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JIAN LIU
1. Introduction
In 1975, M.S.Klamkin [1] established the following inequality: Let ABC be an arbitrary triangle
of sides a, b, c, and let P be an arbitrary point in space, the distances of P from the vertices A, B, C
are R1 , R2 , R3 . x, y, z are real numbers, then
(1.1) (x + y + z)(xR12 + yR22 + zR32 ) > yza2 + zxb2 + xyc2 ,
* * * *
with equality if and only if P lies the plane of 4ABC and x : y : z = S 4P BC : S 4P CA : S 4P AB ( S 4P CA
denote the algebra area, etc.)
Inequality (1.1) is called the polar moment of inertia inequality, it is one of the most important
inequality for the triangle, there exist many consequences and applications for this one, see [1]-[5].
In this paper, we will apply Klamkin’ inequality (1.1) and the inversion transformation to deduce
a new weighted geometric inequality, then we discuss applications of our results. In addition, we
also pose some conjectures.
2. Main Result
In order to prove our new results, first of all, we give the following lemma.
Lemma Let ABC be an arbitrary triangle, and let P be an arbitrary point in the plane of the
triangle ABC, if the following inequality:
(2.1) f (a, b, c, R1 , R2 , R3 ) > 0.
holds, then we have the dual inequality:
(2.2) f (aR1 , bR2 , cR3 , R2 R3 , R3 R1 , R1 R2 ) > 0.
Indeed, the above conclusion can be deduced by using inversion transformation, see[6] or [3], [7].
Now, We state and prove main result.
Theorem Let x, y, z be positive real numbers, then for any triangle ABC and arbitrary point P
in the plane of 4ABC holds the following inequality:
R12 R22 R32 aR1 + bR2 + cR3
(2.3) + + > √ ,
x y z yz + zx + xy
with equality if and only if 4ABC is acute-angled, P coincide with its orthocenter and x : y : z =
cot A : cot B : cot C.
Proof. If P coincide with one of the vertices of 4ABC, for example P ≡ A, then P A = 0, P B =
c, P C = b, (2.3) becomes trivial inequality and we easily to know it is holds true. In this case,
equality in (2.3) obviously cannot occur.
Next assume P does not coincide with the vertices.
2000 Mathematics Subject Classification. Primary 51M16.
Key words and phrases. triangle, point, polar moment of inertia inequality.
This paper was typeset using AMS-LATEX.
1
2 JIAN LIU
If x, y, z are positive real numbers, then by the polar moment of inertia inequality (1.1) we have
µ ¶
2 2 2 1 1 1 a2 b2 c2
(xR1 + yR2 + zR3 ) + + > + + .
yz zx xy x y z
On the other hand, from Cauchy-Schwarz inequality we get
a2 b2 c2 (a + b + c)2
+ + > ,
x y z x+y+z
with equality if and only if x : y : z = a : b : c.
Combining these two above inequalities, for any positive real numbers x, y, z, the following inequal-
ity holds:
µ ¶
2 2 2 1 1 1 (a + b + c)2
(2.4) (xR1 + yR2 + zR3 ) + + > .
yz zx xy x+y+z
and we easily to know that the equality if and only if x : y : z = a : b : c and P is the incenter of
4ABC.
Now, applying the inversion transformation in the lemma to inequality (2.4), we obtain
µ ¶
£ 2 2 2
¤ 1 1 1 (aR1 + bR2 + cR3 )2
x(R2 R3 ) + y(R3 R1 ) + z(R1 R2 ) + + > ,
yz zx xy x+y+z
or equivalently
µ ¶2
(R2 R3 )2 (R3 R1 )2 (R1 R2 )2 aR1 + bR2 + cR3
(2.5) + + > .
yz zx xy x+y+z
where x, y, z are positive numbers.
For x → xR12 , y → yR22 , z → zR32 , then holds
µ ¶
1 1 1 aR1 + bR2 + cR3 2
(2.6) + + > .
yz zx xy xR12 + yR22 + zR32
Remark 2.1. If P does not coincide with the vertices, then inequality (2.4) is equivalent to the
following result in [8]:
r
R2 R3 R3 R1 R1 R2 xyz
(2.8) x +y +z >2 s,
R1 R2 R3 x+y+z
where s is the semi-perimeter of 4ABC, x, y, z are positive real numbers. In [8], the proof of (2.8)
without using the polar moment of inertia inequality, So does not start from Klamkin’s inequality
(1.1) we can deduce the inequality of the theorem.
A WEIGHTED GEOMETRIC INEQUALITY AND ITS APPLICATIONS 3
Applying the inversion transformation of the lemma to the above inequality, then divide both sides
by R1 R2 R3 we get
Corollary 4 If P is arbitrary point which does not coincide with the vertices of 4ABC, then
µ ¶
1 1 1
(3.9) (R2 R3 + R3 R1 + R1 R2 )2 + + > 4s2 .
R2 R3 R3 R1 R1 R2
It is not difficult to know that the above inequality is stronger than the following result which
the author obtained many years ago:
r r r q
R2 R3 R3 R 1 R 1 R2 √
(3.10) + + > 2 3s
R1 R2 R3
Now, Let P be an interior point of the triangle ABC, then we have the well known inequalities(see[13]):
aR1 > br3 + cr2 , bR2 > cr1 + ar3 , cR3 > ar2 + br1 ,
Summing them up, note that a + b + c = 2s and the identity ar1 + br2 + cr3 = 2rs we easily get
(3.11) aR1 + bR2 + cR3 > 2s(r1 + r2 + r3 ) − 2rs.
Multiplying both sides by 2 then adding inequality (3.1) and using ∆ = rs, then
3(aR1 + bR2 + cR3 ) > 4s(r1 + r2 + r3 ),
that is
aR1 + bR2 + cR3 4
(3.12) > s.
r 1 + r2 + r 3 3
According to this and the equivalent form (2.5) of inequality (2.3), we get immediately the result:
Corollary 5 Let P be an interior point of the triangle ABC, then
(R2 R3 )2 (R3 R1 )2 (R1 R2 )2 16
(3.13) + + > s2 .
r2 r3 r3 r1 r 1 r2 9
From inequality (3.8) and (3.12) we infer that
16 2
(R2 R3 + R3 R1 + R1 R2 )(R1 + R2 + R3 )2 > s (r1 + r2 + r3 )2 ,
9
Note that again 3(R2 R3 + R3 R1 + R1 R2 ) 6 (R1 + R2 + R3 )2 , we get the following inequality:
Corollary 6 Let P be an interior point of triangle ABC, then
(R1 + R2 + R3 )2 4
(3.14) > √ s.
r1 + r2 + r3 3
Letting x = ra , y = rb , z = rc in (2.3) and note that identity rb rc + rc ra + ra rb = s2 , hence
R12 R22 R32 1
(3.15) + + > (aR1 + bR2 + cR3 ).
ra rb rc s
This inequality and (3.12) lead us get the following inequality:
Corollary 7 Let P be an interior point of the triangle ABC, then
R12 R22 R32 4
(3.16) + + > (r1 + r2 + r3 ).
ra rb rc 3
Adding (3.1) and (3.11) then dividing both sides by 2, we have
(3.17) aR1 + bR2 + cR3 > s(r1 + r2 + r3 + r).
From this and (3.15), we get again the following inequality which similar to (3.16):
Corollary 8 Let P be an interior point of the triangle ABC, then
R12 R22 R32
(3.18) + + > r1 + r2 + r3 + r.
ra rb rc
A WEIGHTED GEOMETRIC INEQUALITY AND ITS APPLICATIONS 5
When P locates interior of the triangle ABC, let D, E, F be the feet of the perpendicular from
P to the sides BC, CA, AB respectively. Take x = ar1 , y = br2 , z = cr3 in equivalent form (2.6) of
inequality (2.3), then
µ ¶2
1 1 1 aR1 + bR2 + cR3
+ + > ,
bcr2 r3 car3 r1 abr1 r2 ar1 R1 + br2 R2 + cr3 R3
Using identity ar1 + br2 + cr3 = 2∆ and well known identity(see[7]):
(3.19) ar1 R12 + br2 R22 + cr3 R32 = 8R2 ∆p ,
(where ∆p is the area of pedal triangle DEF )We get
abcr1 r2 r3 (aR1 + bR2 + cR3 )2 6 64∆R4 ∆2p .
Let sp , rp denote the semi-perimeter of the triangle DEF and radius of incircle respectively. Note
that ∆p = rp sp , aR1 + bR2 + cR3 = 4Rsp , from the above inequality we obtain the following
inequality which established by the author in [14]:
Corollary 9 Let P be an interior point of the triangle ABC, then
r1 r2 r3
(3.20) 6 2R,
rp2
equality holds if and only if P is the orthocenter of the triangle ABC.
It is well known, the inequality related to a triangle and two points is very few. Several years
ago, the author guessed the following inequality holds:
R12 R22 R32
(3.21) + + > 4(r1 + r2 + r3 ),
d1 d2 d3
where d1 , d2 , d3 denote the distances from an interior point Q to the sides of 4ABC.
Inequality (3.21) is very interesting, the author have been trying to prove it. In what follows, we
will prove the stronger result. To do so, we need a corollary of the following conclusion(see[15]):
Let Q be an interior point of 4ABC, t1 , t2 , t3 denote the bisector of ∠BQC, ∠CQA, ∠AQB
respectively, 4A0 B 0 C 0 is an arbitrary triangle, then
1
(3.22) t2 t3 sin A0 + t3 t1 sin B 0 + t1 t2 sin C 0 6 ∆,
2
0 0 0
with equality if and only if 4A B C ∼ 4ABC, and Q is the circumcentre of 4ABC.
In (3.22), letting 4ABC be equilateral, then we immediately get
1
(3.23) t2 t3 + t3 t1 + t1 t2 6 √ ∆.
3
2
√
From this and the simple inequality s > 3 3∆, we have
1
(3.24) t2 t3 + t3 t1 + t1 t2 6 s2 .
9
According to the inequality (2.3) of the theorem and (3.24), we can see that
R12 R22 R32 3
(3.25) + + > (aR1 + bR2 + cR3 ).
t1 t2 t3 s
By using inequality (3.12), we obtain the following stronger of inequality (3.21)
Corollary 10 Let P and Q be two interior points of the 4ABC, then
R12 R22 R32
(3.26) + + > 4(r1 + r2 + r3 ).
t1 t2 t3
with equality if and only if 4ABC is equilateral, P and Q both are its center.
Analogously, from inequality (3.17) and inequality (3.25) we get
Corollary 11 Let P and Q be two interior points of 4ABC, then
R12 R22 R32
(3.27) + + > 3(r1 + r2 + r3 + r).
t1 t2 t3
6 JIAN LIU
with equality if and only if 4ABC is equilateral, and P and Q both are its center.
4. Some conjectures
In this section, we will state some conjectures in allusion to the inequality which appeared in
this paper.
Inequality (3.8) is equivalent to
µ ¶
aR1 + bR2 + cR3 2
(4.1) R2 R3 + R3 R1 + R1 R2 > .
R1 + R2 + R3
Enlightened by this one and the well known inequality:
(4.2) R2 R3 + R3 R1 + R1 R2 > 4(w2 w3 + w3 w1 + w1 w2 ),
We pose the following
Conjecture 1 Let P be an arbitrary interior point of the triangle ABC, then
µ ¶
aR1 + bR2 + cR3 2
(4.3) > 4(w2 w3 + w3 w1 + w1 w2 ).
R1 + R 2 + R3
We consider the stronger of corollary 5, the author posed these two conjectures:
Conjecture 2 Let P be an arbitrary interior point of the triangle ABC, then
(R2 R3 )2 (R3 R1 )2 (R1 R2 )2 4
(4.4) + + > (a2 + b2 + c2 ).
w2 w3 w3 w1 w1 w2 3
Conjecture 8 Let 4ABC be acute-angled and P is an arbitrary point in its interior, then
aR1 + bR2 + cR3
(4.11) > 2s.
w1 + w2 + w3 + r
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(Jian Liu) East China Jiaotong University, Nanchang City, Jxian Province, 330013, China
E-mail address: liujian99168@yahoo.com.cn