aa r X i v : . [ m a t h . G M ] S e p New Inequalities and Applications
Daiyuan Zhang
September 6, 2019
College of ComputerNanjing University of Posts and TelecommunicationsNanjing, P.R. [email protected], [email protected]
Abstract
This paper presents some new inequalities, the most important ofwhich is the inequality given in Theorem 2.1. It can solve a class ofinequalities by a unified method. An important application of the in-equality given in Theorem 2.1 is to derive another new general formof inequality. The famous Nesbitt’s inequality is a special case of thisgeneral form of inequality when n = 3.The new inequality in Theorem 2.1 proposed in this paper is easy touse and expand, and many new inequalities can be derived and obtainedby direct calculation, so it has a wide range of applications.Many known inequalities can also be directly calculated by the in-equalities proposed in this paper, and the calculation is simple andconvenient. Keywords: inequality, Nesbitt’s inequality, generalization of Nesbitt’s in-equality, application.
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This is my original work. Prohibit any copying or plagiarism.
As we all know, inequalities are widely used in mathematics and natural sci-ence. Compared with other branches of mathematics, inequality develops rel-atively late. It is believed that Hardy, Littlewood and Polya are the first onesto systematically discuss inequalities (see Ref. [1]). In the following decades,documents [2] and [3] also systematically discuss inequalities.1
Daiyuan Zhang
However, the proof and application of inequalities often require relativelyhigh skills. To exaggerate, the problem of inequalities is basically a problem-by-problem solution, which is too ”concrete analysis of specific problems”,and lacks a unified way of thinking and systematic methods. Because of therelative independence of inequality methods and the seemingly lack of intrinsiccorrelation, researchers and applicants will often encounter a lot of confusion.Obviously, it is impossible to prove and apply all inequalities in one way,but people hope to provide as systematic a method as possible, so that thismethod can solve a class of problems or a large class of problems, so as toavoid the unrelated ”concrete analysis of specific problems” as far as possible.This paper makes an attempt to link some seemingly unrelated inequalitiesand deal with them in a unified way. For example, the following inequalitiesseem to be independent of each other (where x i ∈ R + , if and only if each x i isequal to each other, the following equalities hold): x x + x + x x + x + x x + x ≥
32 ;1 x + x + 1 x + x + 1 x + x ≥
92 ( x + x + x ) ;11 − x + 11 − x ≥
41 + 2 x x , x + x = 1;1 x + x + 1 x + x + 1 x + x + 1 x + x + 1 x + x + 1 x + x ≥ x + x + x + x . However, through this study, we know that they all originate from oneinequality (or that these inequalities have the same ”father”), which is theinequality I put forward in this paper.Among those inequalities mentioned above, the first two are well known,and the first one is the famous Nesbitt’s inequality. However, the last twoinequalities may be rarely known. I have proved the following theorem.
Theorem 2.1.
Let S be a constant, S − x i > , i = 1 , , · · · , n , ¯ x = n P i =1 x i /n , then the following inequality holds n X i =1 S − x i ≥ nS − ¯ x (2.1) ew Inequalities and Applications Equality holds iff x = x = · · · = x n . Inequality (2.1) given in theorem 2.1 is very important and has many applica-tions. In this section, I first generalize Nesbitt’s inequality to a general formby using inequality (2.1) that I proposed. Other applications are given laterin section 3.It is well known that the following inequality holds for any given positivereal number x , x and x x x + x + x x + x + x x + x ≥ x = x = x .The above inequality is called Nesbitt’s inequality. Nesbitt’s inequality hasalso been discussed in some literatures (e.g. [4], [5], [6], [7], [8], etc.), but theyare basically limited to Nesbitt’s inequality of three quantities, four quantitiesand six quantities. Although the proof of Nesbitt’s inequality is given in thoseliteratures (e.g. [4], [5], [6], [7], [8], etc.), the proof methods are individualized,lack generality, and are not easy to generalize to the general form.If any given n positive real numbers x , x ,, x n , How to determine the lowerbound of the following expression? x x + x + · · · + x n + x x + x + · · · + x n + · · · + x n x + x + · · · + x n − This requires that the Nesbitt’s inequality be extended to the general formof n variables. The result is given in this paper. Moreover, under certainconditions, the upper bound of the above expression is also given, see theorem2.3. Theorem 2.2.
Let x i ∈ R + , i = 1 , , · · · , n , then the following inequalityholds n X i =1 x in P j =1 x j − x i ≥ nn − Equality holds iff x = x = · · · = x n . Proof.
Let S = n X j =1 x j (2.3) Daiyuan Zhang
According to inequality (2.1), we have n X i =1 n P j =1 x jn P j =1 x j − x i ≥ n n − n X i =1 n P j =1 x j − x i + x in P j =1 x j − x i ≥ n n − n X i =1 x in P j =1 x j − x i ≥ n n − n X i =1 n X i =1 x in P j =1 x j − x i ≥ n n − n + n X i =1 x in P j =1 x j − x i ≥ n n − n X i =1 x in P j =1 x j − x i ≥ nn − n = 3, From inequality (2.4), we get Nesbitt’s inequality. Nesbitt’sinequality is a special case of inequality (2.4), or inequality (2.4) proposed inthis paper extends Nesbitt’s inequality to general form. Inequality (2.4) gives the lower bound. A very direct question is what is theupper bound? Obviously, generally speaking, the upper bound does not exist,or the upper bound can tend to infinity. This is easy to understand. Forexample, you can choose that x is large and the rest of x i ( i = 1) is small, sothat the left side of inequality (2.4) becomes very large and can be larger thanany given large number. ew Inequalities and Applications x i is constrained by some other conditions besides being a positive number.As for the upper bound, I give the following theorem. Theorem 2.3. x i ∈ R + , i = 1 , , · · · , n , and n X j =1 x j − x i > x i (2.5) then n X i =1 x in P j =1 x j − x i < Theorem 2.4.
Let x i ∈ R + , i = 1 , , · · · , n , and n P j =1 x j − x i > x i , then nn − ≤ n X i =1 x in P j =1 x j − x i < Equality holds iff x = x = · · · = x n . Nesbitt’s inequality has been generalized by using theorem 2.1 proposedabove. Moreover, there are many other applications of theorem 2.1. Here arejust a few examples.
Example 3.1
Let n = 3, S = x + x + x , by substituting the inequality (2.1)of theorem 2.1, we obtain:1 x + x + 1 x + x + 1 x + x = X i =1 S − x i ≥ S − ¯ x = 3 x + x + x − ( x + x + x )= 92 ( x + x + x )Equality holds iff x = x = x . Daiyuan Zhang
Example 3.2
Let n = 3, S = x + x + x , a = x + x , a = x + x , a = x + x , by substituting inequality (2.1) of theorem 2.1, We get a well-known result as follows:1 x + 1 x + 1 x = X i =1 S − a i ≥ S − ¯ a = 3 x + x + x − ( a + a + a )= 3 x + x + x − ( x + x + x )= 9 x + x + x Equality holds iff x = x = x .This is a well-known result, but it can be obtained directly, convenientlyand quickly by using theorem 2.1 of this paper.Because the conditions of the theorem (2.1) are easily satisfied, many in-equalities can be easily constructed. Here are some examples, in the followingexamples, we assume that x i ∈ R + . Example 3.3
Let n = 3, S = x + x + x , by substituting inequality (2.1) oftheorem 2.1, We have:1 x + x + 1 x + x = X i =1 S − x i ≥ S − ¯ x = 2( x + x + x ) − ( x + x )= 42 ( x + x + x ) − ( x + x )= 4 x + x + 2 x Equality holds iff x = x = x . Example 3.4
Let n = 4, S = x + x + x + x , a = x + x , a = x + x , a = x + x , a = x + x , a = x + x , a = x + x , by substituting inequality ew Inequalities and Applications x + x + 1 x + x + 1 x + x + 1 x + x + 1 x + x + 1 x + x = X i =1 S − a i ≥ S − ¯ a = 6 x + x + x + x − ( a + a + a + a + a + a )= 6 x + x + x + x − ( x + x + x + x )= 12 x + x + x + x Equality holds iff a = a = a = a = a = a , then we can deduce that theequality holds iff x = x = x = x . Example 3.5
Let n = 4, S = x + x + x + x , by substituting inequality (2.1)of theorem 2.1, We have:1 x + x + x + 1 x + x + x + 1 x + x + x + 1 x + x + x = X i =1 S − x i ≥ S − ¯ x = 4 x + x + x + x − ( x + x + x + x )= 163 ( x + x + x + x )Equality holds iff x = x = x = x . Example 3.6
Let n = 2, S = ( x + x ) , a = x , a = x , by substitutinginequality (2.1) of theorem 2.1, We have:1( x + x ) − x + 1( x + x ) − x = X i =1 S − a i ≥ S − ¯ a = 2( x + x ) − ( a + a ) = 2( x + x ) − ( x + x )= 4( x + x ) + 2 x x Daiyuan Zhang i.e. 1(2 x + x ) x + 1(2 x + x ) x ≥ x + x ) + 2 x x Equality holds iff a = a , then we can deduce that the equality holds iff x = x .If let x + x = 1, then the above inequality becomes11 − x + 11 − x ≥
41 + 2 x x Equality holds iff x = x . Example 3.7 Application in Geometry
The above results can be directly appliedto geometry. Suppose x , x ,, x n are the lengths of the sides of a given polygon.Obviously, they satisfy the conditions of theorem 2.4. By using theorem 2.4,the following results can be obtained directly: nn − ≤ x x + x + · · · + x n + x x + x + · · · + x n + x + · · · + x n x + x + · · · + x n − < x = x = · · · = x n .The above results can be directly applied to triangles. Assuming that x , x and x are the lengths of the sides of a given triangle, using theorem 2.4,let n = 3, we can get the following corollary directly. Corollary 3.1.
Assuming that x , x and x are the lengths of the sides ofa given triangle, we have: ≤ x x + x + x x + x + x x + x < Equality holds iff x = x = x . This is a well-known result.
The inequalities presented in this paper can solve a class of lower or up-per bounds of fractional functions with n variables such as n P i =1 (1 /S − x i ) or n P i =1 x i / n P j =1 x j − x i ! . ew Inequalities and Applications S − x i > i = 1 , , · · · , n ) can be easily satisfied,many new inequalities can be easily created by using inequality (2,1) proposedin this paper. The generalization of Nesbitt’s inequality to the general form of n variables is a good example.The inequalities presented in this paper can be easily used to prove someknown results, and the calculation is simple, fast, easy to remember and easyto use.Up to now, there are tens of thousands of inequalities (some of them aregiven in [9]). It is impossible for people to use them all, let alone memorizethem, and there is no need to do so.The author believes that although it is impossible to deal with all inequal-ities in unique unified way, we should try our best to establish the internalrelationship between seemingly different inequalities and adopt a unified wayof thinking to deal with inequalities as much as possible, which is an importantresearch field. The result of this paper is only the tip of the iceberg, and thereis still much work to be done.I hope that the proposed theorems in this paper can be applied more widely. References [1] G. H. Hardy, J. E. Littlewood, and G. Polya,
Inequalities.
CambridgeUniversity Press, 1934.[2] E. F. Beckenbach and R. Bellman,
Inequalities.
Berlin-Gottingen-Heidelberg: Springer-Verlag, 1961.[3] D. S. Mitrinovic, Analytic
Inequalities.
Berlin-New York: Springer-Verlag,1970.[4] J. M. Steele,
The CauchySchwarz Master Class An Introduction to theArt of Mathematical Inequalities.
New York: Cambridge University Press,2004.[5] A. W. Marshall, I. Olkin, and B. C. Arnold,
Inequalities: Theory of Ma-jorization and Its Applications, Second Edition.
LLC, 233 Spring Street,New York: Springer Science+Business Media, 2010.[6] Z. Cvetkovski,
Inequalities Theorems, Techniques and Selected Problems.
Berlin Heidelberg: Springer-Verlag, 2012.[7] P. K. Hung,
Secrets in Inequalities - advanced inequalities, Vol.2.
GILPublishing House, 2008.0
Daiyuan Zhang [8] D. S. Mitrinovic,
Analytic Inequalities.
Berlin Heidelberg: Springer-Verlag, 1970.[9] B. G. Pachpatte,