INTRODUCTION ANDMAIN RESULTS LetP,Qbe two arbitrary interior points in4ABC, and leta,b,cbe the lengths of its sides, S the area, Rthe circumradius andr the inradius, respectively

A GEOMETRIC INEQUALITY OF THE GENERALIZED ERDÖS-MORDELL TYPE

YU-DONG WU, CHUN-LEI YU, AND ZHI-HUA ZHANG DEPARTMENT OFMATHEMATICS

ZHEJIANGXINCHANGHIGHSCHOOL

SHAOXING312500, ZHEJIANG

PEOPLE’SREPUBLIC OFCHINA

yudong.wu@yahoo.com.cn seetill@126.com DEPARTMENT OFMATHEMATICS

SHILISENIORHIGHSCHOOL INZIXING

CHENZHOU423400, HUNAN

PEOPLE’SREPUBLIC OFCHINA

zxzh1234@163.com

Received 20 April, 2009; accepted 09 October, 2009 Communicated by S.S. Dragomir

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

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.

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

2000 Mathematics Subject Classification. Primary 51M16.

1. INTRODUCTION ANDMAIN RESULTS

LetP,Qbe two arbitrary interior points in4ABC, and leta,b,cbe the lengths of its sides, S the area, Rthe circumradius andr the inradius, respectively. Denote byR₁, R₂, R₃ andr₁, r₂, r₃ the distances fromP to the verticesA, B, C and the sidesBC, CA, AB, respectively.

For the interior pointQ, defineD₁,D₂,D₃ andd₁,d₂,d₃similarly (see Figure 1.1).

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

The authors would like to thank the anonymous referee for his very careful reading and some valuable suggestions.

107-09

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 1.1:

concerningR_i andr_i(i= 1,2,3)is called the Erdös-Mordell inequality. Inequality (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 1.1. For a triangleABCand two arbitrary interior pointsP,Q, prove or disprove that

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

We will solve Problem 1.1 in this paper.

From inequality (1.5), we get

R₁D₁ +R₂D₂+R₃D₃ ≥4(d₂d₃+d₃d₁+d₁d₂).

Hence, we obtain the following result.

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).

2. PRELIMINARYRESULTS

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 ≥h_a, R₂ +r₂ ≥h_b, R3+r3 ≥hc.

Lemma 2.2 ([4, 5]). For real numbersx₁, x₂, x₃, y₁, y₂, y₃such 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 arbitrary point P, we have

(2.5) R₁R₂

ab + R₂R₃

bc +R₃R₁ ca ≥1.

Equality holds if and only if P is the orthocenter of the acute triangle ABC 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).

Proof. Suppose that the actual barycentric coordinates of P are (x, y, z), Then x = area of 4P 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 Lemma 2.4 by taking (u, v, w) =

1 x,1

y,1 z

.

This completes the proof of Lemma 2.5.

3. SOLUTION OF PROBLEM 1.1 Proof. In view of Lemmas 2.1 – 2.3 and 2.5, 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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