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volume 7, issue 3, article 108, 2006.

Received 18 February, 2006;

accepted 23 March, 2006.

Communicated by:A. Lupa¸s

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Journal of Inequalities in Pure and Applied Mathematics

A CLASS OF INEQUALITIES RELATED TO THE ANGLE BISECTORS AND THE SIDES OF A TRIANGLE

SHAN-HE WU AND ZHI-HUA ZHANG

Department of Mathematics Longyan College

Longyan Fujian 364012 People’s Republic of China.

EMail:wushanhe@yahoo.com.cn Zixing Educational Research Section Chenzhou, Hunan 423400

People’s Republic of China.

EMail:zxzh1234@163.com

c

2000Victoria University ISSN (electronic): 1443-5756 046-06

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A Class of Inequalities Related to the Angle Bisectors and the

Sides of a Triangle Shan-He Wu and Zhi-Hua Zhang

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J. Ineq. Pure and Appl. Math. 7(3) Art. 108, 2006

http://jipam.vu.edu.au

Abstract

In the present paper a class of geometric inequalities concerning the angle bisectors and the sides of a triangle are established. Moreover an interesting open problem is proposed.

2000 Mathematics Subject Classification:26D15.

Key words: Inequality; Triangle; Angle bisector; Cyclic sum; Best coefficient.

The research was supported by the Natural Science Foundation of Fujian Province Education Department of China under grant No. JA05324.

1. Introduction and Main Results

For a given triangle ABC we assume that A, B, C denote its angles, a, b, c denote the lengths of its corresponding sides,wa, wb, wc denote respectively the bisectors of angles A, B, C. Let R, r and s be the circumradius, the inradius and the semi-perimeter of a triangle respectively. In addition we will customarily use the symbolsP

(cyclic sum) andQ

(cyclic product), such as Xf(a) = f(a) +f(b) +f(c), Y

f(a) =f(a)f(b)f(c).

The angle bisectors of triangles have many interesting properties. In par- ticular, inequalities for angle bisectors is a very attractive subject and plays an important role in the study of geometry. A large number of related results can be found in the well-known monographs [1] – [3]. In recent years, we have given considerable attention to these inequalities (see [4] – [8]). Recently, the following interesting double inequality concerning the angle bisectors and the sides, which was presented by X.-Zh. Yang, T.-Y. Ma and W.-L. Dong in [9,10], has come to our attention:

(1.1) 3√ 3 2 + 8

3 − 3√ 3 2

! 1− 2r

R

≤Xw_a

a ≤ 3√ 3 2 + 2√

3 R

2r −1

.

The above result also motivates us to investigate some similar inequalities.

We give here sharp lower and upper bounds for the sum P _a

wa. Moreover, in Section 3 the obtained result will be used for establishing an analogue of inequality (1.1).

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Theorem 1.1. In any triangleABCthe following double inequalities hold

(1.2) 1

2 s

r +√ 3

≤X a w_a ≤

√2 2

s r + 2√

6−3√ 3

,

with equality if and only if the triangle is an equilateral. Furthermore, ¹₂ and

√2

2 are the best coefficients in (1.2).

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2. Proof of Theorem 1.1

To prove Theorem 1.1, we shall use the following known results [2, p. 3, p.

241] (see also [1])

Lemma 2.1. In any triangleABC we have the following inequalities (2.1) s⁴ ≤s²(4R²+ 20Rr−2r²)−r(4R+r)³,

(2.2) 2R²+ 10Rr−r²−2(R−2r)√

R²−2Rr

≤s² ≤2R²+ 10Rr−r²+ 2(R−2r)√

R²−2Rr, with equality if and only if the triangle is isosceles.

(2.3) s≤ 1

√3(4R+r), with equality if and only if the triangle is equilateral.

In any acute triangleABC we have

(2.4) s² ≥4R²+ 4Rr+r²,

with equality if and only if the triangle is equilateral.

Proof of Theorem1.1. By the formula for angle bisector of triangleABC w_a =

2bc

b+ccos^A₂, we have X a

w_a =X

(cscB+ cscC) sinA 2

=X

cscA

XsinA 2

−1 2

XsecA 2.

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Based on the above result, it follows from the identity Q

cot^A₂ = _r^s that the inequality (1.2) is equivalent to the following inequality

1 2

YcotA 2 +

√3

2 ≤X

cscA

XsinA 2

−1 2

XsecA (2.5) 2

≤

√2 2

Ycot A 2 +4√

3−3√ 6

2 .

Using a substitutionA → π−2A,B →π −2B, C → π−2Cin (2.5), then the inequality (2.5) can be translated to

1 2

YtanA+

√3

2 ≤X

csc 2A X

cosA

− 1 2

XcscA (2.6)

≤

√2 2

YtanA+4√

3−3√ 6

2 .

Now, in order to prove the inequality (2.5), it is enough to prove that the inequality (2.6) to be valid for any acute triangle.

Note that the following known identities for a triangle [2, p. 55-60]:

Xcsc 2A= s⁴+s²(2r²−8Rr−4R²) + 16R³r+ 20R²r²+ 8Rr³+r⁴ 4rs(s²−4R²−4Rr−r²) , YtanA= 2rs

s²−4R²−4Rr−r², XcscA= s²+ 4Rr+r²

2rs ,

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XcosA= R+r R ,

taking these identities into (2.6), we find that the inequality (2.6) is equivalent to

(2.7) 4Rr²s²+ 2√

3 s² −4R²−4Rr−r² srR

≤H ≤4√

2Rr²s²+ (8√

3−6√

6) s²−4R²−4Rr−r² srR, where

H = (R+r)

s⁴+s²(2r²−8Rr−4R²)

+ 16R³r+ 20R²r²+ 8Rr³+r⁴

−R s²+ 4Rr+r²

s²−4R²−4Rr−r² . Let us now prove the inequality (2.7) to be valid for any acute triangle.

Using the inequalities (2.1), (2.3) and (2.4), we have H−4√

2Rr²s² − 8√

3−6√ 6

s²−4R²−4Rr−r² srR (2.8)

≤(R+r)

s²(4R²+ 20Rr−2r²)−r(4R+r)³

+s²(2r²−8Rr−4R²) + 16R³r+ 20R²r²+ 8Rr³+r⁴

−R s²+ 4Rr+r²

s²−4R²−4Rr−r²

−4√

2Rr²s²

−(8−6√

2) s²−4R²−4Rr−r²

(4R+r)rR

=s²h

−40 + 24√ 2

R+

6 + 2√ 2

ri Rr +

160−96√ 2

R⁴r+

152−120√ 2

R³r²

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+

52−48√ 2

R²r³+

6−6√ 2

Rr⁴

=F(s, R, r).

From Euler’s inequality R ≥ 2r, we observe that (−40 + 24√

2)R + (6 + 2√

2)r <0.

Case 1. WhenR >(√

2 + 1)r, by inequality (2.4) we have

F(s, R, r)≤(4R²+ 4Rr+r²)h

−40 + 24√ 2

R+

6 + 2√ 2

ri Rr +

160−96√ 2

R⁴r+ (152−120√ 2)R³r² +

52−48√ 2

R²r³+

6−6√ 2

Rr⁴

=

16−16√ 2

R³r²+

36−16√ 2

R²r³+

12−4√ 2

Rr⁴

=−4Rr²h

R−√ 2 + 1

ri h 4√

2−4 R+

4√ 2−5

ri

<0.

Case 2. When2r≤R ≤(√

2 + 1)r, by inequality (2.2) we get F(s, R, r)≤h

2R² + 10Rr−r²−2(R−2r)√

R² −2Rri

×h

−40 + 24√ 2

R+

6 + 2√ 2

ri

Rr+

160−96√ 2

R⁴r +

152−120√ 2

R³r²+

52−48√ 2

R²r³+

6−6√ 2

Rr⁴

= 4Rr³(R−2r)[F1(R, r) +F2(R, r)],

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where

F₁(R, r) =

20−12√ 2

R r

−3−√ 2

s R

r R

r −2

,

F₂(R, r) =

20−12√ 2

R r

2

+

−19 + 7√ 2

R r

+√

2.

We deduce from2≤R/r≤√

2 + 1that F₁(R, r)≤h

20−12√ 2 √

2 + 1

−3−√

2ir√

2 + 1 √

2 + 1−2

= 7√ 2−7,

F₂(R, r)≤

20−12√ 2 √

2 + 12

+

−19 + 7√ 2 √

2 + 1 +√

2

= 7−7√ 2, which leads toF(s, R, r)≤0.

Consequently (2.9) H−4√

2Rr²s²− 8√

3−6√ 6

s²−4R²−4Rr−r²

srR≤0.

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On the other hand, utilizing the inequalities (2.3) and (2.4), we have H−4Rr²s²−2√

3 s²−4R² −4Rr−r² srR

≥(R+r)

s⁴+s²(2r²−8Rr−4R²) + 16R³r+ 20R²r² + 8Rr³+r⁴

−R s²+ 4Rr+r²

s²−4R² −4Rr−r²

−4√

2Rr²s²−2 s²−4R²−4Rr−r²

(4R+r)R

=r s²−4R²−4Rr−3r²2

+ 4r(R+r)F(s, R, r), where

F(s, R, r) = −s²(3R−2r) + 12R³+ 4R²r−Rr²−2r³.

By Euler’s inequality R ≥ 2r, we conclude that 3R −2r > 0. Using the inequality (2.2) yields

F(s, R, r)≥ −(3R−2r)h

2R² + 10Rr−r²+ 2(R−2r)√

R²−2Rri + 12R³+ 4R²r−Rr²−2r³

= 2(R−2r)h

R(3R−5r) +r²−(3R−2r)√

R²−2Rri

= 2(R−2r) [Rr²(3R−2r) +r⁴] R(3R−5r) +r²+ (3R−2r)√

R²−2Rr

≥0.

Consequently

(2.10) H−4Rr²s²−2√

3 s²−4R²−4Rr−r²

srR≥0.

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Combining (2.9) and (2.10) yields the inequality (2.7), then from (2.7), the inequality (1.2) follows immediately. Moreove, from the process of proving inequality (1.2), it is easy to observe that the equalities hold in (1.2) if and only if the triangle is equilateral.

Next, we need to show that the coefficients¹₂ and

√ 2

2 in (1.2) are best possible in the strong sense.

Consider the inequality (1.2) in a general form as

(2.11) λ s

r + 2√ 3 λ −3√

3

!

≤X a wa

≤k s r +2√

3 k −3√

3

! .

Puttinga= 1, b = 1, c= 2tin (2.11) yields that λ(1 +t)+

2√

3−3√ 3λ

t

r1−t 1 +t (2.12)

≤ 4t²+ (1 + 2t)√ 2−2t 2 + 2t

≤k(1 +t) + 2√

3−3√ 3k

t

r1−t 1 +t.

In (2.12), passing the limit ast→0andt→1respectively, we find thatλ≤ ¹₂ andk ≥

√ 2

2 . Thus the best possible values forλandkin (2.11) is thatλ_max= ¹₂, k_min=

√2

2 . This completes the proof of Theorem1.1.

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3. An Application

As an application of Theorem 1.1, we establish an analogue of the inequality (1.1), as follows.

Theorem 3.1. In any triangleABCthe following double inequalities hold

(3.1) 2√ 3 + 3

2

1− 2r R

≤X a

w_a ≤2√

3 + 2√ 2

R 2r −1

,

with equality if and only if the triangle is equilateral. Furthermore,2√

2is the best coefficient in the right-hand side of inequality (3.1).

Proof. Applying Theorem1.1and Blundon’s inequality [11]s ≤2R+ (3√ 3− 4)r, it follows that

X a w_a ≤

√2 2

s r + 2√

6−3√ 3

≤2√

3 + 2√ 2

R 2r −1

.

On the other hand, by using Theorem1.1 and Gerretsen’s inequality [12]s² ≥ 16Rr−5r², we have

X a wa

−2√ 3−3

2

1− 2r R

≥ 1 2

s r +√

3

−2√ 3−3

2

1− 2r R

= 1

2Rr[sR−3(R−2r)r−6√ 3Rr]

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≥ 1 2Rr

h R√

16Rr−5r²−3(R−2r)r−3√ 3Rr

i

= R²(16Rr−5r²)−

3(R−2r)r+ 3√ 3Rr2

2Rr R√

16Rr−5r²+ 3(R−2r)r+ 3√ 3Rr

= 16(R−2r)³ + (55−18√

3)(R−2r)²+ (64−36√

3)(R−2r) 2R

R√

16Rr−5r²+ 3(R−2r)r+ 3√ 3Rr

≥0, so that

X a

w_a ≥2√ 3 + 3

2

1− 2r R

.

The inequality (3.1) is proved. It follows directly from Theorem 1.1 that the equalities hold in (3.1) if and only if the triangle is equilateral.

Let us now show that the coefficient2√

2in the right-hand side of inequality (3.1) is best possible.

Consider the inequality (3.1) in a general form as

(3.2) X a

w_a ≤2√ 3 +µ

R 2r −1

.

Puttinga= 1, b= 1, c= 2tin (3.2) yields that (3.3) 4t²+ (1 + 2t)√

2−2t

2 + 2t ≤ µ

4√

1−t² + 2√

3−µ t

r1−t 1 +t. Passing the limit as t → 0 in (3.3), we get µ ≥ 2√

2. Thus the best possible value for µin (3.2) should be µ_min = 2√

2. The proof of Theorem 3.1 is complete.

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It is worth noticing that the coefficient ³₂ is not best possible for the left-hand side of inequality (3.1), this may lead us to further discussion of the following significant problem.

Open Problem. Determine the best coefficientµfor which the inequality below holds

(3.4) X a

w_a ≥2√ 3 +µ

1− 2r

R

.

It seems that the problem is complicated and difficult. Indeed, it is unable to be solved in a same way as the foregoing technique.

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References

[1] O. BOTTEMA, R.Z. DJORDJEVI Ć, R.R. JANI Ć, D.S. MITRI- NOVI Ć ANDP.M. VASI ´C, Geometric Inequalities., Groningen, Wolters- Noordhoff, 1969.

[2] D.S. MITRINOVI ´C, J.E. PE ˘CARI ´CANDV. VOLENEC, Recent Advances in Geometric Inequalities, Dordrecht, Netherlands, Kluwer Academic Publishers, 1989.

[3] D.S. MITRINOVI ´C, J.E. PE ˘CARI ´C, V. VOLENEC AND J. CHEN, Ad- denda to the monograph: Recent Advances in Geometric Inequalities (I), Journal of Ningbo University, 4(2) (1991), 1–145.

[4] SH.-H. WU, An inverse inequality for the angle bisectors and its analogue.

Journal of Suzhou Railway Teachers College, 17(8) (2000), 42–44.(in Chi- nese)

[5] SH.-H. WU, A strengthened inequality for the angle bisectors and its ana- logue. Journal of Longyan Teachers College, 20(10) (2002), 98–99.(in Chinese)

[6] SH.-H. WU, Some inequalities related to the angle bisectors and the radius of escribed circle, Journal of Guizhou Educational College, 13(8) (2002), 16–18. (in Chinese)

[7] SH.-H. WU, An inverse inequality for the angle bisectors, Maths Teaching in Middle School, 6 (2001), 62–63. (in Chinese)

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[8] SH.-H. WU, A simple proof of an inequality concerning the angle bisec- tors, Fujian High-School Mathematics, 3 (1997), 19–20. (in Chinese) [9] X.-ZH. YANG, Sharpness of an inequality involving the angle bisectors,

Middle School Mathematics, 12 (1995), 20–22. (in Chinese)

[10] T.-Y. MA AND W.-L. DONG, The inverse version of an inequality con- cerning the angle bisectors, Bulletin of Maths (Wuhan), 11 (1995), 19–20.

(in Chinese)

[11] W.J. BLUNDON, Inequalities associated with the triangle, Canad. Math.

Bull., 8 (1965), 615–626.

[12] J.C. GERRETSEN, Ongelijkheden in de Driehoek, Nieuw Tijdschr. Wisk., 41 (1953), 1–7.