Inequalities on the Generalized ABC Index

: In this work, we obtained new results relating the generalized atom-bond connectivity index with the general Randi´c index. Some of these inequalities for ABC α improved, when α = 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.


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: 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: for any α ∈ R \ {0}. Note that ABC 1/2 = √ 2 2 ABC and ABC −3 is the augmented Zagreb index. There are many papers that have studied the ABC and ABC α 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 ABC α improved, when α = 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.
The next result relates the ABC α and R β indices.
Theorem 1. Let G be a graph with maximum degree ∆ and minimum degree δ and α > 0, β ∈ R \ {0}. Denote by m 2 the cardinality of the set of isolated edges in G.
(1) If β/α ≤ −1 and δ > 1, then: The equality in each bound is attained if and only if G is a regular graph.
(2) If β/α ≤ −1 and δ = 1, then: 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.
(4) If −1 < β/α ≤ −1/2 and δ = 1, then: 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 β > 0 and δ > 1, then: The equality in each bound is attained if and only if G is a regular graph.
(6) If β > 0, δ = 1 and 1 + α/β ≥ ∆, then: 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 ∆+1 and m 2 isolated edges.
(7) If β > 0, δ = 1 and 1 + α/β ≤ 2, then: 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 β > 0, δ = 1 and 2 < 1 + α/β < ∆, then: 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 α/β ∈ Z + and G is a union of star graphs S α/β+2 and m 2 isolated edges.
Proof. First of all, note that ABC α (P 2 ) = 0 and R β (P 2 ) = 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, ∆ ≥ 2.
If λ = β/α, then: for every uv ∈ E(G) and, consequently, The previous argument shows that the equality in the upper bound is attained if and only if d u = d v = ∆ for every uv ∈ E(G), i.e., G is regular. If δ > 1, then the equality in the lower bound is attained if and only if d u = d v = δ for every uv ∈ E(G), i.e., G is regular.
If λ = β/α and δ = 1, then: for every uv ∈ E(G) and, consequently, The equality in this bound is attained if and only if {d u , d v } = {1, 2} for every uv ∈ E(G), i.e., G is a union of path graphs P 3 .
(3) and (4). In what follows, by symmetry, we can assume that x ≤ y. We have: We have: Since 2λ + 2 > 0 and −2λ − 1 ≥ 0, we have: Thus, g(x) ≥ g(δ) and: If λ = β/α and δ > 1, then: for every uv ∈ E(G) and, consequently, The previous argument shows that the equality in this bound is attained if and only if d u = d v = δ for every uv ∈ E(G), i.e., G is regular. If λ = β/α, then: for every uv ∈ E(G) and, consequently, The equality in this bound is attained if and only if {d u , d v } = {1, 2} for every uv ∈ E(G), i.e., G is a union of path graphs P 3 .
If β > 0 and λ = β/α, then we obtain: The equality in the lower bound is attained if and only if d u = d v = ∆ for every uv ∈ E(G), i.e., G is regular. The equality in the upper bound is attained if and only if {d u , d v } = {1, ∆} for every uv ∈ E(G), i.e., G is a union of star graphs S ∆+1 .
If β > 0 and λ = β/α, then we obtain: and we have for every uv ∈ E(G): The equality in the lower bound is attained if and only if d u = d v = ∆ for every uv ∈ E(G), i.e., G is regular. The equality in the upper bound is attained if and only if α/β ∈ Z + and {d u , d v } = {1, 1 + α/β} for every uv ∈ E(G), i.e., G is a union of star graphs S α/β+2 . Note that ABC α (G) is not well defined if α < 0 and G has an isolated edge. The argument in the proof of Theorem 1 gives directly the following result for α < 0. Theorem 2. Let G be a graph without isolated edges, with maximum degree ∆ and minimum degree δ, and α < 0, β ∈ R \ {0}.
(1) If β/α ≤ −1 and δ > 1, then: The equality in each bound is attained if and only if G is a regular graph.
(2) If β/α ≤ −1 and δ = 1, then: 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 .
(4) If −1 < β/α ≤ −1/2 and δ = 1, then: The equality in the bound is attained if and only if G is a union of path graphs P 3 .
(5) If β < 0 and δ > 1, then: The equality in each bound is attained if and only if G is a regular graph.
(6) If β < 0, δ = 1 and 1 + α/β ≥ ∆, then: The equality in the lower bound is attained if and only if G is a union of star graphs S ∆+1 . The equality in the upper bound is attained if and only if G is a regular graph.
(7) If β < 0, δ = 1 and 1 + α/β ≤ 2, then: 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 β < 0, δ = 1 and 2 < 1 + α/β < ∆, then: The equality in the lower bound is attained if and only if α/β ∈ Z + and G is a union of star graphs S α/β+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: Theorem 1 has the following consequence.

Corollary 1.
Let G be a graph with minimum degree δ and m 2 isolated edges.
(1) If δ > 1, then: The equality in the bound is attained if and only if G is a regular graph.
(2) If δ = 1, then 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.
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, µ) be a measure space, f , g : X → R measurable functions, and 1 < p, q < ∞ with 1/p + 1/q = 1. If there exist positive constants a, b with a|g| q ≤ | f | p ≤ b|g| q µ-a.e., then: with: If these norms are finite, the equality in the bound is attained if and only if a = b and | f | p = a|g| q µ-a.e. or f = g = 0 µ-a.e. Remark 1. Since: , Theorem 3 generalizes Lemma 1 (note that a = ω 2 and b = Ω 2 ).
Proof. If p = 2, then Lemma 1 (with ω = a 1/2 and Ω = b 1/2 ) gives the result. Assume now p = 2, and let us define: We will check at the end of the proof that k p (a, b) = K p (a, b). Let us consider t ∈ (0, 1) and define: G t is strictly decreasing on (0, 1) and strictly increasing on (1, ∞). Thus, if 0 < s ≤ S are two constants and we consider s ≤ x ≤ S, then: If x 1 , x 2 > 0 and sx 2 ≤ x 1 ≤ Sx 2 , then: By continuity, this last inequality holds for every x 1 , x 2 ≥ 0 with sx 2 ≤ x 1 ≤ Sx 2 . If the equality is attained for some x 1 , x 2 ≥ 0 with sx 2 ≤ x 1 ≤ Sx 2 , then x 1 = sx 2 or x 1 = Sx 2 (the cases x 1 = 0 and x 2 = 0 are direct).
Choose t = 1/p (thus, Thus, for every x, y ≥ 0 with sy q ≤ x p ≤ Sy q . If the equality is attained for some x, y ≥ 0 with sy q ≤ x p ≤ Sy q , then x p = sy q or x p = Sy q . If f p = 0 or g q = 0, then a|g| q ≤ | f | p ≤ b|g| q µ-a.e. gives f p = g q = 0, and the equality in (7) holds. Assume now that f p = 0 = g q .
Finally, assume that the equality in (7) is attained. Seeking for a contradiction, assume that a = b. The previous argument gives that: µ-a.e. Since we proved G 1/p (a/b) 1/2 = G 1/p (b/a) 1/2 (recall that p = 2 and a < b), we can conclude that: Hence, Since the equality in Young's inequality gives f p p = h q q , we obtain a = b, a contradiction. Therefore, a = b and | f | p = h q µ-a.e. Hence, | f | p = a |g| q µ-a.e. Theorem 3 has the following consequence. Corollary 2. If 1 < p, q < ∞ with 1/p + 1/q = 1, x j , y j ≥ 0 and ay q j ≤ x p j ≤ by q j for 1 ≤ j ≤ k and some positive constants a, b, then: where K p (a, b) is the constant in Theorem 3. If x j > 0 for some 1 ≤ j ≤ k, then the equality in the bound is attained if and only if a = b and x p j = ay q j for every 1 ≤ j ≤ k.
(1) Then: 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.
Hölder's inequality gives: If G is a regular or biregular graph with m edges, then: 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: for every uv ∈ E(G). Note that the function H : is increasing in each variable. If uv, uw ∈ E(G), then: Thus, for each vertex u ∈ V(G), 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.

Corollary 2 gives:
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: 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 δ > 1, then: The equality in this bound is attained if and only if G is regular.
(3) If δ = 1, then: Theorem 5. If G is a graph with m edges and m 2 isolated edges and α ∈ R, then: 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 ABC α (P 2 ) = 0 and R β (P 2 ) = 1, it suffices to prove the theorem for the case m 2 = 0, i.e., when G is a graph without isolated edges.
If α < 0, then we obtain the converse inequality.