An inequality on the medians of a triangle
Vasile JIGLĂU1
Abstract. In this Note a proof of the geometric inequality (1) below is provided.
Keywords: median, circumradius, inradius, Brocard′ s angle.
MSC 2010: 51M04.
In this Note we shall give a proof of the following geometric inequality:
R
m2a + m2b + m2c
≤
2(ma mb + mb mc + cc ma ) − (m2a + m2b + m2c )
2r
(1)
with the usual notations in a triangle.
1. Preliminaries. First, we establish some results to be used later.
Let ABC be a triangle. Denote by ω its Brocard angle. The following properties
of the angle ω are well-known:
(2) (i) ctg ω =
and ω =
[2].
a2 + b 2 + c2
,
4S
2S
,
(ii) sin ω = √
2
2
a b + b 2 c2 + c2 a2
(iii) ω ≤
π
,
6
π
for the equilateral triangles. For details we refer to the Chapter XVII in
6
Lemma 1. In any triangle we have
P 2
»
a
P
P 2 = t + t(t + 1),
(3)
2 ab − a
where
t = ctg2 ω/(4
R
+ 1).
r
Proof. Taking into account (2(i)) as well as the formulas
X
X
a2 = 2(p2 − r2 − 4Rr),
ab = p2 + r2 + 4Rr
one obtains that (3) is successively equivalent to:
P 2
P
P 2 P
a
r
r
( a2 )2
a
( a2 )2
r
.
=
+
(
+ 1),
2
2
4r(4R + r)
16S
4R + r
4S
4R + r 16S 4R + r
P 2
r
p
a
(p2 − r2 − 4Rr)2
,
=
+
+
4R + r
4p(4R + r)
4R + r
(4R + r)2 .4p2
»
X
4p2 =
a2 + 2 4rp2 (4R + r) + (p2 − r2 − 4Rr)2 ,
»
X
X
(
a2 )2 −
a2 = 2 (p2 + r2 + 4Rr)2 ,
X
ab = p2 + r2 + 4Rr.
1 [email protected]
13
Hence (3) holds.
Lemma 2. In any triangle the following inequalities hold:
1
4R2 − 3Rr + 6r2
R2
3R 3
≤
=
−
+ ,
2
2
2
4r
r
4r
2
sin ω
2
2
2
R
R
3
1
4R
−
3Rr
+
2r
= 2 −
+ .
ctg2 ω ≤
4r2
r
4r
2
(4)
(5)
Proof. Taking into account that
X
X
X
a2 b 2 = (
ab)2 − 2abc
a = (p2 + r2 + 4Rr)2 − 16Rrp2
one gets
(6)
1
(p2 + r2 + 4Rr)2 − 16rp2
.
=
2
4r2 p2
sin ω
It is clear that (4) follows from the double inequality that will be proved below.
(7)
4R2 − 3Rr + 6r2
1
p2 + 3r2 − 7Rr
≤
.
≤
4r2
4r2
sin2 ω
Indeed, according to (6), the first inequality (7) reduces to (p2 + r2 + 4Rr)2 −
16rp2 ≤ p2 (p2 + 3r2 − 7Rr) and this in turn becomes r3 + 16R2r + 8Rr2 ≤ p2 (r + R).
Since by the second Gerretsen′ s inequality 16Rr − 5r2 ≤ p2 [3], it suffices to show that
r3 + 16R2 r + 8Rr2 ≤ (r + R)(16Rr − 5r2 ). By some algebra this inequality reduces
to the Euler inequality 2r ≤ R. Thus the first inequality (7) holds.
The second inequality (7) easily follows by using the first Gerretsen′ s inequality:
2
p ≤ 4R2 + 4Rr + 3r2 [3]. Indeed, p2 + 3r2 − 7Rr ≤ (4R2 + 4Rr + 3r2 ) + 3r2 − 7Rr =
4R2 − 3Rr + 6r2 . Concluding, the proof of the inequality (4) is complete.
1
The identity ctg2 ω =
− 1 help us to easily derive the inequality (5) from
sin2 ω
(4).
Remark. A remarkable consequence of the preceding results is the well-known
inequalities
(8)
2≤
R
1
≤ .
sin ω
r
(the left-side inequality follows from (2(iii)) and the right-side inequality follows from
4R2
4R2 − 3r(R − 2r)
1
≤
etc.). In [1] one finds a refinement of the
≤
(4):
2
2
4r
4r2
sin ω
right-side inequality (8).
2. Proof of the inequality (1). Let Am Bm Cm be the triangle whose sides
have the lengths respectively equal to the lengths ma , mb , mc of the medians of the
triangle ABC and let Rm , rm , Sm , ωm be entities easy to understand associated to it.
14
A well-known and veryP
useful property says that ω = ωm . Indeed, by (2(i)) we
P 2
3
a2
ma
have ctg ωm =
= ctg ω and (2(iii)) implies ω = ωm .
= 4 3
4Sm
4. 4 S
Applying the Lemma 1 to the triangle Am Bm Cm one obtains
P 2
»
ma
P
P 2 = tm + tm (tm + 1),
2 ma mb − ma
where
Rm
Rm
+ 1) = ctg2 ω/(4
+ 1).
rm
rm
Therefore, the inequality (1) reduces to
»
R
(9)
tm (tm + 1) ≤
− tm .
2r
tm = ctg2 ωm /(4
Å
ã
Rm
R
R
2
4
+ 1 or
− tm ≥ 0 or, equivalently, ctg ω ≤
We need to show that
2r
2r
rm
1R
R Rm
1
+
+ 1.
≤2 .
2
r
r
2
r
sin ω
m
R
1
1
Rm
≥
≥
and
r
sin ω
rm
sin ω
1
1 1
1
1
+
+ 1.
≤2
and it is obvious that we have
2
sin ω sin ω 2 sin ω
sin ω
Squaring, (9) can be written in the following equivalent forms:
Å
ã
R
R
1 R2
1 R2
2
2
−
.
t
+
t
⇔
+
1
t
≤
tm + tm ≤
m
m
m
4 r2
r
r
4 r2
The last inequality is true since from (8) it follows that
Thus the inequality (1) is equivalent to
Å
ãï 2 Å
ãò
1
Rm
R
R
2
(10)
ctg ω ≤
4
+1
/
+1 .
4
rm
r2
r
R
Rm
. Obviously, we have also
≥
r
rm
Å
ã
ã
Å
2
R
Rm
Rm
R2
/
+1 .
+1 ≥ 2 /
r2
r
rm
rm
Case 1. Assume that
Thus we can write
Å
Å
ãï 2 Å
ãï 2 Å
ãò
ãò
Rm
Rm
1
R
R
Rm
Rm
1
4
4
+1
/
+
1
/
+
1
.
+
1
≥
(11)
2
4
rm
r2
r
4
rm
rm
rm
On the other hand, according to (5), we have
(12)
ctg2 ω = ctg2 ωm ≤
15
2
Rm
3 Rm
1
−
+ .
2
rm
4 rm
2
By (10),(11) and (12) it follows that the inequality (10) is true if the following inequality holds:
ãï 2 Å
ãò
Å
2
Rm
Rm
Rm
1
3 Rm
1
Rm
(13)
+ ≤
+1
+1 .
−
/
4
2
2
rm
4 rm
2
4
rm
rm
rm
Setting
(14)
Rm
= u, (13) takes the form
rm
Å
ã
3
1
1
2
u − u+
(u + 1) ≤ (4u + 1)u2 .
4
2
4
Next, (14) reduces to u ≥ 2, that is Rm ≥ 2rm (Euler). Thus, the inequality (1) is
true in the considered case.
Rm
R
. Then
≤
Case 2. Assume that
r
rm
Å
Å
ãï 2 Å
ãò
ãï 2 Å
ãò
1
Rm
R
1
R
R
R
R
4
4
+1
/
/
+
1
≥
+
1
+
1
4
rm
r2
r
4
r
r2
r
and from (5) it follows that
ctg2 ω ≤
3R 1
R2
−
+ .
2
r
4r
2
The inequality (10), hence and (1) will be true if we show that
ï 2 Å
ãò
3R 1
1 R
R
R
R2
−
+ ≤ (4 + 1) 2 /
+1 .
(15)
r2
4r
2
4 r
r
r
We notice that (15) is just the inequality (13) written for the triangle ABC. Thus
it holds good and the inequality (1) is completely proved.
References
1. V. Jiglău – An inequality involving medians and altitudes (in Romanian), Recreaţii
matematice, nr.2/2016, 112-114 (in Romanian).
2. T. Lalescu – Geometry of the Triangle (in Romanian), Editura Tineretului,
Bucharest, 1958.
3. D.S. Mitrinović, J.E. Pec̆arić, V. Volenec – Recent Advances in Geometric
Inequalities, Kluwer Academic Publishers, Dordrecht/Boston/London, 1989.
16