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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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THE BEST CONSTANTS FOR A DOUBLE INEQUALITY IN A TRIANGLE

YU-DONG WU NU-CHUN HU

Department of Mathematics Department of Mathematics Zhejiang Xinchang High School Zhejiang Normal University Shaoxing 312500, Zhejiang Jinhua 321004, Zhejiang People’s Republic of China. People’s Republic of China.

EMail:yudong.wu@yahoo.com.cn EMail:nuchun@zjnu.cn

WEI-PING KUANG

Department of Mathematics Huaihua University Huaihua 418008, Hunan People’s Republic of China.

EMail:sy785153@126.com

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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Close 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.

Dedicatory: Dedicated to Professor Bi-Cheng Yang on the occasion of his 63rd birth- day.

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Contents

1 Introduction and Main Results 4

2 Preliminary Results 6

3 The Proof of Theorem 1.1 13

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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 semi- perimeter,Rthe circumradius andrthe inradius, respectively.

In 1957, Kooistra (see [1]) built the following double inequality for any triangle:

(1.1) 2<cosA

2 + cosB

2 + cosC

2 ≤ 3√ 3 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 constantµsuch that

(1.2) 3√

3 2

!µ

·s R

1−µ

≤cosA

2 + cosB

2 + cosC 2 holds for any4ABC.

In this short note, we solve the above problem and obtain the following result.

Theorem 1.1. Let

λ ≥λ0 = 1 and

µ≤µ0 = 2 ln (2−√

2) + ln 2

4 ln 2−3 ln 3 ≈0.7194536993.

Then the double inequality

(1.3) 3√ 3 2

!µ

·s R

1−µ

≤cosA

2 + cosB

2 + cosC

2 ≤ 3√ 3 2

!λ

·s R

1−λ

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holds for any4ABC, while the constantsλ0 andµ0 are both the best constant for inequality (1.3).

Remark 1. Whenλ0 = 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:









3

3 2

µ0

· Rs1−µ0

<2

0< Rs <21−µ10

2 3

3

1−µµ0

0

,

3 3 2

µ0

· Rs1−µ0

≥2

21−µ10

2 3

3

1−µµ0

0Rs3

3 2

.

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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 inequalitys ≥(>)f(R, r)holds for any isosceles triangle whose top-angle is greater than or equal to π3, then the inequalitys ≥ (>

)f(R, r)holds for any triangle.

Lemma 2.2 (see [2,3]). The homogeneous inequality

(2.1) s ≥(>)f(R, r)

holds for any acute-angled triangle if and only if it holds for any acute isosceles triangle whose top-angle A ∈ π

3,π2

with 2r ≤ R < (√

2 + 1)r and any right- angled triangle withR≥(√

2 + 1)r.

For the convenience of our readers, we give below the proof by Chen in [2,3].

Proof. Let J

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≥2rfor any acute-angled triangle. So we consider the following two cases:

(i) When2r ≤R <(√

2+1)r:In this case , we can construct an isosceles triangle A1B1C1 whose circumcircle is alsoJ

O and the top-angle of 4A1B1C1(see Figure1) is

A1 = 2 arcsin1 2 1 +

r 1−2r

R

! .

It is easy to see that (see [4,5]):

R1 =R, r1 =r, s1 ≤s and π

3 ≤A1 < π 2.

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I2

C1

A1

O

C

A B

I1 B1

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- angleA ∈π

3,π2 . (ii) When R ≥ (√

2 + 1)r: In this case we can construct a right-angled triangle A2B2C2 whose inscribed circle is alsoJ

I and the length of its hypotenuse is c2 = 2R(see Figure2). This implies that

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A

C

B

O A2

B2 C2

I

Figure 2:

r2 = 1

2(a2+b2−c2) =r, R2 = 1

2c2 =R, s2 = 1

2(a2+b2+c2) = 2R2 +r2 = 2R+r < s.

Thus we have the inequality (2.2) since the inequality (2.1) holds for any right- angled triangle.

Lemma 2.3 (see [2, 3]). The homogeneous inequality (2.1) holds for any acute- angled triangle if and only if

(2.3) p

(1−x)(3 +x)3 ≥(>)f(2,1−x2)

0≤x <√

2−1 ,

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and

(2.4) 5−x2 ≥(>)f(2,1−x2) √

2−1≤x <1 .

Proof. Since the inequality (2.1) is homogeneous, we may assumeR = 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-angleA∈[π3,π2). Let

t = sinA 2 ∈

"

1 2,

√2 2

! . Then we have (see [4,5])

(2.5) r= 4t(1−t) and s= 4(1 +t)√ 1−t2. Letx= 2t−1. Then the inequality (2.1) is just the inequality (2.3).

(ii) WhenR ≥(√

2 + 1)r: We only need to consider a right-angled triangle. Let r= 2r

R = 4t(1−t)∈ 0,√

2−1

√2

2 ≤t <1

! .

Thus we have

(2.6) s = 2R+r = 4 + 4t(1−t).

Letx= 2t−1. Then the inequality (2.1) is just the inequality (2.4).

This completes the proof Lemma2.3.

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Lemma 2.4 ([3,4,5]). The homogeneous inequality

(2.7) s ≤(<)f(R, r)

holds for any triangle if and only if it holds for any isosceles triangle whose top- angleA∈(0,π3], or the following inequality holds

(2.8) p

(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 acute- angled 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, Lemma2.5straightforwardly follows from Lemma2.4 and Lemma2.1.

Lemma 2.6. Define

(2.9) G1(x) := 2 ln (1−x) + 2 ln (1 +x)

3 ln (1−x) + 3 ln (3 +x) + 2 ln (1 +x)−3 ln 3. ThenG1 is decreasing on(−1,√

2−1), and (2.10) lim

x→( 2−1)

G1(x) = 2 ln (2−√

2) + ln 2

4 ln 2−3 ln 3 < G1(x)<1 = lim

x→−1+G1(x).

Proof. LetG01be the derivative ofG1. 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)

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with

g1(x) := (x−1) ln (1−x)−3(x+3)[ln (3+x)−ln 3]+2(x+1) ln (1+x).

(2.12)

Moreover, we know that

(2.13) g10(x) = ln (1−x)−3 ln (3 +x) + 2 ln (1 +x) + 3 ln 3 and

(2.14) g001(x) = −8x

(1−x2)(3 +x). Now we show thatG1 is decreasing on −1,√

2−1 .

(i) It is easy to see that g100(x) ≥ 0 when −1 < x ≤ 0, and g10 is increasing on (−1,0]. Thus,g10(x) ≤g01(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].

(ii) It is easy to see thatg100(x)≤0when0≤x <√

2−1, andg10 is decreasing on 0,√

2−1

. Thus, g10(x) ≤ g10(0) = 0, andg1 is decreasing on 0,√

2−1 . Therefore,g1(x) ≤g1(0) = 0, andG01(x)≤ 0. This means thatG1is decreas- ing on

0,√ 2−1

.

Combining (i) and (ii), it follows thatG1is 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.

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ThenG2 is increasing on

2−1,1 , and (2.16) G2(x)≥G2(√

2−1) = 2 ln (2−√

2) + ln 2 4 ln 2−3 ln 3 . Proof. LetG02be the derivative ofG2. It is easy to see that

(2.17) G02(x) = 4xg2(x)

(2 ln (5−x2) + 2 ln (1−x2)−3 ln 3)2(1−x2)(5−x2), g20(x) =−2xh2(x),

and

(2.18) h02(x) = −16x

(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

(2.19) h2(x) = 2 ln 1−x2

−2 ln 5−x2

+ 3 ln 3.

Thus it follows thath02(x) < 0when√

2−1 ≤ x < 1, andh2 is decreasing on √

2−1,1 , and

h2(x)≤h2(√

2−1) = ln 27 34 + 24√

2 <0.

Thereforeg20(x)>0, andg2 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 thatG02(x)> 0, andG2 is increasing on√

2−1,1

, and (2.16) holds.

The proof of Lemma2.7is thus completed.

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3. The Proof of Theorem 1.1

Proof. (i) The first inequality of (1.3) is equivalent to (3.1) cosA

2 + cosB

2 + cosC

2 ≥ 3√ 3 2

!µ

·(sinA+ sinB+ sinC)1−µ with application to the well known identity

sinA+ sinB+ sinC = s R. Taking

A →π−2A, B →π−2B and C →π−2C, then inequality (3.1) is equivalent to

(3.2) sinA+ sinB+ sinC ≥ 3√ 3 2

!µ

·(sin 2A+ sin 2B+ sin 2C)1−µ for an acute-angled triangleABC.

By the well known identities

sin 2A+ sin 2B + sin 2C= 4 sinAsinBsinC, and

sinAsinBsinC = rs 2R2, the inequality (3.2) can be written as follows:

(3.3) s

R ≥ 3√ 3 2

!µ

· 2rs

R2 1−µ

⇐⇒s R

µ

≥ 3√ 3 2

!µ

· 2r

R 1−µ

.

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Furthermore, by Lemma2.3, the inequality (3.3) holds if and only if the following two inequalities

(3.4)

p(1−x)(3 +x)3 2

!µ

≥ 3√ 3 2

!µ

1−x21−µ

0≤x <√ 2−1

and (3.5)

5−x2 2

µ

≥ 3√ 3 2

!µ

1−x21−µ

2−1≤x <1

hold. In other words,

(3.6) µ≤ min

0≤x<1G(x) where

(3.7) G(x) =

(G1(x) 0≤x <√ 2−1

, G2(x) √

2−1≤x <1 , whileG1(x)andG2(x)are defined by (2.9) and (2.15) respectively.

By Lemma2.6and Lemma2.7, it follows that

0≤x<1min G(x) = G√

2−1 .

Thus the first inequality of (1.3) holds, and the best constantµfor inequality (1.3) is µ0 = 2 ln 2−√

2 + ln 2 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

(3.8)

s R

λ

≤ 3√ 3 2

!λ

· 2r

R 1−λ

.

By Lemma 2.5, the inequality (3.8) holds if and only if the following inequality holds:

(3.9)

p(1−x)(3 +x)3 2

!λ

≤ 3√ 3 2

!λ

(1−x2)1−λ (−1< x≤0),

or equivalently,

(3.10) λ≥ sup

−1<x≤0

G1(x), whereG1(x)is given by (2.9).

By Lemma 2.6, it follows that λ ≥ 1. Moreover, the second inequality of (1.3) holds whenλ0 = 1. Thus the second inequality of (1.3) holds and the best constant λfor inequality (1.3) isλ0 = 1. The proof of Theorem1.1is hence completed.

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References

[1] O. BOTTEMA, R. ˘Z. DJORDJEVI ´C, R.R. JANI ´C, D.S. MITRINOVI ´C, AND

P.M. VASI ´C. 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)

[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)

[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) [5] Y.-D. WU. The best constant for a geometric inequality, J. Inequal. Pure

Appl. Math., 6(4) (2005), Art. 111. [ONLINEhttp://jipam.vu.edu.au/

article.php?sid=585].

[6] X.-Z. YANGANDH.-Y. YIN, The comprehensive investigations of trigonomet- ric 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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