THE BEST CONSTANTS FOR A DOUBLE
INEQUALITY IN A TRIANGLE
YU-DONG WU
NU-CHUN HU
Department of Mathematics
Zhejiang Xinchang High School
Shaoxing 312500, Zhejiang
People’s Republic of China.
EMail: yudong.wu@yahoo.com.cn
Department of Mathematics
Zhejiang Normal University
Jinhua 321004, Zhejiang
People’s Republic of China.
EMail: nuchun@zjnu.cn
WEI-PING KUANG
Department of Mathematics
Huaihua University
Huaihua 418008, Hunan
People’s Republic of China.
Double Inequality in a
Triangle
Yu-Dong Wu, Nu-Chun Hu
and Wei-Ping Kuang
vol. 10, iss. 1, art. 29, 2009
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EMail: sy785153@126.com
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Received:
08 August, 2008
Accepted:
02 February, 2009
Communicated by:
S.S. Dragomir
2000 AMS Sub. Class.:
51M16
Key words:
Inequality; Best Constant; Triangle.
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Abstract:
In this short note, by using some of Chen’s theorems and classic analysis,
we obtain a double inequality for triangle and give a positive answer to a
problem posed by Yang and Yin [6].
Acknowledgment:
The authors would like to thank Prof. Zhi-Hua Zhang and Dr. Zhi-Gang
Wang for their careful reading and making some valuable suggestions in
the preparation of this paper.
Double Inequality in a
Triangle
Yu-Dong Wu, Nu-Chun Hu
and Wei-Ping Kuang
vol. 10, iss. 1, art. 29, 2009
Dedicatory:
Dedicated to Professor Bi-Cheng Yang on the occasion of his 63rd birthday.
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Contents
1
Introduction and Main Results
4
2
Preliminary Results
6
3
The Proof of Theorem 1.1
13
Double Inequality in a
Triangle
Yu-Dong Wu, Nu-Chun Hu
and Wei-Ping Kuang
vol. 10, iss. 1, art. 29, 2009
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1.
Introduction and Main Results
For 4ABC, let a, b, c denote the side-lengths, A, B, C the angles, s the semiperimeter, R the circumradius and r the inradius, respectively.
In 1957, Kooistra (see [1]) built the following double inequality for any triangle:
√
B
C
3 3
A
(1.1)
2 < cos + cos + cos ≤
.
2
2
2
2
In 2000, Yang and Yin [6] considered a new bound of inequality (1.1) and posed
a problem as follows:
Problem 1. Determine the best constant µ such that
√ !µ s 1−µ
A
B
C
3 3
·
≤ cos + cos + cos
(1.2)
2
R
2
2
2
holds for any 4ABC.
In this short note, we solve the above problem and obtain the following result.
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λ ≥ λ0 = 1
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√
µ ≤ µ0 =
vol. 10, iss. 1, art. 29, 2009
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Theorem 1.1. Let
and
Double Inequality in a
Triangle
Yu-Dong Wu, Nu-Chun Hu
and Wei-Ping Kuang
2 ln (2 − 2) + ln 2
≈ 0.7194536993.
4 ln 2 − 3 ln 3
Then the double inequality
√ !µ 3 3
s 1−µ
A
B
C
·
≤ cos + cos + cos ≤
(1.3)
2
R
2
2
2
√ !λ 3 3
s 1−λ
·
2
R
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holds for any 4ABC, while the constants λ0 and µ0 are both the best constant for
inequality (1.3).
Remark 1. When λ0 = 1, the right hand of inequality (1.3) is just the right hand of
inequality (1.1).
Remark 2. It is not difficult to demonstrate that:
 √ µ
µ0 1−µ
0
1
0

3 3
s 1−µ0
s
2
1−µ0

√
<2 0< R <2
,
· R

2
3
3



√ µ0



 323
·
s 1−µ0
R
≥2 2
1
1−µ0
2
√
3 3
µ0
1−µ
0
≤
s
R
≤
√
3 3
2
Double Inequality in a
Triangle
Yu-Dong Wu, Nu-Chun Hu
and Wei-Ping Kuang
vol. 10, iss. 1, art. 29, 2009
.
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2.
Preliminary Results
In order to establish our main theorem, we shall require the following lemmas.
Lemma 2.1 (see [3, 4, 5]). If the inequality s ≥ (>)f (R, r) holds for any isosceles
triangle whose top-angle is greater than or equal to π3 , then the inequality s ≥ (>
)f (R, r) holds for any triangle.
Lemma 2.2 (see [2, 3]). The homogeneous inequality
s ≥ (>)f (R, r)
(2.1)
vol. 10, iss. 1, art. 29, 2009
holds for any acute-angled triangle
if
√ for any acute isosceles
and only if it holds
triangle whose top-angle A ∈ π3 , π2 with 2r ≤ R < ( 2 + 1)r and any right√
angled triangle with R ≥ ( 2 + 1)r.
For the convenience of our readers, we give below the proof by Chen in [2, 3].
J
Proof. Let
O denote the circumcircle of 4ABC. Necessity is obvious from
Lemma 2.1. Thus we only need to prove the sufficiency. It is well known that
R ≥ 2r for any acute-angled triangle. So we consider the following two cases:
√
(i) When 2r ≤ R < ( 2+1)r : In this case
J , we can construct an isosceles triangle
A1 B1 C1 whose circumcircle is also
O and the top-angle of 4A1 B1 C1 (see
Figure 1) is
!
r
1
2r
A1 = 2 arcsin
1+ 1−
.
2
R
It is easy to see that (see [4, 5]):
R1 = R,
r1 = r,
s1 ≤ s
and
Double Inequality in a
Triangle
Yu-Dong Wu, Nu-Chun Hu
and Wei-Ping Kuang
π
π
≤ A1 < .
3
2
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C
C1
I1
O
A
B1
I2
B
A1
Double Inequality in a
Triangle
Yu-Dong Wu, Nu-Chun Hu
and Wei-Ping Kuang
vol. 10, iss. 1, art. 29, 2009
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Figure 1:
Thus we have
(2.2)
s ≥ s1 ≥ f (R1 , r1 ) = f (R, r).
because the inequality
(2.1) holds for any acute isosceles triangle whose top
angle A ∈ π3 , π2 .
√
(ii) When R ≥ ( 2 + 1)r: In this case we
Jcan construct a right-angled triangle
A2 B2 C2 whose inscribed circle is also
I and the length of its hypotenuse is
c2 = 2R (see Figure 2). This implies that
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A2
C
A
I
O
C2
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Yu-Dong Wu, Nu-Chun Hu
and Wei-Ping Kuang
B2
vol. 10, iss. 1, art. 29, 2009
B
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Figure 2:
1
1
r2 = (a2 + b2 − c2 ) = r, R2 = c2 = R,
2
2
1
s2 = (a2 + b2 + c2 ) = 2R2 + r2 = 2R + r < s.
2
Thus we have the inequality (2.2) since the inequality (2.1) holds for any rightangled triangle.
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Lemma 2.3 (see [2, 3]). The homogeneous inequality (2.1) holds for any acuteangled triangle if and only if
p
√
2
3
(2.3)
(1 − x)(3 + x) ≥ (>)f (2, 1 − x )
0≤x< 2−1 ,
and
5 − x2 ≥ (>)f (2, 1 − x2 )
(2.4)
√
2−1≤x<1 .
Proof. Since the inequality (2.1) is homogeneous, we may assume R = 2 without
losing generality.
√
(i) When 2r ≤ R < ( 2 + 1)r: By Lemma 2.2, we only need to consider the
isosceles triangle whose top-angle A ∈ [ π3 , π2 ). Let
" √ !
A
1 2
t = sin ∈
,
.
2
2 2
r = 4t(1 − t) and
√
s = 4(1 + t) 1 − t2 .
Let x = 2t − 1. Then the inequality (2.1) is just the inequality (2.3).
√
(ii) When R ≥ ( 2 + 1)r: We only need to consider a right-angled triangle. Let
!
√
√
2r
2
r=
= 4t(1 − t) ∈ 0, 2 − 1
≤t<1 .
R
2
Thus we have
(2.6)
vol. 10, iss. 1, art. 29, 2009
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Then we have (see [4, 5])
(2.5)
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Yu-Dong Wu, Nu-Chun Hu
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s = 2R + r = 4 + 4t(1 − t).
Let x = 2t − 1. Then the inequality (2.1) is just the inequality (2.4).
This completes the proof Lemma 2.3.
Lemma 2.4 ([3, 4, 5]). The homogeneous inequality
s ≤ (<)f (R, r)
(2.7)
holds for any triangle if and only if it holds for any isosceles triangle whose topangle A ∈ (0, π3 ], or the following inequality holds
p
(2.8)
(1 − x)(3 + x)3 ≤ (<)f 2, 1 − x2
(−1 < x ≤ 0) .
Lemma 2.5 (see [2, 3]). The homogeneous inequality (2.7) holds for any acuteangled triangle if and only if it holds for any isosceles triangle whose top-angle
A ∈ (0, π3 ], or the inequality (2.8) holds.
Proof. As acute-angled triangles include all isosceles triangles whose top-angle is
less than or equal to π3 , Lemma 2.5 straightforwardly follows from Lemma 2.4 and
Lemma 2.1.
Lemma 2.6. Define
2 ln (1 − x) + 2 ln (1 + x)
(2.9)
G1 (x) :=
.
3 ln (1 − x) + 3 ln (3 + x) + 2 ln (1 + x) − 3 ln 3
√
Then G1 is decreasing on (−1, 2 − 1), and
√
2 ln (2 − 2) + ln 2
(2.10)
lim
G1 (x) =
< G1 (x) < 1 = lim + G1 (x).
√
x→−1
4 ln 2 − 3 ln 3
x→( 2−1)−
Proof. Let G01 be the derivative of G1 . It is easy to see that
(2.11)
G01 (x) =
4xg1 (x)
(3 ln (1−x)+3 ln (3+x)+2 ln (1+x)−3 ln 3)2 (1−x2 )(3+x)
Double Inequality in a
Triangle
Yu-Dong Wu, Nu-Chun Hu
and Wei-Ping Kuang
vol. 10, iss. 1, art. 29, 2009
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with
(2.12)
g1 (x) := (x−1) ln (1−x)−3(x+3)[ln (3+x)−ln 3]+2(x+1) ln (1+x).
Moreover, we know that
(2.13)
g10 (x) = ln (1 − x) − 3 ln (3 + x) + 2 ln (1 + x) + 3 ln 3
and
−8x
.
(1 − x2 )(3 + x)
√
Now we show that G1 is decreasing on −1, 2 − 1 .
g100 (x) =
(2.14)
(i) It is easy to see that g100 (x) ≥ 0 when −1 < x ≤ 0, and g10 is increasing on
(−1, 0]. Thus, g10 (x) ≤ g10 (0) = 0, and g1 is decreasing on (−1, 0]. Therefore,
g1 (x) ≥ g1 (0) = 0, and G01 (x) ≤ 0. This means that G1 is decreasing on
(−1, 0].
√
00
(ii) It
is
easy
to
see
that
g
(x)
≤
0
when
0
≤
x
<
2 − 1, and g10 is decreasing
on
1
√
√
0
0
0, 2 − 1 . Thus, g1 (x) ≤ g1 (0) = 0, and g1 is decreasing on 0, 2 − 1 .
0
Therefore,
g1 (x) ≤
g1 (0) = 0, and G1 (x) ≤ 0. This means that G1 is decreas √
ing on 0, 2 − 1 .
√
Combining (i) and (ii), it follows that G1 is decreasing on (−1, 2−1) and (2.10)
holds, and hence the proof is complete.
Lemma 2.7. Define
(2.15)
G2 (x) :=
2 ln (1 − x2 )
.
2 ln (5 − x2 ) + 2 ln (1 − x2 ) − 3 ln 3
Double Inequality in a
Triangle
Yu-Dong Wu, Nu-Chun Hu
and Wei-Ping Kuang
vol. 10, iss. 1, art. 29, 2009
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Then G2 is increasing on
√
2 − 1, 1 , and
√
√
2 ln (2 − 2) + ln 2
(2.16)
G2 (x) ≥ G2 ( 2 − 1) =
.
4 ln 2 − 3 ln 3
Proof. Let G02 be the derivative of G2 . It is easy to see that
4xg2 (x)
(2.17)
G02 (x) =
,
2
(2 ln (5 − x ) + 2 ln (1 − x2 ) − 3 ln 3)2 (1 − x2 )(5 − x2 )
g20 (x) = −2xh2 (x),
and
−16x
(2.18)
h02 (x) =
;
(1 − x2 ) (5 − x2 )
where
g2 (x) := 2 1 − x2 [ln (1 − x)+ln (1 + x)]+2 x2 − 5 ln 5 − x2 +3 5 − x2 ln 3,
and
2
− 2 ln 5 − x + 3 ln 3.
√
0
Thus
it
follows
that
h
(x)
<
0
when
2 − 1 ≤ x < 1, and h2 is decreasing on
2
√
2 − 1, 1 , and
√
27
√ < 0.
h2 (x) ≤ h2 ( 2 − 1) = ln
34 + 24 2
√
Therefore g20 (x) > 0, and g2 is increasing on 2 − 1, 1 , and
√
√
√
√
g2 (x) ≥ g2 ( 2 − 1) = 6( 2 + 1) ln 3 − 8 ln 2 − 8 2 ln
2 + 1 > 0.
√
This means that G02 (x) > 0, and G2 is increasing on
2 − 1, 1 , and (2.16) holds.
The proof of Lemma 2.7 is thus completed.
(2.19)
h2 (x) = 2 ln 1 − x
2
Double Inequality in a
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Yu-Dong Wu, Nu-Chun Hu
and Wei-Ping Kuang
vol. 10, iss. 1, art. 29, 2009
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3.
The Proof of Theorem 1.1
Proof. (i) The first inequality of (1.3) is equivalent to
√ !µ
A
B
C
3 3
(3.1)
cos + cos + cos ≥
· (sin A + sin B + sin C)1−µ
2
2
2
2
Double Inequality in a
Triangle
Yu-Dong Wu, Nu-Chun Hu
and Wei-Ping Kuang
with application to the well known identity
sin A + sin B + sin C =
s
.
R
vol. 10, iss. 1, art. 29, 2009
Taking
A → π − 2A,
B → π − 2B
and C → π − 2C,
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then inequality (3.1) is equivalent to
(3.2)
sin A + sin B + sin C ≥
√ !µ
3 3
· (sin 2A + sin 2B + sin 2C)1−µ
2
for an acute-angled triangle ABC.
By the well known identities
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sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C,
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and
sin A sin B sin C =
rs
,
2R2
the inequality (3.2) can be written as follows:
√ !µ 1−µ
s µ
s
3 3
2rs
≥
·
⇐⇒
≥
(3.3)
R
2
R2
R
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√ !µ 1−µ
3 3
2r
·
.
2
R
Furthermore, by Lemma 2.3, the inequality (3.3) holds if and only if the following
two inequalities
!µ
p
√ !µ
√
(1 − x)(3 + x)3
3 3
2 1−µ
≥
1−x
0≤x< 2−1
(3.4)
2
2
and
(3.5)
5 − x2
2
µ
≥
√ !µ
1−µ
3 3
1 − x2
2
√
2−1≤x<1
Double Inequality in a
Triangle
Yu-Dong Wu, Nu-Chun Hu
and Wei-Ping Kuang
vol. 10, iss. 1, art. 29, 2009
hold. In other words,
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µ ≤ min G(x)
(3.6)
0≤x<1
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where
(3.7)
G(x) =
(
G1 (x)
G2 (x)
√
0≤x< 2−1 ,
√
2−1≤x<1 ,
while G1 (x) and G2 (x) are defined by (2.9) and (2.15) respectively.
By Lemma 2.6 and Lemma 2.7, it follows that
√
min G(x) = G
2−1 .
0≤x<1
Thus the first inequality of (1.3) holds, and the best constant µ for inequality (1.3) is
√ 2 ln 2 − 2 + ln 2
µ0 =
.
4 ln 2 − 3 ln 3
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(ii) By applying a similar method to (i), it follows that the second inequality of (1.3)
is equivalent to
√ !λ 1−λ
s λ
3 3
2r
(3.8)
≤
·
.
R
2
R
By Lemma 2.5, the inequality (3.8) holds if and only if the following inequality
holds:
!λ
p
√ !λ
(1 − x)(3 + x)3
3 3
(1 − x2 )1−λ
(−1 < x ≤ 0) ,
(3.9)
≤
2
2
vol. 10, iss. 1, art. 29, 2009
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or equivalently,
(3.10)
Double Inequality in a
Triangle
Yu-Dong Wu, Nu-Chun Hu
and Wei-Ping Kuang
λ ≥ sup G1 (x),
−1<x≤0
where G1 (x) is given by (2.9).
By Lemma 2.6, it follows that λ ≥ 1. Moreover, the second inequality of (1.3)
holds when λ0 = 1. Thus the second inequality of (1.3) holds and the best constant
λ for inequality (1.3) is λ0 = 1. The proof of Theorem 1.1 is hence completed.
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References
[1] O. BOTTEMA, R.Z̆. DJORDJEVIĆ, R.R. JANIĆ, D.S. MITRINOVIĆ, AND
P.M. VASIĆ. Geometric Inequality. Wolters-Noordhoff Publishing, Groningen,
The Netherlands, 1969.
[2] S.-L. CHEN, Inequalities involving R, r, s in acute-angled triangle, Geometric
Inequalties in China, Jiangsu Educational Press, Nanjing (1996), No. 72-81. (in
Chinese)
Double Inequality in a
Triangle
Yu-Dong Wu, Nu-Chun Hu
and Wei-Ping Kuang
[3] S.-L. CHEN, The simplified method to prove inequalities in triangle, Studies of
Inequalities, Tibet People’s Press, Lhasa (2000), 3–8. (in Chinese)
vol. 10, iss. 1, art. 29, 2009
[4] S.-L. CHEN, A new method to prove one kind of inequalities-equate substitution
method, Fujian High-School Mathematics, No. 20-23, 1993(3). (in Chinese)
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[5] Y.-D. WU. The best constant for a geometric inequality, J. Inequal. Pure
Appl. Math., 6(4) (2005), Art. 111. [ONLINE http://jipam.vu.edu.au/
article.php?sid=585].
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[6] X.-Z. YANG AND H.-Y. YIN, The comprehensive investigations of trigonometric inequalities for half-angles of triangle in China, Studies of Inequalities, Tibet
People’s Press, Lhasa (2000), No.123–174. (in Chinese)
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