Inequalities on the Generalized ABC Index

Abstract

In this work, we obtained new results relating the generalized atom-bond connectivity index with the general Randić index. Some of these inequalities for $AB{C}_{\alpha }$ improved, when $\alpha =1/2$, known results on the $ABC$ index. Moreover, in order to obtain our results, we proved a kind of converse Hölder inequality, which is interesting on its own.

1. Introduction

Mathematical inequalities are at the basis of the processes of approximation, estimation, dimensioning, interpolation, monotonicity, extremes, etc. In general, inequalities appear in models for the study or approach to a certain reality (either objective or subjective). This reason makes it clear that when working with mathematical inequalities, we can essentially find relationships and approximate values of the magnitudes and variables that are associated with one or another practical problem.

In mathematical chemistry, a topological descriptor is a function that associates each molecular graph with a real value; if it correlates well with some chemical property, it is called a topological index. For additional information see [1], for application examples see [2,3,4,5,6,7].

The atom-bond connectivity index of a graph G was defined in [8] as:

$$ABC\left(G\right)=\sum _{uv\in E\left(G\right)}\sqrt{\frac{2(d_u+d_v-2)}{d_u{d}_v}}=\sqrt{2}\sum _{uv\in E\left(G\right)}\sqrt{\frac{d_u+d_v-2}{d_u{d}_v}},$$

where $uv$ denotes the edge of the graph G connecting the vertices u and v and $d_u$ is the degree of the vertex u.

The generalized atom-bond connectivity index was defined in [9] as:

$$AB{C}_{\alpha }\left(G\right)=\sum _{uv\in E\left(G\right)}{\left(\frac{d_u+d_v-2}{d_u{d}_v}\right)}^{\alpha }.$$

for any $\alpha \in \mathbb{R}\setminus \left\{0\right\}$. Note that $AB{C}_{1/2}=\frac{\sqrt{2}}{2}ABC$ and $AB{C}_{-3}$ is the augmented Zagreb index.

There are many papers that have studied the $ABC$ and $AB{C}_{\alpha }$ indices (see, e.g., [9,10,11,12,13,14,15]). In this paper, we obtained new inequalities relating these indices with the general Randić index. Some of these inequalities for $AB{C}_{\alpha }$ improved, when $\alpha =1/2$, known results on the $ABC$ index. In order to obtain our results, we proved a kind of converse Hölder inequality, Theorem 3, which is interesting on its own.

Throughout this work, a path graph $P_n$ is a tree with n vertices and maximum degree two and a star graph $S_n$ is a tree with n vertices and maximum degree $n-1$.

2. Inequalities Involving ${\mathit{ABC}}_{\mathit{\alpha }}$

In 1998, Bollobás and Erdos [16] generalized the Randić index for $\alpha \in \mathbb{R}\setminus \left\{0\right\}$,

$$R_{\beta }\left(G\right)=\sum _{uv\in E\left(G\right)}{\left(d_u{d}_v\right)}^{\beta }.$$

The general Randić index, also called the variable Zagreb index in 2004 by Miličević and Nikolić [17], was extensively studied in [18,19,20].

The next result relates the $AB{C}_{\alpha }$ and $R_{\beta }$ indices.

Theorem 1.

Let G be a graph with maximum degree Δ and minimum degree δ and $\alpha >0,\beta \in \mathbb{R}\setminus \left\{0\right\}$. Denote by $m_2$ the cardinality of the set of isolated edges in G.

(1) If $\beta /\alpha \le -1$ and $\delta >1$, then:

$${(2\delta -2)}^{\alpha }{\delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right)\le AB{C}_{\alpha }\left(G\right)\le {(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right).$$

The equality in each bound is attained if and only if G is a regular graph.

(2) If $\beta /\alpha \le -1$ and $\delta =1$, then:

$$2^{-\alpha -\beta }\left(R_{\beta }\left(G\right)-m_2\right)\le AB{C}_{\alpha }\left(G\right)\le {(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }\left(R_{\beta }\left(G\right)-m_2\right).$$

The equality in the lower bound is attained if and only if G is a union of path graphs $P_3$ and $m_2$ isolated edges. The equality in the upper bound is attained if and only if G is a union of a regular graph and $m_2$ isolated edges.

(3) If $-1<\beta /\alpha \le -1/2$ and $\delta >1$, then:

$${(2\delta -2)}^{\alpha }{\delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right)\le AB{C}_{\alpha }\left(G\right).$$

The equality in the bound is attained if and only if G is a regular graph.

(4) If $-1<\beta /\alpha \le -1/2$ and $\delta =1$, then:

$$2^{-\alpha -\beta }\left(R_{\beta }\left(G\right)-m_2\right)\le AB{C}_{\alpha }\left(G\right).$$

The equality in the bound is attained if and only if G is a union of path graphs $P_3$ and $m_2$ isolated edges.

(5) If $\beta >0$ and $\delta >1$, then:

$${(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right)\le AB{C}_{\alpha }\left(G\right)\le {(2\delta -2)}^{\alpha }{\delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right).$$

The equality in each bound is attained if and only if G is a regular graph.

(6) If $\beta >0$, $\delta =1$ and $1+\alpha /\beta \ge \Delta$, then:

$$\begin{array}{cc}\hfill {(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }\left(R_{\beta }\left(G\right)-m_2\right)& \le AB{C}_{\alpha }\left(G\right)\le {(\Delta -1)}^{\alpha }{\Delta }^{-\alpha -\beta }\left(R_{\beta }\left(G\right)-m_2\right).\hfill \end{array}$$

The equality in the lower bound is attained if and only if G is a union of a regular graph and $m_2$ isolated edges. The equality in the upper bound is attained if and only if G is a union of star graphs $S_{\Delta +1}$ and $m_2$ isolated edges.

(7) If $\beta >0$, $\delta =1$ and $1+\alpha /\beta \le 2$, then:

$$\begin{array}{cc}\hfill {(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }\left(R_{\beta }\left(G\right)-m_2\right)& \le AB{C}_{\alpha }\left(G\right)\le 2^{-\alpha -\beta }\left(R_{\beta }\left(G\right)-m_2\right).\hfill \end{array}$$

The equality in the lower bound is attained if and only if G is a union of a regular graph and $m_2$ isolated edges. The equality in the upper bound is attained if and only if G is a union of path graphs $P_3$ and $m_2$ isolated edges.

(8) If $\beta >0$, $\delta =1$ and $2<1+\alpha /\beta <\Delta$, then:

$$\begin{array}{cc}\hfill {(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }\left(R_{\beta }\left(G\right)-m_2\right)& \le AB{C}_{\alpha }\left(G\right)\le \frac{{\alpha }^{\alpha }{\beta }^{\beta }}{{(\alpha +\beta )}^{\alpha +\beta }}\left(R_{\beta }\left(G\right)-m_2\right).\hfill \end{array}$$

The equality in the lower bound is attained if and only if G is a union of a regular graph and $m_2$ isolated edges. The equality in the upper bound is attained if and only if $\alpha /\beta \in {\mathbb{Z}}^{+}$ and G is a union of star graphs $S_{\alpha /\beta +2}$ and $m_2$ isolated edges.

Proof. 

First of all, note that $AB{C}_{\alpha }\left(P_2\right)=0$ and $R_{\beta }\left(P_2\right)=1$. Therefore, it suffices to prove the theorem for the case $m_2=0$, i.e., when G is a graph without isolated edges. Hence, $\Delta \ge 2$.

We computed the extremal values (for fixed $\lambda \in \mathbb{R}$) of the function $f:[\delta ,\Delta ]\times \left(\right[\delta ,\Delta ]\setminus [1,2\left)\right)⟶\mathbb{R}$ given by:

$$f(x,y)=(x+y-2){\left(xy\right)}^{-\lambda -1}.$$

$\left(1\right)$ and $\left(2\right)$. If $\lambda \le -1$, then $-\lambda -1\ge 0$ and f is a strictly increasing function in each variable, and so,

$$(2\delta -2){\delta }^{-2\lambda -2}\le f(x,y)\le (2\Delta -2){\Delta }^{-2\lambda -2}.$$

The equality in the lower (respectively, upper) bound is attained if and only if $(x,y)=(\delta ,\delta )$ (respectively, $(x,y)=(\Delta ,\Delta )$).

If $\delta =1$, then $f(x,y)\ge f(1,2)=2^{-\lambda -1}$, since $x\in [1,\Delta ]$ and $y\in [2,\Delta ]$, and the equality in this inequality is attained if and only if $(x,y)=(1,2)$.

If $\lambda =\beta /\alpha$, then:

$${(2\delta -2)}^{\alpha }{\delta }^{-2\beta -2\alpha }{\left(d_u{d}_v\right)}^{\beta }\le {\left(\frac{d_u+d_v-2}{d_u{d}_v}\right)}^{\alpha }\le {(2\Delta -2)}^{\alpha }{\Delta }^{-2\beta -2\alpha }{\left(d_u{d}_v\right)}^{\beta }$$

for every $uv\in E\left(G\right)$ and, consequently,

$${(2\delta -2)}^{\alpha }{\delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right)\le AB{C}_{\alpha }\left(G\right)\le {(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right).$$

The previous argument shows that the equality in the upper bound is attained if and only if $d_u=d_v=\Delta$ for every $uv\in E\left(G\right)$, i.e., G is regular. If $\delta >1$, then the equality in the lower bound is attained if and only if $d_u=d_v=\delta$ for every $uv\in E\left(G\right)$, i.e., G is regular.

If $\lambda =\beta /\alpha$ and $\delta =1$, then:

$$2^{-\beta -\alpha }{\left(d_u{d}_v\right)}^{\beta }\le {\left(\frac{d_u+d_v-2}{d_u{d}_v}\right)}^{\alpha }$$

for every $uv\in E\left(G\right)$ and, consequently,

$$2^{-\alpha -\beta }{R}_{\beta }\left(G\right)\le AB{C}_{\alpha }\left(G\right).$$

The equality in this bound is attained if and only if $\{d_u,d_v\}=\{1,2\}$ for every $uv\in E\left(G\right)$, i.e., G is a union of path graphs $P_3$.

$\left(3\right)$ and $\left(4\right)$. In what follows, by symmetry, we can assume that $x\le y$. We have:

$$\begin{array}{cc}\hfill \frac{\partial f}{\partial y}(x,y)& =x^{-\lambda -1}\left(y^{-\lambda -1}+(x+y-2)(-\lambda -1)y^{-\lambda -2}\right)\hfill \\ & =x^{-\lambda -1}{y}^{-\lambda -2}\left(y+(x+y-2)(-\lambda -1)\right).\hfill \end{array}$$

If $-1<\lambda \le -1/2$, then $-\lambda -1\ge -1/2$, and so,

$$\begin{array}{cc}\hfill \frac{\partial f}{\partial y}(x,y)& \ge x^{-\lambda -1}{y}^{-\lambda -2}\left(y-\frac{x+y-2}{2}\right)\hfill \\ & =x^{-\lambda -1}{y}^{-\lambda -2}\frac{y-x+2}{2}\ge x^{-\lambda -1}{y}^{-\lambda -2}>0.\hfill \end{array}$$

Hence,

$$f(x,y)\ge f(x,x)=(2x-2)x^{-2\lambda -2}=g\left(x\right).$$

We have:

$$\begin{array}{cc}\hfill g^{\prime }\left(x\right)& =2x^{-2\lambda -2}+(2x-2)(-2\lambda -2)x^{-2\lambda -3}\hfill \\ & =2x^{-2\lambda -3}\left(x+(x-1)(-2\lambda -2)\right)\hfill \\ & =2x^{-2\lambda -3}\left((-2\lambda -1)x+2\lambda +2\right).\hfill \end{array}$$

Since $2\lambda +2>0$ and $-2\lambda -1\ge 0$, we have:

$$\begin{array}{cc}\hfill g^{\prime }\left(x\right)& =2x^{-2\lambda -3}\left((-2\lambda -1)x+2\lambda +2\right)\hfill \\ & \ge 2x^{-2\lambda -3}(2\lambda +2)>0.\hfill \end{array}$$

Thus, $g\left(x\right)\ge g\left(\delta \right)$ and:

$$\begin{array}{cc}\hfill f(x,y)& \ge g\left(x\right)\ge (2\delta -2){\delta }^{-2\lambda -2},\hfill \end{array}$$

if $\delta \ge 2$.

If $\lambda =\beta /\alpha$ and $\delta >1$, then:

$${(2\delta -2)}^{\alpha }{\delta }^{-2\beta -2\alpha }{\left(d_u{d}_v\right)}^{\beta }\le {\left(\frac{d_u+d_v-2}{d_u{d}_v}\right)}^{\alpha }$$

for every $uv\in E\left(G\right)$ and, consequently,

$${(2\delta -2)}^{\alpha }{\delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right)\le AB{C}_{\alpha }\left(G\right).$$

The previous argument shows that the equality in this bound is attained if and only if $d_u=d_v=\delta$ for every $uv\in E\left(G\right)$, i.e., G is regular.

Assume that $\delta =1$. We proved that $f(x,y)\ge g\left(x\right)\ge g\left(2\right)=2^{-2\lambda -1}$ for every $x,y\in [2,\Delta ]$. Since $\partial f/\partial y(1,y)>0$ for every $y\in [2,\Delta ]$, we have $f(1,y)\ge f(1,2)=2^{-\lambda -1}$ for every $y\in [2,\Delta ]$. Since $\lambda <0$, we have $2^{-2\lambda -1}>2^{-\lambda -1}$ and $f(x,y)\ge 2^{-\lambda -1}$ for every $x\in [1,\Delta ]\cap \mathbb{Z},y\in [2,\Delta ]\cap \mathbb{Z}$. Furthermore, the equality in this bound is attained if and only if $(x,y)=(1,2)$.

If $\lambda =\beta /\alpha$, then:

$$2^{-\beta -\alpha }{\left(d_u{d}_v\right)}^{\beta }\le {\left(\frac{d_u+d_v-2}{d_u{d}_v}\right)}^{\alpha }$$

for every $uv\in E\left(G\right)$ and, consequently,

$$2^{-\alpha -\beta }{R}_{\beta }\left(G\right)\le AB{C}_{\alpha }\left(G\right).$$

The equality in this bound is attained if and only if $\{d_u,d_v\}=\{1,2\}$ for every $uv\in E\left(G\right)$, i.e., G is a union of path graphs $P_3$.

$\left(5\right)$. Assume now that $\lambda >0$. Thus, $-\lambda -1<-1$ and:

$$\begin{array}{cc}\hfill \frac{\partial f}{\partial y}(x,y)& =x^{-\lambda -1}{y}^{-\lambda -2}\left(y+(x+y-2)(-\lambda -1)\right)\hfill \\ & <x^{-\lambda -1}{y}^{-\lambda -2}(2-x),\hfill \end{array}$$

and:

$$\begin{array}{cc}\hfill \frac{\partial f}{\partial x}(x,y)& <y^{-\lambda -1}{x}^{-\lambda -2}(2-y).\hfill \end{array}$$

If $\delta >1$, then f is a strictly decreasing function in each variable, and so,

$$(2\Delta -2){\Delta }^{-2\lambda -2}\le f(x,y)\le (2\delta -2){\delta }^{-2\lambda -2}.$$

(1)

The equality in the lower (respectively, upper) bound is attained if and only if $(x,y)=(\Delta ,\Delta )$ (respectively, $(x,y)=(\delta ,\delta )$).

If $\beta >0$ and $\lambda =\beta /\alpha$, then:

$${(2\Delta -2)}^{\alpha }{\Delta }^{-2\beta -2\alpha }{\left(d_u{d}_v\right)}^{\beta }\le {\left(\frac{d_u+d_v-2}{d_u{d}_v}\right)}^{\alpha }\le {(2\delta -2)}^{\alpha }{\delta }^{-2\beta -2\alpha }{\left(d_u{d}_v\right)}^{\beta }$$

for every $uv\in E\left(G\right)$ and, consequently,

$${(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right)\le AB{C}_{\alpha }\left(G\right)\le {(2\delta -2)}^{\alpha }{\delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right).$$

The equality in the lower bound is attained if and only if $d_u=d_v=\Delta$ for every $uv\in E\left(G\right)$, i.e., G is regular. Furthermore, the equality in the upper bound is attained if and only if $d_u=d_v=\delta$ for every $uv\in E\left(G\right)$, i.e., G is regular.

$\left(6\right)$. Note that:

$${\left(\frac{{\Delta }^2}{2}\right)}^{\lambda +1}>\frac{{\Delta }^2}{2}\ge 2\Delta -2\Rightarrow 2^{-\lambda -1}>(2\Delta -2){\Delta }^{-2\lambda -2}.$$

(2)

We also have:

$${\Delta }^{\lambda +1}>\Delta \ge 2\Rightarrow (\Delta -1){\Delta }^{-\lambda -1}>(2\Delta -2){\Delta }^{-2\lambda -2}.$$

(3)

Assume that $\delta =1$. If $2\le x,y\le \Delta$, then $f(x,y)\le f(2,2)=2^{-2\lambda -1}$. This inequality and the lower bound in (1) give:

$$(2\Delta -2){\Delta }^{-2\lambda -2}\le f(x,y)\le 2^{-2\lambda -1},$$

(4)

for every $2\le x,y\le \Delta$.

Let us consider the function $h\left(y\right)=f(1,y)=(y-1)y^{-\lambda -1}$ with $2\le y\le \Delta$. We have:

$$h^{\prime }\left(y\right)=-\lambda y^{-\lambda -1}+(\lambda +1)y^{-\lambda -2}=y^{-\lambda -2}(-\lambda y+\lambda +1),$$

and so, h strictly increases on $(0,1+1/\lambda )$ and strictly decreases on $(1+1/\lambda ,\infty )$.

If $1+1/\lambda \ge \Delta$, then h strictly increases on $(0,\Delta ]$ and:

$$2^{-\lambda -1}=h\left(2\right)\le h\left(y\right)\le h(\Delta )=(\Delta -1){\Delta }^{-\lambda -1},$$

for every $2\le y\le \Delta$. These inequalities and Equation (4) give:

$$min\left\{2^{-\lambda -1},(2\Delta -2){\Delta }^{-2\lambda -2}\right\}\le f(x,y)\le max\left\{(\Delta -1){\Delta }^{-\lambda -1},2^{-2\lambda -1}\right\}.$$

for every $x\in [1,\Delta ]\cap \mathbb{Z},y\in [2,\Delta ]\cap \mathbb{Z}$. Since we have in this case $2^{-\lambda -1}=h\left(2\right)\le h(\Delta )=(\Delta -1){\Delta }^{-\lambda -1}$, we conclude:

$$\begin{array}{cc}\hfill (\Delta -1){\Delta }^{-\lambda -1}& \le max\left\{(\Delta -1){\Delta }^{-\lambda -1},2^{-2\lambda -1}\right\}\hfill \\ & \le max\left\{(\Delta -1){\Delta }^{-\lambda -1},2^{-\lambda -1}\right\}=(\Delta -1){\Delta }^{-\lambda -1}.\hfill \end{array}$$

Equation (2) gives:

$$min\left\{2^{-\lambda -1},(2\Delta -2){\Delta }^{-2\lambda -2}\right\}=(2\Delta -2){\Delta }^{-2\lambda -2}.$$

Hence,

$$(2\Delta -2){\Delta }^{-2\lambda -2}\le f(x,y)\le (\Delta -1){\Delta }^{-\lambda -1},$$

for every $x\in [1,\Delta ]\cap \mathbb{Z},y\in [2,\Delta ]\cap \mathbb{Z}$. The equality in the lower (respectively, upper) bound is attained if and only if $(x,y)=(\Delta ,\Delta )$ (respectively, $(x,y)=(1,\Delta )$).

If $\beta >0$ and $\lambda =\beta /\alpha$, then we obtain:

$$\begin{array}{cc}\hfill {(2\Delta -2)}^{\alpha }{\Delta }^{-2\beta -2\alpha }{\left(d_u{d}_v\right)}^{\beta }& \le {\left(\frac{d_u+d_v-2}{d_u{d}_v}\right)}^{\alpha }\le {(\Delta -1)}^{\alpha }{\Delta }^{-\beta -\alpha }{\left(d_u{d}_v\right)}^{\beta },\hfill \\ \hfill {(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right)& \le AB{C}_{\alpha }\left(G\right)\le {(\Delta -1)}^{\alpha }{\Delta }^{-\alpha -\beta }{R}_{\beta }\left(G\right).\hfill \end{array}$$

The equality in the lower bound is attained if and only if $d_u=d_v=\Delta$ for every $uv\in E\left(G\right)$, i.e., G is regular. The equality in the upper bound is attained if and only if $\{d_u,d_v\}=\{1,\Delta \}$ for every $uv\in E\left(G\right)$, i.e., G is a union of star graphs $S_{\Delta +1}$.

$\left(7\right)$. If $1+1/\lambda \le 2$, then h strictly decreases on $[2,\Delta ]$ and:

$$(\Delta -1){\Delta }^{-\lambda -1}=h(\Delta )\le h\left(y\right)\le h\left(2\right)=2^{-\lambda -1},$$

for every $2\le y\le \Delta$. These inequalities and Equation (4) give:

$$min\left\{(\Delta -1){\Delta }^{-\lambda -1},(2\Delta -2){\Delta }^{-2\lambda -2}\right\}\le f(x,y)\le max\left\{2^{-\lambda -1},2^{-2\lambda -1}\right\},$$

for every $x\in [1,\Delta ]\cap \mathbb{Z},y\in [2,\Delta ]\cap \mathbb{Z}$. Equation (3) gives:

$$(2\Delta -2){\Delta }^{-2\lambda -2}\le f(x,y)\le 2^{-\lambda -1},$$

for every $x\in [1,\Delta ]\cap \mathbb{Z},y\in [2,\Delta ]\cap \mathbb{Z}$. The equality in the lower (respectively, upper) bound is attained if and only if $(x,y)=(\Delta ,\Delta )$ (respectively, $(x,y)=(1,2)$).

If $\beta >0$ and $\lambda =\beta /\alpha$, then we obtain for every $uv\in E\left(G\right)$:

$$\begin{array}{cc}\hfill {(2\Delta -2)}^{\alpha }{\Delta }^{-2\beta -2\alpha }{\left(d_u{d}_v\right)}^{\beta }& \le {\left(\frac{d_u+d_v-2}{d_u{d}_v}\right)}^{\alpha }\le 2^{-\beta -\alpha }{\left(d_u{d}_v\right)}^{\beta },\hfill \\ \hfill {(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right)& \le AB{C}_{\alpha }\left(G\right)\le 2^{-\alpha -\beta }{R}_{\beta }\left(G\right).\hfill \end{array}$$

The equality in the lower bound is attained if and only if $d_u=d_v=\Delta$ for every $uv\in E\left(G\right)$, i.e., G is regular. The equality in the upper bound is attained if and only if $\{d_u,d_v\}=\{1,2\}$ for every $uv\in E\left(G\right)$, i.e., G is a union of path graphs $P_3$.

$\left(8\right)$. If $2<1+1/\lambda <\Delta$, then:

$$h\left(y\right)\ge min\left\{h\left(2\right),h(\Delta )\right\}=min\left\{2^{-\lambda -1},(\Delta -1){\Delta }^{-\lambda -1}\right\},$$

for every $2\le y\le \Delta$. Furthermore,

$$h\left(y\right)\le h(1+1/\lambda )=\frac{1}{\lambda }{\left(\frac{\lambda +1}{\lambda }\right)}^{-\lambda -1}=\frac{{\lambda }^{\lambda }}{{(\lambda +1)}^{\lambda +1}},$$

for every $2\le y\le \Delta$. These facts and (4) give:

$$\begin{array}{cc}\hfill min\left\{2^{-\lambda -1},(\Delta -1){\Delta }^{-\lambda -1},(2\Delta -2){\Delta }^{-2\lambda -2}\right\}& \le f(x,y)\hfill \\ & \le max\left\{\frac{{\lambda }^{\lambda }}{{(\lambda +1)}^{\lambda +1}},2^{-2\lambda -1}\right\}\hfill \end{array}$$

for every $x\in [1,\Delta ]\cap \mathbb{Z},y\in [2,\Delta ]\cap \mathbb{Z}$.

Equations (2) and (3) give:

$$min\left\{2^{-\lambda -1},(\Delta -1){\Delta }^{-\lambda -1},(2\Delta -2){\Delta }^{-2\lambda -2}\right\}=(2\Delta -2){\Delta }^{-2\lambda -2}.$$

Since $h\left(2\right)\le h(1+1/\lambda )$, we obtain:

$$\begin{array}{cc}\hfill 2^{-2\lambda -1}& <2^{-\lambda -1}\le \frac{{\lambda }^{\lambda }}{{(\lambda +1)}^{\lambda +1}},\hfill \end{array}$$

and so,

$$\begin{array}{cc}\hfill (2\Delta -2){\Delta }^{-2\lambda -2}& \le f(x,y)\le \frac{{\lambda }^{\lambda }}{{(\lambda +1)}^{\lambda +1}}\hfill \end{array}$$

for every $x\in [1,\Delta ]\cap \mathbb{Z},y\in [2,\Delta ]\cap \mathbb{Z}$. The equality in the lower (respectively, upper) bound is attained if and only if $(x,y)=(\Delta ,\Delta )$ (respectively, $(x,y)=(1,1+1/\lambda )$).

If $\beta >0$ and $\lambda =\beta /\alpha$, then we obtain:

$${\left(\frac{{\lambda }^{\lambda }}{{(\lambda +1)}^{\lambda +1}}\right)}^{\alpha }=\frac{{(\beta /\alpha )}^{\beta }}{{(\beta /\alpha +1)}^{\beta +\alpha }}=\frac{{\alpha }^{\alpha }{\beta }^{\beta }}{{(\alpha +\beta )}^{\alpha +\beta }},$$

and we have for every $uv\in E\left(G\right)$:

$$\begin{array}{cc}\hfill {(2\Delta -2)}^{\alpha }{\Delta }^{-2\beta -2\alpha }{\left(d_u{d}_v\right)}^{\beta }& \le {\left(\frac{d_u+d_v-2}{d_u{d}_v}\right)}^{\alpha }\le \frac{{\alpha }^{\alpha }{\beta }^{\beta }}{{(\alpha +\beta )}^{\alpha +\beta }}{\left(d_u{d}_v\right)}^{\beta },\hfill \\ \hfill {(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right)& \le AB{C}_{\alpha }\left(G\right)\le \frac{{\alpha }^{\alpha }{\beta }^{\beta }}{{(\alpha +\beta )}^{\alpha +\beta }}{R}_{\beta }\left(G\right).\hfill \end{array}$$

The equality in the lower bound is attained if and only if $d_u=d_v=\Delta$ for every $uv\in E\left(G\right)$, i.e., G is regular. The equality in the upper bound is attained if and only if $\alpha /\beta \in {\mathbb{Z}}^{+}$ and $\{d_u,d_v\}=\{1,1+\alpha /\beta \}$ for every $uv\in E\left(G\right)$, i.e., G is a union of star graphs $S_{\alpha /\beta +2}$. □

Note that $AB{C}_{\alpha }\left(G\right)$ is not well defined if $\alpha <0$ and G has an isolated edge. The argument in the proof of Theorem 1 gives directly the following result for $\alpha <0$.

Theorem 2.

Let G be a graph without isolated edges, with maximum degree Δ and minimum degree δ, and $\alpha <0,\beta \in \mathbb{R}\setminus \left\{0\right\}$.

(1) If $\beta /\alpha \le -1$ and $\delta >1$, then:

$${(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right)\le AB{C}_{\alpha }\left(G\right)\le {(2\delta -2)}^{\alpha }{\delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right).$$

The equality in each bound is attained if and only if G is a regular graph.

(2) If $\beta /\alpha \le -1$ and $\delta =1$, then:

$${(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right)\le AB{C}_{\alpha }\left(G\right)\le 2^{-\alpha -\beta }{R}_{\beta }\left(G\right).$$

The equality in the lower bound is attained if and only if G is a regular graph. The equality in the upper bound is attained if and only if G is a union of path graphs $P_3$.

(3) If $-1<\beta /\alpha \le -1/2$ and $\delta >1$, then:

$$AB{C}_{\alpha }\left(G\right)\le {(2\delta -2)}^{\alpha }{\delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right).$$

The equality in the bound is attained if and only if G is a regular graph.

(4) If $-1<\beta /\alpha \le -1/2$ and $\delta =1$, then:

$$AB{C}_{\alpha }\left(G\right)\le 2^{-\alpha -\beta }{R}_{\beta }\left(G\right).$$

The equality in the bound is attained if and only if G is a union of path graphs $P_3$.

(5) If $\beta <0$ and $\delta >1$, then:

$${(2\delta -2)}^{\alpha }{\delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right)\le AB{C}_{\alpha }\left(G\right)\le {(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right).$$

The equality in each bound is attained if and only if G is a regular graph.

(6) If $\beta <0$, $\delta =1$ and $1+\alpha /\beta \ge \Delta$, then:

$$\begin{array}{cc}\hfill {(\Delta -1)}^{\alpha }{\Delta }^{-\alpha -\beta }{R}_{\beta }\left(G\right)& \le AB{C}_{\alpha }\left(G\right)\le {(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right).\hfill \end{array}$$

The equality in the lower bound is attained if and only if G is a union of star graphs $S_{\Delta +1}$. The equality in the upper bound is attained if and only if G is a regular graph.

(7) If $\beta <0$, $\delta =1$ and $1+\alpha /\beta \le 2$, then:

$$\begin{array}{cc}\hfill 2^{-\alpha -\beta }{R}_{\beta }\left(G\right)& \le AB{C}_{\alpha }\left(G\right)\le {(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right).\hfill \end{array}$$

The equality in the lower bound is attained if and only if G is a union of path graphs $P_3$. The equality in the upper bound is attained if and only if G is a regular graph.

(8) If $\beta <0$, $\delta =1$ and $2<1+\alpha /\beta <\Delta$, then:

$$\begin{array}{cc}\hfill \frac{{\left|\alpha \right|}^{\alpha }{\left|\beta \right|}^{\beta }}{{|\alpha +\beta |}^{\alpha +\beta }}{R}_{\beta }\left(G\right)& \le AB{C}_{\alpha }\left(G\right)\le {(2\Delta -2)}^{\alpha }{\Delta }^{-2\alpha -2\beta }{R}_{\beta }\left(G\right).\hfill \end{array}$$

The equality in the lower bound is attained if and only if $\alpha /\beta \in {\mathbb{Z}}^{+}$ and G is a union of star graphs $S_{\alpha /\beta +2}$. The equality in the upper bound is attained if and only if G is a regular graph.

Note that Theorems 1 and 2 generalize the classical inequalities:

$$2\sqrt{\delta -1}R\left(G\right)\le ABC\left(G\right)\le 2\sqrt{\Delta -1}R\left(G\right).$$

(5)

Theorem 1 has the following consequence.

Corollary 1.

Let G be a graph with minimum degree δ and $m_2$ isolated edges.

(1) If $\delta >1$, then:

$$2\sqrt{1-\frac{1}{\delta }}{R}_{-1/4}\left(G\right)\le ABC\left(G\right).$$

The equality in the bound is attained if and only if G is a regular graph.

(2) If $\delta =1$, then

$$2^{1/4}\left(R_{-1/4}\left(G\right)-m_2\right)\le ABC\left(G\right).$$

The equality in the bound is attained if and only if G is a union of path graphs $P_3$ and $m_2$ isolated edges.

Corollary 1 improves the inequality:

$$2\left(1-\frac{1}{\sqrt{\delta }}\right)R_{-1/4}\left(G\right)\le ABC\left(G\right)$$

in ([21], Theorem 2.5).

In [22], Lemma 4, the following result appeared.

Lemma 1.

Let $(X,\mu )$ be a measure space and $f,g:X\to \mathbb{R}$ measurable functions. If there exist positive constants $\omega ,\mathrm{\Omega }$ with $\omega \left|g\right|\le \left|f\right|\le \mathrm{\Omega }\left|g\right|$μ-a.e., then:

$${\parallel f\parallel }_2{\parallel g\parallel }_2\le \frac{1}{2}\left(\sqrt{\frac{\mathrm{\Omega }}{\omega }}+\sqrt{\frac{\omega }{\mathrm{\Omega }}}\right){\parallel fg\parallel }_1.$$

(6)

If these norms are finite, the equality in the bound is attained if and only if $\omega =\mathrm{\Omega }$ and $\left|f\right|=\omega \left|g\right|$ μ-a.e. or $f=g=0$ μ-a.e.

We need the following converse Hölder inequality, which is interesting on its own. This result generalizes Lemma 1 and improves the inequality in [23] (Theorem 2).

Theorem 3.

Let $(X,\mu )$ be a measure space, $f,g:X\to \mathbb{R}$ measurable functions, and $1<p,q<\infty$ with $1/p+1/q=1$. If there exist positive constants $a,b$ with ${a\left|g\right|}^q\le {\left|f\right|}^p\le b{\left|g\right|}^q$ μ-a.e., then:

$${\parallel f\parallel }_p{\parallel g\parallel }_q\le K_p(a,b){\parallel fg\parallel }_1,$$

(7)

with:

$$K_p(a,b)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{p}{\left(\frac{a}{b}\right)}^{1/\left(2q\right)}+\frac{1}{q}{\left(\frac{b}{a}\right)}^{1/\left(2p\right)},}\hfill & \mathrm{if}1<p<2,\hfill \\ \hfill & \hfill \\ {\displaystyle \frac{1}{p}{\left(\frac{b}{a}\right)}^{1/\left(2q\right)}+\frac{1}{q}{\left(\frac{a}{b}\right)}^{1/\left(2p\right)},}\hfill & \mathrm{if}p\ge 2.\hfill \end{array}\right.$$

If these norms are finite, the equality in the bound is attained if and only if $a=b$ and ${\left|f\right|}^p=a{\left|g\right|}^q$ μ-a.e. or $f=g=0$ μ-a.e.

Remark 1.

Since:

$$K_2(a,b)=\frac{1}{2}{\left(\frac{b}{a}\right)}^{1/4}+\frac{1}{2}{\left(\frac{a}{b}\right)}^{1/4},$$

Theorem 3 generalizes Lemma 1 (note that $a={\omega }^2$ and $b={\mathrm{\Omega }}^2$).

Proof. 

If $p=2$, then Lemma 1 (with $\omega =a^{1/2}$ and $\Omega =b^{1/2}$) gives the result. Assume now $p\ne 2$, and let us define:

$$k_p(a,b)=max\left\{\frac{1}{p}{\left(\frac{a}{b}\right)}^{1/\left(2q\right)}+\frac{1}{q}{\left(\frac{b}{a}\right)}^{1/\left(2p\right)},\frac{1}{p}{\left(\frac{b}{a}\right)}^{1/\left(2q\right)}+\frac{1}{q}{\left(\frac{a}{b}\right)}^{1/\left(2p\right)}\right\}.$$

We will check at the end of the proof that $k_p(a,b)=K_p(a,b)$.

Let us consider $t\in (0,1)$ and define:

$$G_t\left(x\right):=t{x}^{1-t}+(1-t)x^{-t}$$

for $x>0$. Since:

$$G_t^{\prime }\left(x\right)=t(1-t)x^{-t}-t(1-t)x^{-t-1}=t(1-t)x^{-t-1}(x-1),$$

$G_t$ is strictly decreasing on $(0,1)$ and strictly increasing on $(1,\infty )$. Thus, if $0<s\le S$ are two constants and we consider $s\le x\le S$, then:

$$G_t\left(x\right)\le max\{G_t\left(s\right),G_t\left(S\right)\}=:A,$$

and if $G_t\left(x\right)=A$ for some $s\le x\le S$, then $x=s$ or $x=S$.

Note that if $G_t\left(s\right)\ne G_t\left(S\right)$, the following facts hold: if $G_t\left(s\right)>G_t\left(S\right)$ and $G_t\left(x\right)=A=G_t\left(s\right)$, then $x=s$; if $G_t\left(s\right)<G_t\left(S\right)$ and $G_t\left(x\right)=A=G_t\left(S\right)$, then $x=S$.

If $x_1,x_2>0$ and $s{x}_2\le x_1\le S{x}_2$, then:

$$\begin{array}{cc}\hfill t{\left(\frac{x_1}{x_2}\right)}^{1-t}+(1-t){\left(\frac{x_2}{x_1}\right)}^t& \le A,\hfill \\ \hfill t{x}_1+(1-t)x_2& \le A{x}_1^t{x}_2^{1-t}.\hfill \end{array}$$

By continuity, this last inequality holds for every $x_1,x_2\ge 0$ with $s{x}_2\le x_1\le S{x}_2$. If the equality is attained for some $x_1,x_2\ge 0$ with $s{x}_2\le x_1\le S{x}_2$, then $x_1=s{x}_2$ or $x_1=S{x}_2$ (the cases $x_1=0$ and $x_2=0$ are direct).

Choose $t=1/p$ (thus, $1-t=1/q$), $x=x_1^t=x_1^{1/p}$ and $y=x_2^{1-t}=x_2^{1/q}$. Thus,

$$\frac{x^p}{p}+\frac{y^q}{q}\le Axy$$

(8)

for every $x,y\ge 0$ with $s{y}^q\le x^p\le S{y}^q$. If the equality is attained for some $x,y\ge 0$ with $s{y}^q\le x^p\le S{y}^q$, then $x^p=s{y}^q$ or $x^p=S{y}^q$.

If ${\parallel f\parallel }_p=0$ or ${\parallel g\parallel }_q=0$, then ${a\left|g\right|}^q\le {\left|f\right|}^p\le b{\left|g\right|}^q$ $\mu$-a.e. gives ${\parallel f\parallel }_p={\parallel g\parallel }_q=0$, and the equality in (7) holds. Assume now that ${\parallel f\parallel }_p\ne 0\ne {\parallel g\parallel }_q$.

Let us define the function:

$$h:={\left(ab\right)}^{1/\left(2q\right)}\left|g\right|.$$

We have:

$$\sqrt{\frac{a}{b}}{h}^q={a\left|g\right|}^q,\sqrt{\frac{b}{a}}{h}^q={b\left|g\right|}^q,\sqrt{\frac{a}{b}}{h}^q\le {\left|f\right|}^p\le \sqrt{\frac{b}{a}}{h}^q.$$

If $x=\left|f\right|$, $y=h$, $s={(a/b)}^{1/2}$, and $S={(b/a)}^{1/2}$, then $s{h}^q\le {\left|f\right|}^p\le S{h}^q$ and (8) gives:

$$\begin{array}{cc}\hfill \frac{1}{p}{\left|f\right|}^p+\frac{1}{q}{h}^q& \le A\left|f\right|h.\hfill \end{array}$$

If the equality in this inequality is attained at some point, then:

$${\left|f\right|}^p=\sqrt{\frac{a}{b}}{h}^q\mathrm{or}{\left|f\right|}^p=\sqrt{\frac{b}{a}}{h}^q$$

at that point.

Note that:

$$G_{1/p}\left(x\right)=\frac{1}{p}{x}^{1/q}+\frac{1}{q}{\left(\frac{1}{x}\right)}^{1/p}$$

and so,

$$\begin{array}{cc}\hfill A& =max\{G_t\left(s\right),G_t\left(S\right)\}=max\left\{G_{1/p}\left({(a/b)}^{1/2}\right),G_{1/p}\left({(b/a)}^{1/2}\right)\right\}=k_p(a,b).\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill \frac{1}{p}{\left|f\right|}^p+\frac{1}{q}{h}^q& \le k_p(a,b)\left|f\right|h,\hfill \\ \hfill \frac{1}{p}{\parallel f\parallel }_p^p+\frac{1}{q}{\parallel h\parallel }_q^q& \le k_p(a,b){\parallel fh\parallel }_1.\hfill \end{array}$$

Recall that these norms are well defined, although they can be infinite.

If these norms are finite and the equality in the last inequality is attained, then:

$${\left|f\right|}^p=\sqrt{\frac{a}{b}}{h}^q\mathrm{or}{\left|f\right|}^p=\sqrt{\frac{b}{a}}{h}^q$$

$\mu$-a.e. Young’s inequality states that:

$$xy\le \frac{x^p}{p}+\frac{y^q}{q}$$

for every $x,y\ge 0$, and the equality holds if and only if $x^p=y^q$. Thus,

$$\begin{array}{cc}\hfill {\parallel f\parallel }_p{\parallel h\parallel }_q& \le \frac{1}{p}{\parallel f\parallel }_p^p+\frac{1}{q}{\parallel h\parallel }_q^q\le k_p(a,b){\parallel fh\parallel }_1.\hfill \end{array}$$

Therefore, by homogeneity, we conclude:

$$\begin{array}{cc}\hfill {\parallel f\parallel }_p{\parallel g\parallel }_q& \le k_p(a,b){\parallel fg\parallel }_1.\hfill \end{array}$$

Let us prove now that $k_p(a,b)=K_p(a,b)$. Consider the function $H_t\left(x\right):=G_t\left(x\right)-G_t(1/x)$ for $t\in (0,1)$ and $x\in (0,1]$. We have:

$$\begin{array}{cc}\hfill H_t^{\prime }\left(x\right)& =G_t^{\prime }\left(x\right)+\frac{1}{x^2}{G}_t^{\prime }\left(\frac{1}{x}\right)\hfill \\ & =t(1-t)x^{-t-1}(x-1)+t(1-t)\frac{1}{x^2}{x}^{t+1}\left(\frac{1}{x}-1\right)\hfill \\ & =t(1-t)x^{-t-1}(x-1)+t(1-t)x^{t-2}(1-x)\hfill \\ & =t(1-t)(1-x)x^{-t-1}(x^{2t-1}-1).\hfill \end{array}$$

If $t\in (0,1/2)$, then $2t-1<0$ and $H_t^{\prime }\left(x\right)>0$ for every $x\in (0,1)$, and so, $H_t\left(x\right)<H_t\left(1\right)=0$ for every $x\in (0,1)$. Hence, $G_t\left(x\right)<G_t(1/x)$ for every $x\in (0,1)$. If $p>2$ and $a<b$, then $G_{1/p}\left({(a/b)}^{1/2}\right)<G_{1/p}\left({(b/a)}^{1/2}\right)$, and:

$$k_p(a,b)=\frac{1}{p}{\left(\frac{b}{a}\right)}^{1/\left(2q\right)}+\frac{1}{q}{\left(\frac{a}{b}\right)}^{1/\left(2p\right)}.$$

If $t\in (1/2,1)$, then $2t-1>0$ and $H_t^{\prime }\left(x\right)<0$ for every $x\in (0,1)$, and so, $H_t\left(x\right)>H_t\left(1\right)=0$ for every $x\in (0,1)$. Hence, $G_t\left(x\right)>G_t(1/x)$ for every $x\in (0,1)$. If $1<p<2$ and $a<b$, then $G_{1/p}\left({(a/b)}^{1/2}\right)>G_{1/p}\left({(b/a)}^{1/2}\right)$, and:

$$k_p(a,b)=\frac{1}{p}{\left(\frac{a}{b}\right)}^{1/\left(2q\right)}+\frac{1}{q}{\left(\frac{b}{a}\right)}^{1/\left(2p\right)}.$$

Therefore, $k_p(a,b)=K_p(a,b)$.

If $a=b$ and ${\left|f\right|}^p=a{\left|g\right|}^q$ $\mu$-a.e. or $f=g=0$ $\mu$-a.e., then a computation gives that the equality in (7) is attained.

Finally, assume that the equality in (7) is attained. Seeking for a contradiction, assume that $a\ne b$. The previous argument gives that:

$${\left|f\right|}^p=\sqrt{\frac{a}{b}}{h}^q\mathrm{or}{\left|f\right|}^p=\sqrt{\frac{b}{a}}{h}^q$$

$\mu$-a.e. Since we proved $G_{1/p}\left({(a/b)}^{1/2}\right)\ne G_{1/p}\left({(b/a)}^{1/2}\right)$ (recall that $p\ne 2$ and $a<b$), we can conclude that:

$${\left|f\right|}^p=\sqrt{\frac{a}{b}}{h}^q\mu -\mathrm{a}.\mathrm{e}.\mathrm{or}{\left|f\right|}^p=\sqrt{\frac{b}{a}}{h}^q\mu -\mathrm{a}.\mathrm{e}.$$

Hence,

$${\parallel f\parallel }_p^p=\sqrt{\frac{a}{b}}{\parallel h\parallel }_q^q{\mathrm{or}\parallel f\parallel }_p^p=\sqrt{\frac{b}{a}}{\parallel h\parallel }_q^q.$$

Since the equality in Young’s inequality gives ${\parallel f\parallel }_p^p={\parallel h\parallel }_q^q$, we obtain $a=b$, a contradiction. Therefore, $a=b$ and ${\left|f\right|}^p=h^q$ $\mu$-a.e. Hence, ${\left|f\right|}^p=a{\left|g\right|}^q$ $\mu$-a.e. □

Theorem 3 has the following consequence.

Corollary 2.

If $1<p,q<\infty$ with $1/p+1/q=1$, $x_j,y_j\ge 0$ and $a{y}_j^q\le x_j^p\le b{y}_j^q$ for $1\le j\le k$ and some positive constants $a,b,$ then:

$${\left(\sum _{j=1}^k{x}_j^p\right)}^{1/p}{\left(\sum _{j=1}^k{y}_j^q\right)}^{1/q}\le K_p(a,b)\sum _{j=1}^k{x}_j{y}_j,$$

where $K_p(a,b)$ is the constant in Theorem 3. If $x_j>0$ for some $1\le j\le k$, then the equality in the bound is attained if and only if $a=b$ and $x_j^p=a{y}_j^q$ for every $1\le j\le k$.

The Platt number is defined (see, e.g., [24]) as:

$$F\left(G\right)=\sum _{uv\in E\left(G\right)}\left(d_u+d_v-2\right).$$

Theorem 4.

Let G be a graph with $m_2$ isolated edges and $0<\alpha <1$.

(1) Then:

$$\begin{array}{c}\hfill AB{C}_{\alpha }\left(G\right)\le F{\left(G\right)}^{\alpha }{\left(R_{-\alpha /(1-\alpha )}\left(G\right)-m_2\right)}^{1-\alpha }.\end{array}$$

The equality in this bound is attained for the union of any regular or biregular graph and $m_2$ isolated edges; if G is the union of a connected graph and $m_2$ isolated edges, then the equality in this bound is attained if and only if G is the union of any regular or biregular connected graph and $m_2$ isolated edges.

(2) If $\delta >1$, then:

$$\begin{array}{c}\hfill AB{C}_{\alpha }\left(G\right)\ge \frac{{(\Delta -1)}^{\alpha /2}{\Delta }^{{\alpha }^2/(1-\alpha )}{(\delta -1)}^{(1-\alpha )/2}{\delta }^{\alpha }F{\left(G\right)}^{\alpha }{R}_{-\alpha /(1-\alpha )}{\left(G\right)}^{1-\alpha }}{\alpha {(\Delta -1)}^{1/2}{\Delta }^{\alpha /(1-\alpha )}+(1-\alpha ){(\delta -1)}^{1/2}{\delta }^{\alpha /(1-\alpha )}},\end{array}$$

if $\alpha \in (0,1/2]$, and:

$$\begin{array}{c}\hfill AB{C}_{\alpha }\left(G\right)\ge \frac{{(\delta -1)}^{\alpha /2}{\delta }^{{\alpha }^2/(1-\alpha )}{(\Delta -1)}^{(1-\alpha )/2}{\Delta }^{\alpha }F{\left(G\right)}^{\alpha }{R}_{-\alpha /(1-\alpha )}{\left(G\right)}^{1-\alpha }}{\alpha {(\delta -1)}^{1/2}{\delta }^{\alpha /(1-\alpha )}+(1-\alpha ){(\Delta -1)}^{1/2}{\Delta }^{\alpha /(1-\alpha )}},\end{array}$$

if $\alpha \in (1/2,1)$. The equality in these bounds is attained if and only if G is regular.

(3) If $\delta =1$, then:

$$\begin{array}{c}\hfill AB{C}_{\alpha }\left(G\right)\ge \frac{2^{\alpha }{(\Delta -1)}^{\alpha /2}{\Delta }^{{\alpha }^2/(1-\alpha )}F{\left(G\right)}^{\alpha }{\left(R_{-\alpha /(1-\alpha )}\left(G\right)-m_2\right)}^{1-\alpha }}{\alpha {(2\Delta -2)}^{1/2}{\Delta }^{\alpha /(1-\alpha )}+(1-\alpha )2^{\alpha /(2-2\alpha )}},\end{array}$$

if $\alpha \in (0,1/2]$, and:

$$\begin{array}{c}\hfill AB{C}_{\alpha }\left(G\right)\ge \frac{2^{{\alpha }^2/(2-2\alpha )}{\Delta }^{\alpha }{(2\Delta -2)}^{(1-\alpha )/2}F{\left(G\right)}^{\alpha }{\left(R_{-\alpha /(1-\alpha )}\left(G\right)-m_2\right)}^{1-\alpha }}{\alpha 2^{\alpha /(2-2\alpha )}+(1-\alpha ){(2\Delta -2)}^{1/2}{\Delta }^{\alpha /(1-\alpha )}},\end{array}$$

if $\alpha \in (1/2,1)$.

Proof. 

Since $AB{C}_{\alpha }\left(P_2\right)=0$ and $R_{\beta }\left(P_2\right)=1$, it suffices to prove the theorem for the case $m_2=0$, i.e., when G is a graph without isolated edges. Hence, $\Delta \ge 2$.

Hölder’s inequality gives:

$$\begin{array}{cc}\hfill AB{C}_{\alpha }\left(G\right)& =\sum _{uv\in E\left(G\right)}{\left(\frac{d_u+d_v-2}{d_u{d}_v}\right)}^{\alpha }\hfill \\ & \le {\left(\sum _{uv\in E\left(G\right)}{\left({(d_u+d_v-2)}^{\alpha }\right)}^{1/\alpha }\right)}^{\alpha }{\left(\sum _{uv\in E\left(G\right)}{\left(\frac{1}{{\left(d_u{d}_v\right)}^{\alpha }}\right)}^{1/(1-\alpha )}\right)}^{1-\alpha }\hfill \\ & ={\left(\sum _{uv\in E\left(G\right)}\left(d_u+d_v-2\right)\right)}^{\alpha }{\left(\sum _{uv\in E\left(G\right)}{\left(d_u{d}_v\right)}^{-\alpha /(1-\alpha )}\right)}^{1-\alpha }\hfill \\ & =F{\left(G\right)}^{\alpha }{R}_{-\alpha /(1-\alpha )}{\left(G\right)}^{1-\alpha }.\hfill \end{array}$$

If G is a regular or biregular graph with m edges, then:

$$\begin{array}{cc}\hfill F{\left(G\right)}^{\alpha }{R}_{-\alpha /(1-\alpha )}{\left(G\right)}^{1-\alpha }& ={\left((\Delta +\delta -2)m\right)}^{\alpha }{\left({(\Delta \delta )}^{-\alpha /(1-\alpha )}m\right)}^{1-\alpha }\hfill \\ & =\frac{{(\Delta +\delta -2)}^{\alpha }}{{(\Delta \delta )}^{\alpha }}m=AB{C}_{\alpha }\left(G\right).\hfill \end{array}$$

Assume that G is connected and that the equality in the first inequality is attained. Hölder’s inequality gives that there exists a constant c with:

$$d_u+d_v-2=c{\left(d_u{d}_v\right)}^{-\alpha /(1-\alpha )}$$

for every $uv\in E\left(G\right)$. Note that the function $H:[1,\infty )\times [1,\infty )\to [0,\infty )$ given by $H(x,y)=(x+y-2){\left(xy\right)}^{\alpha /(1-\alpha )}$ is increasing in each variable. If $uv,uw\in E\left(G\right)$, then:

$$c=(d_u+d_v-2){\left(d_u{d}_v\right)}^{\alpha /(1-\alpha )}=(d_u+d_w-2){\left(d_u{d}_w\right)}^{\alpha /(1-\alpha )}$$

implies $d_w=d_v$. Thus, for each vertex $u\in V\left(G\right)$, every neighbor of u has the same degree. Since G is a connected graph, this holds if and only if G is regular or biregular.

Assume now that $\delta >1$. If $\alpha \in (0,1/2]$, then:

$$\begin{array}{cc}& K_{1/\alpha }\left((2\delta -2){\delta }^{2\alpha /(1-\alpha )},(2\Delta -2){\Delta }^{2\alpha /(1-\alpha )}\right)\hfill \\ & =\alpha {\left(\frac{\Delta -1}{\delta -1}\right)}^{(1-\alpha )/2}{\left(\frac{\Delta }{\delta }\right)}^{\alpha }+(1-\alpha ){\left(\frac{\delta -1}{\Delta -1}\right)}^{\alpha /2}{\left(\frac{\delta }{\Delta }\right)}^{{\alpha }^2/(1-\alpha )}\hfill \\ & =\frac{\alpha {(\Delta -1)}^{(1-\alpha )/2}{\Delta }^{\alpha }{(\Delta -1)}^{\alpha /2}{\Delta }^{{\alpha }^2/(1-\alpha )}+(1-\alpha ){(\delta -1)}^{\alpha /2}{\delta }^{{\alpha }^2/(1-\alpha )}{(\delta -1)}^{(1-\alpha )/2}{\delta }^{\alpha }}{{(\Delta -1)}^{\alpha /2}{\Delta }^{{\alpha }^2/(1-\alpha )}{(\delta -1)}^{(1-\alpha )/2}{\delta }^{\alpha }}\hfill \\ & =\frac{\alpha {(\Delta -1)}^{1/2}{\Delta }^{\alpha /(1-\alpha )}+(1-\alpha ){(\delta -1)}^{1/2}{\delta }^{\alpha /(1-\alpha )}}{{(\Delta -1)}^{\alpha /2}{\Delta }^{{\alpha }^2/(1-\alpha )}{(\delta -1)}^{(1-\alpha )/2}{\delta }^{\alpha }}.\hfill \end{array}$$

If $\alpha \in (1/2,1)$, then a similar computation gives:

$$\begin{array}{cc}\hfill K_{1/\alpha }& \left((2\delta -2){\delta }^{2\alpha /(1-\alpha )},(2\Delta -2){\Delta }^{2\alpha /(1-\alpha )}\right)\hfill \\ & =\frac{\alpha {(\delta -1)}^{1/2}{\delta }^{\alpha /(1-\alpha )}+(1-\alpha ){(\Delta -1)}^{1/2}{\Delta }^{\alpha /(1-\alpha )}}{{(\delta -1)}^{\alpha /2}{\delta }^{{\alpha }^2/(1-\alpha )}{(\Delta -1)}^{(1-\alpha )/2}{\Delta }^{\alpha }}.\hfill \end{array}$$

Since:

$$\begin{array}{cc}\hfill (2\delta -2){\delta }^{2\alpha /(1-\alpha )}& \le (d_u+d_v-2){\left(d_u{d}_v\right)}^{\alpha /(1-\alpha )}=\frac{d_u+d_v-2}{{\left(d_u{d}_v\right)}^{-\alpha /(1-\alpha )}}\hfill \\ & \le (2\Delta -2){\Delta }^{2\alpha /(1-\alpha )},\hfill \end{array}$$

Corollary 2 gives:

$$\begin{array}{cc}\hfill AB{C}_{\alpha }\left(G\right)& =\sum _{uv\in E\left(G\right)}{\left(\frac{d_u+d_v-2}{d_u{d}_v}\right)}^{\alpha }\hfill \\ & \ge \frac{{\left({\sum }_{uv\in E\left(G\right)}\left(d_u+d_v-2\right)\right)}^{\alpha }{\left({\sum }_{uv\in E\left(G\right)}{\left(d_u{d}_v\right)}^{-\alpha /(1-\alpha )}\right)}^{1-\alpha }}{K_{1/\alpha }\left((2\delta -2){\delta }^{2\alpha /(1-\alpha )},(2\Delta -2){\Delta }^{2\alpha /(1-\alpha )}\right)}\hfill \\ & =\frac{F{\left(G\right)}^{\alpha }{R}_{-\alpha /(1-\alpha )}{\left(G\right)}^{1-\alpha }}{K_{1/\alpha }\left((2\delta -2){\delta }^{2\alpha /(1-\alpha )},(2\Delta -2){\Delta }^{2\alpha /(1-\alpha )}\right)}.\hfill \end{array}$$

This gives the second and third inequalities.

If the graph is regular, then:

$$\begin{array}{cc}& \frac{F{\left(G\right)}^{\alpha }{R}_{-\alpha /(1-\alpha )}{\left(G\right)}^{1-\alpha }}{K_{1/\alpha }\left((2\delta -2){\delta }^{2\alpha /(1-\alpha )},(2\Delta -2){\Delta }^{2\alpha /(1-\alpha )}\right)}\hfill \\ & =\frac{{\left((2\delta -2)m\right)}^{\alpha }{\left({\delta }^{-2\alpha /(1-\alpha )}m\right)}^{1-\alpha }}{K_{1/\alpha }\left((2\delta -2){\delta }^{2\alpha /(1-\alpha )},(2\delta -2){\delta }^{2\alpha /(1-\alpha )}\right)}\hfill \\ & =\frac{{(2\delta -2)}^{\alpha }}{{\delta }^{2\alpha }}m=AB{C}_{\alpha }\left(G\right).\hfill \end{array}$$

If we have the equality in the second or third inequality, then Corollary 2 gives $(2\delta -2){\delta }^{2\alpha /(1-\alpha )}=(2\Delta -2){\Delta }^{2\alpha /(1-\alpha )}$. Since the function $h\left(t\right)=(2t-2)t^{2\alpha /(1-\alpha )}$ is strictly increasing on $[1,\infty )$, we conclude that $\delta =\Delta$ and G is regular.

Finally, assume that $\delta =1$. If $\alpha \in (0,1/2]$, then:

$$\begin{array}{cc}& K_{1/\alpha }\left(2^{\alpha /(1-\alpha )},(2\Delta -2){\Delta }^{2\alpha /(1-\alpha )}\right)\hfill \\ & =\alpha {(2\Delta -2)}^{(1-\alpha )/2}{\left(\frac{\Delta }{2^{1/2}}\right)}^{\alpha }+(1-\alpha ){\left(\frac{1}{2\Delta -2}\right)}^{\alpha /2}{\left(\frac{2^{1/2}}{\Delta }\right)}^{{\alpha }^2/(1-\alpha )}\hfill \\ & =\frac{\alpha {(2\Delta -2)}^{(1-\alpha )/2}{\Delta }^{\alpha }{(2\Delta -2)}^{\alpha /2}{\Delta }^{{\alpha }^2/(1-\alpha )}+(1-\alpha )2^{{\alpha }^2/(2-2\alpha )}{2}^{\alpha /2}}{{(2\Delta -2)}^{\alpha /2}{\Delta }^{{\alpha }^2/(1-\alpha )}{2}^{\alpha /2}}\hfill \\ & =\frac{\alpha {(2\Delta -2)}^{1/2}{\Delta }^{\alpha /(1-\alpha )}+(1-\alpha )2^{\alpha /(2-2\alpha )}}{2^{\alpha }{(\Delta -1)}^{\alpha /2}{\Delta }^{{\alpha }^2/(1-\alpha )}}.\hfill \end{array}$$

If $\alpha \in (1/2,1)$, then a similar computation gives:

$$\begin{array}{cc}\hfill K_{1/\alpha }& \left(2^{\alpha /(1-\alpha )},(2\Delta -2){\Delta }^{2\alpha /(1-\alpha )}\right)\hfill \\ & =\frac{\alpha 2^{\alpha /(2-2\alpha )}+(1-\alpha ){(2\Delta -2)}^{1/2}{\Delta }^{\alpha /(1-\alpha )}}{2^{{\alpha }^2/(2-2\alpha )}{\Delta }^{\alpha }{(2\Delta -2)}^{(1-\alpha )/2}}.\hfill \end{array}$$

Since:

$$\begin{array}{cc}\hfill 2^{\alpha /(1-\alpha )}& \le (d_u+d_v-2){\left(d_u{d}_v\right)}^{\alpha /(1-\alpha )}=\frac{d_u+d_v-2}{{\left(d_u{d}_v\right)}^{-\alpha /(1-\alpha )}}\hfill \\ & \le (2\Delta -2){\Delta }^{2\alpha /(1-\alpha )},\hfill \end{array}$$

Corollary 2 gives:

$$\begin{array}{cc}\hfill AB{C}_{\alpha }\left(G\right)& =\sum _{uv\in E\left(G\right)}{\left(\frac{d_u+d_v-2}{d_u{d}_v}\right)}^{\alpha }\hfill \\ & \ge \frac{{\left({\sum }_{uv\in E\left(G\right)}\left(d_u+d_v-2\right)\right)}^{\alpha }{\left({\sum }_{uv\in E\left(G\right)}{\left(d_u{d}_v\right)}^{-\alpha /(1-\alpha )}\right)}^{1-\alpha }}{K_{1/\alpha }\left(2^{\alpha /(1-\alpha )},(2\Delta -2){\Delta }^{2\alpha /(1-\alpha )}\right)}\hfill \\ & =\frac{F{\left(G\right)}^{\alpha }{R}_{-\alpha /(1-\alpha )}{\left(G\right)}^{1-\alpha }}{K_{1/\alpha }\left(2^{\alpha /(1-\alpha )},(2\Delta -2){\Delta }^{2\alpha /(1-\alpha )}\right)}.\hfill \end{array}$$

This gives the fourth and fifth inequalities. □

Theorem 4 has the following consequence.

Corollary 3.

Let G be a graph with $m_2$ isolated edges.

(1) Then:

$$\begin{array}{c}\hfill ABC\left(G\right)\le \sqrt{2F\left(G\right)\left(R_{-1}\left(G\right)-m_2\right)}.\end{array}$$

The equality in this bound is attained for the union of any regular or biregular graph and $m_2$ isolated edges; if G is the union of a connected graph and $m_2$ isolated edges, then the equality in this bound is attained if and only if G is the union of any regular or biregular connected graph and $m_2$ isolated edges.

(2) If $\delta >1$, then:

$$\begin{array}{c}\hfill ABC\left(G\right)\ge \frac{2\sqrt{2\Delta \delta }{(\Delta -1)}^{1/4}{(\delta -1)}^{1/4}F{\left(G\right)}^{1/2}{R}_{-1}{\left(G\right)}^{1/2}}{\Delta \sqrt{\Delta -1}+\delta \sqrt{\delta -1}}.\end{array}$$

The equality in this bound is attained if and only if G is regular.

(3) If $\delta =1$, then:

$$\begin{array}{c}\hfill ABC\left(G\right)\ge \frac{2\sqrt{2\Delta }{(\Delta -1)}^{1/4}F{\left(G\right)}^{1/2}{\left(R_{-1}\left(G\right)-m_2\right)}^{1/2}}{\Delta \sqrt{\Delta -1}+1}.\end{array}$$

Theorem 5.

If G is a graph with m edges and $m_2$ isolated edges and $\alpha \in \mathbb{R}$, then:

$$\begin{array}{cc}\hfill AB{C}_{\alpha }\left(G\right)& \le {(m-m_2-1)}^{\alpha }\left(R_{-\alpha }\left(G\right)-m_2\right),\mathrm{if}\alpha >0,\hfill \\ \hfill AB{C}_{\alpha }\left(G\right)& \ge {(m-1)}^{\alpha }{R}_{-\alpha }\left(G\right),\mathrm{if}\alpha <0\mathrm{and}{m}_2=0.\hfill \end{array}$$

The equality in the first bound is attained if and only if G is the union of a star graph and $m_2$ isolated edges. The equality in the second bound is attained if and only if G is a star graph.

Proof. 

Since $AB{C}_{\alpha }\left(P_2\right)=0$ and $R_{\beta }\left(P_2\right)=1$, it suffices to prove the theorem for the case $m_2=0$, i.e., when G is a graph without isolated edges.

In any graph, the inequality $d_u+d_v\le m+1$ holds for every $uv\in E\left(G\right)$. If $\alpha >0$, then:

$$\begin{array}{cc}\hfill \frac{{\left(\frac{d_u+d_v-2}{d_u{d}_v}\right)}^{\alpha }}{{\left(\frac{1}{d_u{d}_v}\right)}^{\alpha }}& ={(d_u+d_v-2)}^{\alpha }\le {(m-1)}^{\alpha },\hfill \\ \hfill {\left(\frac{d_u+d_v-2}{d_u{d}_v}\right)}^{\alpha }& \le {(m-1)}^{\alpha }{\left(d_u{d}_v\right)}^{-\alpha },\hfill \\ \hfill AB{C}_{\alpha }\left(G\right)& \le {(m-1)}^{\alpha }{R}_{-\alpha }\left(G\right).\hfill \end{array}$$

If $\alpha <0$, then we obtain the converse inequality.

If G is a star graph, then $d_u+d_v=m+1$ for every $uv\in E\left(G\right)$, and the equality is attained for every $\alpha$.

If the equality is attained in some inequality, then the previous argument gives that $d_u+d_v=m+1$ for every $uv\in E\left(G\right)$. In particular, G is a connected graph. If $m=2$, then $\{d_u,d_v\}=\{1,2\}$ for every $uv\in E\left(G\right)$, and so, $G=P_3=S_3$. Assume now $m\ge 3$. Seeking for a contradiction, assume that $\{d_u,d_v\}\ne \{m,1\}$ for some $uv\in E\left(G\right)$. Since $d_u+d_v=m+1$, we have $2\le d_u,d_v\le m-1$, and so, there exist two different vertices $u^{\prime },v^{\prime }\in V\left(G\right)\setminus \{u,v\}$ with $u{u}^{\prime },v{v}^{\prime }\in E\left(G\right)$. Since $v{v}^{\prime }$ is not incident on u and $u^{\prime }$, we have $d_u+d_{u^{\prime }}<m+1$, a contradiction. Hence, $\{d_u,d_v\}=\{m,1\}$ for every $uv\in E\left(G\right)$, and so, G is a star graph. □

Corollary 4.

If G is a graph with m edges and $m_2$ isolated edges, then:

$$\begin{array}{c}\hfill ABC\left(G\right)\le \sqrt{2(m-m_2-1)}\left(R\left(G\right)-m_2\right),\end{array}$$

and the equality is attained if and only if G is the union of a star graph and $m_2$ isolated edges.

Note that Theorem 5 (and Corollary 4) improves Items $\left(1\right)$ and $\left(2\right)$ in Theorems 1 and 2 for many graphs (when $m<2\Delta -1$).

3. Conclusions

Topological indices have become a useful tool for the study of theoretical and practical problems in different areas of science. An important line of research associated with topological indices is to find optimal bounds and relations between known topological indices, in particular to obtain bounds for the topological indices associated with invariant parameters of a graph (see [1]).

From the theoretical point of view in this research, a new type of Hölder converse inequality was proposed (Theorem 3 and Corollary 2). From the practical point of view, this inequality was successfully applied to establish new relationships of the generalizations of the indexes $ABC$ and R; in particular, it was applied to prove Theorem 4 and Corollary 3. In addition, other new relationships were obtained between these indices (Theorems 1, 2, and 5) that generalized and improved already known results.