A GEOMETRIC INEQUALITY OF THE GENERALIZED ERDÖS-MORDELL TYPE

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Erdös-Mordell Type Geometric Inequality Yu-Dong Wu, Chun-Lei Yu

and Zhi-Hua Zhang vol. 10, iss. 4, art. 106, 2009

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A GEOMETRIC INEQUALITY OF THE GENERALIZED ERDÖS-MORDELL TYPE

YU-DONG WU, CHUN-LEI YU ZHI-HUA ZHANG

Department of Mathematics Department of Mathematics

Zhejiang Xinchang High School Shili Senior High School in Zixing

Shaoxing 312500, Zhejiang Chenzhou 423400, Hunan

People’s Republic of China People’s Republic of China EMail:yudong.wu@yahoo.com.cn EMail:zxzh1234@163.com

seetill@126.com

Received: 20 April, 2009

Accepted: 09 October, 2009 Communicated by: S.S. Dragomir 2000 AMS Sub. Class.: Primary 51M16.

Key words: Geometric inequality, triangle, Erdös-Mordell inequality, Hayashi’s inequality, Klamkin’s inequality.

Abstract: In this short note, we solve an interesting geometric inequality problem relating to two points in triangle posed by Liu [7], and also give two corollaries.

Acknowledgements: Dedicated to Mr. Ting-Feng Dong on the occasion of his 55th birthday.

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1. Introduction and Main Results

LetP,Qbe two arbitrary interior points in4ABC, and leta,b,cbe the lengths of its sides,Sthe area,R the circumradius andr the inradius, respectively. Denote by R₁,R₂,R₃andr₁,r₂,r₃ the distances fromP to the verticesA, B,C and the sides BC,CA, AB, respectively. For the interior point Q, define D1, D2, D3andd1,d2, d3 similarly (see Figure1).

D1

D₂

D₃ R₁

R2 R

r₁ 3

r3

r2

d1

d₃

d₂ G

H

F

M N

L A

B C

P Q

Figure 1:

The following well-known and elegant result (see [1, Theorem 12.13, pp.105]) (1.1) R₁+R₂+R₃ ≥2(r₁+r₂+r₃)

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concerningR_iandr_i (i= 1,2,3)is called the Erdös-Mordell inequality. Inequal- ity (1.1) was generalized as follows [9, Theorem 15, pp. 318]:

(1.2) R₁x²+R₂y²+R₃z² ≥2(r₁yz+r₂zx+r₃xy) for allx, y, z≥0.

And the special casen= 2of [9, Theorem 8, pp. 315-316] states that

(1.3) p

R₁D₁+p

R₂D₂ +p

R₃D₃ ≥2p

r₁d₁+p

r₂d₂+p r₃d₃

, which also extends (1.1).

Recently, for allx, y, z≥0, J. Liu [8, Proposition 2] obtained

(1.4) p

R₁D₁x²+p

R₂D₂y²+p

R₃D₃z²

≥2p

r1d1yz+p

r2d2zx+p r3d3xy

which generalizes inequality (1.3).

In 2008, J. Liu [7] posed the following interesting geometric inequality problem.

Problem 1. For a triangle ABC and two arbitrary interior points P, Q, prove or disprove that

(1.5) R₁D₁+R₂D₂+R₃D₃ ≥4(r₂r₃+r₃r₁+r₁r₂).

We will solve Problem1in this paper.

From inequality (1.5), we get

R₁D₁+R₂D₂+R₃D₃ ≥4(d₂d₃+d₃d₁+d₁d₂).

Hence, we obtain the following result.

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Corollary 1.1. For any4ABC and two interior pointsP,Q, we have (1.6) R₁D₁+R₂D₂+R₃D₃ ≥4p

(r₂r₃+r₃r₁+r₁r₂)(d₂d₃+d₃d₁+d₁d₂).

From inequality (1.5), and by making use of an inversion transformation [2, pp.48-49] (see also [3, pp.108-109]) in the triangle, we easily get the following result.

Corollary 1.2. For any4ABC and two interior pointsP,Q, we have

(1.7) D₁

R₁r₁ + D₂

R₂r₂ + D₃

R₃r₃ ≥4· |P Q| · 1

R₁R₂ + 1

R₂R₃ + 1 R₃R₁

. Remark 1. With one of Liu’s theorems [8, Theorem 3], inequality (1.2) implies (1.4).

However, we cannot determine whether inequalities (1.1) and (1.3) imply inequality (1.5) or inequality (1.6), or inequalities (1.5) and (1.3) imply inequality (1.1).

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2. Preliminary Results

Lemma 2.1. We have for any4ABC and an arbitrary interior pointP that

(2.1) aR₁ ≥br₂+cr₃,

(2.2) bR₂ ≥cr₃+ar₁,

(2.3) cR₃ ≥ar₁+br₂.

Proof. Inequalities (2.1) – (2.3) directly follow from the obvious fact ar₁+br₂+cr₃ = 2S,

the formulas of the altitude h_a = 2S

a , h_b = 2S

b , h_c = 2S c , and the evident inequalities [11]

R1+r1 ≥ha, R₂+r₂ ≥h_b, R₃ +r₃ ≥h_c.

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Lemma 2.2 ([4,5]). For real numbersx₁, x₂, x₃, y₁, y₂, y₃ such that x1x2+x2x3+x3x1 ≥0

and

y₁y₂+y₂y₃+y₃y₁ ≥0, the inequality

(2.4) (y₂ +y₃)x₁+ (y₃+y₁)x₂+ (y₁+y₂)x₃

≥2p

(x₁x₂+x₂x₃+x₃x₁)(y₁y₂+y₂y₃+y₃y₁) holds, with equality if and only if ^x_y¹

1 = ^x_y²

2 = ^x_y³

3.

Lemma 2.3 (Hayashi’s inequality, [9, pp.297, 311]). For any 4ABC and an ar- bitrary pointP, we have

(2.5) R₁R₂

ab +R₂R₃

bc +R₃R₁ ca ≥1.

Equality holds if and only ifP is the orthocenter of the acute triangleABC or one of the vertexes of triangleABC.

Lemma 2.4 (Klamkin’s inequality, [6,10]). LetA, B, C be the angles of4ABC.

For positive real numbersu, v, w, the inequality (2.6) usinA+vsinB +wsinC ≤ 1

2(uv+vw+wu)

ru+v+w uvw holds, with equality if and only ifu=v =wand4ABC is equilateral.

Lemma 2.5. For any4ABC and an arbitrary interior pointP, we have

(2.7) p

abr₁r₂+bcr₂r₃+car₃r₁ ≥2(r₂r₃+r₃r₁+r₁r₂).

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Proof. Suppose that the actual barycentric coordinates ofP are(x, y, z), Thenx = area of4P BC, and therefore

x

x+y+z = area(4P BC)

S = r₁a

bcsinA = 2r₁ bc · a

2 sinA = 2Rr₁ bc . Therefore

r₁ = bc

2R · x x+y+z, r₂ = ca

2R · y x+y+z, r₃ = ab

2R · z x+y+z. Thus, inequality (2.7) is equivalent to

(2.8) abc

2R(x+y+z)

√xy+yz+zx

≥ abc R(x+y+z)²

a

2Ryz+ b

2Rzx+ c 2Rxy

or

(2.9) 1

2(x+y+z)√

xy+yz+zx≥yzsinA+zxsinB +xysinC.

Inequality (2.9) follows from Lemma2.4by taking (u, v, w) =

1 x,1

y,1 z

. This completes the proof of Lemma2.5.

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3. Solution of Problem 1

Proof. In view of Lemmas2.1–2.3and2.5, we have that R₁D₁ +R₂D₂+R₃D₃

=aR₁· D₁

a +bR₂·D₂

b +cR₃· D₃ c

≥(br₂+cr₃)· D₁

a + (cr₃+ar₁)·D₂

b + (ar₁+br₂)·D₃ c

≥2 s

(abr₁r₂+bcr₂r₃+car₃r₁)

D₁D₂

ab +D₂D₃

bc + D₃D₁ ca

≥2p

abr₁r₂+bcr₂r₃ +car₃r₁

≥4(r₂r₃+r₃r₁+r₁r₂).

The proof of inequality (1.5) is thus completed.

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References

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[10] W.-X. SHEN, Introduction to Simplices, Hunan Normal University Press, Changsha, 2000, 179. (in Chinese)

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