Volume 10 (2009), Issue 3, Article 79, 8 pp.
A GEOMETRIC INEQUALITY INVOLVING A MOBILE POINT IN THE PLACE
OF THE TRIANGLE
XIAO-GUANG CHU AND YU-DONG WU
S UZHOU H ENGTIAN T RADING C O . LTD .,
S UZHOU 215128, J IANGSU
P EOPLE ’ S R EPUBLIC OF C HINA
srr345@163.com
D EPARTMENT OF M ATHEMATICS
Z HEJIANG X INCHANG H IGH S CHOOL
S HAOXING 312500, Z HEJIANG
P EOPLE ’ S R EPUBLIC OF C HINA
yudong.wu@yahoo.com.cn
Received 04 March, 2008; accepted 27 August, 2009
Communicated by S.S. Dragomir
Dedicated to Professor Lu Yang on the occasion of his 73rd birthday.
A BSTRACT. By using Bottema’s inequality and several identities in triangles, we prove a
weighted inequality concerning the distances between a mobile point P and three vertexes
A, B, C of 4ABC. As an application, a conjecture with regard to Fermat’s sum P A+P B+P C
is proved.
Key words and phrases: Bottema’s inequality, Euler’s inequality, Fermat’s sum, triangle, mobile point.
2000 Mathematics Subject Classification. Primary 51M16; Secondary 51M25, 52A40.
1. I NTRODUCTION AND M AIN R ESULTS
For 4ABC, let a, b, c denote the side-lengths, A, B, C the angles, ∆ the area, p the semiperimeter, R the circumradius and r the inradius, respectively. In addition, supposing that P is
a mobile point in the plane containing 4ABC, let P A, P B, P C denote the distances between
P
P and A, B, C, respectively. We will customarily use the cyclic symbol, that is:
f (a) =
P
Q
f (a) + f (b) + f (c), f (a, b) = f (a, b) + f (b, c) + f (c, a), f (a) = f (a)f (b)f (c), etc.
The authors would like to thank Dr. Zhi-Gang Wang and Zhi-Hua Zhang for their enthusiastic help.
065-08
2
X.-G. C HU AND Y.-D. W U
The following inequality can be easily proved by making use of Bottema’s inequality:
A
B
C
p2 + 2Rr + r2
+ (P C + P A) cos + (P A + P B) cos ≥ p ·
.
2
2
2
4R2
Here we choose to omit the details. From inequality (1.1) and the following known inequality
(1.1) (P B + P C) cos
(1.2) and identity (1.3) (see [3, 4, 6]):
P A cos
(1.2)
B
C
A
+ P B cos + P C cos ≥ p,
2
2
2
and
(1.3)
q1 = cos2
C −A
A−B
p2 + 4R2 + 2Rr + r2
B−C
+ cos2
+ cos2
=
,
2
2
2
4R2
we easily get
(1.4)
PA + PB + PC ≥ p ·
cos2
B−C
2
+ cos2 C−A
+ cos2
2
cos A2 + cos B2 + cos C2
A−B
2
.
Considering the refinement of inequality (1.4), Chu [2] posed a conjecture as follows.
Conjecture 1.1. For any 4ABC,
(1.5)
PA + PB + PC ≥ p ·
cos B−C
+ cos C−A
+ cos A−B
2
2
2
.
cos A2 + cos B2 + cos C2
The main object of this paper is to prove Conjecture 1.1, which is easily seen to follow from
the following stronger result.
Theorem 1.2. In 4ABC, we have
(1.6) (P B + P C) cos
A
B
C
+ (P C + P A) cos + (P A + P B) cos
2
2
2
B−C
C −A
A−B
≥ p · cos
+ cos
+ cos
−1 .
2
2
2
2. P RELIMINARY R ESULTS
In order to prove our main result, we shall require the following four lemmas.
Lemma 2.1. In 4ABC, we have that
(2.1)
q2 = cos
B−C
C −A
A−B
p2 + 2Rr + r2
· cos
· cos
=
,
2
2
2
8R2
cos
(2.2)
(2.3)
A
B
C
p
· cos · cos =
,
2
2
2
4R
B
C
B−C 2
cos cos
b + c 2 − a2
2
2
2
4
2
p + 2Rrp − r (2R + r) (4R + r)2
=
,
4R2
q3 =
X
cos
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 79, 8 pp.
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A G EOMETRIC I NEQUALITY
3
1X
2 B
2 C
q4 =
b 2 + c 2 − a2
cos
+ cos
2
2
2
(2.4)
(2R + 3r) p2 − r (4R + r)2
=
,
2R
and
X
C −A
A−B
B−C
Q=
− cos
cos
cos
2
2
2
Q
2
4
p∆
(b − c)
= q1 − 3q2 − 2 2 2 Q
,
abc
(X + x)
(2.5)
where
p
X = a bc (p − b) (p − c),
p
Z = c ab (p − a) (p − b),
(2.6)
Y =b
p
ca (p − c) (p − a),
and
x = (b + c) (p − b) (p − c) ,
(2.7)
y = (c + a) (p − c) (p − a) ,
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
r
r
p (p − a)
A
A
(p − b) (p − c)
cos =
,
sin =
,
2
bc
2
bc
r
B−C
b + c (p − b) (p − c)
cos
=
,
2
a
bc
and (see [5, pp.52])
Y
a = 4Rrp,
X
a = 2p
and
X
bc = p2 + 4Rr + r2 ,
we get that
q3 = p
X (b + c) (p − b) (p − c)
b 2 + c 2 − a2
a2 bc
X
p
= 2 2 2
bc (b + c) (c + a − b) (a + b − c) b2 + c2 − a2
4a b c
p
= 2 2 2 [6(ab + bc + ca)2 (a + b + c)3 − 8(ab + bc + ca)3 (a + b + c)
4a 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)a2 b2 c2 ]
p4 + 2Rrp2 − r (2R + r) (4R + r)2
=
4R2
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 79, 8 pp.
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4
X.-G. C HU AND Y.-D. W U
and
1X
2 B
2 C
cos
+ cos
b 2 + c 2 − a2
2
2
2
X
A
=
a2 cos2
2
p X 3 X 4
=
p
a −
a
abc p 5
(ab + bc + ca)(a + b + c)2 − 2(ab + bc + ca)2
=
abc 2
5
1
4
− (a + b + c) − (a + b + c)abc
2
2
=
(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
−p2 + 8R2 − 2Rr − r2
8R2
X
1
= 2 2 2
x [bc (b + c) − (c + a) (a + b) (s − a)],
abc
q1 − 3q2 =
and
X
B−C
C −A
A−B
Q=
cos
− cos
cos
2
2
2
P
X [bc (b + c) − (c + a) (a + b) (s − a)]
=
.
a2 b 2 c 2
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
a2 b2 c2 [Q − (q1 − 3q2 )]
X
=
[bc(b + c) − (c + a)(a + b)(p − a)](X − x)
X
∆2 (b − c)2
=
[bc(b + c) − (c + a)(a + b)(p − a)]
X +x
2
X
(a
−
b)(a
−
c)(b
−
c)
=
p∆2
.
(X + x)
Therefore,
(a − b)(a − c)(b − c)2
a2 b2 c2 (X + x)
p∆2 (a − b)(a − c)(b − c) X b − c
=
,
a2 b 2 c 2
X +x
Q − (q1 − 3q2 ) =
X
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 79, 8 pp.
p∆2
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A G EOMETRIC I NEQUALITY
5
where
c−a
a−b
b−c
+
+
X +x Y +y Z +z
(b − c) (Y − X + y − x) (a − b) (Y − Z + y − z)
=
+
(X + x) (Y + y)
(Z + z) (Y + y)
p (p − c) (b − c) (b − a) pabc (p − c) (b − c) (b − a)
=
+
(X + x) (Y + y)
(X + x) (Y + y) (X + Y )
p (p − a) (a − b) (b − c) pabc (p − a) (a − b) (b − c)
+
+
(Z + z) (Y + y)
(Z + z) (Y + y) (Z + Y )
p (b − c) (a − b)
= Q
[(p − a) (X + x) − (p − c) (Z + z)]
(X + x)
pabc (b − c) (a − b)
Q
+
(X + Y ) (Y + Z) (X + x)
× [(p − a) (X + x) (X + Y ) − (p − c) (Z + z) (Y + Z)]
−∆2 (b − c) (a − b) (a − c)
Q
=
(X + x)
i
Y
p (b − c) (a − b) (a − c) h
Q
abc
(p − a) − ca (p − b) Y
+
(Z + X) (X + x)
Y
Y
pabc (b − c) (a − b) (a − c) h
Q
abc
(p − a) − Y
(p − a)
+
(X + Y ) (Z + Y ) (X + x)
Q
abc (Y − abc) (p − a)
+
Z +X
Q
Q
abc (pb + ca) (p − b) (p − a) − ca (p − b) Y (p − a)
+
Z +X
2
pabc(b − c)(a − b)(a − c)
−∆ (b − c)(a − b)(a − c)
Q
Q
+
=
(X + x)
(X + Y )(Z + Y ) (X + x)
nhY
i
p
·
(p − a) − (p − b) ca(p − c)(p − a) (X + Y )(Y + Z)
h
i
Y
p
+ abc
(p − a) Y − abc + pb(p − b) + ca(p − b) − (p − b) ca(p − c)(p − a)
o
Y
+(Z + X)(abc − Y )
(p − a)
=
−∆2 (b − c)(a − b)(a − c)
Q
,
(X + x)
which implies the assertion (2.5).
Lemma 2.2. For any 4ABC,
s
(2.8)
A
B
C
cos cos cos
2
2
2
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 79, 8 pp.
A
B
C
cos + cos + cos
2
2
2
≥
p
.
2R
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6
X.-G. C HU AND Y.-D. W U
Proof. From Euler’s inequality R ≥ 2r, abc = 4Rrp, a + b + c = 2p and the law of sines, we
obtain that
(2.9)
2R2 p ≥ 4Rrp ⇐⇒ R2 (a + b + c) ≥ abc
⇐⇒ sin A + sin B + sin C ≥ 4 sin A sin B sin C.
Taking
A→
π−A
,
2
B→
π−B
,
2
and
C→
π−C
,
2
we easily get
cos
(2.10)
B
C
A
B
C
A
+ cos + cos ≥ 4 cos cos cos .
2
2
2
2
2
2
Inequality (2.8) follows immediately in view of (2.10) and (2.2).
Lemma 2.3. In 4ABC, we have
(2.11)
X
p4 + 2Rrp2 − r (2R + r) (4R + r)2
B
C 2
2
2
cos cos
b +c −a ≥
.
2
2
4R2
Proof. By employing (2.3) and the formulas of half-angles, inequality (2.11) is equivalent to
X
X
B
C 2
B
C
B−C 2
(2.12)
cos cos
b + c 2 − a2 ≥
cos cos cos
b + c 2 − a2 ,
2
2
2
2
2
or
(2.13)
i
X (b2 + c2 − a2 ) h p
a bc (s − b) (s − c) − (b + c) (s − b) (s − c) ≥ 0,
a2 bc
that is
(2.14)
X ∆2 (b2 + c2 − a2 )
·
(b − c)2 ≥ 0,
abc
a (X + x)
where X, Y, Z and x, y, z are given, just as in the proof of Lemma 2.1, by (2.6) and (2.7),
respectively.
Without loss of generality, we can assume that a ≥ b ≥ c to obtain
a (X + x) ≥ b (Y + y) ≥ c (Z + z) ,
and
(a − c)2 ≥ (b − c)2 ,
and thus
(b − c)2
(c − a)2
≤
.
a (X + x)
b (Y + y)
Hence, in order to prove inequality (2.14), we only need to prove that
∆2 (b2 + c2 − a2 )
(c2 + a2 − b2 )
2
2
(2.15)
(b − c) +
(c − a) ≥ 0.
abc
a(X + x)
b(Y + y)
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 79, 8 pp.
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A G EOMETRIC I NEQUALITY
7
We readily arrive at the following result for a ≥ b ≥ c,
(c2 + a2 − b2 )
∆2 (b2 + c2 − a2 )
2
2
(b − c) +
(c − a)
abc
a(X + x)
b(Y + y)
(c − a)2
∆2 2
b + c 2 − a2 + c 2 + a2 − b 2
≥
abc
b (Y + y)
∆2 (c − a)2
= 2c ·
·
≥ 0.
abc b (Y + y)
2
This shows that the inequality (2.15) or (2.11) holds true. The proof of Lemma 2.3 is thus
complete.
Lemma 2.4 (Bottema’s inequality, see [1, pp. 118, Theorem 12.56]). Let ∆0 denote the area of
4A0 B 0 C 0 , and a0 , b0 , c0 the side-lengths of 4A0 B 0 C 0 , respectively. Then
(2.16)
2
(a0 P A + b0 P B + c0 P C)
1
≥ [a02 (b2 + c2 − a2 ) + b02 (c2 + a2 − b2 ) + c02 (a2 + b2 − c2 )] + 8∆∆0 .
2
3. T HE P ROOF OF T HEOREM 1.2
Proof. It is easy to show that
B
C
+ cos ,
2
2
A
B
c0 = cos + cos
2
2
a0 = cos
b0 = cos
C
A
+ cos ,
2
2
and
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
r
8∆
Y
2
AX
A 1X
B
C
cos
cos +
cos + cos
b 2 + c 2 − a2
2
2
2
2
2
X
2
B−C
2
≥p
cos
−1
2
or
(3.1)
r
Y
AX
A
cos
cos + q4
8∆
2
2
X
B
C 2
+
cos cos
b + c2 − a2 + 2p2 Q ≥ p2 (q1 + 1) .
2
2
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 79, 8 pp.
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8
X.-G. C HU AND Y.-D. W U
With identities (1.3), (2.4), (2.5), together with Lemma 2.2 and Lemma 2.3, in order to prove
inequality (3.1), we only need to prove that
(2R + 3r) p2 − r (4R + r)2 p4 + 2Rrp2 − r (2R + r) (4R + r)2
p
8∆ ·
+
+
2R
2R
4R2
"
#
Q
2
4
2
2
2
p∆
(b
−
c)
−p
+
8R
−
2Rr
−
r
+ 2p2
− 2 2 2Q
8R2
abc
(X + x)
2
p + 4R2 + 2Rr + r2
2
≥p
+1
4R2
or
Q
2p3 ∆4 (b − c)2
−p4 + (4R2 + 20Rr − 2r2 ) p2 − r (4R + r)3
≥ 2 2 2Q
.
4R2
abc
(X + x)
(3.2)
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
Y
(3.3)
(X + x) ≥ 2r4 p5 .
For X ≥ x, and with the following two known identities (see [5, pp.53])
Y
Y
(b + c) = 2p(p2 + 2Rr + r2 ),
(p − a) = r2 p,
we obtain
Y
(X + x) ≥ 8
Y
x=8
Y
(b + c)
Y
(p − a)2
= 16r4 p3 (p2 + 2Rr + r2 ) > 16r4 p5 > 2r4 p5 .
Therefore, inequality (3.3) holds. This completes the proof of Theorem 1.2.
4. R EMARKS
Remark 1. From inequalities (1.2) and (1.6), it is easy to see that inequality (1.5) holds.
Remark 2. In view of
X
X
B−C
p2 + 4R2 + 2Rr + r2
2 B −C
cos
≥
cos
=
2
2
4R2
X
B−C
p2 + 2Rr + r2
⇐⇒
cos
−1≥
,
2
4R2
it follows that inequality (1.6) is a refinement of inequality (1.1).
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 79, 8 pp.
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A G EOMETRIC I NEQUALITY
9
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J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 79, 8 pp.
http://jipam.vu.edu.au/