{\beta – 1} \Delta^{2\alpha + \beta}. \end{aligned}\]
From the conditions in Theorem 1.2, we have that \[\begin{aligned} M_{\alpha, \beta}(G) =& (k + 2)(n – k – 2)^{\alpha + 1} \Delta^{\alpha} (n – k – 2 + \Delta)^{\beta}\\ &+ \, (n – k – 2)(n – k – 3)2^{\beta – 1} \Delta^{2\alpha + \beta}. \end{aligned}\]
Thus \[\begin{aligned} S_1 =& (k + 2)(n – k – 2)^{\alpha + 1} \Delta^{\alpha} (n – k – 2 + \Delta)^{\beta},\\ S_2 =& (n – k – 2)(n – k – 2)2^{\beta – 1} \Delta^{2\alpha + \beta}. \end{aligned}\]
Therefore \(d(x) = n – k – 2\) for each \(x \in I\), \(G[V – I]\) is a complete graph of order \((n – k – 2)\), and \(d(y) = \Delta = n – 1\) for each \(y \in V – I\). Since \(n \geq 2k + 2\), we have that \(|V – I| \geq k\). If \(|V – I| \geq (k + 1)\), then \(G\) is traceable, a contradiction. Thus \(|V – I| = k\) and \(G\) is \(K_{k + 2}^{c} \vee K_{k}\). ◻
Proof of Theorem 1.3. Suppose \(G\) is a graph with \(n\) vertices and \(\delta \geq 1\). Let \(I\) be an independent set such that \(|I| = \gamma\). Then \(1 \leq \gamma \leq n – 1\) and \(1 \leq n – \gamma \leq n – 1\). Clearly, \(\delta \leq d(x) \leq n – \gamma\) for each \(x \in I\) and \(\delta \leq d(y) \leq \Delta\) for each \(y \in V – I\). Using the arguments similar to the ones in Proof of Theorem 1.1, we have that \[\begin{aligned} S_1 :=& \sum\limits_{uv \in E, u \in I, v \in V – I}(d(u) d(v))^{\alpha}(d(u) + d(v))^{\beta}\\ \leq& \gamma (n – \gamma)^{\alpha + 1} \Delta^{\alpha} (n – \gamma + \Delta)^{\beta},\\ S_2 :=& \sum\limits_{uv \in E, u \in V – I, v \in V – I, u \neq v } (d(u) d(v))^{\alpha}(d(u) + d(v))^{\beta}\\ \leq& (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \Delta^{2\alpha + \beta}. \end{aligned}\]
Therefore \[\begin{aligned} M_{\alpha, \, \beta}(G) =& S_1 + S_2\\ \leq& \gamma (n – \gamma)^{\alpha + 1} \Delta^{\alpha} (n – \gamma + \Delta)^{\beta} + (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \Delta^{2\alpha + \beta}. \end{aligned}\]
If \[\begin{aligned} M_{\alpha, \, \beta}(G) =& \gamma (n – \gamma)^{\alpha + 1} \Delta^{\alpha} (n – \gamma + \Delta)^{\beta} + (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \Delta^{2\alpha + \beta}, \end{aligned}\] then \[\begin{aligned} S_1 &= \gamma (n – \gamma)^{\alpha + 1} \Delta^{\alpha} (n – \gamma + \Delta)^{\beta},\\ S_2 &= (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \Delta^{2\alpha + \beta}. \end{aligned}\]
In reviewing all the above arguments in this proof, we have that
\(d(x) = n – \gamma\) for each \(x \in I\), \(G[V
– I]\) is a complete graph of order \((n – \gamma)\), and \(d(y) = \Delta = n – 1\) for each \(y \in V – I\). Therefore \(G\) is \(K_{\gamma}^{c} \vee K_{n – \gamma}\). If
\(G\) is \(K_{\gamma}^{c} \vee K_{n – \gamma}\), a
simple computation verifies that
\[\begin{aligned}
M_{\alpha, \, \beta}(G) =& \gamma (n – \gamma)^{\alpha + 1}
\Delta^{\alpha} (n – \gamma + \Delta)^{\beta} +
(n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \Delta^{2\alpha +
\beta}.
\end{aligned}\] ◻
Proof of Theorem 1.4. Suppose \(G\) is a graph with \(n\) vertices and \(\delta \geq 1\). Let \(I\) be an independent set such that \(|I| = \gamma\). Then \(1 \leq \gamma \leq n – 1\) and \(1 \leq n – \gamma \leq n – 1\). Clearly, \(\delta \leq d(x) \leq n – \gamma\) for each \(x \in I\) and \(\delta \leq d(y) \leq \Delta\) for each \(y \in V – I\). Using the arguments similar to the ones in Proof of Theorem 1.1, we have that \[\begin{aligned} S_1 &:= \sum\limits_{uv \in E, u \in I, v \in V – I}(d(u) d(v))^{\alpha}(d(u) + d(v))^{\beta}\\ &\leq \sum\limits_{u \in I, v \in V – I} \left((n – \gamma)^{\alpha} \Delta^{\alpha} (\delta + \delta)^{\beta} \right) \\ &= \gamma (n – \gamma)^{\alpha + 1} (2 \delta)^{\beta} \Delta^{\alpha},\\ S_2 &:= \sum\limits_{uv \in E, u \in V – I, v \in V – I, u \neq v } (d(u) d(v))^{\alpha}(d(u) + d(v))^{\beta}\\ &\leq \sum\limits_{u \in V – I, v \in V – I, u \neq v }\left(\Delta^{\alpha} \Delta^{\alpha} (\delta + \delta)^{\beta})\right)\\ &= (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \delta^{\beta} \Delta^{2\alpha}. \end{aligned}\]
Therefore \[\begin{aligned} M_{\alpha, \, \beta}(G) =& S_1 + S_2\\ \leq& \gamma (n – \gamma)^{\alpha + 1} (2 \delta)^{\beta} \Delta^{\alpha} + (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \delta^{\beta} \Delta^{2\alpha}. \end{aligned}\]
If \[\begin{aligned} M_{\alpha, \, \beta}(G) = \gamma (n – \gamma)^{\alpha + 1} (2 \delta)^{\beta} \Delta^{\alpha} + (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \delta^{\beta} \Delta^{2\alpha}, \end{aligned}\] then \[\begin{aligned} S_1 &= \gamma (n – \gamma)^{\alpha + 1} (2 \delta)^{\beta} \Delta^{\alpha},\\ S_2 &= (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \delta^{\beta} \Delta^{2\alpha}. \end{aligned}\]
In reviewing all the above arguments in this proof, we have that \(d(x) = n – \gamma\) for each \(x \in I\), \(G[V – I]\) is a complete graph of order \((n – \gamma)\), and \(d(y) = \delta = \Delta = n – 1\) for each \(y \in V – I\). Therefore \(G\) is \(K_{n}\). If \(G\) is \(K_{n}\), a simple computation verifies that \[\begin{aligned} M_{\alpha, \, \beta}(G) =& \gamma (n – \gamma)^{\alpha + 1} (2 \delta)^{\beta} \Delta^{\alpha} + (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \delta^{\beta} \Delta^{2\alpha}. \end{aligned}\] ◻
Proof of Theorem 1.5. Suppose \(G\) is a graph with \(n\) vertices and \(\delta \geq 1\). Let \(I\) be an independent set such that \(|I| = \gamma\). Then \(1 \leq \gamma \leq n – 1\) and \(1 \leq n – \gamma \leq n – 1\). Clearly, \(\delta \leq d(x) \leq n – \gamma\) for each \(x \in I\) and \(\delta \leq d(y) \leq \Delta\) for each \(y \in V – I\). Using the arguments similar to the ones in Proof of Theorem 1.1, we have that \[\begin{aligned} S_1 &:= \sum\limits_{uv \in E, u \in I, v \in V – I}(d(u) d(v))^{\alpha}(d(u) + d(v))^{\beta}\\ &\leq \sum\limits_{u \in I, v \in V – I} \left(\delta^{\alpha} \delta^{\alpha} (n – \gamma + \Delta)^{\beta} \right) \\ &= \gamma (n – \gamma) \delta^{2\alpha}(n – \gamma + \Delta)^{\beta},\\ S_2 &:= \sum\limits_{uv \in E, u \in V – I, v \in V – I, u \neq v } (d(u) d(v))^{\alpha}(d(u) + d(v))^{\beta}\\ &\leq \sum\limits_{u \in V – I, v \in V – I, u \neq v }\left(\delta^{\alpha} \delta^{\alpha} (\Delta + \Delta)^{\beta})\right)\\ &= (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \delta^{2 \alpha} \Delta^{\beta}. \end{aligned}\]
Therefore \[\begin{aligned} M_{\alpha, \, \beta}(G) =& S_1 + S_2\\ \leq& \gamma (n – \gamma) \delta^{2\alpha}(n – \gamma + \Delta)^{\beta} + (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \delta^{2 \alpha} \Delta^{\beta}. \end{aligned}\]
If \[\begin{aligned} M_{\alpha, \, \beta}(G) =& \gamma (n – \gamma) \delta^{2\alpha}(n – \gamma + \Delta)^{\beta} + (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \delta^{2 \alpha} \Delta^{\beta}, \end{aligned}\] then \[\begin{aligned} S_1 =& \gamma (n – \gamma) \delta^{2\alpha}(n – \gamma + \Delta)^{\beta},\\ S_2 =& (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \delta^{2 \alpha} \Delta^{\beta}. \end{aligned}\]
In reviewing all the above arguments in this proof, we have that \(d(x) = n – \gamma = \delta\) for each \(x \in I\), \(G[V – I]\) is a complete graph of order \((n – \gamma)\), and \(d(y) = \delta = \Delta = n – 1\) for each \(y \in V – I\). Therefore \(G\) is \(K_{n}\). If \(G\) is \(K_{n}\), a simple computation verifies that \[\begin{aligned} M_{\alpha, \, \beta}(G) =& \gamma (n – \gamma) \delta^{2\alpha}(n – \gamma + \Delta)^{\beta} + (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \delta^{2 \alpha} \Delta^{\beta}. \end{aligned}\] ◻
Proof of Theorem 1.6. Suppose \(G\) is a graph with \(n\) vertices and \(\delta \geq 1\). Let \(I\) be an independent set such that \(|I| = \gamma\). Then \(1 \leq \gamma \leq n – 1\) and \(1 \leq n – \gamma \leq n – 1\). Clearly, \(\delta \leq d(x) \leq n – \gamma\) for each \(x \in I\) and \(\delta \leq d(y) \leq \Delta\) for each \(y \in V – I\). Using the arguments similar to the ones in Proof of Theorem 1.1, we have that \[\begin{aligned} S_1 :=& \sum\limits_{uv \in E, u \in I, v \in V – I}(d(u) d(v))^{\alpha}(d(u) + d(v))^{\beta}\\ \leq& \sum\limits_{u \in I, v \in V – I} \left(\delta^{\alpha} \delta^{\alpha} (\delta + \delta)^{\beta} \right) \\ =& \gamma (n – \gamma) 2^{\beta} \delta^{2\alpha + \beta},\\ S_2 :=& \sum\limits_{uv \in E, u \in V – I, v \in V – I, u \neq v } (d(u) d(v))^{\alpha}(d(u) + d(v))^{\beta}\\ \leq& \sum\limits_{u \in V – I, v \in V – I, u \neq v }\left(\delta^{\alpha} \delta^{\alpha} (\delta + \delta)^{\beta})\right)\\ =& (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \delta^{2 \alpha + \beta}. \end{aligned}\]
Therefore \[\begin{aligned} M_{\alpha, \, \beta}(G) = S_1 + S_2 \leq (n – \gamma)(n + \gamma – 1) 2^{\beta – 1} \delta^{2 \alpha + \beta}. \end{aligned}\]
If \[\begin{aligned} M_{\alpha, \, \beta}(G) = (n – \gamma)(n + \gamma – 1) 2^{\beta – 1} \delta^{2 \alpha + \beta}, \end{aligned}\] then \[\begin{aligned} S_1 =& \gamma (n – \gamma) 2^{\beta} \delta^{2\alpha + \beta},\\ S_2 =& (n – \gamma)(n – \gamma – 1) 2^{\beta – 1} \delta^{2 \alpha + \beta}. \end{aligned}\]
In reviewing all the above arguments in this proof, we have that \(d(x) = n – \gamma = \delta\) for each \(x \in I\), \(G[V – I]\) is a complete graph of order \((n – \gamma)\), and \(d(y) = \delta = n – 1\) for each \(y \in V – I\). Therefore \(G\) is \(K_{n}\). If \(G\) is \(K_{n}\), a simple computation verifies that \[\begin{aligned} M_{\alpha, \, \beta}(G) = (n – \gamma)(n + \gamma – 1) 2^{\beta – 1} \delta^{2 \alpha + \beta}. \end{aligned}\] ◻
The author would like to thank the referees for their suggestions or comments which improve the initial version of the paper.