Received09October,2006;accepted04April,2007CommunicatedbyB.Yang x = t + t + t + ··· + t , ·················· = t + t + t , = t + t ,x x = t ,x  x ≤ x ≤ x ≤···≤ x with n ∈ N ,thenweset(1.1) In[1,2],L.YangsuggestedtheuseofDifferenceSubstitution

PROVING INEQUALITIES IN ACUTE TRIANGLE WITH DIFFERENCE SUBSTITUTION

YU-DONG WU, ZHI-HUA ZHANG, AND YU-RUI ZHANG XINCHANGHIGHSCHOOL

XINCHANGCITY

ZHEJIANGPROVINCE312500 P. R. CHINA.

zjxcwyd@tom.com

ZIXINGEDUCATIONALRESEARCHSECTION

CHENZHOUCITY, HUNANPROVINCE423400 P. R. CHINA

zxzh1234@163.com

URL:http://www.zxslzx.com/zzhweb/

XINCHANGHIGHSCHOOL

XINCHANGCITYZHEJIANGPROVINCE312500 P. R. CHINA.

xczxzyr@163.com

Received 09 October, 2006; accepted 04 April, 2007 Communicated by B. Yang

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.

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

2000 Mathematics Subject Classification. 26D15.

1. INTRODUCTION

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

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

(1.1)















x₁ =t₁, x₂ =t₁+t₂, x₃ =t₁+t₂+t₃,

· · · ·

x_n=t₁+t₂+t₃+· · ·+t_n,

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

253-06

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 the n-variant polynomials, there are n! 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, wherev ≥0,w≥0.

Sequentially, fory ≤ 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 by DS(F). If all the coefficients of these members DS(F) are nonnegative, then F ≥ 0 whenever 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 denote A, B, C as the interior angles, a, b, c as the side-lengths, S as the area, s as the semi-perimeter, R as the circumradius, r as the inradius, h_a, h_b, h_c as the altitudes, m_a, m_b, m_c as the medians, andr_a, r_b, r_c as the radii of the described circles of triangleABC respectively. Moreover, we will customarily use the cyclic sum symbol, that is:

Pf(a) =f(a) +f(b) +f(c),andP

f(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.

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 forx, 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) =P(u, v, w) (1.9)

= 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.

Obviously F(x, y, z) = P(u, v, w) ≥ 0 from (1.8) andu > 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.

CalculatingDS(F), it consists of 3 polynomials withu >0, v ≥0, w ≥0as 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, 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 P_i(u, v, w) ≥ 0 for1 ≤ i ≤ 3. Hence, inequality (1.10) is proved.

2. SOME PROBLEMS AND THEIRPROOFS

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

Problem 2.1. Let4ABCbe 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 + 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) =−4576x⁷z⁵ −5590x⁶z⁶−116x¹⁰z²−2453x⁴z⁸ −2453x⁸z⁴ (2.4)

−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 + 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 forx, y, z, there is no harm in supposing thatx ≤ y ≤ z. Using the transformation (1.2), thenF(x, y, z)in (2.4) becomes

F(x, y, z) =P(u, v, w) (2.5)

=(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 + 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 sin^2α₂, 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 + 54a²b²c⁴−13a⁴b⁴−13a⁴c⁴+ 12a²b⁶+ 6a⁶b²+ 16b⁷c+ 16c⁷b

−64b³c³a²+ 48b⁵c³+ 48b³c⁵ ≥0.

Considering (1.6), inequality (2.7) is transformed into

F(x, y, z) =x⁸+ (4z+ 4y)x⁷+ (2z²+ 40yz+ 2y²)x⁶ (2.8)

+ (−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.

CalculatingDS(F), it consists of 3 polynomials withu >0, v ≥0, w ≥0as 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+ 1272v²w⁴] + (7344v⁷+ 432w⁵v²+ 25704v⁶w+ 33912v⁵w²+ 20520v⁴w³

+ 5400v³w⁴)u+ 20772v⁷w+ 5193v⁸+ 36w⁶v² + 1332w⁵v³+ 32418v⁶w²+ 24552v⁵w³+ 9009v⁴w⁴

forx≤y≤z,

P₂(u, v, w) =(2u²−v²−vw)[(−384vw+ 192v²+ 192w²)u⁴ (2.10)

+ (−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) =384u⁶v²+ 11072w²u²v⁴+ 20992w²u³v³+ 19552w²u⁴v² (2.11)

+ 8832w²u⁵v+ 2008w⁴uv³+ 5296w⁴u²v²+ 5376w⁴u³v + 36w²v⁶+ 1536w²u⁶+ 2792w³uv⁴+ 10400w³u²v³ + 15744w³u³v²+ 10816w³u⁴v+ 2368w²uv⁵+ 840w⁵uv² + 1344w⁵u²v+ 2816w³u⁵+ 132w³v⁵+ 1888w⁴u⁴ + 193w⁴v⁴ + 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 thatP₁(u, v, w)≥0andP₃(u, v, w)≥0becauseu >0, v ≥0, w ≥0 and2u²−v²−vw >0.

In order to proveP2(u, v, w)≥0, we only need prove the following inequality p(u, v, w) =(−384vw+ 192v²+ 192w²)u⁴

(2.12)

+ (−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.

(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 that24u⁴+ 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.

2.4. Remarks.

Remark 2.1. By the same argument as above, we also prove the following inequalities conjec- tures [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 2.2. The operations in this paper were performed using mathematical software Maple 9.0.

3. GENERALIZATION OF THEMETHOD

In fact, Difference Substitution can go even further. Now, we consider the following inequal- ity [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 withx, 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,

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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