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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


Academic year: 2022

Ossza meg "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"


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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 to120in the triangle.

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

2000 Mathematics Subject Classification. 26D15.


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








x1 =t1, x2 =t1+t2, x3 =t1+t2+t3,

· · · ·

xn=t1+t2+t3+· · ·+tn,

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



whereti ≥0for2≤i≤nandi∈N.

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

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




x=u, y=u+v, z =u+v+w, wherev ≥0,w≥0.

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




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 membersPi(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, ha, hb, hc as the altitudes, ma, mb, mc as the medians, andra, rb, rc 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,


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


+a5c−ab4c+a2b4−b6−c6−a6 −abc4 ≥0.







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]


= 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)≤ ha

ma. 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 Pi(u, v, w) ≥ 0 for1 ≤ i ≤ 3. Hence, inequality (1.10) is proved.


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


≤ 3 4, and

(2.2) sinA

2 ≤

√mbmc 2ma . 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


−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


−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 ≤ m2bm2c 16m4a.

By using the formulacosα= 1−2 sin2α2, 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)


=(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



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


+ (−192vw2+ 896v3−960v2w+ 256w3)u3

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

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


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

m2ah2a ≥X m2ara2,

(2.16) X


cos8 A 2,

(2.17) X

(b−c)2 ≥X a b+c


(rb−rc)2, and

(2.18) X

(hb +hc −ha)3 ≥3mambmc.

Remark 2.2. The operations in this paper were performed using mathematical software Maple 9.0.



In fact, Difference Substitution can go even further. Now, we consider the following inequal- ity [12]. In4ABC, ifmax (A, B, C)≤ 3 , then

(3.1) s2 ≥R2+ 10Rr+ 3r2.

Utilizing the known formulasR = abc4S, r = Ss 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)


=(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)≤ 3 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


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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