A GEOMETRIC INEQUALITY INVOLVING A MOBILE POINT IN THE PLACE OF THE TRIANGLE

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A Geometric Inequality Xiao-Guang Chu and Yu-Dong Wu

vol. 10, iss. 3, art. 79, 2009

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A GEOMETRIC INEQUALITY INVOLVING A MOBILE POINT IN THE PLACE OF THE TRIANGLE

XIAO-GUANG CHU YU-DONG WU

Suzhou Hengtian Trading Co. Ltd., Department of Mathematics

Suzhou 215128, Jiangsu Zhejiang Xinchang High School

People’s Republic of China Shaoxing 312500, Zhejiang, China

EMail:srr345@163.com EMail:yudong.wu@yahoo.com.cn

Received: 04 March, 2008

Accepted: 27 August, 2009

Communicated by: S.S. Dragomir

2000 AMS Sub. Class.: Primary 51M16; Secondary 51M25, 52A40.

Key words: Bottema’s inequality, Euler’s inequality, Fermat’s sum, triangle, mobile point.

Abstract: By using Bottema’s inequality and several identities in triangles, we prove a weighted inequality concerning the distances between a mobile pointPand three vertexes A, B, C of4ABC. As an application, a conjecture with regard to Fermat’s sumP A+P B+P Cis proved.

Acknowledgements: The authors would like to thank Dr. Zhi-Gang Wang and Zhi-Hua Zhang for their enthusiastic help.

Dedicatory: Dedicated to Professor Lu Yang on the occasion of his 73rd birthday.

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1. Introduction and Main Results

For 4ABC, let a, b, c denote the side-lengths, A, B, C the angles, ∆ the area, p the semi-perimeter,Rthe circumradius andrthe inradius, respectively. In addition, supposing thatP is a mobile point in the plane containing4ABC, letP A, P B, P C denote the distances betweenP andA, B, C, respectively. We will customarily use the cyclic symbol, that is: P

f(a) = f(a) +f(b) +f(c), P

f(a, b) = f(a, b) + f(b, c) +f(c, a),Q

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

The following inequality can be easily proved by making use of Bottema’s in- equality:

(1.1) (P B+P C) cosA

2 + (P C+P A) cosB

2 + (P A+P B) cosC 2

≥p· p²+ 2Rr+r² 4R² . Here we choose to omit the details. From inequality (1.1) and the following known inequality (1.2) and identity (1.3) (see [3,4,6]):

(1.2) P AcosA

2 +P BcosB

2 +P CcosC 2 ≥p, and

(1.3) q₁ = cos² B −C

2 + cos² C−A

2 + cos² A−B

2 = p²+ 4R²+ 2Rr+r²

4R² ,

we easily get

(1.4) P A+P B+P C ≥p· cos² ^B−C₂ + cos² ^C−A₂ + cos² ^A−B₂ cos^A₂ + cos^B₂ + cos^C₂ .

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Considering the refinement of inequality (1.4), Chu [2] posed a conjecture as follows.

Corollary 1.1. For any4ABC,

(1.5) P A+P B+P C ≥p·cos^B−C₂ + cos^C−A₂ + cos^A−B₂ cos^A₂ + cos^B₂ + cos^C₂ .

The main object of this paper is to prove Conjecture1.1, which is easily seen to follow from the following stronger result.

Theorem 1.2. In4ABC, we have (1.6) (P B+P C) cosA

2 + (P C+P A) cosB

2 + (P A+P B) cosC 2

≥p·

cosB−C

2 + cosC−A

2 + cosA−B 2 −1

.

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2. Preliminary Results

In order to prove our main result, we shall require the following four lemmas.

Lemma 2.1. In4ABC, we have that (2.1) q₂ = cosB−C

2 ·cosC−A

2 ·cosA−B

2 = p²+ 2Rr+r² 8R² ,

(2.2) cosA

2 ·cosB

2 ·cosC 2 = p

4R, q₃ =X

cosB 2 cosC

2 cosB−C

2 b²+c²−a² (2.3)

= p⁴+ 2Rrp² −r(2R+r) (4R+r)²

4R² ,

q₄ = 1 2

X cos² B

2 + cos² C 2

b²+c²−a² (2.4)

= (2R+ 3r)p²−r(4R+r)²

2R ,

and

Q=X

cosB−C

2 −cosC−A

2 cosA−B 2

(2.5)

=q₁ −3q₂− p∆⁴Q

(b−c)² a²b²c²Q

(X+x),

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where

X =ap

bc(p−b) (p−c), Y =bp

ca(p−c) (p−a), (2.6)

Z =cp

ab(p−a) (p−b), and

x= (b+c) (p−b) (p−c), y= (c+a) (p−c) (p−a), (2.7)

z = (a+b) (p−a) (p−b).

Proof. The proofs of identities (2.1) and (2.2) were given in [6]. Now, we present the proofs of identities (2.3) – (2.5). By utilizing the formulas

cosA 2 =

rp(p−a)

bc , sinA 2 =

r(p−b) (p−c)

bc ,

cosB−C

2 = b+c a

r(p−b) (p−c)

bc ,

and (see [5, pp.52])

Ya = 4Rrp, X

a= 2p and X

bc=p²+ 4Rr+r², we get that

q₃ =pX(b+c) (p−b) (p−c)

a²bc b²+c²−a²

= p

4a²b²c²

Xbc(b+c) (c+a−b) (a+b−c) b²+c²−a²

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= p

4a²b²c²[6(ab+bc+ca)²(a+b+c)³−8(ab+bc+ca)³(a+b+c)

−(ab+bc+ca)(a+b+c)⁵−2abc(ab+bc+ca)(a+b+c)² + 8abc(ab+bc+ca)²−abc(a+b+c)⁴−4(a+b+c)a²b²c²]

= p⁴+ 2Rrp²−r(2R+r) (4R+r)² 4R²

and

1 2

X cos² B

2 + cos² C 2

b²+c²−a²

=X

a²cos² A 2

= p abc

pX

a³ −X a⁴

= p abc

5

2(ab+bc+ca)(a+b+c)²−2(ab+bc+ca)²

−1

2(a+b+c)⁴− 5

2(a+b+c)abc

= (2R+ 3r)p²−r(4R+r)²

2R .

Thus, identities (2.3) and (2.4) hold true.

With (1.3), (2.1) and the formulas of half-angles, we obtain that q₁−3q₂ = −p²+ 8R²−2Rr−r²

8R²

= 1

a²b²c²

Xx[bc(b+c)−(c+a) (a+b) (s−a)],

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and

Q=X

cosB−C

2 −cosC−A

2 cosA−B 2

=

PX[bc(b+c)−(c+a) (a+b) (s−a)]

a²b²c² . It is easy to see that

X−x= ∆²(b−c)²

X+x , Y −y= ∆²(c−a)²

Y +y , and Z−z = ∆²(a−b)² Z+z . Then

a²b²c²[Q−(q1−3q2)]

=X

[bc(b+c)−(c+a)(a+b)(p−a)](X−x)

=X

[bc(b+c)−(c+a)(a+b)(p−a)]∆²(b−c)² X+x

=X

p∆²(a−b)(a−c)(b−c)² (X+x) . Therefore,

Q−(q1−3q2) =X

p∆²(a−b)(a−c)(b−c)² a²b²c²(X+x)

= p∆²(a−b)(a−c)(b−c) a²b²c²

X b−c X+x,

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where b−c

X+x + c−a

Y +y + a−b Z+z

= (b−c) (Y −X+y−x)

(X+x) (Y +y) +(a−b) (Y −Z +y−z) (Z+z) (Y +y)

= p(p−c) (b−c) (b−a)

(X+x) (Y +y) +pabc(p−c) (b−c) (b−a) (X+x) (Y +y) (X+Y) +p(p−a) (a−b) (b−c)

(Z +z) (Y +y) + pabc(p−a) (a−b) (b−c) (Z +z) (Y +y) (Z +Y)

= p(b−c) (a−b)

Q(X+x) [(p−a) (X+x)−(p−c) (Z+z)]

+ pabc(b−c) (a−b) (X+Y) (Y +Z)Q

(X+x)

×[(p−a) (X+x) (X+Y)−(p−c) (Z +z) (Y +Z)]

= −∆²(b−c) (a−b) (a−c) Q(X+x)

+p(b−c) (a−b) (a−c) (Z+X)Q

(X+x) h

abcY

(p−a)−ca(p−b)Y i

+ pabc(b−c) (a−b) (a−c) (X+Y) (Z +Y)Q

(X+x) h

abcY

(p−a)−Y Y

(p−a) +abc(Y −abc)Q

(p−a) Z +X

+ abc(pb+ca) (p−b)Q

(p−a)−ca(p−b)Y Q

(p−a) Z +X

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= −∆²(b−c)(a−b)(a−c)

Q(X+x) + pabc(b−c)(a−b)(a−c) (X+Y)(Z +Y)Q

(X+x)

·nhY

(p−a)−(p−b)p

ca(p−c)(p−a)i

(X+Y)(Y +Z) +abcY

(p−a)h

Y −abc+pb(p−b) +ca(p−b)−(p−b)p

ca(p−c)(p−a)i +(Z+X)(abc−Y)Y

(p−a)o

= −∆²(b−c)(a−b)(a−c) Q(X+x) , which implies the assertion (2.5).

Lemma 2.2. For any4ABC, (2.8)

s cosA

2 cosB 2 cosC

2

cosA

2 + cosB

2 + cosC 2

≥ p 2R.

Proof. From Euler’s inequalityR ≥ 2r, abc= 4Rrp, a+b+c = 2pand the law of sines, we obtain that

2R²p≥4Rrp⇐⇒R²(a+b+c)≥abc (2.9)

⇐⇒sinA+ sinB+ sinC ≥4 sinAsinBsinC.

Taking

A→ π−A

2 , B → π−B

2 , and C → π−C 2 , we easily get

(2.10) cosA

2 + cosB

2 + cosC

2 ≥4 cosA 2 cosB

2 cosC 2.

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Inequality (2.8) follows immediately in view of (2.10) and (2.2).

Lemma 2.3. In4ABC, we have (2.11) X

cosB 2 cosC

2 b² +c²−a²

≥ p⁴+ 2Rrp²−r(2R+r) (4R+r)²

4R² .

Proof. By employing (2.3) and the formulas of half-angles, inequality (2.11) is equiv- alent to

(2.12) X

cosB 2 cosC

2 b²+c²−a²

≥X cosB

2 cosC

2 cosB−C

2 b²+c²−a² ,

or

(2.13) X(b²+c²−a²) a²bc

h ap

bc(s−b) (s−c)−(b+c) (s−b) (s−c) i

≥0, that is

(2.14) X ∆²

abc · (b²+c²−a²)

a(X+x) (b−c)² ≥0,

whereX, Y, Z andx, y, z are given, just as in the proof of Lemma2.1, by (2.6) and (2.7), respectively.

Without loss of generality, we can assume thata≥b≥cto obtain a(X+x)≥b(Y +y)≥c(Z+z),

and

(a−c)² ≥(b−c)²,

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and thus

(b−c)²

a(X+x) ≤ (c−a)² b(Y +y).

Hence, in order to prove inequality (2.14), we only need to prove that

(2.15) ∆²

abc

(b²+c²−a²)

a(X+x) (b−c)²+(c²+a²−b²)

b(Y +y) (c−a)²

≥0.

We readily arrive at the following result fora≥b ≥c,

∆² abc

(b²+c²−a²)

a(X+x) (b−c)²+(c²+a²−b²)

b(Y +y) (c−a)²

≥ ∆²

abc b²+c²−a²+c²+a²−b² (c−a)² b(Y +y)

= 2c² · ∆²

abc · (c−a)² b(Y +y) ≥0.

This shows that the inequality (2.15) or (2.11) holds true. The proof of Lemma2.3 is thus complete.

Lemma 2.4 (Bottema’s inequality, see [1, pp. 118, Theorem 12.56]). Let ∆⁰ de- note the area of4A⁰B⁰C⁰, and a⁰, b⁰, c⁰ the side-lengths of 4A⁰B⁰C⁰, respectively.

Then

(2.16) (a⁰P A+b⁰P B+c⁰P C)²

≥ 1

2[a⁰²(b²+c²−a²) +b⁰²(c²+a²−b²) +c⁰²(a² +b²−c²)] + 8∆∆⁰.

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3. The Proof of Theorem 1.2

Proof. It is easy to show that a⁰ = cosB

2 + cosC

2, b⁰ = cosC

2 + cosA

2, and c⁰ = cosA

2 + cosB 2

are three side-lengths of a certain triangle. By using Bottema’s inequality (2.16), in order to prove inequality (1.6), we only need to prove that

8∆

r

YcosA 2

XcosA 2 + 1

2 X

cosB

2 + cosC 2

2

b²+c²−a²

≥p²

XcosB−C 2 −1

2

or

(3.1) 8∆

r

YcosA 2

XcosA 2 +q₄ +X

cosB 2 cosC

2 b²+c²−a²

+ 2p²Q≥p²(q₁+ 1). With identities (1.3), (2.4), (2.5), together with Lemma2.2and Lemma2.3, in order to prove inequality (3.1), we only need to prove that

8∆· p

2R +(2R+ 3r)p²−r(4R+r)²

2R + p⁴ + 2Rrp²−r(2R+r) (4R+r)² 4R²

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+ 2p²

"

−p²+ 8R²−2Rr−r²

8R² − p∆⁴Q

(X+x)

#

≥p²

p²+ 4R²+ 2Rr+r²

4R² + 1

or

(3.2) −p⁴+ (4R² + 20Rr−2r²)p² −r(4R+r)³

4R² ≥ 2p³∆⁴Q

(X+x). From the known identities (see [5])

∆ =rp and

(b−c)²(c−a)²(a−b)² = 4r²[−p⁴+ (4R²+ 20Rr−2r²)p²−r(4R+r)³], inequality (3.2) is equivalent to

(3.3) Y

(X+x)≥2r⁴p⁵.

ForX ≥x, and with the following two known identities (see [5, pp.53]) Y(b+c) = 2p(p²+ 2Rr+r²), Y

(p−a) =r²p, we obtain

Y(X+x)≥8Y

x= 8Y

(b+c)Y

(p−a)²

= 16r⁴p³(p²+ 2Rr+r²)>16r⁴p⁵ >2r⁴p⁵. Therefore, inequality (3.3) holds. This completes the proof of Theorem1.2.

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4. Remarks

Remark 1. From inequalities (1.2) and (1.6), it is easy to see that inequality (1.5) holds.

Remark 2. In view of XcosB−C

2 ≥X

cos² B−C

2 = p²+ 4R²+ 2Rr+r² 4R²

⇐⇒X

cosB−C

2 −1≥ p²+ 2Rr+r² 4R² , it follows that inequality (1.6) is a refinement of inequality (1.1).

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References

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