PROVING INEQUALITIES IN ACUTE TRIANGLE WITH DIFFERENCE SUBSTITUTION

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Proving Inequalities with Difference Substitution Yu-Dong Wu, Zhi-Hua Zhang

and Yu-Rui Zhang vol. 8, iss. 3, art. 81, 2007

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PROVING INEQUALITIES IN ACUTE TRIANGLE WITH DIFFERENCE SUBSTITUTION

YU-DONG WU ZHI-HUA ZHANG

Xinchang High School Zixing Educational Research Section

Xinchang City, Zhejiang Province 312500 Chenzhou City, Hunan Province 423400

P.R. China. P. R. China.

EMail:zjxcwyd@tom.com EMail:zxzh1234@163.com

YU-RUI ZHANG

Xinchang High School

Xinchang City, Zhejiang Province 312500 P. R. China.

EMail:xczxzyr@163.com

Received: 09 October, 2006

Accepted: 04 April, 2007

Communicated by: B. Yang 2000 AMS Sub. Class.: 26D15.

Key words: Inequalities, Acute Triangle, Difference Substitution, Linear transformation.

Abstract: In this paper, we prove several inequalities in the acute triangle by means of so- called Difference Substitution. As generalization of the method, we also consider an example that the greatest interior angle is less than or equal to120^◦in the triangle.

Acknowledgements: The authors would like to thank Professor Lu Yang for his enthusiastic help.

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1. Introduction

In [1, 2], L. Yang suggested the use of Difference Substitution to prove asymmetric polynomial inequalities, as it had been used previously to deal with symmetric ones.

Ifx₁ ≤x₂ ≤x₃ ≤ · · · ≤x_nwithn∈N^∗, then we set

(1.1)











x₁ =t₁, x2 =t1+t2, x3 =t1+t2+t3,

· · · ·

x_n=t₁+t₂+t₃+· · ·+t_n, wheret_i ≥0for2≤i≤nandi∈N^∗.

The expansion (1.1) is so-called a “splitting” transformation, and{t₁, t₂, . . . , t_n} is simply the difference sequence of{x₁, x₂, . . . , x_n}.

In general, for then-variant polynomials, there aren!different orders of{x₁, x₂, . . . , x_n}, sorting by size. In the instance ofn= 3, we letx≤y ≤z, and take

(1.2)







x=u, y=u+v, z =u+v+w, wherev ≥0,w≥0.

Analogously, ify≤x≤z, then its “splitting” transformation is

(1.3)







y=u, x=u+v, z =u+v+w,

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wherev ≥0,w≥0.

Sequentially, for y ≤ z ≤ xorz ≤ x ≤ yorz ≤ y ≤ xorx ≤ z ≤ y, we set four similar linear transformations.

For a 3-variant polynomial F(x, y, z), by using the six linear transformations above, we obtain 6 membersP_i(u, v, w)with1≤i≤6, and call the set{P₁, P₂, . . . , P₆} the Difference Substitution of F(x, y, z) and denote this byDS(F). If all the co- efficients of these membersDS(F)are nonnegative, then F ≥ 0whenever x, y, z all are nonnegative. In other words, F is positive semi-definite on R³+. Difference substitution is a very valid method for proving inequalities. For more information on Difference Substitution, please refer to [3] and [4].

In this paper, by using Difference Substitution, the authors prove several inequalities in acute triangles.

Throughout the paper we denoteA, B, C as the interior angles,a, b, cas the side- lengths, S as the area, sas the semi-perimeter, R as the circumradius, r as the in- radius, h_a, h_b, h_c as the altitudes, m_a, m_b, m_c as the medians, and r_a, r_b, r_c as the radii of the described circles of triangleABC respectively. Moreover, we will cus- tomarily use the cyclic sum symbol, that is: P

f(a) = f(a) +f(b) +f(c), and Pf(a, b) =f(a, b) +f(b, c) +f(c, a), etc.

Let us begin with the well-known Walker’s inequality [5]. In the acute triangle, show that

(1.4) s² ≥2R²+ 8Rr+ 3r²,

or

(1.5) −2a³b³+a⁴b²−a⁴bc+a⁵b+ab⁵+b⁵c+b⁴c²

−2b³c³+b²c⁴ −2c³a³+c⁴a²+c⁵a+c⁵b+c²a⁴

+a⁵c−ab⁴c+a²b⁴−b⁶−c⁶−a⁶−abc⁴ ≥0.

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Let

(1.6)











x= ^b+c−a₂ >0, y = ^c+a−b₂ >0, z = ^a+b−c₂ >0.

Then inequality (1.4) or (1.5) is equivalent to

(1.7) F(x, y, z) = 6xyz⁴+ 2xy²z³+ 2xy³z² + 6xy⁴z+ 2x²yz³+ 2x²y³z + 2x³yz²+ 2x³y²z+ 6x⁴yz−x⁴y²−x²z⁴ −2x³z³−x⁴z²

−2x³y³−y⁴z²−y⁴x²−2y³z³−y²z⁴−18x²y²z² ≥0.

There is no harm in supposingx ≤ y ≤ z since inequality (1.7) is symmetric for x, y, z. Then, by using (1.2), for the acute triangle, it follows that

b²+c²−a² = (z+x)²+ (x+y)²−(y+z)² = 2[x²+ (y+z)x−yz]

(1.8)

= 2{u²+ [(u+v) + (u+v+w)]u−(u+v)(u+v+w)}

= 2(2u²−v² −vw)>0, andF(x, y, z)in (1.7) is transformed into

F(x, y, z) (1.9)

=P(u, v, w)

= 2u²−v²−vw 4v²+ 4w²+ 4vw u² + 8v³+ 20vw²+ 12v²w+ 8w³

u+ 4v⁴+ 8v³w+ 2w⁴ +18vw³+ 22v²w²

+ 24v³w²+ 36v²w³+ 12vw⁴ u + 34v³w³+ 19v²w⁴+ 2vw⁵+ 17v⁴w².

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Obviously F(x, y, z) = P(u, v, w) ≥ 0 from (1.8) and u > 0, v ≥ 0, w ≥ 0, i.e., inequality (1.4) or (1.5) is true.

Now, let us consider another semi-symmetric inequality [6] in the acute triangle

(1.10) cos(B −C)≤ h_a

m_a. It is equivalent to

(1.11) −a⁴+ 3b²+ 3c²

a²−2 (b−c)²(b+c)² ≥0, and from (1.6), this equals

(1.12) F(x, y, z)

= −y²−z²+ 14yz

x²−(y+z) z²−14yz+y²

x+yz(y+z)² ≥0.

Calculating DS(F), it consists of 3 polynomials with u > 0, v ≥ 0, w ≥ 0 as follows

(1.13) P1(u, v, w)

= 40u⁴+ 112u³v + 108u²v²+ 56u³w+ 14u²w² + 40uv³+ 20uvw² + 60uv²w+ 108u²vw+ 8v³w+ 5v²w²+vw³+ 4v⁴,

(1.14) P2(u, v, w)

= 2u²−v²−vw

20u²+ (24w+ 52v)u+ 53v²+ 6w²+ 52vw + 72v³+ 36vw²+ 108v²w

u+ 57v²w² + 51v⁴+ 6vw³+ 102v³w,

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and

(1.15) P₃(u, v, w)

= 2u²−v²−vw

20u²+ (52v + 28w)u+ 53v²+ 54vw+ 7w² + 72v³+ 36vw²+ 108v²w

u+ 57v²w² + 51v⁴+ 6vw³+ 102v³w.

By (1.8), we immediately obtain Pi(u, v, w) ≥ 0for1 ≤ i ≤ 3. Hence, inequality (1.10) is proved.

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2. Some Problems and their Proofs

2.1. The Problems

In 2004-2005, J. Liu [7,8] posed the following conjectures for the inequality in the acute triangle.

Problem 1. Let4ABC be an acute triangle. Prove the following inequalities

(2.1) X

sin 2A sinB+ sinC

2

≤ 3 4, and

(2.2) sinA

2 ≤

√m_bm_c 2m_a . 2.2. The Proof of Inequality (2.1)

Proof. Usingsin 2α= 2 sinαcosα, we find that inequality (2.1) is equivalent to 4a¹⁰b² −10a⁵b⁵c² −24b⁶c⁵a+ 16a⁹b³+ 4a⁸b⁴−5b⁴c⁶a²

(2.3)

−8a¹¹b−5a⁴b⁶c²+ 8a⁹c²b+ 4a⁸b²c²−16a¹⁰cb+ 8a⁹cb²−5a⁶b⁴c² + 32a⁸b³c+ 8a⁷b⁴c+ 8a⁷c⁴b−5a⁶c⁴b²+ 32a⁸c³b+ 4a¹⁰c²−8a¹¹c

−4a¹²−4b¹²−4c¹²+ 4a⁸c⁴+ 16a⁹c³−8a⁷b⁵−8a⁶b⁶ −8a⁶c⁶

−8a⁷c⁵−8a⁵b⁷ + 4a⁴b⁸−24a⁶b⁵c−24a⁵b⁶c+ 2a⁵b³c⁴ + 6a⁴b⁴c⁴ + 4c¹⁰b²−26a⁶b³c³+ 2a⁵b⁴c³−24a⁵c⁶b−5a⁴c⁶b²−24a⁶c⁵b

−10a⁵c⁵b²+ 8a⁴b⁷c−8c¹¹b+ 32b⁸a³c+ 8b⁹a²c+ 4b⁸a²c² + 2b⁵c³a⁴−26b⁶c³a³+ 2b⁵c⁴a³−5b⁶c⁴a²−16b¹⁰ca+ 8b⁹c²a

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+ 32b⁸c³a+ 8b⁷c⁴a−10b⁵c⁵a²+ 16b⁹a³+ 4b¹⁰a²−8b¹¹a

−8b¹¹c+ 4b¹⁰c²+ 16b⁹c³+ 4b⁸c⁴−8b⁷c⁵−8b⁶c⁶+ 4a⁴c⁸

−8a⁵c⁷−24b⁵c⁶a+ 8a⁴c⁷b+ 2c⁵b⁴a³−26c⁶b³a³+ 2c⁵b³a⁴

+ 4c⁸b²a²+ 8c⁹a²b+ 32c⁸a³b+ 8b⁴c⁷a−8c¹¹a+ 32c⁸b³a+ 8c⁹b²a

−16c¹⁰ab+ 4c¹⁰a²+ 16c⁹a³−8b⁵c⁷+ 4b⁴c⁸+ 16c⁹b³ ≥0.

From (1.6), inequality (2.3) equals F(x, y, z)

(2.4)

=−4576x⁷z⁵−5590x⁶z⁶−116x¹⁰z²−2453x⁴z⁸−2453x⁸z⁴

−4576x⁵z⁷−788x³z⁹−788x⁹z³−2453x⁸y⁴−4576y⁷z⁵

−2453y⁸z⁴−2453x⁴y⁸ −788x⁹y³−788x³y⁹−5590x⁶y⁶

−4576x⁷y⁵−4576x⁵y⁷ −2453y⁴z⁸−4576y⁵z⁷−5590y⁶z⁶

−116y¹⁰z²−788y⁹z³−116y¹⁰x²−788y³z⁹−116y²z¹⁰ + 13448x⁶y⁵z+ 8176x⁷yz⁴+ 13448x⁶yz⁵ + 13448x⁵yz⁶ + 8176x⁴yz⁷+ 6448x²y³z⁷+ 1220xy⁹z²+ 6448x²y⁷z³ + 6448x³y⁷z²+ 1220x⁹yz²+ 10862x⁴y⁶z²+ 14288x²y⁵z⁵ + 10862x²y⁴z⁶+ 14288x⁵y²z⁵+ 10862x⁶y²z⁴+ 6448x⁷y²z³

−28248x³y⁵z⁴−28248x⁴y⁵z³+ 14288x⁵y⁵z²−57474x⁴y⁴z⁴

−28248x³y⁴z⁵−8672x³y³z⁶+ 6448x³y²z⁷+ 10862x²y⁶z⁴

−8672x³y⁶z³−28248x⁵y⁴z³+ 10862x⁶y⁴z²−28248x⁴y³z⁵

−28248x⁵y³z⁴−8672x⁶y³z³+ 6448x⁷y³z²+ 10862x⁴y²z⁶

+ 3420x⁸yz³+ 3420x⁸y³z+ 1220x²yz⁹+ 280xy¹⁰z+ 4066x²y²z⁸

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+ 3420x³yz⁸+ 3420x³y⁸z+ 4066x⁸y²z²+ 4066x²y⁸z² + 1220xy²z⁹+ 8176x⁷y⁴z+ 3420xy³z⁸+ 3420xy⁸z³

+ 1220x²y⁹z+ 280xyz¹⁰+ 280x¹⁰yz+ 1220x⁹y²z+ 8176xy⁴z⁷ + 13448xy⁵z⁶+ 13448xy⁶z⁵+ 8176x⁴y⁷z+ 8176xy⁷z⁴

+ 13448x⁵y⁶z−116x¹⁰y² −116x²z¹⁰≥0.

Since (2.4) is symmetric for x, y, z, there is no harm in supposing that x ≤ y ≤ z.

Using the transformation (1.2), thenF(x, y, z)in (2.4) becomes F(x, y, z)

(2.5)

=P(u, v, w)

=(2u² −v²−vw)[(180224w²+ 180224v²+ 180224vw)u⁸ + (1794048v²w+ 1810432vw²+ 606208w³+ 1196032v³)u⁷ + (4360192vw³+ 7030784v³w+ 771072w⁴+ 7875584v²w² + 3515392v⁴)u⁶+ (6049280v⁵+ 520704w⁵+ 19394048v³w² + 13967872v²w³+ 4689152vw⁴+ 15123200v⁴w)u⁵

+ (2838144vw⁵+ 6838400v⁶+ 12647648v²w⁴ + 30324704v⁴w² + 210048w⁶+ 26457408v³w³+ 20515200v⁵w)u⁴ + (19291776v⁶w + 52480w⁷ + 1074176vw⁶+ 5511936v⁷+ 32787968v⁵w²

+ 33740480v⁴w³+ 20662912v³w⁴+ 6899776v²w⁵)u³

+ (32727200v⁵w³+ 7968w⁸+ 2528912w⁶v²+ 24395856v⁴w⁴ + 27385760v⁶w²+ 14122880v⁷w+ 268096w⁷v+ 3530720v⁸

+ 10723072v³w⁵)u²+ (9558576v⁸w+ 4185944v³w⁶+ 13383144v⁴w⁵ + 676240v²w⁷ + 2124128v⁹+ 672w⁹+ 24737624v⁵w⁴+ 45200vw⁸

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+ 20832112v⁷w²+ 28305704v⁶w³)u+ 15686836v⁵w⁵+ 6092840v⁴w⁶ + 1326664v³w⁷+ 139150v²w⁸+ 24651416v⁶w⁴+ 24921352v⁷w³ + 1378920v¹⁰+ 6894600v⁹w+ 16572238v⁸w²+ 5112vw⁹+ 24w¹⁰] + (27659640v⁹w²+ 10558592v¹⁰w+ 689380v³w⁸+ 4642800v⁴w⁷ + 36001700v⁶w⁵+ 45278940v⁸w³+ 16715660v⁵w⁶ + 720vw¹⁰ + 49540936v⁷w⁴+ 1919744v¹¹+ 44048v²w⁹)u+ 5020v²w¹⁰ + 49008067v⁸w⁴+ 142314v³w⁹+ 23121662v¹⁰w²+ 8144784v¹¹w + 1451049v⁴w⁸+ 40947790v⁹w³ + 24vw¹¹+ 7353016v⁵w⁷

+ 1357464v¹²+ 39938152v⁷w⁵+ 21582818v⁶w⁶.

This impliesF(x, y, z) =P(u, v, w)≥0from (1.8). Hence, inequality (2.1) holds.

The proof is completed.

2.3. The Proof of Inequality (2.2) Proof. Inequality (2.2) is equivalent to

(2.6) sin⁴ A

2 ≤ m²_bm²_c 16m⁴_a.

By using the formulacosα = 1−2 sin²^α₂, the law of cosines and the formulas of the medians, we find that (2.6) is simply the following inequality

(2.7) −a⁸−4b⁸ + 6a⁶c²−34b²c⁶+ 20b³a⁴c+ 12a²c⁶−32b⁵a²c−32c⁵a²b

−34b⁶c²−51b⁴c⁴−4c⁸−4a⁶bc+ 20c³a⁴b−26a⁴b²c²+ 54a²b⁴c²

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+ 54a²b²c⁴−13a⁴b⁴−13a⁴c⁴ + 12a²b⁶ + 6a⁶b²+ 16b⁷c+ 16c⁷b

−64b³c³a²+ 48b⁵c³+ 48b³c⁵ ≥0.

Considering (1.6), inequality (2.7) is transformed into F(x,y, z)

(2.8)

=x⁸+ (4z+ 4y)x⁷+ (2z²+ 40yz+ 2y²)x⁶ + (−8z³+ 84yz² −8y³ + 84y²z)x⁵

+ (−20y²z²+ 76yz³+ 76y³z−7z⁴−7y⁴)x⁴

+ (4z⁵+ 48y⁴z−248y³z²−248y²z³+ 4y⁵ + 48yz⁴)x³

+ (4y⁶−234y⁴z²−234y²z⁴+ 28y⁵z+ 4z⁶ + 28yz⁵+ 32y³z³)x² + (−84z⁵y²+ 8z⁶y−84y⁵z²+ 256y³z⁴+ 256y⁴z³+ 8y⁶z)x + 84z⁵y³−63z⁴y⁴−12y⁶z²−12z⁶y²+ 84y⁵z³ ≥0.

It it easy to see that inequality (2.8) is symmetric fory, z. Therefore, we only need to prove that inequality (2.8) holds whenx≤y≤z,y≤x≤zandy≤z ≤x.

Calculating DS(F), it consists of 3 polynomials with u > 0, v ≥ 0, w ≥ 0 as follows

P₁(u, v, w) (2.9)

=(2u²−v²−vw)[(192w²+ 768v²+ 768vw)u⁴ + (256w³+ 2112vw²+ 4800v²w+ 3200v³)u³

+ (5808v⁴+ 80w⁴+ 7376v²w²+ 11616v³w+ 1568vw³)u²

+ (6336v⁵+ 15840v⁴w+ 13440v³w²+ 16w⁵+ 4320v²w³+ 416vw⁴)u + 5112v⁶+ 15336v⁵w+ 48vw⁵+ 16560v⁴w²+ 7560v³w³

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+ 1272v²w⁴] + (7344v⁷+ 432w⁵v²+ 25704v⁶w+ 33912v⁵w² + 20520v⁴w³+ 5400v³w⁴)u+ 20772v⁷w+ 5193v⁸+ 36w⁶v² + 1332w⁵v³+ 32418v⁶w²+ 24552v⁵w³+ 9009v⁴w⁴

forx≤y≤z, P₂(u, v, w) (2.10)

=(2u²−v²−vw)[(−384vw+ 192v²+ 192w²)u⁴ + (−192vw²+ 896v³−960v²w+ 256w³)u³

+ (−976v²w²+ 1776v⁴+ 80w⁴+ 224vw³−288v³w)u²

+ (2032v⁵−480v³w²+ 16w⁵+ 1328v⁴w+ 128v²w³+ 240vw⁴)u + 1640v⁶+ 2128v⁵w+ 544v⁴w²+ 328v³w³+ 416v²w⁴+ 80vw⁵] + (2064v⁵w²+ 32w⁶v+ 4176v⁶w+ 776v⁴w³+ 2320v⁷

+ 416v²w⁵+ 968v³w⁴)u+ 1640v⁸+ 2708w²v⁶ + 817w⁴v⁴ + 524w⁵v³+ 956w³v⁵+ 84w⁶v²+ 3768wv⁷

fory≤x≤z, and P3(u,v, w) (2.11)

=384u⁶v²+ 11072w²u²v⁴+ 20992w²u³v³+ 19552w²u⁴v² + 8832w²u⁵v+ 2008w⁴uv³+ 5296w⁴u²v²+ 5376w⁴u³v + 36w²v⁶+ 1536w²u⁶ + 2792w³uv⁴+ 10400w³u²v³ + 15744w³u³v²+ 10816w³u⁴v+ 2368w²uv⁵+ 840w⁵uv² + 1344w⁵u²v+ 2816w³u⁵+ 132w³v⁵+ 1888w⁴u⁴+ 193w⁴v⁴

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+ 184w⁶uv+ 144w⁵v³+ 640w⁵u³+ 1200wuv⁶+ 13120wu³v⁴ + 6256wu²v⁵+ 13824wu⁴v³+ 58w⁶v² + 128w⁶u²+ 7296wu⁵v² + 3360u⁴v⁴+ 1792u⁵v³+ 288uv⁷+ 1504u²v⁶+ 3168u³v⁵ + 1536wu⁶v+ 12w⁷v+w⁸+ 16w⁷u

fory≤z ≤x.

It is not difficult to see that P₁(u, v, w) ≥ 0 and P₃(u, v, w) ≥ 0 because u >

0, v ≥0, w ≥0and2u² −v²−vw >0.

In order to proveP₂(u, v, w)≥0, we only need prove the following inequality p(u,v, w)

(2.12)

=(−384vw+ 192v² + 192w²)u⁴

+ (−192vw²+ 896v³−960v²w+ 256w³)u³

+ (−976v²w²+ 1776v⁴+ 80w⁴+ 224vw³−288v³w)u²

+ (2032v⁵−480v³w² + 16w⁵+ 1328v⁴w+ 128v²w³+ 240vw⁴)u + 1640v⁶+ 2128v⁵w+ 544v⁴w²+ 328v³w³+ 416v²w⁴+ 80vw⁵

≥0,

whereu >0,v ≥0andw≥0.

(i)Foru >0, v ≥w≥0, takingv =w+twitht ≥0, then we have

p(u, v, w) =192t²u⁴+ (576tw²+ 1728t²w+ 896t³)u³+ (816w⁴+ 4512w³t + 8816w²t²+ 6816wt³+ 1776t⁴)u²+ (2032t⁵+ 11488wt⁴ + 3264w⁵+ 14528w⁴t+ 26976w³t²+ 25152w²t³)u+ 50544w⁴t²

+ 56584w³t³+ 5136w⁶+ 24552w⁵t+ 1640t⁶+ 35784w²t⁴+ 11968wt⁵. It obviously follows thatp(u, v, w)≥0, i.e., inequality (2.12) holds.

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(ii)Whenu >0, w ≥v ≥0, settingw=v+tfort≥0, we get p(u, v, w) =(2u+ 10v)t⁵ + (10u²+ 40uv+ 102v²)t⁴

+ (156uv²+ 32u³+ 349v³+ 68u²v)t³

+ (24u⁴ + 188uv³+ 22u²v²+ 72u³v + 603v⁴)t² +v²(783v³−72u³+ 224uv²−156u²v)t

+ 6v⁴(17u² + 68uv+ 107v²)

=p₁(u, v, t) +p₂(u, v, t), where

p1(u, v, t) = (2u+ 10v)t⁵+ (10u²+ 40uv+ 102v²)t⁴

+ (156uv²+ 32u³+ 349v³+ 68u²v)t³ ≥0, and

(2.13) p₂(u, v, t) = (24u⁴+ 188uv³+ 22u²v² + 72u³v+ 603v⁴)t² +v²(783v³−72u³+ 224uv²−156u²v)t

+ 6v⁴(17u²+ 68uv+ 107v²).

It is easy to see that 24u⁴ + 188uv³ + 22u²v² + 72u³v + 603v⁴ > 0, and the discriminant of the quadratic function (2.13) with respect totis

(2.14) ∆(u, v) =−v⁴(935415v⁶+ 480144u³v³+ 1116096uv⁵

+ 803456u²v⁴+ 4608u⁶+ 196032u⁴v²+ 46080u⁵v)≤0.

This is to say thatp₂(u, v, t)≥0.

Hence,P₂(u, v, w)≥0. From the proof above, the required result (2.6) is proved.

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2.4. Remarks

Remark 1. By the same argument as above, we also prove the following inequalities conjectures [9,10,11] in the acute triangle

(2.15) X

m²_ah²_a ≥X m²_ar_a²,

(2.16) X

sin⁸A≥X

cos⁸ A 2,

(2.17) X

(b−c)² ≥X a b+c

2

(r_b−r_c)², and

(2.18) X

(h_b+h_c −h_a)³ ≥3m_am_bm_c.

Remark 2. The operations in this paper were performed using mathematical software Maple 9.0.

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3. Generalization of the Method

In fact, Difference Substitution can go even further. Now, we consider the following inequality [12]. In4ABC, ifmax (A, B, C)≤ ^2π₃ , then

(3.1) s² ≥R²+ 10Rr+ 3r².

Utilizing the known formulasR = ^abc_4S, r = ^S_s andS = p

s(s−a)(s−b)(s−c), from (1.6), inequality (3.1) is equivalent to

(3.2) 3c²a²b²−a²bc³−a³bc²−a³b²c−a²b³c−ab²c³−ab³c² −2b³c³

−2a³c³+c⁴b²−2a³b³+c⁵b+a⁴c²+b⁴c²+b⁵a+a⁵c+c⁵a

−a⁶+a⁴b²+a⁵b−b⁶−c⁶+a²b⁴+b⁵c+a²c⁴ ≥0, or

(3.3) F(x, y, z)

=−42x²y²z²+ 14y⁴zx+ 14xyz⁴+ 2xy²z³ + 2x²y³z+ 2xy³z² + 14x⁴yz+ 2x³yz²+ 2x²yz³+ 2x³y²z−x⁴y²−x²z⁴

−2x³z³−x⁴z²−2x³y³−y⁴z² −y⁴x²−2y³z³−y²z⁴ ≥0, wherex >0, y >0, z >0.

Since inequality (3.3) is symmetric with x, y, z, there is no harm in supposing thatx≤y≤z. From (1.2),F(x, y, z)in (3.3) is transformed into

F(x, y, z) =P(u, v, w) (3.4)

=(8u²+ 4uv+ 2uw−v²−vw)(8uvw²+ 12uv²w+ 4u²vw + 4v⁴+ 4u²v²+ 2uw³+ 8uv³+ 8v³w+ 7v²w²+ 3vw³) + 2w²(v+ 2u)²(v+ 2u+w)² ≥0,

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and formax (A, B, C)≤ ^2π₃ andy = cosxdecreasing inx∈(0, π), we have b²+c²+bc−a² =b²+c²− 1

2bccos2π 3 −a² (3.5)

= 3x²+ 3(y+z)x−yz

= 8u²+ 4uv+ 2uw−v²−vw

≥b²+c²−1

2bccosA−a² = 0.

SinceF(x, y, z) = P(u, v, w) ≥ 0foru > 0, v ≥ 0andw ≥ 0, inequality (3.1) is obtained.

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References

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Guangzhou Univ. (Natural Sciences Edition), 5(2) (2006), 1–7. (in Chinese) [2] L. YANG, Solving harder problems with lesser mathematics, Proceedings of

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[6] X.-G. CHUAND J. LIU, Some semi-symmetric inequalities in triangle, High- School Mathematics Monthly, 3(1999), 23–25. (in Chinese)

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[10] B.-Q. LIU, 110 interesting inequalities problems, Research in Inequalities. Ti- bet People’s Press, 2000, 389–405. (in Chinese)

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[11] B.-Q. LIU, Private Communication, 2003.

[12] X.-G. CHU, Two Geometric inequalities with the Fermat problem, J. Beijing Union Univ. (Natural Sciences), 18(3) (2004), 62–64. (in Chinese)