PROVING INEQUALITIES IN ACUTE TRIANGLE
WITH DIFFERENCE SUBSTITUTION
YU-DONG WU
ZHI-HUA ZHANG
Xinchang High School
Xinchang City, Zhejiang Province 312500
P.R. China.
EMail: zjxcwyd@tom.com
Zixing Educational Research Section
Chenzhou City, Hunan Province 423400
P. R. China.
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 socalled Difference Substitution. As generalization of the method, we also consider
an example that the greatest interior angle is less than or equal to 120◦ in the
triangle.
Acknowledgements:
The authors would like to thank Professor Lu Yang for his enthusiastic help.
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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Contents
1
Introduction
2
Some Problems and their Proofs
2.1 The Problems . . . . . . . .
2.2 The Proof of Inequality (2.1)
2.3 The Proof of Inequality (2.2)
2.4 Remarks . . . . . . . . . . .
3
Generalization of the Method
3
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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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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.
If x1 ≤ x2 ≤ x3 ≤ · · · ≤ xn with n ∈ N ∗ , then we set

x1 = t1 ,





 x2 = t1 + t2 ,
(1.1)
x3 = t1 + t2 + t3 ,



··················



xn = t1 + t2 + t3 + · · · + tn ,
where ti ≥ 0 for 2 ≤ i ≤ n and i ∈ 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 of n = 3, we let x ≤ y ≤ z, and take


 x = u,
(1.2)
y = u + v,

 z = u + v + w,
where v ≥ 0, w ≥ 0.
Analogously, if y ≤ x ≤ z, then its “splitting” transformation is


 y = u,
(1.3)
x = u + v,

 z = u + v + w,
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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where v ≥ 0, w ≥ 0.
Sequentially, for y ≤ z ≤ x or z ≤ x ≤ y or z ≤ y ≤ x or x ≤ 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 members Pi (u, v, w) with 1 ≤ 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 sidelengths, 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, and ra , rb , rc as the
radii of the described circles of triangle ABC P
respectively. Moreover, we will customarily
use
the
cyclic
sum
symbol,
that
is:
f (a) = f (a) + f (b) + f (c), and
P
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 ,
and Yu-Rui Zhang
vol. 8, iss. 3, art. 81, 2007
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or
(1.5)
Proving Inequalities with
Difference Substitution
Yu-Dong Wu, Zhi-Hua Zhang
− 2 a3 b3 + a4 b2 − a4 bc + a5 b + ab5 + b5 c + b4 c2
− 2 b 3 c 3 + b 2 c 4 − 2 c 3 a3 + c 4 a2 + c 5 a + c 5 b + c 2 a4
+ a5 c − ab4 c + a2 b4 − b6 − c6 − a6 − abc4 ≥ 0.
Let


x=




y=




z =
(1.6)
b+c−a
2
> 0,
c+a−b
2
> 0,
a+b−c
2
> 0.
Proving Inequalities with
Difference Substitution
Yu-Dong Wu, Zhi-Hua Zhang
Then inequality (1.4) or (1.5) is equivalent to
(1.7) F (x, y, z) = 6 xyz 4 + 2 xy 2 z 3 + 2 xy 3 z 2 + 6 xy 4 z + 2 x2 yz 3 + 2 x2 y 3 z
+ 2 x3 yz 2 + 2 x3 y 2 z + 6 x4 yz − x4 y 2 − x2 z 4 − 2 x3 z 3 − x4 z 2
− 2 x3 y 3 − y 4 z 2 − y 4 x2 − 2 y 3 z 3 − y 2 z 4 − 18 x2 y 2 z 2 ≥ 0.
and Yu-Rui Zhang
vol. 8, iss. 3, art. 81, 2007
Title Page
There is no harm in supposing x ≤ y ≤ z since inequality (1.7) is symmetric for
x, y, z. Then, by using (1.2), for the acute triangle, it follows that
(1.8)
b2 + c2 − a2 = (z + x)2 + (x + y)2 − (y + z)2 = 2[x2 + (y + z)x − yz]
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= 2{u2 + [(u + v) + (u + v + w)]u − (u + v)(u + v + w)}
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2
2
= 2(2u − v − vw) > 0,
Page 5 of 20
and F (x, y, z) in (1.7) is transformed into
(1.9)
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F (x, y, z)
= P (u, v, w)
2
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2
= 2 u − v − vw
2
2
2
4 v + 4 w + 4 vw u
+ 8 v + 20 vw2 + 12 v 2 w + 8 w3 u + 4 v 4 + 8 v 3 w + 2 w4
+18 vw3 + 22 v 2 w2 + 24 v 3 w2 + 36 v 2 w3 + 12 vw4 u
3
+ 34 v 3 w3 + 19 v 2 w4 + 2 vw5 + 17 v 4 w2 .
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
cos(B − C) ≤
(1.10)
ha
.
ma
Proving Inequalities with
Difference Substitution
Yu-Dong Wu, Zhi-Hua Zhang
It is equivalent to
−a4 + 3 b2 + 3 c2 a2 − 2 (b − c)2 (b + c)2 ≥ 0,
(1.11)
and Yu-Rui Zhang
vol. 8, iss. 3, art. 81, 2007
and from (1.6), this equals
(1.12) F (x, y, z)
Title Page
= −y 2 − z 2 + 14 yz x2 − (y + z) z 2 − 14 yz + y
2
2
x + yz (y + z) ≥ 0.
Calculating DS(F ), it consists of 3 polynomials with u > 0, v ≥ 0, w ≥ 0 as
follows
(1.13) P1 (u, v, w)
4
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Page 6 of 20
3
2 2
3
2
2
3
= 40 u + 112 u v + 108 u v + 56 u w + 14 u w + 40 uv + 20 uvw
2
2
3
2
2
3
2
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4
+ 60 uv w + 108 u vw + 8 v w + 5 v w + vw + 4 v ,
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(1.14) P2 (u, v, w)
= 2 u2 − v 2 − vw
20 u2 + (24 w + 52 v) u + 53 v 2 + 6 w2 + 52 vw
+ 72 v 3 + 36 vw2 + 108 v 2 w u + 57 v 2 w2 + 51 v 4 + 6 vw3 + 102 v 3 w,
and
(1.15) P3 (u, v, w)
= 2 u2 − v 2 − vw
20 u2 + (52 v + 28 w) u + 53 v 2 + 54 vw + 7 w2
+ 72 v 3 + 36 vw2 + 108 v 2 w u + 57 v 2 w2 + 51 v 4 + 6 vw3 + 102 v 3 w.
By (1.8), we immediately obtain Pi (u, v, w) ≥ 0 for 1 ≤ i ≤ 3. Hence, inequality
(1.10) is proved.
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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2.
2.1.
Some Problems and their Proofs
The Problems
In 2004-2005, J. Liu [7, 8] posed the following conjectures for the inequality in the
acute triangle.
Problem 1. Let 4ABC be an acute triangle. Prove the following inequalities
2
X
3
sin 2A
(2.1)
≤ ,
sin B + sin C
4
and
A
sin ≤
2
(2.2)
√
mb mc
.
2ma
Proving Inequalities with
Difference Substitution
Yu-Dong Wu, Zhi-Hua Zhang
and Yu-Rui Zhang
vol. 8, iss. 3, art. 81, 2007
Title Page
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Proof. Using sin 2α = 2 sin α cos α, we find that inequality (2.1) is equivalent to
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(2.3) 4 a10 b2 − 10 a5 b5 c2 − 24 b6 c5 a + 16 a9 b3 + 4 a8 b4 − 5 b4 c6 a2
Page 8 of 20
2.2.
The Proof of Inequality (2.1)
− 8 a11 b − 5 a4 b6 c2 + 8 a9 c2 b + 4 a8 b2 c2 − 16 a10 cb + 8 a9 cb2 − 5 a6 b4 c2
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+ 32 a8 b3 c + 8 a7 b4 c + 8 a7 c4 b − 5 a6 c4 b2 + 32 a8 c3 b + 4 a10 c2 − 8 a11 c
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− 4 a12 − 4 b12 − 4 c12 + 4 a8 c4 + 16 a9 c3 − 8 a7 b5 − 8 a6 b6 − 8 a6 c6
7 5
5 7
4 8
6 5
5 6
5 3 4
4 4 4
− 8 a c − 8 a b + 4 a b − 24 a b c − 24 a b c + 2 a b c + 6 a b c
+ 4 c10 b2 − 26 a6 b3 c3 + 2 a5 b4 c3 − 24 a5 c6 b − 5 a4 c6 b2 − 24 a6 c5 b
− 10 a5 c5 b2 + 8 a4 b7 c − 8 c11 b + 32 b8 a3 c + 8 b9 a2 c + 4 b8 a2 c2
+ 2 b5 c3 a4 − 26 b6 c3 a3 + 2 b5 c4 a3 − 5 b6 c4 a2 − 16 b10 ca + 8 b9 c2 a
Close
+ 32 b8 c3 a + 8 b7 c4 a − 10 b5 c5 a2 + 16 b9 a3 + 4 b10 a2 − 8 b11 a
− 8 b11 c + 4 b10 c2 + 16 b9 c3 + 4 b8 c4 − 8 b7 c5 − 8 b6 c6 + 4 a4 c8
− 8 a5 c7 − 24 b5 c6 a + 8 a4 c7 b + 2 c5 b4 a3 − 26 c6 b3 a3 + 2 c5 b3 a4
+ 4 c8 b2 a2 + 8 c9 a2 b + 32 c8 a3 b + 8 b4 c7 a − 8 c11 a + 32 c8 b3 a + 8 c9 b2 a
− 16 c10 ab + 4 c10 a2 + 16 c9 a3 − 8 b5 c7 + 4 b4 c8 + 16 c9 b3 ≥ 0.
Proving Inequalities with
Difference Substitution
Yu-Dong Wu, Zhi-Hua Zhang
From (1.6), inequality (2.3) equals
and Yu-Rui Zhang
(2.4) F (x, y, z)
vol. 8, iss. 3, art. 81, 2007
= − 4576 x7 z 5 − 5590 x6 z 6 − 116 x10 z 2 − 2453 x4 z 8 − 2453 x8 z 4
− 4576 x5 z 7 − 788 x3 z 9 − 788 x9 z 3 − 2453 x8 y 4 − 4576 y 7 z 5
8 4
4 8
9 3
3 9
Title Page
6 6
− 2453 y z − 2453 x y − 788 x y − 788 x y − 5590 x y
− 4576 x7 y 5 − 4576 x5 y 7 − 2453 y 4 z 8 − 4576 y 5 z 7 − 5590 y 6 z 6
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+ 13448 x6 y 5 z + 8176 x7 yz 4 + 13448 x6 yz 5 + 13448 x5 yz 6
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+ 8176 x4 yz 7 + 6448 x2 y 3 z 7 + 1220 xy 9 z 2 + 6448 x2 y 7 z 3
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− 116 y 10 z 2 − 788 y 9 z 3 − 116 y 10 x2 − 788 y 3 z 9 − 116 y 2 z 10
+ 6448 x3 y 7 z 2 + 1220 x9 yz 2 + 10862 x4 y 6 z 2 + 14288 x2 y 5 z 5
+ 10862 x2 y 4 z 6 + 14288 x5 y 2 z 5 + 10862 x6 y 2 z 4 + 6448 x7 y 2 z 3
3 5 4
4 5 3
5 5 2
3 3 6
3 2 7
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4 4 4
− 28248 x y z − 28248 x y z + 14288 x y z − 57474 x y z
3 4 5
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2 6 4
− 28248 x y z − 8672 x y z + 6448 x y z + 10862 x y z
− 8672 x3 y 6 z 3 − 28248 x5 y 4 z 3 + 10862 x6 y 4 z 2 − 28248 x4 y 3 z 5
− 28248 x5 y 3 z 4 − 8672 x6 y 3 z 3 + 6448 x7 y 3 z 2 + 10862 x4 y 2 z 6
+ 3420 x8 yz 3 + 3420 x8 y 3 z + 1220 x2 yz 9 + 280 xy 10 z + 4066 x2 y 2 z 8
Close
+ 3420 x3 yz 8 + 3420 x3 y 8 z + 4066 x8 y 2 z 2 + 4066 x2 y 8 z 2
+ 1220 xy 2 z 9 + 8176 x7 y 4 z + 3420 xy 3 z 8 + 3420 xy 8 z 3
+ 1220 x2 y 9 z + 280 xyz 10 + 280 x10 yz + 1220 x9 y 2 z + 8176 xy 4 z 7
+ 13448 xy 5 z 6 + 13448 xy 6 z 5 + 8176 x4 y 7 z + 8176 xy 7 z 4
+ 13448 x5 y 6 z − 116 x10 y 2 − 116 x2 z 10 ≥ 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), then F (x, y, z) in (2.4) becomes
(2.5) F (x, y, z)
=P (u, v, w)
=(2u2 − v 2 − vw)[(180224 w2 + 180224 v 2 + 180224 vw)u8
2
2
3
3
4
2
+ (4360192 vw + 7030784 v w + 771072 w + 7875584 v w
Contents
2
+ 13967872 v 2 w3 + 4689152 vw4 + 15123200 v 4 w)u5
+ (2838144 vw5 + 6838400 v 6 + 12647648 v 2 w4 + 30324704 v 4 w2
+ 210048 w6 + 26457408 v 3 w3 + 20515200 v 5 w)u4 + (19291776 v 6 w
+ 52480 w7 + 1074176 vw6 + 5511936 v 7 + 32787968 v 5 w2
8
6 2
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+ 33740480 v 4 w3 + 20662912 v 3 w4 + 6899776 v 2 w5 )u3
3
vol. 8, iss. 3, art. 81, 2007
7
+ 3515392 v 4 )u6 + (6049280 v 5 + 520704 w5 + 19394048 v 3 w2
5
and Yu-Rui Zhang
Title Page
3
+ (1794048 v w + 1810432 vw + 606208 w + 1196032 v )u
3
Proving Inequalities with
Difference Substitution
Yu-Dong Wu, Zhi-Hua Zhang
Close
4
+ (32727200 v w + 7968 w + 2528912 w v + 24395856 v w
4
+ 27385760 v 6 w2 + 14122880 v 7 w + 268096 w7 v + 3530720 v 8
+ 10723072 v 3 w5 )u2 + (9558576 v 8 w + 4185944 v 3 w6 + 13383144 v 4 w5
+ 676240 v 2 w7 + 2124128 v 9 + 672 w9 + 24737624 v 5 w4 + 45200 vw8
+ 20832112 v 7 w2 + 28305704 v 6 w3 )u + 15686836 v 5 w5 + 6092840 v 4 w6
+ 1326664 v 3 w7 + 139150 v 2 w8 + 24651416 v 6 w4 + 24921352 v 7 w3
+ 1378920 v 10 + 6894600 v 9 w + 16572238 v 8 w2 + 5112 vw9 + 24 w10 ]
+ (27659640 v 9 w2 + 10558592 v 10 w + 689380 v 3 w8 + 4642800 v 4 w7
+ 36001700 v 6 w5 + 45278940 v 8 w3 + 16715660 v 5 w6 + 720 vw10
+ 49540936 v 7 w4 + 1919744 v 11 + 44048 v 2 w9 )u + 5020 v 2 w10
+ 49008067 v 8 w4 + 142314 v 3 w9 + 23121662 v 10 w2 + 8144784 v 11 w
4
8
9
3
11
5
+ 1451049 v w + 40947790 v w + 24 vw + 7353016 v w
7
Proving Inequalities with
Difference Substitution
Yu-Dong Wu, Zhi-Hua Zhang
and Yu-Rui Zhang
vol. 8, iss. 3, art. 81, 2007
+ 1357464 v 12 + 39938152 v 7 w5 + 21582818 v 6 w6 .
This implies F (x, y, z) = P (u, v, w) ≥ 0 from (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
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m2b m2c
.
16m4a
2 sin2 α2 ,
By using the formula cos α = 1 −
the law of cosines and the formulas of
the medians, we find that (2.6) is simply the following inequality
(2.7)
Title Page
− a8 − 4 b8 + 6 a6 c2 − 34 b2 c6 + 20 b3 a4 c + 12 a2 c6 − 32 b5 a2 c − 32 c5 a2 b
− 34 b6 c2 − 51 b4 c4 − 4 c8 − 4 a6 bc + 20 c3 a4 b − 26 a4 b2 c2 + 54 a2 b4 c2
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+ 54 a2 b2 c4 − 13 a4 b4 − 13 a4 c4 + 12 a2 b6 + 6 a6 b2 + 16 b7 c + 16 c7 b
− 64 b3 c3 a2 + 48 b5 c3 + 48 b3 c5 ≥ 0.
Considering (1.6), inequality (2.7) is transformed into
(2.8) F (x,y, z)
=x8 + (4 z + 4 y)x7 + (2 z 2 + 40 yz + 2 y 2 )x6
Proving Inequalities with
Difference Substitution
Yu-Dong Wu, Zhi-Hua Zhang
+ (−8 z 3 + 84 yz 2 − 8 y 3 + 84 y 2 z)x5
and Yu-Rui Zhang
+ (−20 y 2 z 2 + 76 yz 3 + 76 y 3 z − 7 z 4 − 7 y 4 )x4
vol. 8, iss. 3, art. 81, 2007
+ (4 z 5 + 48 y 4 z − 248 y 3 z 2 − 248 y 2 z 3 + 4 y 5 + 48 yz 4 )x3
+ (4 y 6 − 234 y 4 z 2 − 234 y 2 z 4 + 28 y 5 z + 4 z 6 + 28 yz 5 + 32 y 3 z 3 )x2
5 2
6
5 2
3 4
4 3
Title Page
6
+ (−84 z y + 8 z y − 84 y z + 256 y z + 256 y z + 8 y z)x
+ 84 z 5 y 3 − 63 z 4 y 4 − 12 y 6 z 2 − 12 z 6 y 2 + 84 y 5 z 3 ≥ 0.
It it easy to see that inequality (2.8) is symmetric for y, z. Therefore, we only need
to prove that inequality (2.8) holds when x ≤ y ≤ z, y ≤ x ≤ z and y ≤ z ≤ x.
Calculating DS(F ), it consists of 3 polynomials with u > 0, v ≥ 0, w ≥ 0 as
follows
(2.9) P1 (u, v, w)
2
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2
2
2
=(2u − v − vw)[(192 w + 768 v + 768 vw)u
3
2
2
3
4
+ (256 w + 2112 vw + 4800 v w + 3200 v )u
Close
3
+ (5808 v 4 + 80 w4 + 7376 v 2 w2 + 11616 v 3 w + 1568 vw3 )u2
+ (6336 v 5 + 15840 v 4 w + 13440 v 3 w2 + 16 w5 + 4320 v 2 w3 + 416 vw4 )u
+ 5112 v 6 + 15336 v 5 w + 48 vw5 + 16560 v 4 w2 + 7560 v 3 w3
+ 1272 v 2 w4 ] + (7344 v 7 + 432 w5 v 2 + 25704 v 6 w + 33912 v 5 w2
+ 20520 v 4 w3 + 5400 v 3 w4 )u + 20772 v 7 w + 5193 v 8 + 36 w6 v 2
+ 1332 w5 v 3 + 32418 v 6 w2 + 24552 v 5 w3 + 9009 v 4 w4
for x ≤ y ≤ z,
(2.10) P2 (u, v, w)
Proving Inequalities with
Difference Substitution
Yu-Dong Wu, Zhi-Hua Zhang
=(2u2 − v 2 − vw)[(−384 vw + 192 v 2 + 192 w2 )u4
and Yu-Rui Zhang
+ (−192 vw2 + 896 v 3 − 960 v 2 w + 256 w3 )u3
vol. 8, iss. 3, art. 81, 2007
+ (−976 v 2 w2 + 1776 v 4 + 80 w4 + 224 vw3 − 288 v 3 w)u2
+ (2032 v 5 − 480 v 3 w2 + 16 w5 + 1328 v 4 w + 128 v 2 w3 + 240 vw4 )u
6
5
4
2
3
3
2
4
Title Page
5
+ 1640 v + 2128 v w + 544 v w + 328 v w + 416 v w + 80 vw ]
+ (2064 v 5 w2 + 32 w6 v + 4176 v 6 w + 776 v 4 w3 + 2320 v 7
+ 416 v 2 w5 + 968 v 3 w4 )u + 1640 v 8 + 2708 w2 v 6 + 817 w4 v 4
+ 524 w5 v 3 + 956 w3 v 5 + 84 w6 v 2 + 3768 wv 7
for y ≤ x ≤ z, and
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(2.11) P3 (u,v, w)
=384 u6 v 2 + 11072 w2 u2 v 4 + 20992 w2 u3 v 3 + 19552 w2 u4 v 2
+ 8832 w2 u5 v + 2008 w4 uv 3 + 5296 w4 u2 v 2 + 5376 w4 u3 v
+ 36 w2 v 6 + 1536 w2 u6 + 2792 w3 uv 4 + 10400 w3 u2 v 3
+ 15744 w3 u3 v 2 + 10816 w3 u4 v + 2368 w2 uv 5 + 840 w5 uv 2
+ 1344 w5 u2 v + 2816 w3 u5 + 132 w3 v 5 + 1888 w4 u4 + 193 w4 v 4
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+ 184 w6 uv + 144 w5 v 3 + 640 w5 u3 + 1200 wuv 6 + 13120 wu3 v 4
+ 6256 wu2 v 5 + 13824 wu4 v 3 + 58 w6 v 2 + 128 w6 u2 + 7296 wu5 v 2
+ 3360 u4 v 4 + 1792 u5 v 3 + 288 uv 7 + 1504 u2 v 6 + 3168 u3 v 5
+ 1536 wu6 v + 12 w7 v + w8 + 16 w7 u
for y ≤ 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 ≥ 0 and 2u2 − v 2 − vw > 0.
In order to prove P2 (u, v, w) ≥ 0, we only need prove the following inequality
Proving Inequalities with
Difference Substitution
Yu-Dong Wu, Zhi-Hua Zhang
and Yu-Rui Zhang
vol. 8, iss. 3, art. 81, 2007
(2.12) p(u,v, w)
=(−384 vw + 192 v 2 + 192 w2 )u4
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+ (−192 vw2 + 896 v 3 − 960 v 2 w + 256 w3 )u3
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+ (−976 v 2 w2 + 1776 v 4 + 80 w4 + 224 vw3 − 288 v 3 w)u2
+ (2032 v 5 − 480 v 3 w2 + 16 w5 + 1328 v 4 w + 128 v 2 w3 + 240 vw4 )u
+ 1640 v 6 + 2128 v 5 w + 544 v 4 w2 + 328 v 3 w3 + 416 v 2 w4 + 80 vw5
≥0,
where u > 0, v ≥ 0 and w ≥ 0.
(i) For u > 0, v ≥ w ≥ 0, taking v = w + t with t ≥ 0, then we have
2 4
2
2
3
3
3
4
2
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3
4
+ 8816 w t + 6816 wt + 1776 t )u + (2032 t + 11488 wt
+ 3264 w5 + 14528 w4 t + 26976 w3 t2 + 25152 w2 t3 )u + 50544 w4 t2
+ 56584 w3 t3 + 5136 w6 + 24552 w5 t + 1640 t6 + 35784 w2 t4 + 11968 wt5 .
It obviously follows that p(u, v, w) ≥ 0, i.e., inequality (2.12) holds.
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p(u, v, w) =192 t u + (576 tw + 1728 t w + 896 t )u + (816 w + 4512 w t
2 2
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(ii) When u > 0, w ≥ v ≥ 0, setting w = v + t for t ≥ 0, we get
p(u, v, w) =(2 u + 10 v)t5 + (10 u2 + 40 uv + 102 v 2 )t4
+ (156 uv 2 + 32 u3 + 349 v 3 + 68 u2 v)t3
+ (24 u4 + 188 uv 3 + 22 u2 v 2 + 72 u3 v + 603 v 4 )t2
+ v 2 (783 v 3 − 72 u3 + 224 uv 2 − 156 u2 v)t
Proving Inequalities with
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Yu-Dong Wu, Zhi-Hua Zhang
+ 6 v 4 (17 u2 + 68 uv + 107 v 2 )
=p1 (u, v, t) + p2 (u, v, t),
and Yu-Rui Zhang
vol. 8, iss. 3, art. 81, 2007
where
p1 (u, v, t) = (2 u + 10 v)t5 + (10 u2 + 40 uv + 102 v 2 )t4
+ (156 uv 2 + 32 u3 + 349 v 3 + 68 u2 v)t3 ≥ 0,
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and
(2.13)
4
3
2 2
3
4
2
p2 (u, v, t) = (24 u + 188 uv + 22 u v + 72 u v + 603 v )t
+ v 2 (783 v 3 − 72 u3 + 224 uv 2 − 156 u2 v)t
+ 6 v 4 (17 u2 + 68 uv + 107 v 2 ).
It is easy to see that 24 u4 + 188 uv 3 + 22 u2 v 2 + 72 u3 v + 603 v 4 > 0, and the
discriminant of the quadratic function (2.13) with respect to t is
(2.14) ∆(u, v) = −v 4 (935415 v 6 + 480144 u3 v 3 + 1116096 uv 5
+ 803456 u2 v 4 + 4608 u6 + 196032 u4 v 2 + 46080 u5 v) ≤ 0.
This is to say that p2 (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
X
X
(2.15)
m2a h2a ≥
m2a ra2 ,
X
(2.16)
(2.17)
X
sin8 A ≥
X
cos8
A
,
2
X a 2
(rb − rc )2 ,
(b − c) ≥
b+c
2
and
(2.18)
X
(hb + hc − ha )3 ≥ 3ma mb mc .
Remark 2. The operations in this paper were performed using mathematical software
Maple 9.0.
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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3.
Generalization of the Method
In fact, Difference Substitution can go even further. Now, we consider the following
inequality [12]. In 4ABC, if max (A, B, C) ≤ 2π
, then
3
s2 ≥ R2 + 10Rr + 3r2 .
(3.1)
Utilizing the known formulas R = abc
,r =
4S
from (1.6), inequality (3.1) is equivalent to
S
s
and S =
p
s(s − a)(s − b)(s − c),
(3.2) 3 c2 a2 b2 − a2 bc3 − a3 bc2 − a3 b2 c − a2 b3 c − ab2 c3 − ab3 c2 − 2 b3 c3
− 2 a3 c3 + c4 b2 − 2 a3 b3 + c5 b + a4 c2 + b4 c2 + b5 a + a5 c + c5 a
− a6 + a4 b2 + a5 b − b6 − c6 + a2 b4 + b5 c + a2 c4 ≥ 0,
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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or
Contents
(3.3) F (x, y, z)
2 2 2
4
4
2 3
2 3
3 2
= −42 x y z + 14 y zx + 14 xyz + 2 xy z + 2 x y z + 2 xy z
+ 14 x4 yz + 2 x3 yz 2 + 2 x2 yz 3 + 2 x3 y 2 z − x4 y 2 − x2 z 4
− 2 x3 z 3 − x4 z 2 − 2 x3 y 3 − y 4 z 2 − y 4 x2 − 2 y 3 z 3 − y 2 z 4 ≥ 0,
where x > 0, y > 0, z > 0.
Since inequality (3.3) is symmetric with x, y, z, there is no harm in supposing
that x ≤ y ≤ z. From (1.2), F (x, y, z) in (3.3) is transformed into
(3.4) F (x, y, z) =P (u, v, w)
=(8 u2 + 4 uv + 2 uw − v 2 − vw)(8 uvw2 + 12 uv 2 w + 4 u2 vw
+ 4 v 4 + 4 u2 v 2 + 2 uw3 + 8 uv 3 + 8 v 3 w + 7 v 2 w2 + 3 vw3 )
+ 2 w2 (v + 2 u)2 (v + 2 u + w)2 ≥ 0,
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and for max (A, B, C) ≤
(3.5)
2π
3
and y = cos x decreasing in x ∈ (0, π), we have
1
2π
b2 + c2 + bc − a2 = b2 + c2 − bc cos
− a2
2
3
= 3x2 + 3(y + z)x − yz
= 8u2 + 4uv + 2uw − v 2 − vw
1
≥ b2 + c2 − bc cos A − a2 = 0.
2
Proving Inequalities with
Difference Substitution
Yu-Dong Wu, Zhi-Hua Zhang
and Yu-Rui Zhang
Since F (x, y, z) = P (u, v, w) ≥ 0 for u > 0, v ≥ 0 and w ≥ 0, inequality (3.1) is
obtained.
vol. 8, iss. 3, art. 81, 2007
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References
[1] L. YANG, Difference substitution and automated inequality proving, J.
Guangzhou Univ. (Natural Sciences Edition), 5(2) (2006), 1–7. (in Chinese)
[2] L. YANG, Solving harder problems with lesser mathematics, Proceedings of
the 10th Asian Technology Conference in Mathematics, ATCM Inc., 2005, 37–
46.
[3] B.-Q. LIU, The generating operation and its application in the proof of the
symmetric inequality in n variables, J. Guangdong Edu. Inst., 25(3) (2005),
10–14. (in Chinese)
[4] Y.-D. WU, On one of H.Alzer’s problems, J. Ineq. Pure Appl. Math.,
7(2) (2006), Art. 71. [ONLINE: http://jipam.vu.edu.au/article.
php?sid=688].
[5] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND V. VOLENEC, Recent Advances in
Geometric Inequalities, Acad. Publ., Dordrecht, Boston, London, 1989, 248.
[6] X.-G. CHU AND J. LIU, Some semi-symmetric inequalities in triangle, HighSchool Mathematics Monthly, 3(1999), 23–25. (in Chinese)
[7] J. LIU, CIQ.51, Communication in Research of Inequalities (2003), no. 6, 94.
(in Chinese)
[8] J. LIU, Nine sine inequality, Manuscript, 2005.
[9] X.-G. CHU, On several inequalities with respect to medians in acute triangle,
Research in Inequalities, Tibet People’s Press, 2000, 296. (in Chinese)
[10] B.-Q. LIU, 110 interesting inequalities problems, Research in Inequalities. Tibet People’s Press, 2000, 389–405. (in Chinese)
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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[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)
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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