A Proof of an Inequality Related to Variable Zagreb Indices for Simple Connected Graphs

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1 MATCH Communications in Mathematical and in Computer Chemistry MATCH Commun. Math. Comput. Chem. 66 (011) ISSN A Proof of an Inequality Related to Variable Zagreb Indices for Simple Connected Graphs Sašo Bogoev Ministry of Finance, IT Department Dame Gruev 14, 1000 Skopje, Republic of Macedonia shcaky@gmail.com (Received July 8, 010) Abstract The paper is concerned with the inequality λ M 1 (G)/n λ M (G)/m for the first and second generalized Zagreb indices of (m n) - graphs. It was conjectured that this inequality holds for general graphs, whenever λ is in the interval [0 /], but the same was only partially proved. We prove that the inequality holds for every (m n) - graph and for every λ from the interval (1/ /]. 1 Introduction 1.1 Definitions In this paper graph has the meaning of a simple connected graph. Let G =(V,E) be a graph with n = V vertices and m = E edges, and let denote the degree of a verte v V by d(v). The variable first and second Zagreb indices are defined as λ M 1 (G) = v V (d(v)) λ, and λ M (G) = uv E (d(u)d(v)) λ respectively, where λ R. A question that remains to be answered is whether and when the following Inequality (1) holds? λ M 1 (G) λ M (G) n m. (1) Inequality (1) is known as generalized Zagreb indices inequality. From the definitions above, we can observe that the order of magnitude for λ M 1 (G) iso(n 3 ), while the order of magnitude for λ M (G) iso(mn ). Hence, λ M 1 (G)/n and λ M (G)/m will

2 -648- have magnitudes of same order and, therefore, evaluation of (1) is more reasonable approach than comparing λ M 1 (G) and λ M (G) directly. Assuming λ =1,wetakea step backward from this generalization to the originally stated definitions for the first and second Zagreb indices (introduced by I. Gutman and N. Trinajstić [1] in 197) and the Zagreb indices inequality. In this case, we simply write M 1 (G) andm (G) instead of 1 M 1 (G) and 1 M (G). Now, let s introduce a definition which refers to the term of a chemical graph: A chemical graph is any simple connected graph G with maimum verte degree Δ(G) 4. In order to obtain a clear notion about the herein analyzed problem, we are going to epose several related results to this topic. 1. Known results and an open problem In this section, we aim to give a brief idea about some of the already resolved issues arising from Inequality (1). In sequel, if in a specified case some inequality does not hold, but the reverse one (with reversed inequality sign) does, we may emphasize it by stating opposite inequality, oropposite strict inequality in case of omitted equality sign. Namely, the opposite of Inequality (1) is given by λ M 1 (G) n λ M (G) m. () At first, the AutoGraphiX system [] proposed the following conjecture: Conjecture 1. For all simple connected graphs G and λ =1, Inequality (1) holds, and the bound is tight for complete graphs. In 007, P. Hansen and D. Vukičević [3] disproved this claim for general graphs. Actually, the above conjecture does not refer to the variable Zagreb indices because the generalization happened later, but however, it was underlying motivation for many further results. Remember that when we say graph, we mean a simple connected graph. Notations and all other terms used in the below listed theorems, are in the sense of the classical Graph Theory. Theorem 1. [3] For all chemical graphs G and λ =1, Inequality (1) holds. Theorem. [4] For all trees T and λ =1, Inequality (1) holds.

3 -649- Theorem 3. [5] For all unicyclic graphs G and λ =1, Inequality (1) holds. Theorem 4. [6] For all graphs G such that Δ(G) δ(g) and λ =1, Inequality (1) holds. Theorem 5. [6] For all graphs G such that Δ(G) δ(g) 3 and δ(g) and λ =1, Inequality (1) holds. Theorem 6. [7] For all graphs G =(V,E) such that v V, d(v) {[s, s + s ] :s N} and λ =1, Inequality (1) holds. Theorem 7. [8] For all graphs G =(V,E) such that v V, d(v) {[a, a + s] :a s(s 1) : s N} and λ =1, Inequality (1) holds. The last two theorems slightly differ from their original, and they are adapted here for the purpose of this section. Now, let s move to the generalized variant. Theorem 8. [9] For all chemical graphs G and λ [0, 1], Inequality (1) holds. Theorem 9. [10] For all trees T and λ [0, 1], Inequality (1) holds. Theorem 10. [11] For all unicyclic graphs G and λ [0, 1], Inequality (1) holds. Theorem 11. [1] For all unicyclic graphs G and λ (, 0), Opposite inequality () holds. It has also been proved that for the distinct classes of graphs mentioned in the above theorems, for λ (1, + ) neither (1) nor () holds, i.e., the relationship of numerical value between λ M 1 (G)/n and λ M (G)/m is not defined [1]. Theorem 1. [13] For all graphs G such that Δ(G) δ(g) and λ [0, 1], Inequality (1) holds. Theorem 13. [13] For all graphs G such that Δ(G) δ(g) and λ (, 0), Opposite inequality () holds. Theorem 14. [13] For all graphs G such that Δ(G) δ(g) 3 and δ(g) and λ [0, 1], Inequality (1) holds. Theorem 15. [13] For all graphs G such that Δ(G) δ(g) 3 and δ(g) and λ (, 0), Opposite inequality () holds.

4 -650- Theorem 16. [9] For all complete unbalanced bipartite graphs G and λ R\[0, 1], Opposite strict inequality () holds. Theorem 17. [15] For all graphs G and λ [0, 1 ], Inequality (1) holds. Theorem 18. [9] For all λ (, 1], there is a graph G such that Opposite strict inequality () holds. From the last two theorems, the net question for general graphs makes sense: What happens with Inequality (1) when λ ( 1, ]? In other words, the same question can be restated as in [9]: Open problem 1. [9] Identify λ from the interval ( 1, ] such that Opposite strict inequality () holds. Preliminaries.1 Karamata Inequality Definition 1. Let (a, b) R. A function f :(a, b) R is said to be conve if for each two points, y (a, b) and each two nonnegative real numbers λ, λ y satisfying λ + λ y =1, the following inequality holds: f(λ + λ y y) λ f()+λ y f(y). (3) The function f is concave if the function f is conve. If in the previous inequality (assuming y), the equality takes place only in the case when λ =0or λ y =0, then the function f is said to be strictly conve. The function f is strictly concave if the function f is strictly conve. For concave functions, the net inequality is valid: f(λ + λ y y) λ f()+λ y f(y). Now, let s define a majorization relation for finite sequences of real numbers. Definition.1. Let a =(a i ) n i=1 and b =(b i ) n i=1 be two (finite) sequences of real numbers. We say that the sequence a majorizes the sequence b, and write a b or b a (4) if there eist permutations i 1 i...i n and j 1 j...j n of all indices of the elements of a and b respectively, such that the following three conditions are simultaneously satisfied:

5 -651- i. a i1 a i... a in, and b j1 b j... b jn ; ii. m k=1 a i k m k=1 b j k, m, 1 m n 1; iii. n k=1 a i k = n k=1 b j k. Net, we give a theorem which plays an essential role in the rest of this paper. Theorem.1. [Karamata Inequality] Let (α, β) R. Leta =(a i ) n i=1 and b =(b i ) n i=1 be two (finite) sequences of real numbers such that a i,b i (α, β) for all i =1,,...,n. If f :(α, β) R is a conve function and a b (in other words (3) and (4) hold), then the following inequality is valid: n n f(a i ) f(b i ). (5) i=1 i=1. Redefining Open problem 1 Now, we return to Open problem 1, confining to matters that are related to our approach to this problem. Namely, it has been shown (see [9, 14]) that the difference λ M (G) m λ M 1 (G) n (6) depends on a function in which the degrees of the vertices of a graph play the role of variables. This function also depends on the parameter λ, and our interest is how the function behaves with respect to the mentioned parameter. What is important here is the sign of the epression (6) and, to this end, we are going to present some connections to this difference in the following paragraph. Let G =(V,E) be a (simple connected) graph and D G = {d(v) :v V } be the set of the degrees of the all vertices of G. Letm i,j bethenumberofthoseedgesfrome that connect vertices of degrees i and j. It is clear that i, j N, andm i,j = m j,i for all pairs (i, j). As shown (see [9, 14]), the sign of the difference (6) can be epressed by the following equivalence: [ λ ] [ M (G) λ Sign m M 1 (G) ] = Sign λ f(i, j, k, l)m i,j m k,l, (i, j), (k, l) N. n i j k l

6 -65- In order to single out the parameter λ from the other variables, we write λ f(i, j, k, l) instead of f(i, j, k, l, λ). Since m i,j =0forall(i, j) / D G (the same applies to m k,l), the sum on the right side of the previous equality can be taken over all pairs (i, j) and (k, l) fromn, such that i j and k l. Clearly, m i,j and m k,l are positive integers or 0, thus, it turns out that the function λ f is a key factor for the sign of the difference (6). This function is defined as ( 1 λ f(α) =i λ j λ k + 1 ) ( 1 + k λ l λ l i + 1 ) i λ 1 j λ 1 k λ 1 l λ 1, (7) j where α =(i, j, k, l) N 4 and λ R. In Section 1., we listed a number of theorems that actually refer to some restrictions on the variables i, j, k, l, and the parameter λ, under which the function λ f is nonnegative. Also, it was shown [8] that in general case, whenever λ f is negative for some fied values i g, j g, k g, l g and λ g,wehave a way to construct a graph G g with D Gg = {i g,j g,k g,l g }, such that for λ = λ g, Inequality (1) does not hold. From the previous fact and the list of theorems in Section 1., we may conclude that for all i, j, k, l such that ma{i, j, k, l} 4, or ma{i, j, k, l} min{i, j, k, l}, or ma{i, j, k, l} min{i, j, k, l} 3where min{i, j, k, l}, wehave 1 f 0. What is more important to us, is that we have λ f 0 for all i, j, k, l when λ [0, 1 ]. On the other hand, the last theorem from the list says us that for λ (, 1], we can always find i, j, k, l such that λ f<0. Hence, the interval [0, 1] could be at most etended to the interval [0, ], on which λ f 0 would hold for any four positive integers i, j, k, l, and that is the assumption to be proved. Thus, Open problem 1 is reduced to Problem. Identify λ from the interval ( 1, ] such that λ f is negative for some positive integers i, j, k, l. 3 Solution of Problem Let X be the set (R + ) 4, ω =(, y, p, q) beapointfromx, andλ be a real number. Define a function φ : X R R as ( 1 φ(ω, λ) = λ y λ p + 1 ) ( 1 + p λ q λ q + 1 ) λ 1 y λ 1 p λ 1 q λ 1, (8) y

7 -653- and for the sake of formal similarity with the notation in (7), denote λ φ(ω) =φ(ω, λ). Now, from (7) and (8) it is clear that λ f is a restriction of λ φ on N 4, i.e., λ f = λ φ N 4. (9) Some characteristics of the function λ φ can be immediately perceived: a. 0 φ(ω) =0, ω X ; λ φ is continuous in both variables ω and λ. Further, the function λ φ has an useful symmetry property with respect to the position of, y, p, q, i.e., for a fied value of λ and an arbitrary point ω =(, y, p, q) X, obviously it is valid b. λ φ(, y, p, q) = λ φ(y,, p, q); λ φ(, y, p, q) = λ φ(p, q,, y). In accordance with Problem, our interest is limited in terms of the range of values of λ. The interval which we consider is [ 1 ] I =,. Let s define the sets: X r = {ω X : y, p q, = min{, y, p, q}} ; X ω = {ω X r : q<y} ; Y r = {ω X : y, p q, y =ma{, y, p, q}} ; Y ω = {ω Y r : <p}. From X r X, for the images of X r and X under the function λ φ, we conclude that λ φ(x r ) λ φ(x), and by assertion b, for any point ω X, there is a point ω r X r (just a rearrangement of the values, y, p, q), such that λ φ(ω) = λ φ(ω r ), which means λ φ(x r ) λ φ(x). Further, from X ω X r, for the images of X ω and X r under the function λ φ,wehave λ φ(x ω ) λ φ(x r ). It is easy to see that an analogous inclusion is valid for the sets Y ω, Y r, X, and for their images λ φ(y ω ), λ φ(y r ), λ φ(x). We obtain: λ φ(x ω ) λ φ(x r )= λ φ(x); λ φ(y ω ) λ φ(y r )= λ φ(x). (10) From this point, we consider the restrictions of the function λ φ on the sets X ω, X r, and Y ω, Y r, and according to (10), it is enough to show that for every λ I, λ φ Xr (ω) 0, or λ φ Yr (ω) 0 (11)

8 -654- holds for all ω. The definitions of the sets allow three possible situations for each of the cases ω X r, ω Y r, in terms of the relationship between, y, p, q. Those are: ω =(, y, p, q) X r : ω =(, y, p, q) Y r : 1 y p q 1 p q y p y q p q y 3 p q<y 3 <p q y Note that the set of all points ω from X r which coordinates, y, p, q satisfy 1 or, and do not satisfy 3, is the set X r \X ω. Analogously, the set of all points ω from Y r which coordinates, y, p, q satisfy 1 or, and do not satisfy 3, is the set Y r \Y ω (to see this, it is enough to recall the definitions of these sets). Furthermore, there is an equivalence between the points from X r \X ω and those from Y r \Y ω in the net sense: The point ω =(, y, p, q) X r \X ω if and only if the point ω y =(p, q,, y) Y r \Y ω. Even more, the coordinates of ω satisfy 1 if and only if the coordinates of ω y satisfy 1 and, the coordinates of ω satisfy if and only if the coordinates of ω y satisfy. For the points ω and ω y, the property b implies that λ φ(ω )= λ φ(ω y ) and, hence, the following two claims are equivalent. Here, we set them out in form of lemmas: Lemma 1. Let ω =(, y, p, q) X r \X ω (, y, p, q satisfy 1 or ) and λ [0, 1]. Then λ φ(ω) 0 holds. Lemma. Let ω =(, y, p, q) Y r \Y ω (, y, p, q satisfy 1 or ) and λ [0, 1]. Then λ φ(ω) 0 holds. A proof of Lemma 1 will be given at the end of this paper. So, for now, we assume that in cases 1 and (1 and ), the inequality λ φ 0 is valid for λ [0, 1]. Starting with case 3,orwithcase3, we have the net two choices respectively: To consider λ φ Xω, or to consider λ φ Yω, which is the same in sense of our problem. If we take the first of these two possibilities, the points which coordinates satisfy = p q<yare covered by, and by Lemma, we have λ φ 0 at these points. Otherwise, if we decide to take the second possibility, the points which coordinates satisfy <p q = y are covered by, and by Lemma 1, we have λ φ 0. Ignoring these two sets of points, cases 3 and 3 become identical. In this way, we also avoid some further undefined epressions. Therefore, without loss of generality, we assume <p q<y. (1)

9 -655- Case 3. Let ω =(, y, p, q) X ω and λ R\{0}. Let s fi λ, and make change of the coordinates of ω by introducing a continuous bijective map g λ : X ω R 4 such that ω ω, which is given by g λ (ω) =(g (ω),g y (ω),g p (ω),g q (ω)), where the components g, g y, g p, g q are real functions on X ω defined as g (ω) =λ ln, g y (ω) =λ ln y, g p (ω) =λ ln p, g q (ω) =λ ln q, and for simpler further writing let denote = g (ω), y = g y (ω), p = g p (ω), q = g q (ω). Remember that in this moment λ does not vary. Now, with an analogous notation as above, we define a function λ ψ : g λ (X ω ) R as λ ψ(ω )=e p +q λ + e p +q y λ + e +y p λ + e +y q λ e y y λ e q q λ e p p λ e λ. (13) The Jacobian of the map g λ is positive at every point (J gλ (ω) = λ4 ypq > 0 for all ω X ω and λ 0), which means that we can turn back to the original coordinates at any moment and, hence, by elementary transformation of (13) we may check that λ φ(ω) = λ ψ(g λ (ω)), ω X ω. (14) The last equality suggests to pay attention to the sign of the function λ ψ, which would lead us to the same goal. Before we start, allow λ to vary over the whole range of values. Then, for each separate λ, in the above presented manner, we are able to construct the corresponding map g λ, after what (14) still applies for all (ω, λ, g λ (ω)). Further, simplify the writing in (13), (14) by introducing the following substitutions: a 1 = p + q λ =ln(pq)λ ; a = p + q y λ =ln(pq)λ ; y a 3 = + y p λ =ln(y)λ ; a 4 = + y q p λ =ln(y)λ ; q b 1 =y y λ =lnyλ y ; b =q q λ =lnqλ q ; b 3 =p p λ =lnpλ p ; b 4 = λ =lnλ ; λ φ(ω) = λ ψ(g λ (ω)) = e a1 +e a +e a3 +e a4 e b1 e b e b3 e b4, ω X ω,λ 0. (15) Since ordering (1) holds for the points ω, due to the monotonicity of the logarithm function, <p q <y remains to hold for the points ω. Now, let s split our

10 -656- consideration on three cases by setting the following conditions: pq > y ; pq < y ; pq = y ; λ [ 1, + ) λ [ 1, + ) λ [ 1, + ) Let pq > y; λ [ 1, + ). First, make several direct estimates referring to eponents a i, b i (i =1,, 3, 4), which will be valid whenever is pq > y and λ 1. Since (1) holds, we easily observe that a 1 =ma{a 1,a,a 3,a 4 } ; a 3 a 4 ; b 1 >b b 3 >b 4. Under these conditions, the net three orderings are possible: 1 a 1 a a 3 a 4 a 1 a 3 a a 4 3 a 1 a 3 a 4 a For ordering 1 to be true, an additional constraint is needed on λ, under which a a 3 will hold. Similarly, for ordering, the missing constraints are those under which a 3 a and a a 4 will hold, and for ordering 3 that refers to a 4 a.all these restrictions on λ are given by S 1 : a a 3 λ ln y p ln pq y S 3 : a a 4 λ ln y q ln pq y ; S : a 3 a λ ln y p ln pq y ; S 4 : a 4 a λ ln y q ln pq y (= F 1 ); (= F 4 ), where F 1, F 4 (and F, F 3, F 5 below) are labels for simpler and shorter notation in the sequel. Obviously, the values F 1 and F 4 are always positive. Further, looking to provide the necessary conditions i, ii and iii of Definition.1, we are going to compare the corresponding partial sums for each of the sequences 1, and 3 (which elements

11 -657- appear in the above-specified order), with the partial sums of the sequence b =(b i ) 4 i=1. P 1 : a 1 b 1 ln (pq)λ ln yλ y pq ( ) y λ 1 y, pq thus, we can see for certain that this applies when λ 1; P : a 1 + a b 1 + b ln (pq)λ y ln (qy)λ qy pλ yλ q λ 1 ln q ln y p (= F ); P 3 : a 1 + a 3 b 1 + b ln (pqy)λ p ln (qy)λ qy qy p ( ) λ qy λ 1; p P 4 : a 1 + a + a 3 b 1 + b + b 3 ln (y)λ (pq) λ py ln (pqy)λ pqy λ yλ q λ ln q ln y (= F 3 ); P 5 : a 1 + a 3 + a 4 b 1 + b + b 3 ln (y)λ (pq) λ pq ln (pqy)λ pqy λ (pq)λ y λ ln y ln pq (= F 5 ). Values F, F 3, F 5 are also positive. Since a 1 + a + a 3 + a 4 = b 1 + b + b 3 + b 4 is valid without any restriction on λ, it is not a limiting factor in sense of Definition.1. Now, combining the conditions S 1,...,S 4 and P 1,...,P 5, we can cover all three possibilities 1, and 3 in a way that assumptions i, ii and iii of Definition.1 always remain satisfied. Namely, if we assume that the conditions S 1 P 1 P P 4 are simultaneously satisfied, then ordering 1 becomes true, and also the majorization we want (a 1,a,a 3,a 4 ) (b 1,b,b 3,b 4 ) is achieved. Similarly, if the conditions S S 3 P 1 P 3 P 4 (S 4 P 1 P 3 P 5 ) are simultaneously satisfied, then ordering (ordering 3 )andthesame majorization a =(a i ) 4 i=1 b =(b i ) 4 i=1 are to be valid. The function e r (r R) is strictly conve. Thus, under the previous assumptions, we infer that Theorem.1 can be applied in (15), so by (5) it follows 4 S 1 P 1 P P 4 S S 3 P 1 P 3 P 4 S 4 P 1 P 3 P 5 (e ai e bi ) 0. (16) The conditions from the left side of the previous implication can be overwritten as: i=1

12 -658- Q 1 : S 1 P 1 P P 4 λ A [ 1, 1] where A =[F 1, min{f,f 3 }]; Q : S S 3 P 1 P 3 P 4 λ B [ 1, 1] where B =[F 4, min{f 1,F 3 }]; Q 3 : S 4 P 1 P 3 P 5 λ C [ 1, 1] where C =[1, min{f 4,F 5 }]. The proof of this case shall be successfully accomplished if we can show the following: [ 1 ] I =, ( [ 1 ] ) (A B C), 1. (17) pq>y ω X ω It means that whenever λ I, at least one of the assumptions Q 1, Q, Q 3 becomes true, which implies (16) also to be true. Since in (17) the interval [ 1, 1] has no other impact ecept discarding the numbers greater than 1, we have only to show the net: I A B C, {ω X ω : pq > y}. (18) Now, we analyze the set (A B C) [ 1, 1]. Suppose that F <F 3. Then it follows 1 ln q ln y p < ln q ln y p <y, hence, combining with (1) we have 1 < ln y p ln q p = ln y p ln pq p < ln y p ln pq = F 1. y On the other hand, F 3 < 1 always holds, thus under the previous assumption we get F <F 3 <F 1, which means A =, i.e., B A =[F 4,F 3 ]. Conversely, if F 3 F, then the upper endpoint of A is F 3. The assumption p q implies that we always have F 4 F 1, so if we assume F 1 F 3,thenB A =[F 4,F 1 ] [F 1,F 3 ]=[F 4,F 3 ]. Otherwise, if we now assume F 3 <F 1 to hold, then again B A =[F 4,F 3 ]. In this way, we show that B A =[F 4,F 3 ] in all cases. Net, it will be proved that neither F 5 F 4 <F 3 (F 5 <F 4 F 3 )norf 3 F 4 <F 5 (F 3 <F 4 F 5 ) is possible. In that order, let s take the case F 5 F 4 <F 3.Thenwehave F 5 F 4 ln y pq ln y ln y pq ln q ; F 4 <F 3 ln y q ln y < ln q pq ln y.now,sumtheleftandtherightsides: ln pq ( ln y y ln q ) < ln y ( ln p q +lnq ln y ) ln pq y ln y q < ln y pq ln q y 0 < 0,

13 -659- which is impossible. In the same manner we disprove the other cases, so it remains to have only the net two conjunctions to be true: F 4 F 3 F 4 F 5 ; F 4 >F 3 F 4 >F 5. Remark. From the preceding contradiction, we observe F 4 = F 3 if and only if F 4 = F 5. Further, from the definitions of F 3 and F 5, we note that both are always less than 1. Thus, when the first of these two cases occurs, i.e., when we have the situation F 4 F 3 and F 4 F 5, it follows that (C B A) [ 1, 1] = ([ 1,F 4] [F 4,F 3 ]) [ 1, 1] = [ 1,F 3]. In this case, the important value is F 3. When the second case occurs, i.e., when F 4 >F 3 and F 4 >F 5, it turns out that (C B A) [ 1, 1] = ([ 1,F 5] ) [ 1, 1] = [ 1,F 5]. Up to this point, it is still possible that the result is the empty set, but it does not affect our conclusions. What is important here is that the set (C B A) [ 1, 1] depends solely on the values F 3 and F 5. Thus, we will end up this case by proving the following two implications: a F 4 F 3 F 4 F 5 F 3 b F 4 >F 3 F 4 >F 5 F 5 a Let F 4 F 3 F 4 F 5. As we showed, in this case (A B C) [ 1, 1] = [ 1,F 3]. Suppose that there eist a point (, y, p, q) X ω and α (0, ], such that α>f 3. α>f 3 ln y > 1 α ln q ; ln y q +lnq > 1 α ln q ; ln y ( ) 1 q > α 1 ln q. α>f 4 ln pq y > 1 α ln y q ; ln p +lnq y > 1 α ln y q ; ln p ( ) 1 > α +1 ln y q. Hence, α>f 3 together with α>f 4 imply the following: ln y ( ) 1 q > α 1 ln q ( ) 1 α 1 ln p ( )( ) 1 1 > α 1 α +1 ln y q. Since ln y q > 0, finally we obtain: 1 > 1 α 1; α>.

14 -660- This contradicts the above assumption and, therefore, we conclude that whenever F 4 F 3 holds, then F 3 must also be valid. Hence, the inclusion which we want [ 1 I A B C = 3],F (19) is obtained, and that is what we intended to show. As it can be noted, the condition F 4 F 5 is not used anywhere above, so the same proof can be applied for the net case when F 4 F 3, to bound F 4 from below. Therefore, we infer that the following two implications must always be correct: F 3 F 4 F 3, and F 4 F 3 F 4. (0) b Let F 4 >F 3 F 4 >F 5. Then, we have (A B C) [ 1, 1] = [ 1,F 5], and we now need to show that under these conditions, F 5 is valid. Suppose in contrary that there eist a point (, y, p, q) X ω and α (0, ], such that α>f 5. Since F 4 >F 3, from (0) it is F 4 and, therefore, we have F 4 α. F 4 α ln y ( pq α ln q y = α ln p ) ( ) 1 +lnq ; y α +1 ln y q ln p. α>f 5 ln p +lnq > 1 α ln y = 1 α ( ln y ) q +lnq. Now, replace ln p : ( ) 1 α +1 ln y q +lnq > 1 ( ln y ) α q +lnq ; ln y > 1 α ln q ; q α>ln ln y = F 3. It turns out that the condition F 4 α>f 5 implies α>f 3. If we assume F 3 >F 5, then there must be β such that F 3 >β>f 5,andfromF 4 >β>f 5, as we showed, it follows that β>f 3, which is impossible. Hence, it has to be true that F 5 F 3,and

15 -661- together with the assumption α>f 5, again we arrive at the following contradiction: >F pq 5 ln > ln y ; >F 3 ln y > ln q. If we rewrite the same in one raw, we get pq ln > ln y > ln q, which is clearly impossible, since p q. Thus, under the above conditions, we infer that our starting assumption was wrong and, therefore, F 5 must be valid. From this, we see that the inclusion we want [ 1 I A B C = 5],F holds, and together with (19), implies that we also have (18), (17) and (16) as valid. Before we move to the net case which will be proved in a quite analogous way, let s mention (to avoid confusion), we will keep a part of the previous labeling unchanged. Let pq < y; λ [ 1, + ). Again, make several direct estimates referring to eponents a i, b i (i =1,, 3, 4), which will be valid whenever is pq < y and λ 1. Since (1) holds, we easily observe that a = min{a 1,a,a 3,a 4 } ; a 3 a 4 ; b 1 >b b 3 >b 4. Under these conditions, the net three orderings are possible: 1 a 3 a 4 a 1 a a 3 a 1 a 4 a 3 a 1 a 3 a 4 a For ordering 1 to be true, an additional constraint is needed on λ, under which a 4 a 1 will hold. Similarly, for ordering, the missing constraints are those under which a 3 a 1 and a 1 a 4 will hold, and for ordering 3 that refers to a 1 a 3.All these restrictions on λ are given by S 1 : a 4 a 1 λ ln q ln y pq ; S : a 1 a 4 λ ln q ln y pq (= G 1 ); S 3 : a 3 a 1 λ ln p ln y pq ; S 4 : a 1 a 3 λ ln p ln y pq (= G 4 ),

16 -66- where G 1, G 4 (and G 3, G 5, G, G 6 below) are labels for simpler and shorter notation in the sequel. Clearly, the values G 1 and G 4 are always positive. Further, looking to provide the necessary conditions i, ii and iii of Definition.1, we are going to compare the corresponding partial sums for each of the sequences 1, and 3 (which elements appear in the above-specified order), with the partial sums of the sequence b =(b i ) 4 i=1. P 1 : a 3 b 1 ln (y)λ p ln yλ y y p ( y ) λ λ ln y p ln y (= G 3 ); P : a 1 b 1 ln (pq)λ ln yλ y y ( y pq ) λ λ ln y ln y pq (= G 5 ); P 3 : a 3 + a 4 b 1 + b ln (y)λ pq ln (qy)λ qy y ( q ) λ p 1 λ ln y p ln q (= G ); P 4 : a 1 + a 3 b 1 + b ln (pqy)λ p ln (qy)λ qy qy p ( ) λ qy λ 1; p P 5 : a 1 + a 3 + a 4 b 1 + b + b 3 ln (y)λ (pq) λ pq ln (pqy)λ pqy λ (pq)λ y λ ln y ln pq (= G 6 ). Here, G 3, G 5, G, G 6 are also positive. Since a 1 + a + a 3 + a 4 = b 1 + b + b 3 + b 4 holds without any restriction on λ, it is not a limiting factor in sense of Definition.1. Now, combining the conditions S 1,...,S 4 and P 1,...,P 5, we can cover all three possibilities 1, and 3 in a way that assumptions i, ii and iii of Definition.1 always remain satisfied. Namely, if we assume that the conditions S 1 P 1 P 3 P 5 are simultaneously satisfied, then ordering 1 becomes true, and also the majorization we want (a 3,a 4,a 1,a ) (b 1,b,b 3,b 4 ) is achieved. Similarly, if the conditions S S 3 P 1 P 4 P 5 (S 4 P P 4 P 5 ) are simultaneously satisfied, then ordering (ordering 3 )andthesame majorization a =(a i ) 4 i=1 b =(b i ) 4 i=1 are to be valid. The function e r (r R) is strictly conve. Thus, under the previous assumptions, we infer that Theorem.1 can be applied in (15), so by (5) it follows 4 S 1 P 1 P 3 P 5 S S 3 P 1 P 4 P 5 S 4 P P 4 P 5 (e ai e bi ) 0. (1) i=1

17 -663- Directly, we can observe that whenever P 1 is true or P is true, then P 5 must also be true. Namely, since pq < y, wehave ln pq < ln y, pq y ln < ln pq, which means that it is always G 3 <G 6 and G 5 <G 6. Therefore, P 5 does not have meaning when it is in conjunction with P 1 or P, and will not be taken into account. The conditions from the left side of the implication (1) can be overwritten as follows: Q 1 : S 1 P 1 P 3 λ A [ 1, 1] where A =[G 1, min{g,g 3 }]; Q : S S 3 P 1 P 4 λ B [ 1, 1] where B =[G 4, min{g 1,G 3 }]; Q 3 : S 4 P P 4 λ C [ 1, 1] where C =[1, min{g 4,G 5 }]. The proof of this case shall be successfully accomplished if we can show the following: [ 1 ] I =, ( [ 1 ] ) (A B C), 1. () pq<y ω X ω It means that whenever λ I, at least one of the assumptions Q 1, Q, Q 3 becomes true, which implies (1) also to be true. Since in () the interval [ 1, 1] has no other impact ecept discarding the numbers greater than 1, we have only to show the net: I A B C, {ω X ω : pq < y}. (3) Let s consider the set (A B C) [ 1, 1]. First, suppose G <G 3. Then it follows 1 ln y p ln q < ln y p ln y y < q, hence, combining with (1) we have 1 < ln q ln q p = ln q ln q pq < ln q ln y = G 1. pq On the other hand, G 3 < 1 always holds, thus under the previous assumption we get G <G 3 <G 1, which means A =, i.e., B A =[G 4,G 3 ]. Conversely, if G 3 G, then the upper endpoint of A is G 3. The assumption p q implies that we always have G 4 G 1,soifweassumeG 1 G 3,thenB A =[G 4,G 1 ] [G 1,G 3 ]=[G 4,G 3 ]. Otherwise, if we now assume G 3 <G 1 to hold, then again B A =[G 4,G 3 ]. In this way, we show that B A =[G 4,G 3 ] in all cases. Net, it will be proved that

18 -664- neither G 5 G 4 <G 3 (G 5 <G 4 G 3 )norg 3 G 4 <G 5 (G 3 <G 4 G 5 )is possible. In that order, let s take the case G 5 G 4 <G 3.Thenwehave G 5 G 4 ln y y ln pq ln p y ln pq ; G 4 <G 3 ln p ln y < ln y y ln. Now, sum the left and the right sides: p pq ln y ( ln y pq ln y ) < ln p ( ln y p p +lny q ln y ) ln y pq ln p < ln p y ln pq 0 < 0, which is impossible. In the same manner we disprove the other cases, so it remains to have only the net two conjunctions to be true: G 4 G 3 G 4 G 5 ; G 4 >G 3 G 4 >G 5. Remark. From the preceding contradiction, we see G 4 = G 3 if and only if G 4 = G 5. Further, from the definitions of F 3 and F 5, we note that both are always less than 1. Thus, when the first of these two cases occurs, i.e., when we have the situation F 4 F 3 and F 4 F 5, it follows that (C B A) [ 1, 1] = ([ 1,F 4] [F 4,F 3 ]) [ 1, 1] = [ 1,F 3]. In this case, the important value is F 3. When the second case occurs, i.e., when F 4 >F 3 and F 4 >F 5, it turns out that (C B A) [ 1, 1] = ([ 1,F 5] ) [ 1, 1] = [ 1,F 5]. Up to this point, it is still possible that the result is the empty set, but it does not affect our conclusions. What is important here is that the set (C B A) [ 1, 1] depends solely on the values F 3 and F 5. Thus, we will end up this case by proving the following two implications: c G 4 G 3 G 4 G 5 G 3 d G 4 >G 3 G 4 >G 5 G 5 c Let G 4 G 3 G 4 G 5. As we showed, in this case (A B C) [ 1, 1] = [ 1,G 3]. Suppose that there eist a

19 -665- point (, y, p, q) X ω and α (0, ], such that α>g 3. α>g 3 ln y > 1 α ln y p ; ln y p +lnp > 1 α ln y p ; ln p > ( 1 α 1 ) ln y p. α>g 4 ln y pq > 1 α ln p ; ln p +lny q > 1 α ln p ; ln y ( ) 1 q > α +1 ln p. Hence, α>g 3 together with α>g 4 imply the following: ln p > ( 1 α 1 ) ln y p ( 1 α 1 ) ln y q > ( 1 α 1 )( 1 α +1 ) ln p. Since ln p > 0, finally we obtain: 1 > 1 α 1; α>. This contradicts the above assumption and, therefore, we conclude that whenever G 4 G 3 holds, then G 3 must also be valid. Hence, the inclusion which we want [ 1 I A B C = 3],G (4) is obtained, and that is what we intended to show. As it can be noted, the condition G 4 G 5 is not used anywhere above, so the same proof can be applied for the net case when G 4 G 3, to bound G 4 from below. Therefore, we infer that the following two implications must always be correct: G 3 G 4 G 3, and G 4 G 3 G 4. (5) d Let G 4 >G 3 G 4 >G 5. Then, we have (A B C) [ 1, 1] = [ 1,G 5], and we now need to show that under these conditions, G 5 is valid. Suppose in contrary that there eist a point (, y, p, q) X ω and α (0, ], such that α>g 5. Since G 4 >G 3, from (5) it is G 4 and, therefore, we have G 4 α. G 4 α ln p ( y α ln pq = α ln ) ( ) 1 p +lny ; q α +1 ln p ln y q.

20 -666- α>g 5 ln y p +lny q > 1 α ln y = 1 α ( ln y ) p +lnp. Now, replace ln y q : ln y ( ) 1 p + α +1 ln p > 1 ( ln y ) α p +lnp ; ln y > 1 α ln y y p ; α>ln p ln y = G 3. It turns out that the condition G 4 α>g 5 implies α>g 3. If we assume G 3 >G 5, then there must be β such that G 3 >β>g 5,andfromG 4 >β>g 5, as we showed, it follows that β>g 3, which is impossible. Hence, it has to be true that G 5 G 3,and together with the assumption α>g 5, again we arrive at the following contradiction: >G y 5 ln pq > ln y ; >G 3 ln y > ln y. If we rewrite the same in one raw, we get p y ln > ln y pq > ln y, which is clearly impossible, since p q. p Thus, under the above conditions, we infer that our starting assumption was wrong and, therefore, G 5 must be valid. From this, we see that the inclusion we want [ 1 I A B C = 5],G holds, and together with (4), implies that we also have (3), () and (1) as valid. Let pq = y; λ [ 1, + ). This case is much simpler. According to (1), the following orderings hold for all λ: a 1 >a 3 a 4 >a ; b 1 >b b 3 >b 4. Again, looking for the necessary conditions i, ii and iii of Definition.1, we may use the known relations from the previous two cases, and having pq = y, simplify them: a 1 b 1 holds for λ 1; a 1 + a 3 b 1 + b holds for all λ ; a 1 + a 3 + a 4 b 1 + b + b 3 holds for λ 1; a 1 + a 3 + a 4 + a = b 1 + b + b 3 + b 4 holds for all λ.

21 -667- Therefore, the majorization (a 1,a 3,a 4,a ) (b 1,b,b 3,b 4 ) is achieved when λ [ 1, 1]. Since the function e r (r R) is strictly conve, the conditions under which it is possible Theorem.1 to be applied in (15) are obtained for this interval. From the last and the previous two cases, we conclude that λ φ Xω (ω) 0 is valid for all λ I. This conclusion together with Lemma 1 give us also the validity of (11), which was to be proved. Since in [13], the proof of Lemma 1 is short, we include it for the sake of completeness. Proof of Lemma 1. [13] Let ω =(, y, p, q) X r \X ω. Thus, either ordering 1 or ordering appears as valid. It is easy to verify that the function λ φ defined by (8), can be represented in the following form: λ φ(ω) =( λ 1 y λ 1 )(y λ λ )+(p λ 1 q λ 1 )(q λ p λ )+ + 1 ypq [p(q y)+yq(p )](pλ q λ λ y λ ). Obviously, when λ [0, 1], all differences on the right side in the above equality are nonnegative, which proves the lemma. After all, we are in position to answer the question from which we started: There is no value of λ within the interval I, suchthat λ φ would be negative at some point (, y, p, q) (R + ) 4. Since λ f is given by (9), we now establish the following theorem: Theorem 19. For all graphs G and λ ( 1, ], Inequality (1) holds. Acknowledgement. I thank Dr. R. Škrekovski who introduced me to the problem, and I also want to epress my gratitude to Dr. D. Vukičević for reviewing the results. References [1] I. Gutman, N. Trinajstić, Graph Theory and molecular orbitals. Total π electron energy of alternant hydrocarbons, Chem. Phys. Lett. 17 (197) [] G. Caporossi, P. Hansen, Variable neighborhood search for etremal graphs: 1. the Auto-GraphiX system, Discr. Math. 1 (000) [3] P. Hansen, D. Vukičević, Comparing the Zagreb indices, Croat. Chem. Acta 80 (007)

22 -668- [4] D. Vukičević, A. Graovac, Comparing Zagreb M 1 and M indices for acyclic molecules, MATCH Commun. Math. Comput. Chem. 57 (007) [5] B. Liu, On a conjecture about comparing Zagreb indices, in: I. Gutman, B. Furtula (Eds.), Recent Results in the Theory of Randic Inde, Univ. Kragujevac, Kragujevac, 008, pp [6] L. Sun, T. Chen, Comparing the Zagreb indices for graphs with small difference between the maimum and minimum degrees, Discr. Appl. Math. 157 (009) [7] V. Andova, N. Cohen, R. Škrekovski, Some classes of graphs (dis)satisfying the Zagreb indices inequality, MATCH Commun. Math. Comput. Chem. 65 (011) [8] V. Andova, S. Bogoev, D. Dimitrov, M. Pilipczuk, R. Škrekovski, On the Zagreb inde inequality of graphs with prescribed verte degrees, submitted. [9] D. Vukičević, Comparing variable Zagreb indices, MATCH Commun. Math. Comput. Chem. 57 (007) [10] D. Vukičević, A. Graovac, Comparing variable Zagreb M 1 and M indices for acyclic molecules, MATCH Commun. Math. Comput. Chem. 60 (008) [11] B. Horoldagva, K. C. Das, Comparing variable Zagreb indices for unicyclic graphs, MATCH Commun. Math. Comput. Chem. 6 (009) [1] M. Zhang, B. Liu, On comparing variable Zagreb indices for unicyclic graphs, MATCH Commun. Math. Comput. Chem. 63 (010) [13] B. Liu, M. Zhang, Y. Huang, Comparing the variable Zagreb indices for graphs, MATCH Commun. Math. Comput. Chem. 65 (011) [14] Y. Huang, B. Liu, M. Zhang, On comparing the variable Zagreb indices, MATCH Commun. Math. Comput. Chem. 63 (010) [15] V. Andova, M. Petruševski, Variable Zagreb indices and Karamata s inequality, MATCH Commun. Math. Comput. Chem. 65 (011)