Sequences of Inequalities among Differences of Gini Means and Divergence Measures
aa r X i v : . [ c s . I T ] N ov Sequences of Inequalities among Differencesof Gini Means and Divergence Measures
Inder Jeet Taneja
Departamento de Matem´aticaUniversidade Federal de Santa Catarina88.040-900 Florian´opolis, SC, Brazil. ∼ taneja Abstract
In 1938, Gini [3] studied a mean having two parameters. Later, many authorsstudied properties of this mean. In particular, it contains the famous means asharmonic, geometric, arithmetic, etc. Here we considered a sequence of inequalitiesarising due to particular values of each parameter of Gini’s mean. This sequencegenerates many nonnegative differences. Not all of them are convex. We havestudied here convexity of these differences and again established new sequences ofinequalities of these differences. Considering in terms of probability distributionsthese differences, we have made connections with some of well known divergencemeasures.
Key words:
Arithmetic mean; Geometric Mean; Harmonic Mean; Gini Mean; PowerMean; Differences of Means; Divergence measures
AMS Classification: r and s The Gini [3] mean of order r and s is given by E r,s ( a, b ) = (cid:0) a r + b r a s + b s (cid:1) r − s r = s exp (cid:0) a r ln a + b r ln ba r + b r (cid:1) r = s = 0 √ ab r = s = 0 . (1)In particular when s = 0 in (1), we have E r, ( a, b ) := B r ( a, b ) = ((cid:0) a r + b r (cid:1) r , r = 0 √ ab, r = 0 (2)Again, when s = r − E r,r − ( a, b ) := K s ( a, b ) = a r + b r a r − + b r − , r ∈ R (3)1he expression (2) is famous as mean of order r or power mean . The expression(3) is known as Lehmer mean [4]. Both these means are monotonically increasing in r .Moreover, these two have the following inequality [2] among each other: B r ( a, b ) ( < L r ( a, b ) , r > > L r ( a, b ) , r < E r,s = E s,r , the Gini-mean E r,s ( a, b ) given by (1) is an increasing function in r or s . Using the monotonicity property [1], [7], [5] we have the following inequalities: E − , − ≤ E − , − ≤ E − / , − / ≤ E − , ≤ E − / , ≤ E − / , / ≤ E , / ≤ E , ≤ E , ≤ E , (5)or E − , − ≤ E − , − ≤ E − / , − / ≤ E − , ≤ E − / , ≤ E − / , / ≤ E , / ≤ E , ≤ E / , ≤ E , (6)Equivalently, let us write as K − ≤ K − ≤ K − / ≤ ( K = B − = H ) ≤ B − / ≤ (cid:0) K / = B = G (cid:1) ≤ B / ≤ ( K = B = A ) ≤ ( B = S ) ≤ K , (7)or K − ≤ K − ≤ K − / ≤ ( K = B − = H ) ≤ B − / ≤ (cid:0) K / = B = G (cid:1) ≤ B / ≤ ( K = B = A ) ≤ E / , ≤ K . (8)where H , G , A and S are respectively , the harmonic , geometric , arithmetic and the square-root means. We observe from the expression (7) that the means considered eitherare the particular cases of (2) or of (3), while in the second case we have we have a mean E / , , which is neither a particular case of (2) nor of (3). We can easily find valuesproving that there is no relation between S and E / , . Summarizing the expressions (7)and (8), we can write them in joint form as P ≤ P ≤ P ≤ H ≤ P ≤ G ≤ N ≤ A ≤ ( P or S ) ≤ P , (9)where P = K − , P = K − , P = K − / , P = B − / , N = B / , P = E / , and P = K .In [9, 10], the author studied the following inequalities: H ≤ G ≤ N ≤ N ≤ N ≤ A ≤ S, (10)where N ( a, b ) = √ a + √ b ! r a + b ! N ( a, b ) = a + √ ab + b . The expression N ( a, b ) is famous as Heron’s mean. The inequalities (10) admit manynon-negative differences. Bases on these difference the author [. . . ] proved the followingresult: D SA ≤ D SH ≤ D AH ≤ (cid:26) D N N ≤ D N G ≤ D AG ≤ D AN D SG ≤ D AG D SN ≤ D SG D SN ≤ D AN D SN ≤ D SN , (11)where D SA = S − A , D SH = S − H , etc. Some applications of the inequalities (11) canbe seen in [6], [7].Combining (9) and (10), we have the following sequence of inequalities: P P P H P G N N N A ( P or S ) P . (12)The inequalities (12) admits many nonnegative differences such as D tp ( a, b ) = bg tp (cid:16) ab (cid:17) = b h f t (cid:16) ab (cid:17) − f p (cid:16) ab (cid:17)i , (13)where g tp ( x ) = f t ( x ) − f p ( x ) , f t ( x ) > f p ( x ) , ∀ x > . More precisely, the expression (12) and the function f : (0 , ∞ ) → R given in (13) leadus to the following result: f P ( x ) f P ( x ) f P ( x ) f H ( x ) f P ( x ) f G ( x ) f N ( x ) f N ( x ) f N ( x ) f A ( x ) ( f P ( x ) or f S ( x )) f P ( x ) . (14)Equivalently, x ( x + 1) x + 1 x ( x + 1) x + 1 x ( √ x + 1) x / + 1 x x x ( √ x + 1) √ x (cid:18) √ x + 12 (cid:19) x + √ x + 13 (cid:18) √ x + 12 (cid:19) r x + 12 ! x + 12 (cid:18) x + 1 √ x + 1 (cid:19) or r x + 12 ! x + 1 x + 1 . (15)In this paper our aim is to produce new inequalities for the difference of means arisingdue to inequalities given in (12). In another words we shall improve considerably theinequalities given in (11). For this we need first to know the convexity of the difference ofmeans. In total, we have 77 differences. Some of them are equal to each other with somemultiplicative constants. Some of them are not convex and some of them are convex.3 Convexity of Difference of Means
Let us prove now the convexity of some of the difference of means arising due to inequalities(11). In order to prove it we shall make use of the following lemma (ref. Taneja [. . . ]).
Lemma 2.1.
Let f : I ⊂ R + → R be a convex and differentiable function satisfying f (1) = 0 . Consider a function φ f ( a, b ) = af (cid:18) ba (cid:19) , a, b > , then the function φ f ( a, b ) is convex in R . Additionally, if f ′ (1) = 0 , then the followinginequality hold: φ f ( a, b ) (cid:18) b − aa (cid:19) φ ′ f ( a, b ) . Inequalities appearing in (12) admits 77 nonnegative differences. Within these wehave the following two set of equalities with multiplicative constants:(i) D P A = D P H = D AH .(ii) D P P = D AN = D AN = D AG = 2 D N N = D N G = D N G .Not all the differences of means appearing in (12) are convex. We shall consider onlythose are convex. It is easy to check that in all the cases, f ( · ) (1) = 1, i.e., g tp (1) = 0, t > p .According to Lemma 2.1, it is suficient to show the convexity of the functions g tp ( x ), i.e.,to show that the second order derivative of g tp ( x ) is nonnegative for all x > For D P S ( a , b ) : We can write D P S ( a, b ) = b g P S ( a / b ), where g P S ( x ) = f P ( x ) − f S ( x ) = x + 1 x + 1 − r x + 12 . This gives g ′′ P S ( x ) = 2 h x + 2) / − ( x + 1) i ( x + 1) (2 x + 2) / . (16)Since, S > A , this implies that S > A , i.e., (cid:18)q x +12 (cid:19) − (cid:0) x +12 (cid:1) >
0. This gives2 (2 x + 2) / − ( x + 1) >
0. Thus we have g ′′ P S ( x ) > x > For D P N ( a , b ) : We can write D P N ( a, b ) = b g P N ( a / b ), where g P N ( x ) = f P ( x ) − f N ( x ) = 4( x + 1) − √ x + 2 ( √ x + 1) ( x + 1)4( x + 1) . g ′′ P N ( x ) = ( x + 1) (cid:0) x / + 1 (cid:1) + 16 x / (2 x + 2) / x / ( x + 1) (2 x + 2) / > , ∀ x > . (17)3. For D P N ( a , b ) : We can write D P N ( a, b ) = b g P N ( a / b ), where g P N ( x ) = f P ( x ) − f N ( x ) = 2( x + 1) − √ x ( √ x + 1) x + 1) . This gives g ′′ P N ( x ) = 48 x / + ( x + 1) x / ( x + 1) > , ∀ x > . (18)4. For D P N ( a , b ) : We can write D P N ( a, b ) = b g P N ( a / b ), where g P N ( x ) = f P ( x ) − f N ( x ) = 3( x + 1) − √ x ( x + √ x + 1)4( x + 1) . This gives g ′′ P N ( x ) = 32 x / + ( x + 1) x / ( x + 1) > , ∀ x > . (19)5. For D P G ( a , b ) : We can write D P G ( a, b ) = b g P G ( a / b ), where g P G ( x ) = f P ( x ) − f G ( x ) = x + 1 x + 1 − √ x = ( √ x − ( x + √ x + 1) x + 1 . This gives g ′′ P G ( x ) = 16 x / + ( x + 1) x / ( x + 1) > , ∀ x > . (20)6. For D P P ( a , b ) : We can write D P P ( a, b ) = b g P P ( a / b ), where g P P ( x ) = f P ( x ) − f P ( x ) = x + 1 x + 1 − x ( √ x + 1) = ( x + 1) ( x − x + 1) + 2 √ x ( x + 1)( x + 1) ( √ x + 1) . This gives g ′′ P P ( x ) = 2 [3 ( x + 1) + 17 x ( x + 1) + 2 √ x ( x + 6 x + 1)] √ x ( √ x + 1) ( x + 1) > , ∀ x > . (21)5. For D P P ( a , b ) : We can write D P P ( a, b ) = b g P P ( a / b ), where g P P ( x ) = f P ( x ) − f P ( x ) = ( x − ( x + x + 1)( x + 1) ( x + 1) . This gives g ′′ P P ( x ) = 2 ( x + 15 x + 16 x + 15 x + 1)( x + 1) ( x + 1) > , ∀ x > . (22)8. For D P P ( a , b ) : We can write D P P ( a, b ) = b g P P ( a / b ), where g P P ( x ) = f P ( x ) − f P ( x ) = ( x − ( x + 1)( x + 1) . This gives g ′′ P P ( x ) = 2 ( x + 1) (cid:0) ( x − − x (cid:1) + x ++ ( x − + x ( x − ! + 8 x ( x + 1) > , ∀ x > . (23)9. For D P A ( a , b ) : We can write D P A ( a, b ) = b g P A ( a / b ), where g P A ( x ) = f P ( x ) − f A ( x ) = (cid:18) x + 1 √ x + 1 (cid:19) − x + 12= ( x + 1) ( √ x − √ x + 1) . This gives g ′′ P A ( x ) = 4 √ x h ( √ x − + √ x i + ( x − x / ( √ x + 1) > , ∀ x > . (24)10. For D P N ( a , b ) : We can write D P N ( a, b ) = b g P N ( a / b ), where g P N ( x ) = f P ( x ) − f N ( x ) = (cid:18) x + 1 √ x + 1 (cid:19) − (cid:18) √ x + 12 (cid:19) r x + 12 ! . This gives g ′′ P N ( x ) = 18 ( √ x + 1) x / ( x + 1) √ x + 2 ×× ( √ x + 1) (cid:0) x / + 1 (cid:1) + 4 √ x + 2 ( x + 1) ×× h √ x (cid:16) ( √ x − + √ x (cid:17) + ( x − i ! > , ∀ x > . (25)61. For D P N ( a , b ) : We can write D P N ( a, b ) = b g P N ( a / b ), where g P N ( x ) = f P ( x ) − f N ( x ) = (cid:18) x + 1 √ x + 1 (cid:19) − x + √ x + 13= ( √ x − (2 x + √ x + 2)2 ( √ x + 1) . This gives g ′′ P N ( x ) = 7 ( √ x − ( x + 6 √ x + 1) + 40 x x / ( √ x + 1) > , ∀ x > . (26)12. For D P N ( a , b ) : We can write D P N ( a, b ) = b g P N ( a / b ), where g P N ( x ) = f P ( x ) − f N ( x ) = (cid:18) x + 1 √ x + 1 (cid:19) − (cid:18) √ x + 12 (cid:19) = ( √ x − (3 x + 2 √ x + 3)4 ( √ x + 1) . This gives g ′′ P N ( x ) = 5 ( √ x − ( x + 6 √ x + 1) + 32 x x / ( √ x + 1) > , ∀ x > . (27)13. For D P G ( a , b ) : We can write D P G ( a, b ) = b g P G ( a / b ), where g P G ( x ) = f P ( x ) − f G ( x ) = (cid:18) x + 1 √ x + 1 (cid:19) − √ x = ( √ x − ( x + √ x + 1)( √ x + 1) . This gives g ′′ P G ( x ) = 3 (cid:2) √ x ( x + 1) + ( x − (cid:3) x / ( √ x + 1) > , ∀ x > . (28)14. For D P H ( a , b ) : We can write D P H ( a, b ) = b g P H ( a / b ), where g P H ( x ) = f P ( x ) − f H ( x ) = (cid:18) x + 1 √ x + 1 (cid:19) − xx + 1 . This gives g ′′ P H ( x ) = 12 x / ( x + 1) ( √ x + 1) ×× (cid:18) ( x + 1) (cid:2) ( x − + 16 x (cid:3) ++ 4 √ x (cid:2) ( x + 1) + 2 x ( x + 6 x + 1) (cid:3) (cid:19) > , ∀ x > . (29)75. For D P P ( a , b ) : We can write D P P ( a, b ) = b g P P (cid:0) ab (cid:1) , where g P P = f P ( x ) − f P ( x ) = (cid:18) x + 1 √ x + 1 (cid:19) − x ( √ x + 1) x / + 1= ( √ x − (cid:0) x + x / + 3 x + √ x + 1 (cid:1) ( √ x + 1) ( x / + 1) . This gives g ′′ P P ( x ) = s ( x )4 x / ( x / + 1) ( √ x + 1) , (30)where s ( x ) = (cid:18) x + 2 x / − x + 75 x / − x ++198 x / − x + 75 x / − x + 2 √ x + 2 (cid:19) . Now we shall show that s ( x ) > ∀ x >
0. Let us consider h ( t ) = s ( t ) = (cid:18) t + 2 t − t + 75 t − t ++198 t − t + 75 t − t + 2 t + 2 (cid:19) . The polynomial equation h ( t ) = 0 of 10 th degree admits 10 solutions. Out of them8 are complex and are given by0 . ± . I ; 0 . ± . I ;0 . ± . I ; 1 . ± . I. The real solutions are − . − . t >
0, this means that there are no realpositive solutions of the equation h ( t ) = 0. Thus we conclude that either h ( t ) > h ( t ) <
0, for all t >
0. In order to check it is sufficient to see for any particularvalue of h ( t ), for example when t = 1. This gives h (1) = 56, hereby proving that h ( t ) > t >
0, consequently, s ( x ) >
0, for all x >
0. Finally, we have g ′′ P P ( x ) > ∀ x > For D P P ( a , b ) : We can write D P P ( a, b ) = b g P P (cid:0) ab (cid:1) , where g P P = f P ( x ) − f P ( x ) = (cid:18) x + 1 √ x + 1 (cid:19) − x ( x + 1) x + 1= ( x + 1) (cid:0) x / − (cid:1) ( √ x + 1) ( x + 1) . This gives g ′′ P P ( x ) = s ( x )2 x / ( x + 1) ( √ x + 1) , (31)8here s ( x ) = (cid:18) x + 4 x / − x − x + 14 x + 92 x / ++102 x + 92 x / + 14 x − x − x + 4 √ x + 1 (cid:19) . Now we shall show that s ( x ) > ∀ x >
0. Let us consider h ( t ) = s ( t ) = (cid:18) t + 4 t − t − t + 14 t + 92 t ++102 t + 92 t + 14 t − t − t + 4 t + 1 (cid:19) . The polynomial equation h ( t ) = 0 of 16 th degree admits 16 solutions. Out of them14 are complex (not written here) and two of them are real given by − . − . t >
0, this means that there are no real positive solutions of the equation h ( t ) = 0.Thus we conclude that either h ( t ) > h ( t ) <
0, for all t >
0. In order tocheck it is sufficient to see for any particular value of h ( t ), for example when t = 1.This gives h (1) = 288, hereby proving that h ( t ) > t >
0, consequently, s ( x ) >
0, for all x >
0. Finally, we have g ′′ P P ( x ) > ∀ x > For D P P ( a , b ) : We can write D P P ( a, b ) = b g P P (cid:0) ab (cid:1) , where g P P = f P ( x ) − f P ( x ) = (cid:18) x + 1 √ x + 1 (cid:19) − x ( x + 1) x + 1= (cid:18) x + 2 x / + 4 x + 4 x / ++4 x + 4 x / + 4 x + 2 √ x + 1 (cid:19) ( √ x − ( √ x + 1) ( x / + 1) . This gives g ′′ P P ( x ) = s ( x )2 x / ( x + 1) ( √ x + 1) , (32)where s ( x ) = x + 4 x / − x + 4 x / + x − x / −− x − x / + 30 x + 144 x / + 99 x ++48 x / + 99 x + 144 x / + 30 x − x / −− x − x / + x + 4 x / − x + 4 √ x + 1 . Now we shall show that s ( x ) > ∀ x >
0. Let us consider h ( t ) = s ( t ) = t + 4 t − t + 4 t + t − t −− t − t + 30 t + 144 t + 99 t ++48 t + 99 t + 144 t + 30 t − t −− t − t + t + 4 t − t + 4 t + 1 . h ( t ) = 0 of 22 nd degree admits 22 solutions. Out of them20 are complex (not written here) and two of them are real given by − . − . t >
0, this means that there are no real positive solutions of the equation h ( t ) = 0.Thus we conclude that either h ( t ) > h ( t ) <
0, for all t >
0. In order tocheck it is sufficient to see for any particular value of h ( t ), for example when t = 1.This gives h (1) = 416, hereby proving that h ( t ) > t >
0, consequently, s ( x ) >
0, for all x >
0. Finally, we have g ′′ P P ( x ) > ∀ x > For D SP ( a , b ) : We can write D SP ( a, b ) = b g SP (cid:0) ab (cid:1) , where g SP = f S ( x ) − f P ( x ) = r x + 12 − x ( √ x + 1) . This gives g ′′ SP ( x ) = 2 h √ x ( √ x + 1) + 3 (2 x + 2) / i √ x ( √ x + 1) (2 x + 2) / > , ∀ x > . (33)19. For D AP ( a , b ) : We can write D AP ( a, b ) = b g AP ( a / b ), where g AP ( x ) = f A ( x ) − f P ( x ) = x + 12 − x ( √ x + 1) . This gives g ′′ AP ( x ) = 6 √ x ( √ x + 1) > , ∀ x > . (34)20. For D SA ( a , b ) : We can write D SA ( a, b ) = b g SA ( a / b ), where g SA ( x ) = f S ( x ) − f A ( x ) = r x + 12 − x + 12 . This gives g ′′ SA ( x ) = 1( x + 1) √ x + 2 > , ∀ x > . (35)21. For D SN ( a , b ) : We can write D SN ( a, b ) = b g SN ( a / b ), where g SN ( x ) = f S ( x ) − f N ( x ) = r x + 12 − r x + 12 (cid:18) √ x + 12 (cid:19) . This gives g ′′ SN ( x ) = 18 x / ( x + 1) ( x + 1) √ x + 2 √ x + 2 ×× (cid:18) √ x + 2 (cid:0) x / + 1 (cid:1) ( x + 1) ++8 x / ( x + 1) √ x + 2 (cid:19) , ∀ x > . (36)102. For D SN ( a , b ) : We can write D SN ( a, b ) = b g SN ( a / b ), where g SN ( x ) = f S ( x ) − f N ( x ) = r x + 12 − (cid:18) √ x + 12 (cid:19) . This gives g ′′ SN ( x ) = 8 x / + ( x + 1) √ x + 28 x / ( x + 1) √ x + 2 > , ∀ x > . (37)23. For D AN ( a , b ) : We can write D AN ( a, b ) = b g AN ( a / b ), where g AN ( x ) = f A ( x ) − f N ( x ) = x + 12 − r x + 12 (cid:18) √ x + 12 (cid:19) . This gives g ′′ AN ( x ) = x / + 18 x / ( x + 1) √ x + 2 > , ∀ x > . (38)24. For D AG ( a , b ) : We can write D AG ( a, b ) = b g AG ( a / b ), where g AG ( x ) = f A ( x ) − f G ( x ) = x + 12 − √ x = ( √ x − . This gives g ′′ AG ( x ) = 14 x / > , ∀ x > . (39)25. For D AH ( a , b ) : We can write D AH ( a, b ) = b g AH ( a / b ), where g AH ( x ) = f A ( x ) − f H ( x ) = x + 12 − xx + 1 = ( x − x + 1) . This gives g ′′ AH ( x ) = 4( x + 1) > , ∀ x > . (40)26. For D N N ( a , b ) : We can write D N N ( a, b ) = b g N N ( a / b ), where g N N ( x ) = f N ( x ) − f N ( x ) = r x + 12 − (cid:18) √ x + 12 (cid:19) . This gives g ′′ N N ( x ) = ( x + 1) √ x + 2 − (cid:0) x / + 1 (cid:1) x / ( x + 1) √ x + 2 . (41)11rom (14), we have ( x + 1) > (cid:16) √ x +12 (cid:17) √ x + 2. This gives( x + 1) √ x + 2 − (cid:0) x / + 1 (cid:1) = ( x + 1) √ x + 2 − (cid:0) √ x + 1 (cid:1) (cid:0) x − √ x + 1 (cid:1) > (cid:18) √ x + 12 (cid:19) (cid:2) x + 2 − (cid:0) x − √ x + 1 (cid:1)(cid:3) > √ x (cid:0) √ x + 1 (cid:1) > , ∀ x > . This proves that g ′′ N N ( x ) > ∀ x > For D SG ( a , b ) : We can write D SG ( a, b ) = b g SG ( a / b ), where g S ( x ) = f S ( x ) − f G ( x ) = r x + 12 − √ x. This gives g ′′ AH ( x ) = 4 x / + √ x + 2 ( x + 1)4 x / √ x + 2 ( x + 1) > , ∀ x > . (42)The convexity of the expressions given in parts 20-27 is already given in [11] but wehave written them again because we need them in the next section. In this section we shall bring sequence of inequalities based on the differences arising dueto (8). The results given in this section are based on the applications of the followinglemma [11]:
Lemma 3.1.
Let f , f : I ⊂ R + → R be two convex functions satisfying the assumptions:(i) f (1) = f ′ (1) = 0 , f (1) = f ′ (1) = 0 ;(ii) f and f are twice differentiable in R + ;(iii) there exists the real constants α, β such that α < β and α f ′′ ( x ) f ′′ ( x ) β, f ′′ ( x ) > , for all x > then we have the inequalities: α φ f ( a, b ) φ f ( a, b ) β φ f ( a, b ) , for all a, b ∈ (0 , ∞ ) , where the function ϕ ( · ) ( a, b ) is as defined in Lemma 2.1. emark 3.1. From the above lemma we observe that η ( x ) = β f ( x ) − f ( x ) , ∀ x > . This gives η (1) = η ′ (1) = 0 . In order to obtain β , let us consider η ′′ (1) = 0 . This gives β = f ′′ (1)/ f ′′ (1) . We shall use this argument to prove the results given in the theorembelow. Theorem 3.1.
The following sequences of inequalities hold: D P P D P P D SA D SH D AH D P N (cid:26) D P N D SP (cid:27) (cid:26) D P N D P P D P G (cid:26) D P H D AP (cid:27) D N N D N G D AG D AN D P G D P N D P N D P N D P A , (43) D SA (cid:26) D SN D SN (cid:27) D SN (cid:26) D P G D SG (cid:27) D P H (44) and (cid:26) D P P D P P (cid:27) (cid:26) D P P D P P (cid:27) D P P D P N D P S D AG . (45) Proof.
We shall prove each part separately.1.
For D P P D P P : Let us consider the function g P P P P ( x ) = f ′′ P P ( x ) (cid:14) f ′′ P P ( x ),where f ′′ P P ( x )and f ′′ P P ( x ) are as given by (23) and (22) respectively. This gives β P P P P = g P P P P (1) = f ′′ P P (1) f ′′ P P (1) = 43 . In order to prove this part we shall show that D P P − D P P >
0. We can write43 D P P − D P P = bf P P P P (cid:16) ab (cid:17) , where f P P P P ( x ) = 43 f P P ( x ) − f P P ( x ) = ( x + 1) ( x − ( x + 1) ( x + 1) . Since f P P P P ( x ) > ∀ x > , x = 1, hence proving the required result.13. For D P P SA : By considering the function g P P SA ( x ) = f ′′ P P ( x ) (cid:14) f ′′ SA ( x ),where f ′′ P P ( x ) and f ′′ SA ( x ) are as given by (22) and (35) respectively, we get β P P SA = g P P SA (1) = 6. We can write6 D SA − D P P = b f SA P P (cid:16) ab (cid:17) , where f SA P P ( x ) = 6 f SA ( x ) − f P P ( x ) = k ( x )( x + 1) ( x + 1) , with k ( x ) = 3 √ x + 2 ( x + 1) (cid:0) x + 1 (cid:1) − (cid:0) x + 5 x + 6 x + 5 x + 4 (cid:1) . Now we shall show that k ( x ) > ∀ x > , x = 1. Let us consider h ( x ) = h √ x + 2 ( x + 1) (cid:0) x + 1 (cid:1)i − (cid:0) x + 5 x + 6 x + 5 x + 4 (cid:1) . After simplifications we have h ( x ) = (cid:0) x + 4 x + 3 x + 4 x + 2 (cid:1) ( x − . Since h ( x ) > k ( x ) > ∀ x > , x = 1. Thus we have f P P P P ( x ) > ∀ x > , x = 1, thereby proving the required result. Argument:
Let a and b two positive numbers, i.e., a > and b > . If a − b > ,then we can conclude that a > b because a − b = ( a − b )/( a + b ) . We have used thisargument to prove k ( x ) > , ∀ x > , x = 1 . We shall use frequently this argumentto prove the other parts of the theorem. For D SA D SH : This result is already appearing in (11).4.
For D SH D AH : This result is already appearing in (11).5.
For D AH D P N : By considering the function g AH P N ( x ) = f ′′ AH ( x ) (cid:14) f ′′ P N ( x ),we get β AH P N = g AH P N (1) = , where f ′′ AH ( x ) and f ′′ P N ( x ) are as given by (40)and (17) respectively. We can write89 D P N − D AH = b f P N AH (cid:16) ab (cid:17) , where f P N AH ( x ) = 89 f P N ( x ) − f AH ( x ) = k ( x )18 ( x + 1) , with k ( x ) = 7 x + 18 x + 7 − √ x + 2 (cid:0) √ x + 1 (cid:1) ( x + 1) . k ( x ) > ∀ x > , x = 1. Let us consider h ( x ) = (cid:0) x + 18 x + 7 (cid:1) − h √ x + 2 (cid:0) √ x + 1 (cid:1) ( x + 1) i . After simplifications we have h ( x ) = (cid:0) x + 4 x / + 38 x + 4 √ x + 17 (cid:1) (cid:0) √ x − (cid:1) . Since h ( x ) > k ( x ) > ∀ x > , x = 1. Thus we have f P P AH ( x ) > ∀ x > , x = 1, thereby proving the required result.6. For D AH D P N : By considering the function g AH P N ( x ) = f ′′ AH ( x ) (cid:14) f ′′ P N ( x ),we get β AH P N = g AH P N (1) = , where f ′′ AH ( x ) and f ′′ P N ( x ) are as given by (40)and (18) respectively. We can write67 D P N − D AH = b f P N AH (cid:16) ab (cid:17) , where f P N AH ( x ) = 67 f P N ( x ) − f AH ( x ) = ( √ x −
18 ( x + 1) . Since f P N AH ( x ) > ∀ x > , x = 1, hence proving the required result.7. For D AH D SP : By considering the function g AH SP ( x ) = f ′′ AH ( x ) (cid:14) f ′′ SP ( x ), weget β AH SP = g AH SP (1) = , where f ′′ AH ( x ) and f ′′ SP ( x ) are as given by (40) and(33) respectively. We can write23 D SP − D AH = b f SP AH (cid:16) ab (cid:17) , where f SP AH ( x ) = 23 f SP ( x ) − f AH ( x ) = k ( x )10 ( x + 1) ( √ x + 1) , with k ( x ) = 4 ( x + 1) √ x + 2 (cid:0) √ x + 1 (cid:1) − h x + 10 x / + 17 x + 17 x + 10 x (cid:0) √ x − (cid:1) + 10 √ x + 5 i . Now we shall show that k ( x ) > ∀ x > , x = 1. Let us consider h ( x ) = h x + 1) √ x + 2 (cid:0) √ x + 1 (cid:1) i − h x + 10 x / + 17 x + 17 x + 10 x (cid:0) √ x − (cid:1) + 10 √ x + 5 i . h ( x ) = (cid:0) x + 70 x / + 201 x + 340 x / + 201 x + 70 √ x + 7 (cid:1) (cid:0) √ x − (cid:1) Since h ( x ) > k ( x ) > ∀ x > , x = 1. Thus we have f SP AH ( x ) > ∀ x > , x = 1, thereby proving the required result.8. For D P N D P N : By considering the function g P N P N ( x ) = f ′′ P N ( x ) (cid:14) f ′′ P N ( x ),we get β P N P N = g P N P N (1) = , where f ′′ P N ( x ) and f ′′ P N ( x ) are as given by(18) and (19) respectively. We can write1415 D P N − D P N = b f P N P N (cid:16) ab (cid:17) , where f P N P N ( x ) = 1415 f P N ( x ) − f P N ( x ) = ( √ x −
30 ( x + 1) . Since f P N P N ( x ) > ∀ x > , x = 1, hence proving the required result.9. For D P N D P P : By considering the function g P N P P ( x ) = f ′′ P N ( x ) (cid:14) f ′′ P P ( x ),we get β P N P P = g P N P P (1) = , where f ′′ P N ( x ) and f ′′ P P ( x ) are as given by(18) and (21) respectively. We can write23 D P P − D P N = b f P P P N (cid:16) ab (cid:17) , where f P P P N ( x ) = 23 f P P ( x ) − f P N ( x ) = √ x ( √ x − √ x + 1) ( x + 1) . Since f P P P N ( x ) > ∀ x > , x = 1, hence proving the required result.10. For D SP D P N : By considering the function g SP P N ( x ) = f ′′ SP ( x ) (cid:14) f ′′ P N ( x ),we get β SP P N = g SP P N (1) = , where f ′′ SP ( x ) and f ′′ P N ( x ) are as given by(33) and (19) respectively. We can write45 D P N − D SP = b f P N SP (cid:16) ab (cid:17) , where f P N SP ( x ) = 45 f P N ( x ) − f SP ( x ) = k ( x )4 ( x + 1) ( √ x + 1) , with k ( x ) = 3 x + 4 x / + 9 x + 4 x (cid:0) √ x − (cid:1) + 9 x + 4 √ x + 3 − √ x + 2 ( x + 1) (cid:0) √ x + 1 (cid:1) . k ( x ) > ∀ x > , x = 1. Let us consider h ( x ) = h x + 4 x / + 9 x + 4 x (cid:0) √ x − (cid:1) + 9 x + 4 √ x + 3 i − h √ x + 2 ( x + 1) (cid:0) √ x + 1 (cid:1) i . After simplifications we have h ( x ) = (cid:0) √ x − (cid:1) (cid:18) x ( √ x − + 4 x + 20 x / ++78 x + 20 x / + 4 x + (2 √ x − (cid:19) . Since h ( x ) > k ( x ) > ∀ x > , x = 1. Thus we have f P N SP ( x ) > ∀ x > , x = 1, thereby proving the required result.11. For D SP D P P : By considering the function g SP P P ( x ) = f ′′ SP ( x ) (cid:14) f ′′ P P ( x ),we get β SP P P = g SP P P (1) = , where f ′′ SP ( x ) and f ′′ P P ( x ) are as given by (33)and (21) respectively. We can write57 D P P − D SP = b f P P SP (cid:16) ab (cid:17) , where f P P SP ( x ) = 57 f P P ( x ) − f SP ( x ) = k ( x )14 ( x + 1) ( √ x + 1) , with k ( x ) = 10 x + 20 x / + 26 x + 26 x + 20 √ x + 10 − √ x + 2 (cid:0) √ x + 1 (cid:1) ( x + 1) . Now we shall show that k ( x ) > ∀ x > , x = 1. Let us consider h ( x ) = (cid:0) x + 20 x / + 26 x + 26 x + 20 √ x + 10 (cid:1) − h √ x + 2 (cid:0) √ x + 1 (cid:1) ( x + 1) i . After simplifications we have h ( x ) = 2 (cid:0) √ x − (cid:1) (cid:18) x + 8 x / + 94 x + 264 x / ++386 x + 264 x / + 94 x + 8 √ x + 1 (cid:19) . Since h ( x ) > k ( x ) > ∀ x > , x = 1. Thus we have f P P SP ( x ) > ∀ x > , x = 1, thereby proving the required result.172. For D P N D P G : By considering the function g P N P G ( x ) = f ′′ P N ( x ) (cid:14) f ′′ P G ( x ),we get β P N P G = g P N P G (1) = , where f ′′ P N ( x ) and f ′′ P G ( x ) are as given by(17) and (20) respectively. We can write34 D P G − D P N = b f P G P N (cid:16) ab (cid:17) , where f P G P N ( x ) = 34 f P G ( x ) − f P N ( x ) = k ( x )4 ( x + 1) , with k ( x ) = √ x + 2 (cid:0) √ x + 1 (cid:1) ( x + 1) − (cid:0) x + 1 + 3 x / + 3 √ x (cid:1) . Now we shall show that k ( x ) > ∀ x > , x = 1. Let us consider h ( x ) = h √ x + 2 (cid:0) √ x + 1 (cid:1) ( x + 1) i − (cid:0) x + 1 + 3 x / + 3 √ x (cid:1) . After simplifications we have h ( x ) = (cid:0) √ x − (cid:1) (cid:0) x + 2 x / + x + 2 √ x + 1 (cid:1) . Since h ( x ) > k ( x ) > ∀ x > , x = 1. Thus we have f P G P N ( x ) > ∀ x > , x = 1, thereby proving the required result.13. For D P N D P G : By considering the function g P N P G ( x ) = f ′′ P N ( x ) (cid:14) f ′′ P G ( x ),we get β P N P G = g P N P G (1) = , where f ′′ P N ( x ) and f ′′ P G ( x ) are as given by(19) and (20) respectively. We can write56 D P G − D P N = b f P G P N (cid:16) ab (cid:17) , where f P G P N ( x ) = 56 f P G ( x ) − f P N ( x ) = ( √ x −
12 ( x + 1) . Since f P G P N ( x ) > ∀ x > , x = 1, hence proving the required result.14. For D P P D P G : By considering the function g P P P G ( x ) = f ′′ P P ( x ) (cid:14) f ′′ P G ( x ),we get β P P P G = g P P P G (1) = , where f ′′ P P ( x ) and f ′′ P G ( x ) are as given by (21)and (20) respectively. We can write76 D P G − D P P = b f P G P P (cid:16) ab (cid:17) , where f P G P P ( x ) = 76 f P G ( x ) − f P P ( x ) = ( √ x − h ( √ x − + √ x i
12 ( x + 1) . Since f P G P P ( x ) > ∀ x > , x = 1, hence proving the required result.185. For D P G D P H : By considering the function g P G P H ( x ) = f ′′ P G ( x ) (cid:14) f ′′ P H ( x ),we get β P G P H = g P G P H (1) = , where f ′′ P G ( x ) and f ′′ P H ( x ) are as given by (20)and (29) respectively. We can write65 D P H − D P G = b f P H P G (cid:16) ab (cid:17) , where f P H P G ( x ) = 65 f P H ( x ) − f P G ( x ) = ( √ x − h ( √ x − + √ x i x + 1) . Since f P H P G ( x ) > ∀ x > , x = 1, hence proving the required result.16. For D P G AP : By considering the function g P G AP ( x ) = f ′′ P G ( x ) (cid:14) f ′′ AP ( x ),we get β P G AP = g P G AP (1) = 2, where f ′′ P G ( x ) and f ′′ AP ( x ) are as given by (20)and (34) respectively. We can write2 D AP − D P G = b f AP P G (cid:16) ab (cid:17) , where f AP P G ( x ) = 2 f AP ( x ) − f P G ( x ) = √ x ( √ x − ( √ x + 1) ( x + 1) . Since f AP P G ( x ) > ∀ x > , x = 1, hence proving the required result.17. For D P H N N : By considering the function g P H N N ( x ) = f ′′ P H ( x ) (cid:14) f ′′ N N ( x ),we get β P H N N = g P H N N (1) = 10, where f ′′ P H ( x ) and f ′′ N N ( x ) are as given by(29) and (41) respectively. We can write10 D N N − D P H = b f N N P H (cid:16) ab (cid:17) , where f N N P H ( x ) = 10 f N N ( x ) − f P H ( x ) = k ( x )2 ( √ x + 1) ( x + 1) , with k ( x ) = 5 √ x + 2 (cid:0) √ x + 1 (cid:1) ( x + 1) − (cid:0) x + 20 x / + 37 x + 32 x / + 37 x + 20 √ x + 7 (cid:1) . Now we shall show that k ( x ) > ∀ x > , x = 1. Let us consider h ( x ) = h √ x + 2 (cid:0) √ x + 1 (cid:1) ( x + 1) i − (cid:0) x + 20 x / + 37 x + 32 x / + 37 x + 20 √ x + 7 (cid:1) . h ( x ) = (cid:0) √ x − (cid:1) (cid:18) x + 24 x / + 72 x + 120 x / ++126 x + 120 x / + 72 x + 24 √ x + 1 (cid:19) . Since h ( x ) > k ( x ) > ∀ x > , x = 1. Thus we have f N N P H ( x ) > ∀ x > , x = 1, thereby proving the required result.18. For D AP N N : By considering the function g AP N N ( x ) = f ′′ AP ( x ) (cid:14) f ′′ N N ( x ),we get β AP N N = g AP N N (1) = 6, where f ′′ AP ( x ) and f ′′ N N ( x ) are as given by(34) and (41) respectively. We can write6 D N N − D AP = b f N N AP (cid:16) ab (cid:17) , where f N N AP ( x ) = 6 f N N ( x ) − f AP ( x ) = k ( x )2 ( √ x + 1) , with k ( x ) = 3 √ x + 2 (cid:0) √ x + 1 (cid:1) − (cid:0) x + 7 x / + 6 x + 7 √ x + 2 (cid:1) . Now we shall show that k ( x ) > ∀ x > , x = 1. Let us consider h ( x ) = h √ x + 2 (cid:0) √ x + 1 (cid:1) i − (cid:0) x + 14 x / + 12 x + 14 √ x + 4 (cid:1) After simplifications we have h ( x ) = 2 (cid:0) √ x − (cid:1) (cid:0) x + 2 x / + 2 √ x + 1 (cid:1) . Since h ( x ) > k ( x ) > ∀ x > , x = 1. Thus we have f N N AP ( x ) > ∀ x > , x = 1, thereby proving the required result.19. For D N N D N G : This result is already appearing in (11).20. For D N G D AG : This result is already appearing in (11).21. For D AG AN : This result is already appearing in (11).22. For D AN D P G : By considering the function g AN P G ( x ) = f ′′ AN ( x ) (cid:14) f ′′ P G ( x ),we get β AN P G = g AN P G (1) = , where f ′′ AN ( x ) and f ′′ P G ( x ) are as given by (38)and (28) respectively. We can write34 D P G − D AN = b f AN P G (cid:16) ab (cid:17) , f AN P G ( x ) = 16 f P G ( x ) − f AN ( x ) = k ( x )12 ( √ x + 1) , with k ( x ) = k ( x ) > ∀ x > , x = 1. Thus we have f N N AP ( x ) > ∀ x > , x = 1, thereby proving the required result.23. For D P G D P N : By considering the function g P G P N ( x ) = f ′′ P G ( x ) (cid:14) f ′′ P N ( x ),we get β P G P N = g P G P N (1) = , where f ′′ P G ( x ) and f ′′ P N ( x ) are as given by(28) and (27) respectively. We can write32 D P N − D P G = b f P G P N (cid:16) ab (cid:17) , where f P G P N ( x ) = 32 f P G ( x ) − f P N ( x ) = ( √ x − x + 1) . Since f P G P N ( x ) > ∀ x > , x = 1, hence proving the required result.24. For D P N D P N : By considering the function g P N P N ( x ) = f ′′ P N ( x ) (cid:14) f ′′ P N ( x ),we get β P N P N = g P N P N (1) = , where f ′′ P N ( x ) and f ′′ P N ( x ) are as given by(27) and (26) respectively. We can write65 D P N − D P N = b f P N P N (cid:16) ab (cid:17) , where f P N P N ( x ) = 65 f P N ( x ) − f P N ( x ) = ( √ x −
20 ( x + 1) . Since f P N P N ( x ) > ∀ x > , x = 1, hence proving the required result.25. For D P N D P N : By considering the function g P N P N ( x ) = f ′′ P N ( x ) (cid:14) f ′′ P N ( x ),we get β P N P N = g P N P N (1) = , where f ′′ P N ( x ) and f ′′ P N ( x ) are as given by(26) and (25) respectively. We can write109 D P N − D P N = b f P N P N (cid:16) ab (cid:17) , where f P N P N ( x ) = 109 f P N ( x ) − f P N ( x ) = k ( x )18 ( √ x + 1) , with k ( x ) = 2 (cid:0) x + 9 x / + 14 x + 9 √ x + 4 (cid:1) − √ x + 2 (cid:0) √ x + 1 (cid:1) . k ( x ) > ∀ x > , x = 1. Let us consider h ( x ) = (cid:2) (cid:0) x + 9 x / + 14 x + 9 √ x + 4 (cid:1)(cid:3) − h √ x + 2 (cid:0) √ x + 1 (cid:1) i . After simplifications we have h ( x ) = 2 (cid:0) √ x − (cid:1) (cid:0) x + 22 x / + 32 x + 22 √ x + 7 (cid:1) . Since h ( x ) > k ( x ) > ∀ x > , x = 1. Thus we have f P N P N ( x ) > ∀ x > , x = 1, thereby proving the required result.26. For D P N D P A : By considering the function g P N P A ( x ) = f ′′ P N ( x ) (cid:14) f ′′ P A ( x ),we get β P N P A = g P N P A (1) = , where f ′′ P N ( x ) and f ′′ P A ( x ) are as given by(25) and (24) respectively. We can write32 D P A − D P N = b f P A P N (cid:16) ab (cid:17) , where f P A P N ( x ) = 32 f P A ( x ) − f P N ( x ) = k ( x )4 ( √ x + 1) , with k ( x ) = √ x + 2 (cid:0) √ x + 1 (cid:1) − ( x + 1) (cid:0) x + 6 √ x + 1 (cid:1) . Let us consider h ( x ) = h √ x + 2 (cid:0) √ x + 1 (cid:1) i − (cid:2) ( x + 1) (cid:0) x + 6 √ x + 1 (cid:1)(cid:3) . After simplifications we have h ( x ) = (cid:0) √ x − (cid:1) ( x + 1) (cid:0) x + 4 √ x + 1 (cid:1) . Since h ( x ) > k ( x ) > ∀ x > , x = 1. Thus we have f P A P N ( x ) > ∀ x > , x = 1, thereby proving the required result.Parts 1-26 prove the sequences of inequalities appearing in (43).27. For D SA D SN : This result is already appearing in (11).28. For D SA D SN : This result is already appearing in (11).229. For D SN D SN : By considering the function g SN SN ( x ) = f ′′ SN ( x ) (cid:14) f ′′ SN ( x ),we get β SN SN = g SN SN (1) = , where f ′′ SN ( x ) and f ′′ SN ( x ) are as given by (36)and (37) respectively. We can write56 D SN − D SN = b f SN SN (cid:16) ab (cid:17) , where f SN SN ( x ) = 56 f SN ( x ) − f SN ( x ) = 124 × k ( x ) , with k ( x ) = 6 (cid:0) √ x + 1 (cid:1) √ x + 2 − h √ x + 2 + 5 (cid:0) √ x + 1 (cid:1) i . In order to prove k ( x ) > ∀ x > , x = 1, here we shall apply twice the argumentgiven in part 2. Let us consider h ( x ) = h (cid:0) √ x + 1 (cid:1) √ x + 2 i − h √ x + 2 + 5 (cid:0) √ x + 1 (cid:1) i . = 39 x + 41 x / + 3 √ x (cid:0) √ x − (cid:1) + 41 √ x + 39 − (cid:0) √ x + 1 (cid:1) √ x + 2 . Applying again over h ( x ) the argument given in part 2, we get h a ( x ) = h x + 41 x / + 3 √ x (cid:0) √ x − (cid:1) + 41 √ x + 39 i − h (cid:0) √ x + 1 (cid:1) √ x + 2 i = (cid:0) √ x − (cid:1) (cid:0) x + 3116 x / + 4806 x + 3116 √ x + 721 (cid:1) . Since h a ( x ) > h ( x ) > ∀ x > , x = 1 and consequently, we have k ( x ) > ∀ x > , x = 1. Thus we have f SN SN ( x ) > ∀ x > , x = 1, therebyproving the required result.30. For D SN D SN : This result is already appearing in (11).31. For D SN D P G : By considering the function g SN P G ( x ) = f ′′ SN ( x ) (cid:14) f ′′ P G ( x ),we get β SN P G = g SN P G (1) = , where f ′′ SN ( x ) and f ′′ P G ( x ) are as given by (37)and (20) respectively. We can write12 D P G − D SN = b f P G SN (cid:16) ab (cid:17) , where f P G SN ( x ) = 12 f P G ( x ) − f SN ( x ) = k ( x )4 ( x + 1) , k ( x ) = 3 x + 2 x + 3 − x + 1) √ x + 2 . Now we shall show that k ( x ) > ∀ x > , x = 1. Let us consider h ( x ) = (cid:0) x + 2 x + 3 (cid:1) − h x + 1) √ x + 2 i After simplifications we have h ( x ) = ( x − . Since h ( x ) >
0, giving k ( x ) > ∀ x > , x = 1. Thus we have f P G SN ( x ) > ∀ x > , x = 1, thereby proving the required result.32. For D SN D SG : This result is already appearing in (11).33. For D SG D P H : By considering the function g SG P H ( x ) = f ′′ SG ( x ) (cid:14) f ′′ P H ( x ), weget β SG P H = g SG P H (1) = , where f ′′ SG ( x ) and f ′′ P H ( x ) are as given by (42) and(29) respectively. We can write45 D P H − D SG = b f P H SG (cid:16) ab (cid:17) , where f P H SG ( x ) = 45 f P H ( x ) − f SG ( x ) = k ( x )4 ( x + 1) , with k ( x ) = 2 (cid:16) x + 5 x / + 11 x + 3 x (cid:0) √ x − (cid:1) + 11 x + 5 √ x + 4 (cid:17) − x + 1) (cid:0) √ x + 1 (cid:1) √ x + 2 . Now we shall show that k ( x ) > ∀ x > , x = 1. Let us consider h ( x ) = h (cid:16) x + 5 x / + 11 x + 3 x (cid:0) √ x − (cid:1) + 11 x + 5 √ x + 4 (cid:17)i − h x + 1) (cid:0) √ x + 1 (cid:1) √ x + 2 i . After simplifications we have h ( x ) = 2 (cid:0) √ x − (cid:1) (cid:18) x + 8 x / + 64 x + 120 x / ++242 x + 120 x / + 64 x + 8 √ x + 7 (cid:19) . Since h ( x ) >
0, giving k ( x ) > ∀ x > , x = 1. Thus we have f P H SG ( x ) > ∀ x > , x = 1, thereby proving the required result.Parts 27-33 prove the sequences of inequalities appearing in (44).244. For D P P D P P : By considering the function g P P P P ( x ) = f ′′ P P ( x ) (cid:14) f ′′ P P ( x ),we get β P P P P = g P P P P (1) = , where f ′′ P P ( x ) and f ′′ P P ( x ) are as given by(23) and (31) respectively. We can write169 D P P − D P P = b f P P P P (cid:16) ab (cid:17) , where f P P P P ( x ) = 169 f P P ( x ) − f P P ( x )= ( √ x − (cid:18) x + 10 x / + 23 x + 64 x / + 142 x ++180 x / + 142 x + 64 x / + 23 x + 10 √ x + 7 (cid:19) x + 1) ( x + 1) ( √ x + 1) . Since f P P P P ( x ) > ∀ x > , x = 1, hence proving the required result.35. For D P P D P P : By considering the function g P P P P ( x ) = f ′′ P P ( x ) (cid:14) f ′′ P P ( x ),we get β P P P P = g P P P P (1) = , where f ′′ P P ( x ) and f ′′ P P ( x ) are as given by(31) and (22) respectively. We can write1312 D P P − D P P = b f P P P P (cid:16) ab (cid:17) , where f P P P P ( x ) = 1312 f P P ( x ) − f P P ( x )= ( √ x − (cid:18) x + 30 x / + 89 x + 152 x / + 181 x ++190 x / + 181 x + 152 x / + 89 x + 30 √ x + 1 (cid:19)
12 ( x + 1) ( x + 1) ( √ x + 1) . Since f P P P P ( x ) > ∀ x > , x = 1, hence proving the required result.36. For D P P D P P : By considering the function g P P P P ( x ) = f ′′ P P ( x ) (cid:14) f ′′ P P ( x ),we get β P P P P = g P P P P (1) = , where f ′′ P P ( x ) and f ′′ P P ( x ) are as given by(32) and (31) respectively. We can write139 D P P − D P P = b f P P P P (cid:16) ab (cid:17) , where f P P P P ( x ) = 139 f P P ( x ) − f P P ( x )= ( √ x − (cid:18) x + 16 x / + 44 x + 88 x / + 139 x ++162 x / + 139 x + 88 x / + 16 √ x + 44 x + 4 (cid:19) x + 1) ( x + 1) ( √ x + 1) . Since f P P P P ( x ) > ∀ x > , x = 1, hence proving the required result.257. For D P P D P P : By considering the function g P P P P ( x ) = f ′′ P P ( x ) (cid:14) f ′′ P P ( x ),we get β P P P P = g P P P P (1) = , where f ′′ P P ( x ) and f ′′ P P ( x ) are as given by(22) and (30) respectively. We can write127 D P P − D P P = b f P P P P (cid:16) ab (cid:17) , where f P P P P ( x ) = 169 f P P ( x ) − f P P ( x )= ( √ x − (cid:18) x + x / + 17 x + 22 x / ++38 x + 22 x / + 17 x + √ x + 5 (cid:19) x + 1) ( x + 1) ( √ x + 1) . Since f P P P P ( x ) > ∀ x > , x = 1, hence proving the required result.38. For D P P D P P : By considering the function g P P P P ( x ) = f ′′ P P ( x ) (cid:14) f ′′ P P ( x ),we get β P P P P = g P P P P (1) = , where f ′′ P P ( x ) and f ′′ P P ( x ) are as given by(31) and (30) respectively. We can write97 D P P − D P P = b f P P P P (cid:16) ab (cid:17) , where f P P P P ( x ) = 97 f P P ( x ) − f P P ( x )= ( √ x − (cid:18) x + 6 x / + 16 x ++21 x / + 16 x + 6 √ x + 2 (cid:19) x + 1) ( x − √ x + 1) ( √ x + 1) . Since f P P P P ( x ) > ∀ x > , x = 1, hence proving the required result.39. For D P P D P N : By considering the function g P P P N ( x ) = f ′′ P P ( x ) (cid:14) f ′′ P N ( x ),we get β P P P N = g P P P N (1) = , where f ′′ P P ( x ) and f ′′ P N ( x ) are as given by(30) and (17) respectively. We can write97 D P N − D P P = b f P N P P (cid:16) ab (cid:17) , where f P P P N ( x ) = 149 f P N ( x ) − f P P ( x )= ( √ x − x + 268 x / + 1143 x + 782 x / ++2597 x + 2014 x / + 4478 x + 2014 x / ++2597 x + 782 x / + 1143 x + 268 √ x + 1
18 ( x + 1) ( √ x + 1) . Since f P N P P ( x ) > ∀ x > , x = 1, hence proving the required result.260. For D P N D P S : By considering the function g P N P S ( x ) = f ′′ P N ( x ) (cid:14) f ′′ P S ( x ),we get β P N P S = g P N P S (1) = , where f ′′ P N ( x ) and f ′′ P S ( x ) are as given by(17) and (16) respectively. We can write94 D P S − D P N = b f P S P N (cid:16) ab (cid:17) , where f P S P N ( x ) = 94 f P S ( x ) − f P N ( x ) = k ( x )8 ( x + 1) , with k ( x ) = 2 √ x + 2 (cid:0) √ x + 1 (cid:1) ( x + 1) + 10 x + 10 − x + 1) √ x + 2 . In order to prove k ( x ) > ∀ x > , x = 1, we shall apply twice the argumentgiven in part 2. Let us consider h ( x ) = h √ x + 2 (cid:0) √ x + 1 (cid:1) ( x + 1) + 10 x + 10 i − h x + 1) √ x + 2 i = 40 √ x + 2 (cid:0) √ x + 1 (cid:1) (cid:0) x + 1 (cid:1) ( x + 1) − h x + 8 ( x + 1) (cid:0) √ x − (cid:1) + 260 x + 28 x + 260 x + 46 i . Applying again over h ( x ) the argument given in part 2, we get h a ( x ) = h √ x + 2 (cid:0) √ x + 1 (cid:1) (cid:0) x + 1 (cid:1) ( x + 1) i − h x + 8 ( x + 1) (cid:0) √ x − (cid:1) + 260 x + 28 x + 260 x + 46 i . = 4 (cid:0) √ x − (cid:1)
71 + 2316 √ x + 11960 x / + 4090 x + 11960 x / ++11180 x / + 2316 x / + 11180 x / + 12021 x ++12021 x + 6004 x + 4090 x + 71 x . Since h a ( x ) > h ( x ) > ∀ x > , x = 1 and consequently, we have k ( x ) > ∀ x > , x = 1. Thus we have f P S P N ( x ) > ∀ x > , x = 1, therebyproving the required result.41. For D P S D AG : By considering the function g P S AG ( x ) = f ′′ P S ( x ) (cid:14) f ′′ AG ( x ), weget β P S AG = g P S AG (1) = 1, where f ′′ P S ( x ) and f ′′ AG ( x ) are as given by (16) and(17) respectively. We can write D AG − D P S = b f AG P S (cid:16) ab (cid:17) , where f AG P S ( x ) = f AG ( x ) − f P S ( x ) = k ( x )2 ( x + 1) , k ( x ) = √ x + 2 ( x + 1) − (cid:2) ( x − + 2 ( x + 1) √ x (cid:3) . Now we shall show that k ( x ) > ∀ x > , x = 1. Let us consider h ( x ) = h √ x + 2 ( x + 1) i − (cid:2) ( x − + 2 ( x + 1) √ x (cid:3) After simplifications we have h ( x ) = (cid:0) √ x + 1 (cid:1) (cid:0) √ x − (cid:1) . Since h ( x ) >
0, giving k ( x ) > ∀ x > , x = 1. Thus we have f AG P S ( x ) > ∀ x > , x = 1, thereby proving the required result.Parts 34-41 prove the sequences of inequalities appearing in (45). Remark 3.2. (i) In view of Theorem 3.1, we have the following improvement over theinequalities (11): D SA D SH D AH D N N D N G D AG D AN (cid:26) D SN D SN (cid:27) D SN D SG . (ii) The results appearing Parts 1-41 bring us some very interesting measures given by V k ( a, b ) = b f k ( a / b ) , k = 1 , , , , where f ( x ) = ( x + 1) ( x − ( x + 1) ( x + 1) ,f ( x ) = ( √ x − x + 1 ,f ( x ) = √ x ( √ x − ( x + 1) ( √ x + 1) and f ( x ) = ( √ x − h ( √ x − + √ x i
12 ( x + 1) . The measures V ( a, b ) is due to part 1, the measure V ( a, b ) is due to parts 6, 8, 13,23 and 24, the measure V ( a, b ) is due to parts 9 and 16 and the measure V ( a, b ) is dueto parts 14 and 15. Connections with Information Measures Γ n = ( P = ( p , p , ..., p n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p i > , n X i =1 p i = 1 ) , n > , be the set of all complete finite discrete probability distributions. For all P, Q ∈ Γ n , thefollowing inequalities are already proved by author [9]:12 D AH I D N N D N G D AG D AN J T
116 Ψ , (46)where I ( P || Q ) = 12 " n X i =1 p i ln (cid:18) p i p i + q i (cid:19) + n X i =1 q i ln (cid:18) q i p i + q i (cid:19) , (47) J ( P || Q ) = n X i =1 ( p i − q i ) ln (cid:18) p i q i (cid:19) , (48) T ( P || Q ) = n X i =1 (cid:18) p i + q i (cid:19) ln (cid:18) p i + q i √ p i q i (cid:19) (49)and Ψ( P || Q ) = n X i =1 ( p i − q i ) ( p i + q i ) p i q i . (50)The measures I ( P || Q ), J ( P || Q ) and T ( P || Q ) are the respectively, the well-know Jensen-Shannon divergence , J − divergence and arithmetic and geometric mean diver-gence . The measure Ψ( P || Q ) is symmetric chi-square divergence . For detailed studyon these measures refer to Taneja [8, 9, 11]. Moreover, D AH ( P || Q ) = ∆( P || Q ) and D AG ( P || Q ) = h ( P || Q ), where ∆( P || Q ) and h ( P || Q ) are the well-known triangular’s and Hellinger’s discriminations respectively.In the theorem below we shall make connections of the classical divergence measuresgiven in (47)-(50) with the inequalities given in (43). Moreover the theorem below unifiesthe inequalities (43) and (46).
Theorem 4.1.
The following inequalities hold: (cid:26) D P H D AP I (cid:27) D N N D N G D AG D AN D P G (cid:26) D P N D P N D P N D P A J (cid:27) T
116 Ψ . roof. In view of the inequalities given in (43) and (46), it sufficient to show the followingthree inequalities:(i) D AP I ;(ii) D P G J ;(iii) D P A T. Since in each part of the above expressions we have logarithmic form, we shall applya different approach to show the above three inequalities. (i) For D AP I: Let us consider g AP I ( x ) = f ′′ AP ( x ) f ′′ I ( x ) = 12 x ( x + 1) √ x ( √ x + 1) , x = 1 , x > . Calculating the first order derivative of the function g AP I ( x ) with respect to x , onegets g ′ AP I ( x ) = − (cid:0) x / − x + 3 √ x − (cid:1) √ x ( √ x + 1) = − √ x − √ x ( √ x + 1) ( > , x < < , x > g AP I ( x ) is increasing in x ∈ (0 ,
1) anddecreasing in x ∈ (1 , ∞ ), and hence β AP I = sup x ∈ (0 , ∞ ) g AP I ( x ) = g AP I (1) = 32 . (52)By the application of Lemma 3.1 with (52) we get the required result. (ii) For D P G J: Let us consider g P G J ( x ) = f ′′ P G ( x ) f ′′ J ( x ) = 3 √ x (cid:0) x / + 4 √ x + x − x + 1 (cid:1) x + 1) ( √ x + 1) . Calculating the first order derivative of the function g P G J ( x ) with respect to x , onegets g ′ P G J ( x ) = − √ x − (cid:0) x + 8 x / + 6 x + 8 √ x + 1 (cid:1) √ x ( √ x + 1) ( x + 1) ( > , x < < , x > g P G J ( x ) is increasing in x ∈ (0 ,
1) anddecreasing in x ∈ (1 , ∞ ), and hence β P G J = sup x ∈ (0 , ∞ ) g P G J ( x ) = g P G J (1) = 32 . (54)30y the application of Lemma 3.1 with (54) we get the required result. (iii) For D P A T: Let us consider g P A T ( x ) = f ′′ P A ( x ) f ′′ T ( x ) = 2 √ x (cid:0) x / + 4 √ x + x − x + 1 (cid:1) ( x + 1)( x + 1) ( √ x + 1) . Calculating the first order derivative of the function g P A T ( x ) with respect to x , onegets g ′ P A T ( x ) = − ( √ x − (cid:18) √ x ( x + 1) ( √ x − ++ x + 14 x + 10 x + 14 x + 1 (cid:19) √ x ( √ x + 1) ( x + 1) ( > , x < < , x > g P A T ( x ) is increasing in x ∈ (0 ,
1) anddecreasing in x ∈ (1 , ∞ ), and hence β P A T = sup x ∈ (0 , ∞ ) g P A T ( x ) = g P A T (1) = 12 . (56)By the application of Lemma 3.1 with (56) we get the required result.Parts (i)-(iii) completes the proof of the Theorem 4.1.Here we have referred Lemma 3.1, but its extension for the probability distributionsis already proved in [9, 10]. References [1] P. CZINDER and Z. PALES, Local monotonicity properties of two-variable Ginimeans and the comparison theorem revisited,
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