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# Refined Estimates on Conjectures of Woods and Minkowski-I

Kathuria L* and Raka M

Centre for Advanced Study in Mathematics, Panjab University, Chandigarh-160014, India

*Corresponding Author:
Kathuria L
Centre for Advanced Study in Mathematics
Panjab University, Chandigarh-160014
India
Tel: 08754216121
E-mail: [email protected]

Received Date: February 10, 2015; Accepted Date: March 23, 2015; Published Date: April 15, 2015

Citation: Kathuria L, Raka M (2015) Refined Estimates on Conjectures of Woods and Minkowski-I. J Appl Computat Math 4:209. doi: 10.4172/2168-9679.1000209

Copyright: © 2015 Kathuria L, et al. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

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#### Keywords

Lattice; Covering; Non-homogeneous; Product of linear forms; Critical determinant; Korkine and Zolotare reduction; Hermite's constant; Centre density

#### Introduction

Let ,be n real linear forms in n variables x1; : : : ; xn and having determinant The following conjecture is attributed to H. Minkowski:

Conjecture I: For any given real numbers c1; : : : ; cn, there exists integers x1; : : : ; xn such that

(1.1)

Equality is necessary if and only if after a suitable unimodular transformation the linear forms Li have the form 2cixlfor

This result is known to be true for For a detailed history and the related results,

Minkowski's Conjecture is equivalent to saying that [1]

where is given by

Chebotarev proved the weaker inequality

(1.2)

Since then several authors have tried to improve upon this estimate. The bounds have been obtained in the form

(1.3)

where Clearly by considering the linear forms Li=xi and for During 1949-1986, many authors such as Davenport, Woods, Bombieri, Gruber, Skubenko, Andrijasjan, Il'in and Malyshev obtained Vn for large n. obtained for all n [2-4] improved Mordell's estimates for Hans-Gill et al. [12,14] got improvements on the results of [5-8] for Since recently Vn 9=29/2 has been established by the authors [9], we study Vn for in a series of three papers.

In this paper we obtain improved estimates on Minkowski's Conjecture for n=10; 11 and 12. In next papers [10-12], we shall derive improved estimates on Minkowski's Conjecture for n=13; 14; 15 and for respectively [13-16]. For sake of comparison, we give results by our improved Vn in Table 1.

Estimates by Mordell Estimates by Il’in Estimates by Hans-Gill et al Our improved Estimates
n Vn Vn Vn Vn
10 2.899061 3.47989 24.3627506 27.60348
11 2.973102 3.52291 29.2801145 33.47272
12 3.040525 3.55024 32.2801213 39.59199
13 3.102356 3.57856 34.8475153 45.40041
14 3.159373 3.60209 37.8038391 51.26239
15 3.21218 3.61116 40.905198 57.00375
16 3.261252 3.61908 44.3414913 57.4702
17 3.306972 3.63924 47.2339309 57.67598
18 3.349652 3.66176 46.7645724 57.38876
19 3.389556 3.66734 47.2575897 60.09339
20 3.426907 3.67236 46.8640155 58.48592
21 3.461897 3.67692 46.0522028 56.42571
22 3.494699 3.68408 43.6612034 53.94142
23 3.525464 3.68633 37.8802374 50.98842
24 3.55433 3.68978 32.5852958 47.74632
25 3.581421 3.69295 27.8149432 42.39088
26 3.606852 3.69589 23.0801951 38.8657
27 3.630729 3.70012 17.3895105 31.93316
28 3.653149 3.70263 12.9938763 26.10663
29 3.674203 3.70497 9.5796191 19.96254
30 3.693976 3.70867 6.7664335 16.06884
31 3.712547 3.72558 4.745972 11.23872
32 3.729989     8.325879
33 3.746371     5.411488

Table 1: The weaker result.

We shall follow the Remak-Davenport approach. For the sake of convenience of the reader we give some basic results of this approach. Minkowski's Conjecture can be restated in the terminology of lattices as : Any lattice ^ of determinant d(^) in Rn is a covering lattice for the set

The weaker result (1.3) is equivalent to saying that any lattice ^ of determinant d(^) in Rn is a covering lattice for the set

Define the homogeneous minimum of ^ as

Proposition 1. Suppose that Minkowski Conjecture has been proved for dimensions 1, 2,…., n - 1: Then it holds for all lattices ^ in Rn for which MH(^)=0.

Proposition 2. If ^ is a lattice in Rn for with with MH(^) ≠ 0 then there exists an ellipsoid having n linearly independent points of ^ on its boundary and no point of ^ other than O in its interior.

It is well known that using these results, Minkowski's Conjecture would follow from

Conjecture II. If ^ is a lattice in Rn of determinant 1 and there is a sphere |X| <R which contains no point of ^ other than O in its interior and has n linearly independent points of ^ on its boundary then ^ is a covering lattice for the closed sphere of radius Equivalently, every closed sphere of radius lying in Rn contains a point of ^.

They formulated a conjecture from which Conjecture-II follows immediately. To state Woods' conjecture, we need to introduce some terminology [17,18].

Let L be a lattice in Rn. By the reduction theory of quadratic forms introduced by a cartesian co-ordinate system may be chosen in Rn in such a way that L has a basis of the form [19-22],

where A1;A2; : : : ;An are all positive and further for each i=1; 2; : : : ; n any two points of the lattice in Rn-i+1 with basis

are at a distance atleast Ai apart. Such a basis of L is called a reduced basis [23].

Conjecture III (Woods): If A1A2…An=1 and for each i then any closed sphere in Rn of radius contains a point of L.

Woods [10] proved this conjecture for Hans-Gill et al. [12] gave a unified proof of Woods' Conjecture for Hans-Gill et al. [12,14] proved Woods' Conjecture for n=7 and n=8 and thus completed the proof of Minkowski's Conjecture for n=7 and 8 Woods [10,24] proved Conjecture and hence Minkowski's Conjecture for n=9. With the assumptions as in Conjecture III, a weaker result would be that

If any closed sphere in Rn of radius contains a point of L [25,26].

Hans-Gill et al. [12,14] obtained the estimates wn on Woods' Conjecture for As w9=9 has been established by the authors [17] recently, we study wn for in a series of three papers. In this paper we obtain improved estimates wn on Woods' Conjecture for n=10; 11 and 12. In next papers [18,19], we shall derive improved estimates wnon Woods' Conjecture for n=13; 14; 15 and for respectively. Together with the following result of Hans-Gill et al. [12], we get improvements of wn for

Proposition 3. Let L be a lattice in Rn with A1A2…An=1 and for each i. Let where ln and mn are real numbers. Then L is a covering lattice for the sphere where Wn is defined inductively by

Here we prove

Theorem 1. Let n=10; 11; 12. If d(L)=A1 : : :An=1 and for i=2;….; n, then any closed sphere in Rn of radius contains a point of L, where

The earlier best known values were w10=10:5605061, w11=11:9061976 and w12=13:4499927.

To deduce the results on the estimates of Minkowski's Conjecture we also need the following generalization of Proposition 1

Proposition 4. Suppose that we know

Let vn<min Vk1 Vk2…Vks , where the minimum is taken over all (k1; k2; ; ks) such that n=k1+k2+ : : :+ks, ki positive integers for all i and . Then for all lattices in Rn with homogeneous minimum MH(<)=0, the estimate Vn holds for Minkowski's Conjecture.

Since by arithmetic-geometric inequality the sphere is a subset of Propositions 2 and 4 immediately imply

Theorem 2: The values of Vn for the estimates of Minkowski's Conjecture can be taken as

For these values are listed in Table 1. In Section 2 we state some preliminary results and in Sections 3-5 we prove Theorem 1 for n=10; 11 and 12.

#### Preliminary Results and Plan of the Proof

Let L be a lattice in Rn reduced in the sense of Korkine and Zolotare. Let (Sn) denotes the critical determinant of the unit sphere D Sn with centre O in Rn i.e.

has no point other than O in the interior of Sn}

Let be the Hermite's constant i.e. is the smallest real number such that for any positive de nite quadratic form Q in n variables of determinant D, there exist integers u1; u2;…; un not all zero satisfying

It is well known that We write A2Bi=Bi.

We state below some preliminary lemmas. Lemmas 1 and 2 are due to Woods [25], Lemma 3 is due to Korkine and Zolotare [21] and Lemma 4 is due to Pendavingh and Van Zwam [24]. In Lemma 5, the cases n=2 and 3 are classical results of Lagrange and Gauss; n=4 and 5 are due to Korkine and Zolotare [21] while n=6; 7 and 8 are due to Blichfeldt [3].

Lemma 1. If then any closed sphere of radius

in Rn contains a point of L.

Lemma 2. For a Fixed integer i with denote by L1 the lattice in Ri with reduced basis

and denote by L2 the lattice in Rn-i with reduced basis

If any closed sphere in Ri of radius r1 contains a point of L1 and if any closed sphere in Rn-i of radius r2 contains a point of L2 then any closed sphere in Rn of radius contains a point of L:

Lemma 3. For all relevant i,

(2.1)

Lemma 4. For all relevant i,

(2.2)

Throughout the paper we shall denote 0.46873 by ε .

Lemma 5. for n=2; 3; 4; 5; 6; 7 and 8 respectively:

Lemma 6. For any integer s;

and                  (2.4)

This is Lemma 4 of Hans-Gill et al. [12].

Lemma 7.

(2.5)

This is Lemma 6 of Hans-Gill et al. [14].

Remark 1. Let

=the best centre density of packings of unit spheres in Rn;

=the best centre density of lattice packings of unit spheres in Rn:

Then it is known that

(2.6)

and hence is known for Also =4 has been proved by Cohn and Kumar [6]. For using the bounds on given by Cohn and Elkies [5] and inequality (2.6) we find that , ,,

We assume that Theorem 1 is false and derive a contradiction. Let L be a lattice satisfying the hypothesis of the conjecture. Suppose that there exists a closed sphere of radius in Rn that contains no point of L in Rn.

Since Bi=A2 i and d(L)=1; we have B1B2 : : :Bn=1:

We give some examples of inequalities that arise. Let L1 be a lattice in R4 with basis (A1; 0; 0; 0), (a2;1;A2; 0; 0); (a3;1; a3;2;A3; 0); (a4;1; a4;2; a4;3;A4); and Li for be lattices in R1 with basis (Ai+3). Applying Lemma 2 repeatedly and using Lemma 1 we see that if then any closed sphere of radius

contains a point of L: By the initial hypothesis this radius exceeds Since and this results in the conditional inequality : if then

(2.7)

We call this inequality (4; 1;…; 1); since it corresponds to the ordered partition (4; 1;…; 1) of n for the purpose of applying Lemma 2. Similarly the conditional inequality (1;…; 1; 2; 1;…; 1) corresponding to the ordered partition (1;…; 1; 2; 1;…; 1) is : if then

(2.8)

Since (2.8) gives

One may remark here that the condition is necessary only if we want to use inequality (2.8), but it is not necessary if we want to use the weaker inequality (2.9). This is so because if using the partition (1; 1) in place of (2) for the relevant part, we get the upper bound which is clearly less than 2Bi+1. We shall call inequalities of type (2.9) as weak inequalities and denote it by (1;…; 1; 2; 1;….; 1)w.

If is an ordered partition of n, then the conditional inequality arising from it, by using Lemmas 1 and 2, is also denoted by If the conditions in an inequality are satisfied then we say that holds. Sometimes, instead of Lemma 2, we are able to use induction. The use of this is indicated by putting (*) on the corresponding part of the partition. For example, if for n=10, B5 is larger than each of B6;B7;….;B10, and if the inequality (4; 6*) gives

(2.10)

In particular the inequality ((n-1)*; 1) always holds. This can be written as

(2.11)

Also we have because if B1<1, then for each I contradicting B1B2:::Bn=1.

Using the upper bounds on and the inequality (2.5), we obtain numerical lower and upper bounds on Bn, which we denote by ln and mn respectively. We use the approach of Hans-Gill et al. [14], but our method of dealing with

Is somewhat different. In Sections 3-5 we give proof of Theorem 1 for n=10; 11 and 12 respectively. The proof of these cases is based on the truncation of the interval [ln;mn] from both the sides.

In this paper we need to maximize or minimize frequently functions of several variables. When we say that a given function of several variables in x; y; is an increasing/decreasing function of x; y;…., it means that the concerned property holds when function is considered as a function of one variable at a time, all other variables being fixed.

Proof of Theorem 1 for n=10

Here we have W10=10:3,B1<10<2:2636302. Using (2.5), we have l10=0:4007<B10<1:9770808=m10.

The inequality (9*; 1) gives 9(B10) -1/9 + B10<10:3. But for 0:4398 B10 1:9378, this inequality is not true. Hence we must have either B10<0:4398 or B10>1:9378. We will deal with the two cases 0:4007< B10<0:4398 and 1:9378<B10<1:9770808 separately:

0:4007<B10<0:4398

Using the Lemmas 3 & 4 we have:

Claim(i) B2>1:7046

The inequality (2; 2; 2; 2; 2)w gives 2B2 +2B4 +2B6 +2B8 +2B10>10:3. Using (3.1), we find that this inequality is not true for B2 ≤ 1:7046. Hence we must have B2>1:7046.

Claim(ii)

Suppose then using (3.1) and that we find that and So the inequality (1,4,4,1) holds, i.e.B1+4B2- Applying AM-GM inequality we get Now since , and we find that the left side is a decreasing function of B10 and B6. So replacing B10 by and we get Now the left side is a decreasing function of B2, so replacing B2 by 1.8815 we find that for 1<B1<2:2636302, a contradiction. Hence we must have B2<1:8815.

Claim (iii) B3<1:5652

Suppose From (3.1) we have B4B5B6<1:6524 and B8B9B10 <0:1702, so we find that and for B3>1:49.

Applying AM-GM to inequality (2,4,4) we get Since and we find that left side is a decreasing function of B1 and B7. So we replace B1 by B2, B7 by and get that

But left side is a decreasing function of B3, so replacing B3 by 1.5652 we find that for 1:7046<B2<1:8815, a contradiction. Hence we must have B3<1:5652.

Claim (iv) B1>1:9378

Suppose Using (3.1) and that B3<1:5652, B2>1:7046, we find that B2 is larger than each of B3; B4;…;B10. So the inequality (1; 9,*) holds. This gives which is not true for So we must have B1>1:9378.

Claim (v) B3<1:5485

Suppose We proceed as in Claim(iii) and replace B1 by 1.9378 and B7 by to get that

One easily checks that for 1.5485≤ B3<1:5652 and 1:7046< B2<1:8815. Hence we have B3<1:5485.

Claim (vi) B1<2:0187

Suppose Using (3.1) and Claims (ii), (v) we have B2B3B4<4:11. Therefore As we see using (3.1) that B5 is larger than each of B6;B7,… ;B10. Hence the inequality (4; 6,*) holds. This gives Left side is an increasing function of B2B3B4 and decreasing function of B1. So we can replace B2B3B4 by 4:11 and B1 by 2.0187 to find a contradiction. Hence we have B1<2:0187.

Claim (vii) B4<1:337

Suppose then using (3.1) we get Applying AMGM to inequality (1,2,4,2,1) we have

Since B2>1:7046, and we find that left side is a decreasing function of B2, B8 and B10. So we can replace B2 by 1.7046; B8 by and B10 by to get

Now left side is a decreasing function of B4, replacing B4 by 1:337, we find that for 1<B1<2:0187 and 1<B3<1:5485, a contradiction. Hence we have B4<1:337.

Claim (viii) B5<1:1492

Suppose Using (3.1), we get B6B7B8<0:5445: Therefore Also using Lemma 3 & 4, 2 .So the inequality (4*; 4; 2) holds, i.e.4 Now left side is a decreasing function of B5 and B9. So we replace B5 by 1.1492 and B9 by and get that where x=B6B7B8. Using Lemma 3 & 4 we have x==B6B7B8 and It can be verified that for and giving thereby a contradiction. Hence we must have B5<1:1492.

Claim (ix) B2<1:766.

Suppose We have B3B4B5<2:3793. So Also Therefore B6 is larger than each of Hence the inequality (1; 4; 5,*) holds. This gives Left side is an increasing function of B3B4B5, a decreasing function of B2 and an increasing function of B1. One easily checks that this inequality is not true for B1<2:0187;

and B3B4B5<2:3793: Hence we have B2<1:766.

As 2(B2+B4+B6+B8+B10)<2(1:766+1:337+0:9383+0:6597+0:4398)<10:3, the weak inequality (2; 2; 2; 2; 2)w gives a contradiction.

9378<B10<1:9770808

Here and B2=(B1B3…B10)-1 Which implies i.e. B2<1:75076.

Similarly

These respectively give B3<1:46138, B4<1:22883, B6<0:896058 and B8<0:721763. So we have Also and Applying AMGM to inequality (4,2,2,1,1) we have 4B1 + 4B5 + 4B7 + B9 +B10 We find that left side is a decreasing function of B7 and B5, so can replace B7 by and B5 by then it is a decreasing function of B1, so replacing B1 by B10 we have which is not true for and 1:9378<B10<1:9770808. Hence we get a contradiction.

#### Proof of Theorem 1 for n=11

Here we have w11=11.62, Using (2.5), we have l11=0:3673<B11<2:1016019=m11.

The inequality (10*; 1) gives 10:3 But for this inequality is not true. So we must have either or

#### 0:3673<B11<0:4409

Claim (i) B10<0:4692

Suppose then 2B10>B11, so (9*; 2) holds, i.e. 9 As left side is a decreasing function of B10, we can replace B10 by 0.4692 and find that it is not true for 0:3673<B11<0:4409.

Hence we must have B10<0:4692.

Using Lemmas 3 and 4 we have:

(4.1)

Claim (ii) B2>1:913

The inequality (2; 2; 2; 2; 2; 1) w gives 2B2+2B4+2B6+2B8+2B10+B11 > 11:62. Using (4.1) we find that this inequality is not true for B ≤1.913 so we must have B2>1:913.

Claim(iii) B3<1:761

Suppose then we have and

2. Applying AM-GM to the inequality (2,4,4,1) we get One easily finds that it is not true for and Hence we must have B3<1:761:

Claim (iv) B1<2:2436

Suppose As B2B3B4<2:13557×1:761×1:50151<5:6468, we have Also so B5 is larger than each of B6;B7…;B11. Hence the inequality (4; 7,*) holds. This gives Left side is an increasing function of B2B3B4 and decreasing function of B1. One easily checks that the inequality is not true for B2B3B4<5:6468 and B1 ≥ 2:2436. Hence we have B1<2:2436.

Claim (v) B4<1.4465 and B2>1:9686

Suppose B4 ≥ 1.4465 We have B5B6B7<1:2569 and B9B10B11<0:1295. Therefore for B4>1:36, we have and So the inequality (1,2,4,4) holds. Applying AM-GM to inequality(1,2,4,4), we get A simple calculation shows that this is not true for ,, and Hence we have B4<1:4465.

Further if then 2B2+2B4+2B6+2B8+2B10+B11<11:62. So the inequality (2; 2; 2; 2; 2; 1)w gives a contradiction.

Claim (vi) B4<1:4265 and B2>1:9888

Suppose We proceed as in Claim (v) and get a contradiction with improved bounds on B2 and B4.

Claim (vii) B1<2:2056

Suppose As B3B4B5<1:761 × 1:4265 × 1:3347<3:3529, we have Also so B6 is larger than each of B7;B8,…,B11. Hence the inequality (1; 4; 6*) holds, i.e. B1 + 4B2 -

Claim (ix) B1<2:1669

Suppose We proceed as in Claim(iv) and get a contradiction with improved bounds on B1, B2 and B4.

Claim (x) B4<1:403 and B2>2:012

Suppose We proceed as in Claim(v) and get a contradiction with improved bounds on B2 and B4.

As now B3B4B5<1:761×1:403 1:3347<3:2977, we have for B2>2:012. Also each of B7; B8; B11. Hence the inequality (1; 4; 6) holds. Proceeding as in Claim (viii) we find that this inequality is not true for B1<2:1669; B2>2:012 and B3B4B5<3:2977; giving thereby a contradiction.

#### 2:018<B11<2:1016019

Here Therefore using Lemmas 3 & 4 we have B10=(B1 B9B11)-1

Similarly

which gives B4<1:37661.

Claim (i) B10<0:4402

The inequality (9*; 1; 1) gives But this inequality is not true for and 2:018<B11<2:1016019. Hence we must have B10<0:4402.

Now we have < 0:58694, and

Claim (ii) B7<0:768

Suppose Then so (6*; 4; 1) holds. This gives where x=B1B2 : : :B6. The function has its maximum value at Therefore which is less than 11:62 for 2:018<B11<2:1016019. This gives a contradiction.

Now and

Claim (iii) B2<1:795

Suppose then and Applying AMGM to the inequality (1,4,4,1,1) p we get B1 + 4B2 + 4B6 + B10 + B11 - We find that left side is a decreasing function of B6, so we first replace B6 by εB2 then it is a decreasing function of B2, so we replace B2 by 1.795 and get that

Now so which can be verified to be at most 11.62 for and 2:018<B1<2:393347, giving thereby a contradiction.

Claim (iv) B5<0:98392

Suppose We have and Also Applying AM-GM to the inequality (4; 4; 2; 1) we get One can easily check that left side is a decreasing function of B9 and B1 so we can replace B9 by εB and B1 by B11 to get Now the left side is a decreasing function of B5, so replacing B5 by 0.98392 we see that and 2:018<B11 < 2:1016019, a contradiction.

As in Claim(iv), we have Also each of B6;B7,…,B10. Therefore the inequality (4; 6*; 1) holds, i.e. Left side is an increasing function of B2B3B4 and B11 and decreasing function of B1. Using B5<0:98392, we have and One easily checks that for B2B3B4<1:795 × 1:47588 × 1:311894, B11<2:1016019 and Hence we have a contradiction.

#### Proof of Theorem 1 for n=12

Here we have w12=13, Using (2.5), we have l12 =0:3376<B12<2:2254706=m12 and using (2.3) we have i.e

The inequality (11*; 1) gives 11:62(B12)-1/11 +B12>13. But this is not true for So we must have either B12<0:4165 or B12>2:17.

#### 0:3376<B12<0:4165

Claim (i) B11<0:459

Suppose then and 2B11>B12, so (10*; 2) holds, i.e, Left side is a decreasing function of B11, so we can replace B11 by .459 to find that for 0:34425<B12<0:4165, a contradiction. Hence we have B11<0:459.

Claim (ii) B10<0:5432

Suppose From Lemma 3, and B12.Therefore and so the inequality (9*; 3) holds, i.e. One easily checks that it is not true noting that left side is a decreasing function of B10. Hence we must have B10<0:5432.

Claim (iii) B9<0:6655

Suppose then So the inequality (8*; 4) holds. This gives where x=B1B2 … B8. The function has its maximum value at so for where x=B1B2… B8. The function has its maximum value at so for 0:6655 This gives a contradiction.

Using Lemmas 3 & 4 we have:

Claim (iv) B2>1:828, B4>1:426, B6>1:019 and B8>0:715

Suppose Then 2(B2+B4+B6+B8+B10+B12)<2(1:828+ 1:7384+1:1589+0:8148+0:5432+0:4165)<13, giving thereby a contradiction to the weak inequality (2; 2; 2; 2; 2; 2) w.

Similarly we obtain lower bounds on B4;B6 and B8 using (2; 2; 2; 2; 2; 2)w.

Claim(v) B2>2:0299

Suppose Consider following two cases:

Case (i) B3>B4

We have B3>B4>1:426>each of B5,….,B12. So the inequality (2; 10*) holds, i.e. The left side is a decreasing function of B1, so replacing B1 by B2 we get 2B2 + 10:3 which is not true for

Case (ii)

As B4>1:426>each of B5,….,B12, the inequality (3; 9*) holds, i.e. where X=B2B3<min say. Now is an increasing function of X for and so which can be seen to be less than 13. Hence we have B2>2:0299.

Claim (vi) B1>2:17 and B3<1:9517

Using (2.3) we have Therefore B2>2:0299>each of B3,…,B12. So the inequality (1; 11*) holds, i.e. B1 + 11:62 But this is not true for So we must have B1>2:17: Again using (2.3) we have

Claim (vii) B4<1:646

Suppose From (5.1) and Claims (i)-(iii), we have and Applying AM-GM to the inequality (1,2,4,4,1) we get We find that left side is a decreasing function of B2, B8 and B12. So we can replace B2 by 2:0299, B8 by "B4 and B12 by . Then it turns a decreasing function of e 2B4, so can replace B4 by 1.646 to find that ,a contradiction. Hence we have B4<1:646.

Claim (viii) B1<2:4273

Suppose B1≥2:4273. Consider following two cases:

Case (i) B5>B6

Here B5>each of B6,…,B12 as B5 ≥ B1>1.137 >each of B7,…, B12. Also B2B3B4<2:2254706×1:9517×1:646<7:15. So Hence the inequality (4; 8*) holds. This gives

Left side is an increasing function of B2B3B4 and decreasing function of B1. So we can replace B2B3B4 by 7.15 and B1 by 2.4273 to get a contradiction.

Case (ii)

Using (5.1) we have <1:1589 and so Therefore as B2>2:0299 and B3<1:9517. Also from Claim (iv), B6>1:019>each of B7,….,B12. Hence the inequality (1; 4; 7*) holds. This gives 7(B1B2B3B4B5)-1/7>13: Left side is an increasing function of B3B4B5 and B1 and a decreasing function of B2. One can check that inequality is not true for B3B4B5<1:9517×1:5452 ×1:1589, B1<2:5217871 and for B2>2:0299: Hence we must have B1<2:4273:

Claim (ix) B5<1:396

Suppose From (5.1), B6B7B8<0:925 and B10B11B12<0:104, so we have and Applying AMGM to the inequality (1,2,1,4,4) we get + We find that left side is a decreasing function of B2 and B9. So we replace B2 by 2:0299 and B9 by ε B5. Now it becomes a decreasing function of B5 and an increasing function of B1 so replacing

B5 by 1.396 and B1 by 2.4273, we find that above inequality is not true for 1:522<B3<1:9517 and 1:426<B4<1:646, giving thereby a contradiction. Hence we must have B5<1:396.

Claim (x) B3>1:7855

Suppose We have B4>1:426>each of B5;B6,…,B12, hence the inequality (1; 2; 9*) holds. It gives It is easy to check that left side of above inequality is a decreasing function of B2 and an increasing function of B1 and B3. So replacing B1 by 2.4273, B3 by 1.7855 and B2 by 2.0299 we get -15<13; a contradiction. Hence we have B3>1:7855.

Claim (xi)

Suppose We have B3>1:7855>each of B4;B5,…,B12, hence the inequality (2; 10*) holds. It gives The left side is a decreasing function of B1 and an increasing function of B2, so replacing B1 by 2:17 and B2 by 2.0733 we get a contradiction.

Claim (xii) B7<0:92 and B5<1:38

Suppose Here we have B4B5B6<2:67 and B8B9B10<0:295, so and Also Applying AM-GM to the inequality (2,4,4,2) we get We find that left side is a decreasing function of B1 and B11. So we can replace B1 by 2:17 and B11 by ε B7. Then left side becomes a decreasing function of B7 and an increasing function of B2, so can replace B7 by 0.92 and B2 by 2.2254706 to see that for 1:7855<B3<1:9517 and 0:3376<B12<0:4156, a contradiction. Hence B7<0:92. Further gives B5<1:38.

Claim (xiii) B6<1:097

Suppose Here we have B3B4B5<4:44 and B7B8B9<0:5, so and Also Applying AM-GM to the inequality (1,4,4,2,1) we get We find that left side is a decreasing function of B10, B12 and B11. So we can replace B10 by εB6 and B12 by 0.3376 and B11 by B6. Then left side becomes a decreasing function of B6, so we can replace B6 by 1.097 to find that Hence we must have B6<1:097.

Claim (xiv) B5>B6 and

First suppose B5 ≤ B6, then B4B5B6<1:646 × 1:0972<1:981 and Also each of B8,…,B12. Hence the inequality (2; 4; 6*) holds, i.e. Now the left side is a decreasing function of B1 and B3 as well; also it is an increasing function of B2 and B4B5B6. But one can check that this inequality is not true for B1>2:17, B3>1:7855, B2<2:2254706 and B4B5B6<1:981, giving thereby a contradiction. Further suppose then as B5>B6>1:019>each of B7,…, B12, the inequality (4; 8*) holds. Now working as in Case (i) of Claim (viii) we get contradiction for B1>2:17 and B2B3B4<2:2254706 × 1:9517 × 1:646<7:14934.

Claim (xv) B3<1:9 and B1<2:4056

Suppose ,then for B4B5B6<1:646×1:38×1:097<2:492, Also B7≥ε B3>0:89>each of B8,…,B12. Hence the inequality (2; 4; 6*) holds. Now working as in Claim (xiv) we get contradiction for B1>2:17, B2<2:2254706, B3>1:9 and B4B5B6<2:492. So B3<1:9. Further if B1≥ 2:4056, then contradicting Claim (xiv).

Claim (xvi) B4<1:58 and B1<2:373

Suppose then for B5B6B<1:38 × 1:097 × 0:92<1:393, Also B8 B4>0:74>each of B9,…,B12. Hence the inequality (1; 2; 4; 5*) holds, i.e. -19=B1 + 4B2 - Left side is a decreasing function of B2 and B4.

So we replace B2 by 2.0733 and B4 by 1.58. Then it becomes an increasing function of B1, B3 and B5B6B7. So we replace B1 by 2.4056, B3 by 1.9 and B5B6B7 by 1.393 to find that -19<13, a contradiction. Further if B1 ≥ 2:373, then contradicting Claim (xiv).

We have B3B4B<1:9×1:58×1:38<4:15. Therefore >2 Also B6>1:019>each of B7,…,B12. Hence the inequality (1; 4; 7*) holds. Now we get contradiction working as in Case (ii) of Claim (viii).

5.2 2:17<B12<2:2254706

Here 2.17Using Lemma 3 and 4, we have

B11=(B1B2…B10B12)-1<

Claim (i) Either B11<0:4307 or B11>1:818

Suppose 0:4307 ≤B11≤1.818 The inequality (10*; 1; 1) gives 10:3 which is not true for 0:4307 ≤B11≤1.818 and 2:17<B12<2:2254706. So we must have either B11<0:4307 or B11>1:818

Claim (ii) B11<0:4307

Suppose B11 ≥ 0.4307 then using Claim(i) we have B11>1:818. Now we have using Lemmas 3 & 4,

This gives B2<1:777.

This gives B3<1:487

This gives B4<1:213.

This gives B6<0:826.

This gives B7<0:697.

This gives B8<0:559.

This gives B9<0:478.

Therefore we have and each of B6,…,B10. So the inequality (4; 6*; 1; 1) holds, i.e.4B1 (B1B2B3B4B11B12)-1/6 +B11+B12>13 Now the left side is an increasing function of B2B3B4, B11 and of B12 as well. Also it is a decreasing function of B1. So we replace B2B3B4 by 1:777 × 1:487 × 1:213, B11 by 1.8223, B12 by 2.2254706 and B1 by 2.17 to arrive at a contradiction. Hence we must have B11<0:4307.

Claim (iii) B10<0:445

Suppose then 2B10>B11. So the inequality (9*; 2; 1) holds, i.e. B10 and B12>2:2254706, the left side is an increasing function of B12 and a decreasing function of B10, so replacing B12 by 2.2254706 and B10 by 0.445 we find that ,for 3 4(0:445)<B11<0:4307, a contradiction. Hence we must have B10<0:445.

Using Lemmas 3 and 4 we have:

Claim (iv) B3<1:62

Suppose From (5.2), we have B4B5B6<1:712 and B8B9B10<0:178, so and Applying AM-GM to the inequality (2,4,4,1,1) we get We find that left side is a decreasing function of B1, B7 and B11. So we can replace B1 by B12, B7 by ε B3 and B11 by e 2B3. Then it becomes a decreasing function of B3, so replacing B3 by 1.62 we find that φ 21<13; for 1:6275<B2<2:0255 and 2:17<B12<2:2254706, a contradiction. Hence we must have B3<1:62.

Claim (v) B12>2:196

Suppose From (5.2), we have B2B3B4<4:674 and Also each of B6,…, B11. Therefore the inequality (4; 7*; 1) holds, i.e. Left side is an increasing function of B2B3B4 and of B12 as well. Also it is a decreasing function of B1. So we can replace B2B3B4 by 4.674, B12 by 2.196 and B1 by 2.17 to get Φ 22<13, a contradiction. Hence we must have B12>2:196.

Now we have B1 ≥B12>2:196. We proceed as in Claim(v) and use (4; 7*; 1). Here we replace B2B3B4 by 4.674, B12 by 2.2254706 and B1 by 2.196 to get Φ22<13, a contradiction.

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