Inequalities on the inverse degree index

Abstract

The inverse degree index, also called inverse index, first attracted attention through numerous conjectures generated by the computer programme Graffiti. Since then, its relationship with other graph invariants has been studied by several authors. In this paper, we obtain new inequalities involving the inverse degree index, and we characterize graphs which are extremal with respect to them.

1 Introduction

A topological descriptor is a single number that represents a chemical structure in graph-theoretical terms via the molecular graph. They play a significant role in mathematical chemistry, especially in the QSPR/QSAR investigations. A topological descriptor is called a topological index if it correlates with a molecular property. Topological indices are used to understand physicochemical properties of chemical compounds, since they capture some properties of a molecule in a single number. Hundreds of topological indices have been introduced and studied, starting with the seminal work by Wiener [32].

Topological indices based on vertex degrees have been used over 40 years [12]. Some of those indices have been recognized as useful tools for chemical researches and their mathematical properties have been examined in depth. The best known among such degree-based structure-descriptors are the connectivity indices and the two Zagreb indices; for recent surveys focusing on their mathematical properties, see [2, 13, 18], respectively.

The first and second Zagreb indices, denoted by $M_1$ and $M_2$, are defined as

$$\begin{aligned} M_1(G) = \sum _{uv\in E(G)} (d_u + d_v) = \sum _{u\in V(G)} d_u^2, \qquad M_2(G) = \sum _{uv\in E(G)} d_u d_v, \qquad \end{aligned}$$

where uv denotes the edge of the graph G connecting the vertices u and v.

The inverse degree index${ ID}(G)$ of a graph G is defined by

$$\begin{aligned} { ID}(G) = \sum _{u\in V(G)} \frac{1}{d_u} = \sum _{uv\in E(G)} \left( \frac{1}{d_u^2} + \frac{1}{d_v^2}\right) = \sum _{uv\in E(G)}\frac{d_u^2 + d_v^2}{d_u^2 d_v^2}. \end{aligned}$$

The inverse degree index first attracted attention through numerous conjectures generated by the computer programme Graffiti [11]. Since then, its relationship with other graph invariants, such as diameter, edge-connectivity, matching number, Wiener index has been studied by several authors (see, e.g., [3, 7, 9, 10, 21, 36]).

Miličević and Nikolić defined in [20] the first and second variable Zagreb indices as

$$\begin{aligned} ^{\alpha }M_1(G) = \sum _{u\in V(G)} d_u^{2\alpha }, \qquad ^{\alpha }M_2(G) = \sum _{uv\in E(G)} (d_u d_v)^\alpha , \end{aligned}$$

with $\alpha \in \mathbb {R}$. In [16, 33] the first and second general Zagreb indices are introduced as

$$\begin{aligned} M_1^{\alpha }(G) = \sum _{u\in V(G)} d_u^{\alpha }, \qquad M_2^{\alpha }(G) = \sum _{uv\in E(G)} (d_u d_v)^\alpha , \end{aligned}$$

respectively. It is clear that these indices are equivalent to the previous ones, since $^{\alpha }M_1(G)=M_1^{2\alpha }(G)$ and $^{\alpha }M_2(G)=M_2^{\alpha }(G)$. We prefer to use $M_j^{\alpha }(G)$ instead of $^{\alpha }M_j(G)$, for $j=1,2,$ since the inequalities obtained in this paper become simpler with them.

Note that $M_1^0$ is the number of vertices, $M_1^{1}$ is twice the number of edges, $M_1^{2}$ is the first Zagreb index $M_1$, $M_1^{-1}$ is the inverse index ID, $M_1^{3}$ is the forgotten index F, etc.; also, $M_2^{0}$ is the number of edges, $M_2^{-1/2}$ is the usual Randić index R, $M_2^{1}$ is the second Zagreb index $M_2$, $M_2^{-1}$ is the modified Zagreb index, etc.

The concept of the variable molecular descriptors was proposed as a new way of characterizing heteroatoms in molecules (see [23, 24]), but also to assess the structural differences (e.g., the relative role of carbon atoms of acyclic and cyclic parts in alkylcycloalkanes [25]). The idea behind the variable molecular descriptors is that the variables are determined during the regression so that the standard error of estimate for a studied property is as small as possible.

In the paper of Gutman and Tosovic [14], the correlation abilities of 20 vertex-degree-based topological indices occurring in the chemical literature were tested for the case of standard heats of formation and normal boiling points of octane isomers. It is remarkable to realize that the second general Zagreb index $M_2^\alpha$ with exponent $\alpha = {-}\,1$ (and to a lesser extent with exponent $\alpha = {-}\,2$) performs significantly better than the Randić index ($R=M_2^{-1/2}$).

The second variable Zagreb index is used in the structure-boiling point modeling of benzenoid hydrocarbons [22]. Various properties and relations of these indices are discussed in several papers (see, e.g., [1, 15, 17, 29, 34, 35]).

In this paper we obtain new inequalities involving the inverse degree index, and we characterize graphs which are extremal with respect to them. In order to prove our results, we obtain a kind of converse of Hölder’s inequality, which is interesting by itself.

Throughout this paper, $G=(V (G),E (G))$ denotes a (non-oriented) finite simple (without multiple edges and loops) non-trivial (each vertex belongs to some edge) graph.

2 Inequalities involving Zagreb and sum-connectivity indices

Let us start by recalling two well-known and useful inequalities for the inverse degree index of a graph G with n edges, maximum degree $\varDelta$ and minimum degree $\delta$:

$$\begin{aligned} \frac{n}{\varDelta }\le { ID}(G)\le \frac{n}{\delta }. \end{aligned}$$

(1)

Moreover, both equalities are attained if and only if G is regular.

In order to prove our first result, we need the following useful and well-known inequality (see, e.g., [19, Lemma 3.4] for a proof of the statement of equality).

Lemma 1

If $a_j,b_j\ge 0$ and $M b_j \le a_j \le N b_j$ for $1\le j \le k$ and some positive constants M, N,  then

$$\begin{aligned} \left( \sum _{j=1}^k a_j^2 \right) ^{1/2} \left( \sum _{j=1}^k b_j^2 \right) ^{1/2} \le \frac{1}{2} \left( \sqrt{\frac{N}{M}}+ \sqrt{\frac{M}{N}} \,\right) \sum _{j=1}^k a_jb_j. \end{aligned}$$

If $a_j>0$ for some $1\le j \le k$, then the equality holds if and only if $M=N$ and $a_j=M b_j$ for every $1\le j \le k$.

Theorem 1

If G is a non-trivial graph with n vertices, m edges, minimum degree $\delta$ and maximum degree $\varDelta$, then

$$\begin{aligned} \frac{n^2}{2m} \le ID (G) \le \frac{(\varDelta + \delta )^2n^2}{8\varDelta \delta m}. \end{aligned}$$

The equality in each inequality is attained if and only if G is regular.

Proof

Cauchy–Schwarz inequality gives

$$\begin{aligned} n^2 =\left( \sum _{u\in V(G)} \sqrt{d_u} \,\frac{1}{\sqrt{d_u}} \right) ^2 \le \left( \sum _{u\in V(G)} d_u \right) \left( \sum _{u\in V(G)} \frac{1}{d_u}\right) = 2m\,{ ID}(G). \end{aligned}$$

On the other hand, since

$$\begin{aligned} \delta \le \frac{\sqrt{d_u}}{\frac{1}{\sqrt{d_u}}} = d_u \le \varDelta , \end{aligned}$$

Lemma 1 gives

$$\begin{aligned} \begin{aligned} n^2&=\Big (\sum _{u\in V(G)}\sqrt{d_u} \,\frac{1}{\sqrt{d_u}}\Big )^2 \ge \frac{\Big ( \sum _{u\in V(G)} d_u \Big ) \Big (\sum _{u\in V(G)} \frac{1}{d_u}\Big )}{\frac{1}{4}\Big (\sqrt{\frac{\varDelta }{\delta }}+\sqrt{\frac{\delta }{\varDelta }}\,\Big )^2} \\&= \frac{2m\,{ ID}(G)}{\frac{1}{4} \frac{(\varDelta + \delta )^2}{\varDelta \delta }} =\frac{8m\varDelta \delta \,{ ID}(G)}{(\varDelta + \delta )^2}. \end{aligned} \end{aligned}$$

If the graph is regular, then the lower and upper bound are the same, and they are equal to ${ ID}(G)$.

If the equality is attained in the lower bound, then Cauchy–Schwarz inequality gives that the vectors $\big (d_u^{1/2}\big )_{u \in V(G)}$ and $\big (d_u^{-1/2}\big )_{u \in V(G)}$ are parallel; this is equivalent to $d_u=d_v$ for every $u,v \in V(G)$, and G is regular. If the equality is attained in the upper bound, then Lemma 1 gives $\delta =\varDelta$ and the graph is regular. $\square$

In order to prove Theorem 3 below we need a kind of converse of Hölder’s inequality, which is interesting by itself.

Theorem 2

Let $(X,\mu )$ be a measure space and $f,g : X \rightarrow \mathbb {R}$ non-negative measurable functions, and $1<p,q<\infty$ with $1/p+1/q=1$. If $f \in L^p(X,\mu )$, $g \in L^q(X,\mu )$ and $\omega g^q \le f^p \le \varOmega g^q$$\mu$-a.e. for some positive constants $\omega ,\varOmega ,$ then

$$\begin{aligned} \Big (\int _X f^p \,d\mu \Big )^{1/p} \Big (\int _X g^q \,d\mu \Big )^{1/q} \le c_p(\omega ,\varOmega ) \int _X fg \,d\mu , \end{aligned}$$

(2)

where

$$\begin{aligned} c_p(\omega ,\varOmega )= \max \Big \{ \frac{1}{p} \Big ( \frac{\omega }{\varOmega } \Big )^{1/q} + \frac{1}{q} \Big ( \frac{\varOmega }{\omega } \Big )^{1/p}, \, \frac{1}{p} \Big ( \frac{\varOmega }{\omega } \Big )^{1/q} + \frac{1}{q} \Big ( \frac{\omega }{\varOmega } \Big )^{1/p}\, \Big \}. \end{aligned}$$

The equality is attained if and only if we have $\omega = \varOmega$ and $f^p = \omega g^q$$\mu$-a.e. or $f = g = 0$$\mu$-a.e.

Proof

Fix $\lambda \in (0,1)$ and $0<m \le M$. Let us define $F_\lambda (t):= \lambda t^{1-\lambda } + (1-\lambda ) t^{-\lambda }$ for $t> 0$. Since $F_\lambda '(t)= \lambda (1-\lambda ) t^{-\lambda } -\lambda (1-\lambda ) t^{-\lambda -1}= \lambda (1-\lambda ) t^{-\lambda -1}(t-1)$, we have that $F_\lambda$ is strictly decreasing on (0, 1) and strictly increasing on $(1,\infty )$. Hence, $F_\lambda (t) \le \max \{ F_\lambda (m),\,F_\lambda (M)\}=:D$ for every $m \le t \le M$, and if $F_\lambda (t) = D$ for some $m \le t \le M$, then $t=m$ or $t=M$.

If $x,y>0$ and $m y \le x \le M y$, then

$$\begin{aligned} \begin{aligned} \lambda \Big (\frac{x}{y}\Big )^{1-\lambda } + (1-\lambda ) \Big (\frac{y}{x}\Big )^{\lambda }&\le D,\\ \lambda x + (1-\lambda ) y&\le D x^{\lambda } y^{1-\lambda }. \end{aligned} \end{aligned}$$

Note that, by continuity, this last inequality holds for every $x,y\ge 0$ with $m y \le x \le M y$. If the equality is attained for some $x,y\ge 0$ with $m y \le x \le M y$, then $x = m y$ or $x = M y$ (the cases $x=0$ and $y=0$ are direct).

Consider $\lambda =1/p$ (and so, $1-\lambda =1/q$), $a=x^{\lambda }=x^{1/p}$ and $b=y^{1-\lambda }=y^{1/q}$. Thus,

$$\begin{aligned} \frac{a^p}{p} + \frac{b^q}{q} \le D ab \end{aligned}$$

(3)

for every $a,b\ge 0$ with $m b^q \le a^p \le M b^q$. If the equality is attained for some $a,b\ge 0$ with $m b^q \le a^p \le M b^q$, then $a^p = m b^q$ or $a^p = M b^q$.

Since $\omega g^q \le f^p \le \varOmega g^q$$\mu$-a.e., we have $\omega \Vert g\Vert _q^q \le \Vert f\Vert _p^p \le \varOmega \Vert g\Vert _q^q$. If $\Vert f\Vert _p=0$ or $\Vert g\Vert _q=0$, then $\Vert f\Vert _p=\Vert g\Vert _q=0$ and the equality in (2) holds. Assume now that $\Vert f\Vert _p\ne 0 \ne \Vert g\Vert _q$. Thus,

$$\begin{aligned} \frac{\omega }{\varOmega }\,\frac{g^q}{\Vert g\Vert _q^q} \le \frac{f^p}{\Vert f\Vert _p^p} \le \frac{\varOmega }{\omega }\,\frac{g^q}{\Vert g\Vert _q^q} \quad \mu \text {-a.e.} \end{aligned}$$

If we consider $a=f/\Vert f\Vert _p$ and $b=g/\Vert g\Vert _q$ in (3) and we integrate both sides with respect to $\mu$, then we obtain

$$\begin{aligned} \begin{aligned} 1 = \frac{1}{p}\,\frac{\Vert f\Vert _p^p}{\Vert f\Vert _p^p} + \frac{1}{q}\,\frac{\Vert g\Vert _q^q}{\Vert g\Vert _q^q}&\le c_p(\omega ,\varOmega ) \frac{\Vert fg\Vert _1}{\Vert f\Vert _p\Vert g\Vert _q}, \\ \Vert f\Vert _p\Vert g\Vert _q&\le c_p(\omega ,\varOmega ) \Vert fg\Vert _1. \end{aligned} \end{aligned}$$

If the equality is attained, then

$$\begin{aligned} \frac{f^p}{\Vert f\Vert _p^p} = \frac{\omega }{\varOmega }\,\frac{g^q}{\Vert g\Vert _q^q} \quad \text {or} \quad \frac{f^p}{\Vert f\Vert _p^p} = \frac{\varOmega }{\omega }\,\frac{g^q}{\Vert g\Vert _q^q} \quad \mu \text {-a.e.} \end{aligned}$$

(4)

Assume that the first equality in (4) holds in a set A of positive $\mu$-measure. Therefore, we have both $f^p = \omega g^q$ in A and $\Vert f\Vert _p^p = \varOmega \Vert g\Vert _q^q$. Since $\Vert f\Vert _p\ne 0$, these facts imply $\omega = \varOmega$ and $f^p = \omega g^q$$\mu$-a.e.

If the second equality in (4) holds in a set of positive $\mu$-measure, then a similar argument gives $\omega = \varOmega$ and $f^p = \omega g^q$$\mu$-a.e. $\square$

Theorem 2 has the following consequence.

Corollary 1

If $1<p,q<\infty$, $a_j,b_j\ge 0$ and $\omega b_j^q \le a_j^p \le \varOmega b_j^q$ for $1\le j \le k$ and some positive constants $\omega ,\varOmega ,$ then

$$\begin{aligned} \left( \sum _{j=1}^k a_j^p \right) ^{1/p} \left( \sum _{j=1}^k b_j^q \right) ^{1/q} \le c_p(\omega ,\varOmega ) \sum _{j=1}^k a_jb_j, \end{aligned}$$

where $c_p(\omega ,\varOmega )$ is the constant in Theorem 2. If $a_j>0$ for some $1\le j \le k$, then the equality holds if and only if $\omega =\varOmega$ and $a_j^p=\omega b_j^q$ for every $1\le j \le k$.

Next, we prove three theorems that state several inequalities involving the inverse and the first general Zagreb indices.

Theorem 3

If $\alpha \in \mathbb {R}$ and G is a non-trivial graph with n vertices, m edges, minimum degree $\delta$ and maximum degree $\varDelta$, then

$$\begin{aligned} \begin{array}{l@{\quad }l} 2^{\alpha } m^{\alpha } n^{1-\alpha } \le M_1^{\alpha }(G) \le c_\alpha (\delta ^\alpha , \varDelta ^\alpha )^{\alpha } 2^{\alpha } m^{\alpha } n^{1-\alpha }, &{} \text {if }\, \alpha \ge 1,\\ c_{\frac{1}{\alpha }}(\delta , \varDelta )^{-1} 2^{\alpha } m^{\alpha } n^{1-\alpha } \le M_1^{\alpha }(G) \le 2^{\alpha } m^{\alpha } n^{1-\alpha }, &{} \text {if }\, 0\le \alpha \le 1,\\ c_{-\frac{1}{\alpha }}(\varDelta ^{-1}, \delta ^{-1})^{-1} n^{1+\alpha } \le M_1^{\alpha }(G)\, { ID}(G)^{\alpha } \le n^{1+\alpha }, &{} \text {if }\, -1\le \alpha \le 0,\\ c_{-\alpha }(\varDelta ^{\alpha }, \delta ^{\alpha })^{-1} M_1^\alpha (G)^{\frac{-1}{\alpha }} n^{\frac{\alpha +1}{\alpha }} \le { ID}(G) \le M_1^\alpha (G)^{\frac{-1}{\alpha }} n^{\frac{\alpha +1}{\alpha }}, &{} \text {if }\, \alpha \le -1, \end{array} \end{aligned}$$

where $c_p(\omega ,\varOmega )$ is the constant in Theorem 2 if $1<p<\infty$, and $c_1(\omega ,\varOmega )=c_\infty (\omega ,\varOmega )=1$. If $\alpha \ne -1,0,1,$ then the equality is attained in each inequality if and only if G is regular. If $\alpha \in \{-1,0,1\},$ then the inequalities are equalities for every graph G.

Proof

If $\alpha =1$, then $M_1^{1}(G)=2m$ for every graph G. If $\alpha =0$, then $M_1^{0}(G)=n$ for every graph G. If $\alpha =-1$, then $M_1^{-1}(G)={ ID}(G)$ for every graph G. Thus, in the three cases the inequalities are equalities for every graph G.

If $\alpha > 1$, then Hölder’s inequality gives

$$\begin{aligned} 2m = \sum _{u\in V(G)} d_u \le \left( \sum _{u\in V(G)} d_u^\alpha \right) ^{\frac{1}{\alpha }} \left( \sum _{u\in V(G)} 1^{\frac{\alpha }{\alpha -1}}\right) ^{\frac{\alpha -1}{\alpha }} = M_1^{\alpha }(G)^{\frac{1}{\alpha }} n^{\frac{\alpha -1}{\alpha }}, \end{aligned}$$

and so $M_1^{\alpha }(G)\ge 2^\alpha m^\alpha n^{1-\alpha }$. Since $\delta ^{\alpha } \le d_u^{\alpha }=d_u^{\alpha }/1^{\frac{\alpha }{\alpha -1}} \le \varDelta ^{\alpha }$, Corollary 1 gives

$$\begin{aligned} 2m = \sum _{u\in V(G)} d_u \ge \frac{\Big (\sum _{u\in V(G)} d_u^\alpha \Big )^{\frac{1}{\alpha }} \left( \sum _{u\in V(G)} 1^{\frac{\alpha }{\alpha -1}}\right) ^{\frac{\alpha -1}{\alpha }}}{c_\alpha (\delta ^\alpha , \varDelta ^\alpha )} = \frac{M_1^{\alpha }(G)^{\frac{1}{\alpha }} n^{\frac{\alpha -1}{\alpha }}}{c_\alpha (\delta ^\alpha , \varDelta ^\alpha )}, \end{aligned}$$

If $\alpha <-1$, then

$$\begin{aligned} \begin{aligned} { ID}(G)&=\sum _{u\in V(G)} d_u^{-1} \le \left( \sum _{u\in V(G)} (d_u^{-1})^{-\alpha }\right) ^{\frac{-1}{\alpha }} \left( \sum _{u\in V(G)} 1^{\frac{\alpha }{\alpha +1}}\right) ^{\frac{\alpha +1}{\alpha }}\\&=M_1^\alpha (G)^{\frac{-1}{\alpha }} n^{\frac{\alpha +1}{\alpha }}. \end{aligned} \end{aligned}$$

Since $\varDelta ^{\alpha } \le d_u^{\alpha }=(d_u^{-1})^{-\alpha }/1^{\frac{\alpha }{\alpha +1}} \le \delta ^{\alpha }$, Corollary 1 gives

$$\begin{aligned} \begin{aligned} { ID}(G)&=\sum _{u\in V(G)} d_u^{-1} \ge \frac{\Big (\sum _{u\in V(G)} (d_u^{-1})^{-\alpha }\Big )^{\frac{-1}{\alpha }} \Big (\sum _{u\in V(G)} 1^{\frac{\alpha }{\alpha +1}}\Big )^{\frac{\alpha +1}{\alpha }}}{c_{-\alpha }(\varDelta ^{\alpha }, \delta ^{\alpha })}\\&= \frac{M_1^\alpha (G)^{\frac{-1}{\alpha }} n^{\frac{\alpha +1}{\alpha }}}{c_{-\alpha }(\varDelta ^{\alpha }, \delta ^{\alpha })}. \end{aligned} \end{aligned}$$

If $0<\alpha <1$, then

$$\begin{aligned} M_1^{\alpha }(G)=\sum _{u\in V(G)} d_u^\alpha \le \Big (\sum _{u\in V(G)} (d_u^\alpha )^\frac{1}{\alpha }\Big )^\alpha \Big (\sum _{u\in V(G)} 1^\frac{1}{1-\alpha }\Big )^{1-\alpha } =2^\alpha m^\alpha n^{1-\alpha }. \end{aligned}$$

Since $\delta \le d_u =(d_u^\alpha )^\frac{1}{\alpha }/1^\frac{1}{1-\alpha } \le \varDelta$, Corollary 1 gives

$$\begin{aligned} M_1^{\alpha }(G) \ge \frac{\Big (\sum _{u\in V(G)} (d_u^\alpha )^\frac{1}{\alpha }\Big )^\alpha \Big (\sum _{u\in V(G)} 1^\frac{1}{1-\alpha }\Big )^{1-\alpha }}{c_{\frac{1}{\alpha }}(\delta , \varDelta )} =\frac{2^\alpha m^\alpha n^{1-\alpha }}{c_{\frac{1}{\alpha }}(\delta , \varDelta )}. \end{aligned}$$

If $-1<\alpha <0$, then $\varDelta ^{-1} \le d_u^{-1}=(d_u^\alpha )^{-\frac{1}{\alpha }}/1^{\frac{\alpha }{\alpha +1}} \le \delta ^{-1}$, and Corollary 1 gives

$$\begin{aligned} \begin{aligned} M_1^{\alpha }(G)&=\sum _{u\in V(G)} d_u^\alpha \ge \frac{\Big (\sum _{u\in V(G)} (d_u^\alpha )^{-\frac{1}{\alpha }}\Big )^{-\alpha } \Big (\sum _{u\in V(G)} 1^\frac{1}{1+\alpha }\Big )^{1+\alpha }}{c_{-\frac{1}{\alpha }}(\varDelta ^{-1}, \delta ^{-1})}\\&= \frac{{ ID}(G)^{-\alpha } n^{1+\alpha }}{c_{-\frac{1}{\alpha }}(\varDelta ^{-1}, \delta ^{-1})}. \end{aligned} \end{aligned}$$

Assume that $\alpha \ne {-}\,1,0,1,$ and consider any inequality proved by using Hölder’s inequality. By Hölder’s inequality, the equality is attained if and only if the vectors $(d_u^\beta )_{u \in V(G)}$ (for some constant $\beta \ne 0$ which depends on $\alpha$) and $(1)_{u \in V(G)}$ are parallel; this is equivalent to $d_u=d_v$ for every $u,v \in V(G)$, i.e., G is regular.

Assume that $\alpha \ne -1,0,1,$ and consider any inequality proved by using Corollary 1. By Corollary 1, the equality is attained if and only if $\delta ^\beta =\varDelta ^\beta$ (for some constant $\beta \ne 0$ which depends on $\alpha$), i.e., G is regular. $\square$

Remark 1

Recall that the number $\alpha$ in $M_1^{\alpha }(G)$ is not an exponent, it is a parameter.

Theorem 4

If $\alpha \in \mathbb {R}$ and G is a non-trivial graph with n vertices, minimum degree $\delta$ and maximum degree $\varDelta$, then

$$\begin{aligned} \begin{aligned} {\delta }^{\alpha -1} n^2 \le M_1^{\alpha }(G)\, { ID}(G) \le \frac{\big (\varDelta ^{\alpha } + \delta ^{\alpha } \big )^2n^2}{4\varDelta \delta ^{\alpha }}, \qquad&\text {if }\, \alpha \ge 1, \\ {\varDelta }^{\alpha -1} n^2 \le M_1^{\alpha }(G)\, { ID}(G) \le \frac{\big (\varDelta ^{\alpha } + \delta ^{\alpha } \big )^2n^2}{4\varDelta ^{\alpha } \delta }, \qquad&\text {if }\, \alpha \le 1. \end{aligned} \end{aligned}$$

Any equality is attained for some $\alpha \in \mathbb {R}$ if and only if G is regular.

Proof

If $\alpha \ge 1$, then Cauchy–Schwarz inequality gives

$$\begin{aligned} \begin{aligned} n^2&= \left( \sum _{u\in V(G)} d_u^{\alpha /2} d_u^{-\alpha /2}\right) ^{2} \le \left( \sum _{u\in V(G)} d_u^\alpha \right) \left( \sum _{u\in V(G)} d_u^{-\alpha } \right) \\&= M_1^{\alpha }(G) \sum _{u\in V(G)} d_u^{-\alpha +1} d_u^{-1} \le \delta ^{-\alpha +1} M_1^{\alpha }(G) \, { ID}(G). \end{aligned} \end{aligned}$$

The same argument gives $\varDelta ^{\alpha -1}n^2 \le M_1^{\alpha }(G)\, { ID}(G)$ for $\alpha \le 1$.

On the other hand, since we have for every $\alpha \in \mathbb {R}$,

$$\begin{aligned} \min \big \{ \delta ^{\alpha }, \varDelta ^{\alpha }\big \} \le \frac{d_u^{\alpha /2}}{d_u^{-\alpha /2}} = d_u^{\alpha } \le \max \big \{ \delta ^{\alpha }, \varDelta ^{\alpha }\big \}, \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \left( \sqrt{\frac{\max \big \{ \delta ^{\alpha }, \varDelta ^{\alpha }\big \}}{\min \big \{ \delta ^{\alpha }, \varDelta ^{\alpha }\big \}}} + \sqrt{\frac{\min \big \{ \delta ^{\alpha }, \varDelta ^{\alpha }\big \}}{\max \big \{ \delta ^{\alpha }, \varDelta ^{\alpha }\big \}}} \;\right) ^2&= \left( \frac{\varDelta ^{\alpha /2}}{\delta ^{\alpha /2}} + \frac{\delta ^{\alpha /2}}{\varDelta ^{\alpha /2}} \right) ^2\\&= \frac{\big (\varDelta ^{\alpha } + \delta ^{\alpha } \big )^2}{\varDelta ^{\alpha } \delta ^{\alpha }}, \end{aligned} \end{aligned}$$

Lemma 1 gives

$$\begin{aligned} \begin{aligned} n^2&=\Big (\sum _{u\in V(G)} d_u^{\alpha /2} d_u^{-\alpha /2}\Big )^2 \ge \frac{\Big (\sum _{u\in V(G)} d_u^\alpha \Big ) \Big (\sum _{u\in V(G)} d_u^{-\alpha } \Big )}{\frac{(\varDelta ^{\alpha } + \delta ^{\alpha } )^2}{4\varDelta ^{\alpha } \delta ^{\alpha }}}\\&= \frac{4\varDelta ^{\alpha } \delta ^{\alpha } M_1^{\alpha }(G)\Big (\sum _{u\in V(G)} d_u^{-\alpha +1}d_u^{-1} \Big )}{\big (\varDelta ^{\alpha } + \delta ^{\alpha } \big )^2}. \end{aligned} \end{aligned}$$

If $\alpha \ge 1$, then $d_u^{-\alpha +1} \ge \varDelta ^{-\alpha +1}$ and

$$\begin{aligned} n^2 \ge \frac{4\varDelta \delta ^{\alpha } M_1^{\alpha }(G)\,IG(G)}{\big (\varDelta ^{\alpha } + \delta ^{\alpha } \big )^2}. \end{aligned}$$

If $\alpha \le 1$, then $d_u^{-\alpha +1} \ge \delta ^{-\alpha +1}$ and

$$\begin{aligned} n^2 \ge \frac{4\varDelta ^{\alpha } \delta \,M_1^{\alpha }(G)\,IG(G)}{\big (\varDelta ^{\alpha } + \delta ^{\alpha } \big )^2}. \end{aligned}$$

If the graph is regular, then for each $\alpha \in \mathbb {R}$ both bounds are the same, and they are equal to $M_1^{\alpha }(G)\, { ID}(G)$.

If an equality is attained for some $\alpha \ne 1$, then we have either $d_u=\delta$ for every $u \in V(G)$ or $d_u=\varDelta$ for every $u \in V(G)$, and G is regular in both cases. If the lower bound is attained for $\alpha = 1$, then Cauchy–Schwarz inequality gives that the vectors $(d_u^{1/2})_{u \in V(G)}$ and $(d_u^{-1/2})_{u \in V(G)}$ are parallel; this is equivalent to $d_u=d_v$ for every $u,v \in V(G)$, and G is regular. If the upper bound is attained for $\alpha = 1$, then Lemma 1 gives that $\delta =\varDelta$, and G is regular. $\square$

Theorem 5

If G is a non-trivial graph with n vertices, minimum degree $\delta$ and maximum degree $\varDelta$, then

$$\begin{aligned} { ID}(G)\ge \frac{n}{\varDelta +\delta }+\frac{\varDelta \delta }{\varDelta +\delta } \, M_1^{-2}(G). \end{aligned}$$

The equality is attained if and only if each vertex has degree either $\delta$ or $\varDelta$.

Proof

We have

$$\begin{aligned} \begin{aligned} \Big (\frac{1}{\delta }-\frac{1}{d_u}\Big )\Big (\frac{1}{d_u}-\frac{1}{\varDelta }\Big ) \ge 0,&\qquad \frac{1}{d_u\delta }-\frac{1}{\delta \varDelta }-\frac{1}{d_u^2}+\frac{1}{d_u\varDelta } \ge 0,\\ \frac{1}{d_u}\frac{\varDelta +\delta }{\varDelta \delta } \ge \frac{1}{\varDelta \delta }+\frac{1}{d_u^2},&\qquad \frac{1}{d_u} \ge \frac{1}{\varDelta +\delta }+\frac{\varDelta \delta }{\varDelta +\delta }\,\frac{1}{d_u^2},\\ { ID}(G)&\ge \frac{n}{\varDelta +\delta }+\frac{\varDelta \delta }{\varDelta +\delta }\,M_1^{-2}(G). \end{aligned} \end{aligned}$$

The equality is attained if and only if $(\delta -d_u)(d_u-\varDelta )$ for every $u \in V(G)$, i.e., each vertex has degree either $\delta$ or $\varDelta$. $\square$

With motivation from the first Zagreb and harmonic indices, general sum-connectivity index$\chi _{\alpha }$ was defined by Zhou and Trinajstić in [38] as

$$\begin{aligned} \chi _{\alpha }(G) = \sum _{uv\in E(G)} (d_u + d_v)^{\alpha }, \end{aligned}$$

with $\alpha \in \mathbb {R}$. Note that $\chi _{{}_1}$ is the first Zagreb index $M_1$, $2\chi _{{}_{-1}}$ is the harmonic index H, $\chi _{{}_{-1/2}}$ is the sum-connectivity index, etc. Some mathematical properties of the general sum-connectivity index were given in [8, 28, 37,38,39].

The following results relate the general first Zagreb and the general sum-connectivity indices.

Theorem 6

If $\alpha \in \mathbb {R}$ and G is a non-trivial graph, then

$$\begin{aligned} \begin{aligned} M_1^{\alpha +1}(G) \ge 2^{1-\alpha }\chi _{\alpha } (G), \qquad&\text {if }\, \alpha \ge 1 \, \text { or }\, \alpha \le 0,\\ M_1^{\alpha +1}(G) \le 2^{1-\alpha }\chi _{\alpha } (G), \qquad&\text {if }\, 0\le \alpha \le 1. \end{aligned} \end{aligned}$$

If $\alpha \ne 0,1,$ then the equality is attained in each inequality if and only if every connected component of G is regular. If $\alpha \in \{0,1\},$ then the equality holds for every graph G.

Proof

If $\alpha =0$, then $2\chi _{{}_0}(G)=2\sum _{uv\in E(G)} 1=2m=\sum _{u\in V(G)} d_u=M_1^{1}(G)$.

If $\alpha =1$, then $\chi _{{}_1}(G)=\sum _{uv\in E(G)} (d_u + d_v)=M_1(G)=M_1^{2}(G)$.

Let us consider the function $f(x)=x^\alpha$ for $x>0$, with $\alpha \ne 0,1$.

If $\alpha > 1$ or $\alpha <0$, then f is convex and

$$\begin{aligned} \begin{aligned} \Big (\frac{d_u+d_v}{2}\Big )^\alpha&\le \frac{d_u^\alpha +d_v^\alpha }{2}, \qquad 2^{1-\alpha }(d_u+d_v)^\alpha \le d_u^\alpha +d_v^\alpha ,\\ 2^{1-\alpha }\chi _{\alpha }(G)&= 2^{1-\alpha } \sum _{uv\in E(G)}(d_u+d_v)^\alpha \le \sum _{uv\in E(G)} \big (d_u^\alpha +d_v^\alpha \,\big )\\&= \sum _{u \in V(G)} d_u^{\alpha +1} = M_1^{\alpha +1}(G). \end{aligned} \end{aligned}$$

If $0<\alpha <1,$ then $f(x)=x^\alpha$ is concave and the converse inequality holds.

If we fix $\alpha \ne 0,1,$ then f is either strictly convex or strictly concave. Thus, the equality is attained if and only if $d_u=d_v$ for every $uv \in E(G)$, i.e., each connected component of G is regular. If $\alpha \in \{0,1\},$ then we have proved that the equality holds for every graph G. $\square$

Corollary 2

If G is a non-trivial graph, then

$$\begin{aligned} { ID}(G) \ge 8\chi _{_{-2}}(G). \end{aligned}$$

The equality is attained if and only if every connected component of G is regular.

In [27, Corollary 2.2] appears the following result.

Lemma 2

If $0<a \le x,y \le b$, then

$$\begin{aligned} \frac{2\sqrt{ab}}{a + b} \le \frac{2\sqrt{xy}}{x + y} \le 1. \end{aligned}$$

The equality in the lower bound is attained if and only if either $x=a$ and $y=b$, or $x=b$ and $y=a$, and the equality in the upper bound is attained if and only if $x=y$.

Corollary 3

If $0<\delta \le x,y \le \varDelta$, then

$$\begin{aligned} 2 \le \frac{x^2 + y^2}{x y} \le \frac{\varDelta ^2 + \delta ^2}{\varDelta \delta }. \end{aligned}$$

The equality in the upper bound is attained if and only if either $x=\delta$ and $y=\varDelta$, or $x=\varDelta$ and $y=\delta$, and the equality in the lower bound is attained if and only if $x=y$.

Recall that a biregular graph is a bipartite graph for which any vertex in one side of the given bipartition has degree $\varDelta$ and any vertex in the other side of the bipartition has degree $\delta$. We say that a graph is $(\varDelta ,\delta )$-biregular if we want to write explicitly the maximum and minimum degrees.

The following results relate the second general Zagreb and the inverse indices.

Theorem 7

If G is a non-trivial graph with m edges, minimum degree $\delta$ and maximum degree $\varDelta$, then

$$\begin{aligned} \frac{4\varDelta \delta }{\varDelta ^2 + \delta ^2} \,m^{\frac{1}{2}} M_2^{-2}(G)^{\frac{1}{2}} \le { ID}(G) \le \frac{\varDelta ^2+\delta ^2}{\varDelta \delta } \,m^{\frac{1}{2}} M_2^{-2}(G)^{\frac{1}{2}}. \end{aligned}$$

The equality in the upper bound is attained if and only if G is either regular or $(\varDelta ,\delta )$-biregular. The equality in the lower bound is attained if and only if G is regular.

Proof

Cauchy–Schwarz inequality and Corollary 3 give

$$\begin{aligned} \begin{aligned} { ID}(G)^2&= \left( \sum _{uv\in E(G)}\frac{d_u^2 + d_v^2}{d_u^2 d_v^2} \right) ^2 \le \sum _{uv\in E(G)}\left( \frac{d_u^2 + d_v^2}{d_u d_v}\right) ^2 \sum _{uv\in E(G)} \frac{1}{(d_u d_v)^{2}}\\&\le \sum _{uv\in E(G)}\left( \frac{\varDelta ^2+\delta ^2}{\varDelta \delta } \right) ^2 \sum _{uv\in E(G)} (d_u d_v)^{-2} = \left( \frac{\varDelta ^2+\delta ^2}{\varDelta \delta } \right) ^2 \, m\, M_2^{-2}(G). \end{aligned} \end{aligned}$$

On the other hand, since

$$\begin{aligned} 2\delta ^2 \le \frac{\frac{d_u^2 + d_v^2}{d_ud_v}}{\frac{1}{d_ud_v}} = d_u^2 + d_v^2 \le 2 \varDelta ^2, \quad \text { with } \quad \sqrt{\frac{2 \varDelta ^2}{2 \delta ^2}} =\frac{\varDelta }{\delta }, \end{aligned}$$

Lemma 1 and Corollary 3 give

$$\begin{aligned} \begin{aligned} { ID}(G)^2&=\left( \sum _{uv\in E(G)}\frac{d_u^2 + d_v^2}{d_u^2d_v^2}\,\right) ^2 \ge \frac{\sum _{uv\in E(G)}\left( \frac{d_u^2 + d_v^2}{d_u d_v}\right) ^2 \sum _{uv\in E(G)} (d_u d_v)^{-2}}{\frac{1}{4}\big (\frac{\varDelta }{\delta }+\frac{\delta }{\varDelta }\big )^2}\\&\ge \frac{\sum _{uv\in E(G)} 4\,M_2^{-2}(G)}{\frac{1}{4} \,\frac{(\varDelta ^2+\delta ^2)^2}{\varDelta ^2\delta ^2}} =\frac{16\varDelta ^2\delta ^2m\, M_2^{-2}(G)}{(\varDelta ^2 + \delta ^2)^2}. \end{aligned} \end{aligned}$$

By Cauchy–Schwarz inequality and Corollary 3, the equality in the upper bound is attained if and only if every edge has a vertex with degree $\delta$ and the other vertex with degree $\varDelta$; and this happens if and only if G is either regular (if $\varDelta =\delta$) or $(\varDelta ,\delta )$-biregular (if $\varDelta \ne \delta$).

By Lemma 1 and Corollary 3, the equality is attained in the lower bound if and only if $2\delta ^2=2\varDelta ^2$, i.e., G is regular. $\square$

Theorem 8

Let G be a non-trivial graph with minimum degree $\delta$ and maximum degree $\varDelta$. Then

$$\begin{aligned} \begin{aligned}&2\delta ^{-2\alpha -2}M_2^{\alpha }(G) \le { ID}(G) \le 2\varDelta ^{-2\alpha -2} M_2^{\alpha }(G), \quad \text {if } \, \alpha \le -2\varDelta ^2/(\varDelta ^2+\delta ^2),\\&\min \Big \{ 2\delta ^{-2\alpha -2},\,\frac{2}{-\alpha } \Big (\frac{\alpha +2}{-\alpha }\Big )^{-\alpha /2-1}\varDelta ^{-2\alpha -2}\Big \} M_2^{\alpha }(G) \le { ID}(G)\\&\quad \le \max \big \{ 2\varDelta ^{-2\alpha -2},\, (\varDelta ^2+\delta ^2)(\varDelta \delta )^{-\alpha -2} \big \}M_2^{\alpha }(G),\\&\quad \text {if } \, -2\varDelta ^2/(\varDelta ^2+\delta ^2)< \alpha< -1,\\&2M_2^{-1}(G) \le { ID}(G) \le \frac{\varDelta ^2+\delta ^2}{\varDelta \delta } M_2^{-1}(G), \qquad \text {if } \, \alpha =-1,\\&\min \Big \{ 2\varDelta ^{-2\alpha -2},\,\frac{2}{-\alpha } \Big (\frac{\alpha +2}{-\alpha }\Big )^{-\alpha /2-1}\delta ^{-2\alpha -2}\Big \} M_2^{\alpha }(G) \le { ID}(G)\\&\quad \le \max \big \{ 2\delta ^{-2\alpha -2},\, (\varDelta ^2+\delta ^2)(\varDelta \delta )^{-\alpha -2} \big \}M_2^{\alpha }(G),\\&\qquad \qquad \quad \text {if } \, -1< \alpha < -2\delta ^2/(\varDelta ^2+\delta ^2),\\&2\varDelta ^{-2\alpha -2}M_2^{\alpha }(G) \le { ID}(G) \le 2\delta ^{-2\alpha -2} M_2^{\alpha }(G), \text {if } \, \alpha \ge -2\delta ^2/(\varDelta ^2+\delta ^2). \end{aligned} \end{aligned}$$

Every bound is attained for every regular graph G. Furthermore, in the first and last cases, each inequality is attained if and only if G is regular.

Proof

We are going to compute the maximum and minimum values of the function $f: [\delta ,\varDelta ] \times [\delta ,\varDelta ] \rightarrow \mathbb {R}$ given by

$$\begin{aligned} f(x,y) = \Big ( \frac{1}{x^2} + \frac{1}{y^2}\Big )(xy)^{-\alpha } = (x^2+y^2)x^{-\alpha -2}y^{-\alpha -2}. \end{aligned}$$

By symmetry, we can assume that $x \le y$. We have

$$\begin{aligned} \begin{aligned} \frac{\partial f}{\partial x}(x,y)&= 2x \cdot x^{-\alpha -2}y^{-\alpha -2} + (x^2+y^2)(-\alpha -2) x^{-\alpha -3}y^{-\alpha -2}\\&= x^{-\alpha -3}y^{-\alpha -2} \big ( -\alpha x^2 -(\alpha +2) y^2 \big ),\\ \frac{\partial f}{\partial y}(x,y)&= y^{-\alpha -3}x^{-\alpha -2} \big ( -\alpha y^2 -(\alpha +2) x^2 \big ). \end{aligned} \end{aligned}$$

If $\alpha \le -2$, then $\partial f/\partial x,\partial f/\partial y>0$ and so, f is an increasing function in both variables. Hence,

$$\begin{aligned} \begin{aligned} 2\delta ^{-2\alpha -2}&\le f(x,y) \le 2\varDelta ^{-2\alpha -2},\\ 2\delta ^{-2\alpha -2} (d_u d_v)^\alpha&\le \frac{1}{d_u^2} + \frac{1}{d_v^2} \le 2\varDelta ^{-2\alpha -2} (d_u d_v)^\alpha ,\\ 2\delta ^{-2\alpha -2}M_2^{\alpha }(G)&\le { ID}(G) \le 2\varDelta ^{-2\alpha -2} M_2^{\alpha }(G). \end{aligned} \end{aligned}$$

(5)

If $\alpha \ge 0$, then $\partial f/\partial x,\partial f/\partial y<0$ and

$$\begin{aligned} \begin{aligned} 2\varDelta ^{-2\alpha -2}&\le f(x,y) \le 2\delta ^{-2\alpha -2},\\ 2\varDelta ^{-2\alpha -2} (d_u d_v)^\alpha&\le \frac{1}{d_u^2} + \frac{1}{d_v^2} \le 2\delta ^{-2\alpha -2} (d_u d_v)^\alpha ,\\ 2\varDelta ^{-2\alpha -2}M_2^{\alpha }(G)&\le { ID}(G) \le 2\delta ^{-2\alpha -2} M_2^{\alpha }(G). \end{aligned} \end{aligned}$$

(6)

We deal now with the case $-2< \alpha < 0$. If $\nabla f(x,y)=0$, then

$$\begin{aligned} \begin{aligned} 0&= \alpha x^2 +2 y^2 + \alpha y^2,\\ 0&= -\alpha x^2-2 x^2 -\alpha y^2, \end{aligned} \end{aligned}$$

and we conclude $x=y$. Therefore, the maximum and the minimum values of f are attained on the boundary $\{x=\delta ,\, \delta \le y \le \varDelta \} \cup \{y=\varDelta ,\, \delta \le x \le \varDelta \} \cup \{ \delta \le x=y \le \varDelta \}$.

On the set $\{ \delta \le x=y \le \varDelta \}$ we have $f(x,x)=2x^{-2\alpha -2}$. If $-2< \alpha \le -1$, then $2\delta ^{-2\alpha -2} \le f(x,x) \le 2\varDelta ^{-2\alpha -2}$. If $-1 \le \alpha < 0$, then $2\varDelta ^{-2\alpha -2} \le f(x,x) \le 2\delta ^{-2\alpha -2}$.

We have $\partial f/\partial y (\delta ,y) = 0$ for some $y > 0$ if and only if

$$\begin{aligned} -\alpha y^2 -(\alpha +2) \delta ^2 =0 \Leftrightarrow y=y_0:=\sqrt{\frac{\alpha +2}{-\alpha }} \, \delta . \end{aligned}$$

Thus, $\delta< y_0 < \varDelta$ if and only if

$$\begin{aligned} -1< \alpha < \frac{-2\delta ^2}{\varDelta ^2+\delta ^2}. \end{aligned}$$

(7)

If $-2< \alpha \le -1$, then $\partial f/\partial y (\delta ,y) > 0$ for $\delta< y < \varDelta$, and

$$\begin{aligned} 2\delta ^{-2\alpha -2} = f(\delta ,\delta ) \le f(\delta ,y) \le f(\delta ,\varDelta ) = (\varDelta ^2+\delta ^2)(\varDelta \delta )^{-\alpha -2}. \end{aligned}$$

If $-2\delta ^2/(\varDelta ^2+\delta ^2)\le \alpha <0$, then $\partial f/\partial y (\delta ,y) < 0$ for $\delta< y < \varDelta$, and

$$\begin{aligned} (\varDelta ^2+\delta ^2)(\varDelta \delta )^{-\alpha -2} \le f(\delta ,y) \le 2\delta ^{-2\alpha -2}. \end{aligned}$$

If (7) holds, then $\partial f/\partial y(\delta ,\delta ) = -2\delta ^{-2\alpha -3} ( \alpha +1 )<0$, and $\partial f/\partial y(\delta ,\varDelta ) = \varDelta ^{-\alpha -3}\delta ^{-\alpha -2} \big ( -\alpha \varDelta ^2 -(\alpha +2) \delta ^2 \big )>0$, and we conclude

$$\begin{aligned} \begin{aligned} \frac{2}{-\alpha }&\Big (\frac{\alpha +2}{-\alpha }\Big )^{-\alpha /2-1}\delta ^{-2\alpha -2}= f(\delta ,y_0) \le f(\delta ,y)\\ \le&\max \big \{ f(\delta ,\delta ),\, f(\delta ,\varDelta )\big \} = \max \big \{ 2\delta ^{-2\alpha -2},\, (\varDelta ^2+\delta ^2)(\varDelta \delta )^{-\alpha -2}\big \}. \end{aligned} \end{aligned}$$

We have $\partial f/\partial x (x,\varDelta ) = 0$ for some $x > 0$ if and only if

$$\begin{aligned} -\alpha x^2 -(\alpha +2) \varDelta ^2 =0 \Leftrightarrow x=x_0:=\sqrt{\frac{\alpha +2}{-\alpha }} \, \varDelta . \end{aligned}$$

Thus, $\delta< x_0 < \varDelta$ if and only if

$$\begin{aligned} \frac{-2\varDelta ^2}{\varDelta ^2+\delta ^2}< \alpha < -1. \end{aligned}$$

(8)

If $-1\le \alpha <0$, then $\partial f/\partial x (x,\varDelta ) < 0$ for $\delta< x < \varDelta$, and

$$\begin{aligned} 2\varDelta ^{-2\alpha -2} = f(\varDelta ,\varDelta ) \le f(x,\varDelta ) \le f(\delta ,\varDelta ) = (\varDelta ^2+\delta ^2)(\varDelta \delta )^{-\alpha -2}. \end{aligned}$$

If $-2 < \alpha \le -2\varDelta ^2/(\varDelta ^2+\delta ^2)$, then $\partial f/\partial x (x,\varDelta ) > 0$ for $\delta< x < \varDelta$, and

$$\begin{aligned} (\varDelta ^2+\delta ^2)(\varDelta \delta )^{-\alpha -2} \le f(x,\varDelta ) \le 2\varDelta ^{-2\alpha -2}. \end{aligned}$$

If (8) holds, then $\partial f/\partial x(\delta ,\varDelta ) = \delta ^{-\alpha -3}\varDelta ^{-\alpha -2} \big ( -\alpha \delta ^2 -(\alpha +2) \varDelta ^2 \big ) < 0$, and $\partial f/\partial x(\varDelta ,\varDelta ) = -2\varDelta ^{-2\alpha -3} ( \alpha +1 ) > 0$, and we conclude

$$\begin{aligned} \begin{aligned} \frac{2}{-\alpha }&\Big (\frac{\alpha +2}{-\alpha }\Big )^{-\alpha /2-1}\varDelta ^{-2\alpha -2}= f(x_0,\varDelta ) \le f(x,\varDelta )\\ \le&\max \big \{ f(\delta ,\varDelta ),\, f(\varDelta ,\varDelta )\big \} = \max \big \{ (\varDelta ^2+\delta ^2)(\varDelta \delta )^{-\alpha -2},\,2\varDelta ^{-2\alpha -2} \big \}. \end{aligned} \end{aligned}$$

Hence, if $-2 < \alpha \le -2\varDelta ^2/(\varDelta ^2+\delta ^2)$ or $-2\delta ^2/(\varDelta ^2+\delta ^2)\le \alpha <0$, then (5) or (6) holds, respectively.

If $-2\varDelta ^2/(\varDelta ^2+\delta ^2)< \alpha < -1$, then

$$\begin{aligned} \begin{aligned}&\min \Big \{ 2\delta ^{-2\alpha -2},\,\frac{2}{-\alpha } \Big (\frac{\alpha +2}{-\alpha }\Big )^{-\alpha /2-1}\varDelta ^{-2\alpha -2}\Big \} \le f(x,y)\\&\quad \le \max \big \{ 2\varDelta ^{-2\alpha -2},\, (\varDelta ^2+\delta ^2)(\varDelta \delta )^{-\alpha -2} \big \}. \end{aligned} \end{aligned}$$

If $-1< \alpha < -2\delta ^2/(\varDelta ^2+\delta ^2)$, then

$$\begin{aligned} \begin{aligned}&\min \Big \{ 2\varDelta ^{-2\alpha -2},\,\frac{2}{-\alpha } \Big (\frac{\alpha +2}{-\alpha }\Big )^{-\alpha /2-1}\delta ^{-2\alpha -2}\Big \} \le f(x,y)\\&\quad \le \max \big \{ 2\delta ^{-2\alpha -2},\, (\varDelta ^2+\delta ^2)(\varDelta \delta )^{-\alpha -2} \big \}. \end{aligned} \end{aligned}$$

These inequalities finish the proofs of the bounds if $\alpha \ne -1$. If $\alpha = -1$, then we can obtain the bounds by taking limits on the inequalities when $\alpha > -1$.

In the first and last cases, the properties of the function f give that each inequality is attained if and only if either $d_u=d_v=\delta$ for every $uv \in E(G)$ or $d_u=d_v=\varDelta$ for every $uv \in E(G)$, and this happens if and only if G is regular.

If the graph is regular, then the lower and upper bounds are the same in each case, and they are equal to ${ ID}(G)$ (note that we do not have the second and fourth cases if $\delta =\varDelta$). $\square$

3 Inequalities involving the geometric–arithmetic index

The first geometric–arithmetic index $GA_1$ was introduced in [31] as

$$\begin{aligned} GA_1(G) = \sum _{uv\in E(G)}\frac{\sqrt{d_u d_v}}{\frac{1}{2} (d_u + d_v)}. \end{aligned}$$

Although $GA_1$ was introduced in 2009, there are many papers dealing with this index (see, e.g., [4,5,6, 26, 27, 31] and the references therein).

The following results provide inequalities relating inverse degree and geometric–arithmetic indices.

Theorem 9

If G is a non-trivial graph with m edges and maximum degree $\varDelta$, then

$$\begin{aligned} 2GA_1(G) + \varDelta ^2{ ID}(G) \ge 4m, \end{aligned}$$

and the equality is attained if and only if G is regular.

Proof

The inequality $2xy \le x^2 + y^2$ for every $x,y\in \mathbb {R}$, and the geometric–arithmetic inequality give

$$\begin{aligned} \begin{aligned} \frac{2d_ud_v}{(d_u+d_v)^2}+\frac{d_u^2+d_v^2}{(d_u+d_v)^2} =1,&\quad \frac{1}{2}\,\frac{2\sqrt{d_ud_v}}{d_u+d_v}\,\frac{2\sqrt{d_ud_v}}{d_u+d_v}+\frac{d_u^2+d_v^2}{4 d_u d_v} \ge 1,\\ \frac{1}{2}\,\frac{2\sqrt{d_ud_v}}{d_u+d_v}+\frac{d_u^2+d_v^2}{4\frac{d_u^2d_v^2}{\varDelta ^2}} \ge 1,&\quad 2\,\frac{2\sqrt{d_ud_v}}{d_u+d_v}+\varDelta ^2\frac{d_u^2+d_v^2}{d_u^2d_v^2} \ge 4,\\ 2GA_1(G) + \varDelta ^2{ ID}(G)&\ge 4m. \end{aligned} \end{aligned}$$

If the graph is regular, then $2GA_1(G) + \varDelta ^2{ ID}(G) = 4m$.

If the equality is attained, then $d_ud_v=\varDelta ^2$ for every $uv \in E(G)$, and so $d_u=\varDelta$ for every $u \in V(G)$ and G is regular. $\square$

Theorem 10

Let G be a non-trivial graph with minimum degree $\delta$ and maximum degree $\varDelta$. Then

$$\begin{aligned} { ID}(G) \le \max \Big \{ \frac{2}{\delta ^2}, \, \frac{(\varDelta +\delta )(\varDelta ^2+\delta ^2)}{2(\varDelta \delta )^{5/2}} \Big \} GA_1(G), \end{aligned}$$

and the equality is attained for every regular graph G.

Proof

We are going to compute the maximum value of the function $f: [\delta ,\varDelta ] \times [\delta ,\varDelta ] \rightarrow \mathbb {R}$ given by

$$\begin{aligned} f(x,y) = \Big ( \frac{1}{x^2} + \frac{1}{y^2}\Big )\frac{x+y}{2\sqrt{xy}} = \frac{1}{2} (x+y)(x^2+y^2)x^{-5/2}y^{-5/2}. \end{aligned}$$

By symmetry, we can assume that $x \le y$. We have

$$\begin{aligned} \begin{aligned} \frac{\partial f}{\partial x}(x,y)&= \frac{1}{2} (x^2+y^2)x^{-5/2}y^{-5/2} + (x+y)x \cdot x^{-5/2}y^{-5/2}\\&\quad - \frac{5}{4} (x+y)(x^2+y^2)x^{-7/2}y^{-5/2}\\&= \frac{1}{4}\, x^{-7/2}y^{-5/2} \big ( 2x(x^2+y^2) +4x^2(x+y) - 5(x+y)(x^2+y^2) \big )\\&= \frac{1}{4}\, x^{-7/2}y^{-5/2} \big ( x^3-x^2y -3xy^2-5y^3 \big ),\\ \frac{\partial f}{\partial y}(x,y)&= \frac{1}{4}\, y^{-7/2}x^{-5/2} \big ( y^3-y^2x -3yx^2-5x^3 \big ). \end{aligned} \end{aligned}$$

If we define $g(t):=t^3-t^2-3t-5$, then

$$\begin{aligned} \frac{\partial f}{\partial x}(x,y) = \frac{1}{4}\, x^{-7/2}y^{1/2} g\Big ( \frac{x}{y}\Big ). \end{aligned}$$

Since $g'(t)=3t^2-2t-3=0$ if and only if $t=(1 \pm \sqrt{10}\,)/3$, g is a decreasing function on [0, 1]. Since $g(0)=-5$, we have $g(t)<0$ for every $t\in [0,1]$, and

$$\begin{aligned} \frac{\partial f}{\partial x}(x,y) <0, \qquad \text {if }\, \delta \le x \le y \le \varDelta . \end{aligned}$$

Thus, $f(\delta ,y)> f(x,y) > f(y,y)$ for every $x\in (\delta ,y)$ and so, the maximum value of f is attained on the set $\{x=\delta ,\, \delta \le y \le \varDelta \}$, and the minimum value of f is attained on the set $\{ \delta \le x=y \le \varDelta \}$.

If we define $h(t):=t^3-\delta t^2-3\delta ^2t-5\delta ^3$, then

$$\begin{aligned} \frac{\partial f}{\partial y}(\delta ,y) = \frac{1}{4}\, y^{-7/2}\delta ^{-5/2} h ( y). \end{aligned}$$

Since $h'(t)=3t^2-2\delta t-3\delta ^2=0$ if and only if $t=(1 \pm \sqrt{10}\,)\delta /3$, h is a decreasing function on $[\delta ,(1 + \sqrt{10}\,)\delta /3)$ and increasing on $((1 + \sqrt{10}\,)\delta /3,\infty )$. Since $h(\delta )=-8\delta ^3$, we have $h<0$ on $[\delta ,t_1)$ and $h>0$ on $(t_1,\infty )$ for some $t_1 > \delta$. Thus,

$$\begin{aligned} \begin{aligned} f(x,y)&\le \max _{\delta \le y \le \varDelta } f(\delta ,y) = \max \big \{ f(\delta ,\delta ), \, f(\delta ,\varDelta )\big \}\\&= \max \Big \{ \frac{2}{\delta ^2}, \, \frac{(\delta +\varDelta )(\delta ^2+\varDelta ^2)}{2(\delta \varDelta )^{5/2}} \Big \},\\ \frac{1}{d_u^2} + \frac{1}{d_v^2}&\le \max \Big \{ \frac{2}{\delta ^2}, \, \frac{(\varDelta +\delta )(\varDelta ^2+\delta ^2)}{2(\varDelta \delta )^{5/2}} \Big \}\frac{2\sqrt{d_u d_v}}{d_u+d_v},\\ { ID}(G)&\le \max \Big \{ \frac{2}{\delta ^2}, \, \frac{(\varDelta +\delta )(\varDelta ^2+\delta ^2)}{2(\varDelta \delta )^{5/2}} \Big \} GA_1(G). \end{aligned} \end{aligned}$$

If the graph is regular, then

$$\begin{aligned} \begin{aligned} \max \Big \{ \frac{2}{\delta ^2}, \, \frac{(\varDelta +\delta )(\varDelta ^2+\delta ^2)}{2(\varDelta \delta )^{5/2}} \Big \} GA_1(G)&= \max \Big \{ \frac{2}{\delta ^2}, \, \frac{(2\delta )(2\delta ^2)}{2 \delta ^{5}} \Big \} m \\&= \frac{2m}{\delta ^{2}} = { ID}(G). \end{aligned} \end{aligned}$$

$\square$

4 Conclusions

Although only about 1000 benzenoid hydrocarbons are known, the number of possible benzenoid hydrocarbons is huge. For instance, the number of possible benzenoid hydrocarbons with 35 benzene rings is $5.85\times 10^{21}$ [30]. Therefore, the modeling of their physico-chemical properties is very important in order to predict properties of currently unknown species.

Topological indices are employed in the process of correlating the chemical structures with various characteristics such as boiling points, molar heats of formation...Therefore, they are used in theoretical chemistry for the design of chemical compounds with given physical–chemical properties or given pharmacological and biological activities. In particular, the variable molecular descriptors are important since the variables are determined during the regression so that the standard error of estimate for a studied property is as small as possible.

Hence, an important problem in this field is to find the extremal values of a given index among certain categories of graphs.

In this paper we obtain several inequalities involving the inverse degree, the first and second variable Zagreb, the general sum-connectivity and the geometric–arithmetic indices. All these indices have proved to be useful in QSAR and QSPR studies.

In particular, in Theorem 7 we relate the inverse degree index with the second variable Zagreb index $M_2^{-2}$ (recall that $M_2^{-2}$ performs better than the Randić index for the case of standard heats of formation and normal boiling points of octane isomers). Moreover, all extreme graphs are also determined.

Also, Theorem 1 provides the minimal graphs for the inverse degree index when n and m are fixed, and the maximal graphs when n, m, $\varDelta$ and $\delta$ are fixed; Theorem 3 provides the minimal graphs for $M_1^\alpha$$(\alpha > 1)$ when n and m are fixed, and the maximal graphs when n, m, $\varDelta$ and $\delta$ are fixed; and a similar result if $0< \alpha < 1$.

Note that the results obtained in this paper can be used for every graph since they do not have hypotheses on the graphs.