A REMARKABLE INEQUALITY
DANIEL CAMPOS, ROBERTO BOSCH
Introduction.
The following inequality
(1)
1 − cos A 1 − cos B 1 − cos C
(1 − cos A)(1 − cos B)(1 − cos C)
≥
+
+
cos A cos B cos C
1 + cos A 1 + cos B 1 + cos C
for acute-angled triangles was proposed as problem S99 in this journal by the first author. This
problem remained unsolved for a while, until [3] was published. In [3] two solutions are presented:
the first one rewrites (1) into symmetric functions of x = tan A2 , y = tan B2 , z = tan C2 , and the
second one employs the theory of Schur-concave functions [4]. In this paper we show a new and
more simple solution only using trigonometric identities and Arithmetic Mean - Geometric Mean
inequality.
Main result.
Let ABC be an acute-angled triangle. The following inequality holds
(1 − cos A)(1 − cos B)(1 − cos C)
1 − cos A 1 − cos B 1 − cos C
≥
+
+
cos A cos B cos C
1 + cos A 1 + cos B 1 + cos C
Proof :
Rewrite both sides of the inequality as follows
(1 − cos A)(1 − cos B)(1 − cos C)
cos A cos B cos C
=
=
=
8 sin2 A2 sin2 B2 sin2 C2
cos A cos B cos C
tan A tan B tan C
cot A2 cot B2 cot C2
tan A + tan B + tan C
,
cot A2 + cot B2 + cot C2
and
1 − cos A 1 − cos B 1 − cos C
A
B
C
+
+
= tan2 + tan2 + tan2 .
1 + cos A 1 + cos B 1 + cos C
2
2
2
Multiply by 2 and substract 2 on both sides, and then rewrite as
P
tan A + tan B − 2 cot C2
2(tan A + tan B + tan C)
cyc
−2=
,
A
B
C
cot 2 + cot 2 + cot 2
cot A2 cot B2 cot C2
and
A
B
C
2 tan
+ tan2 + tan2
2
2
2
2
−2
2
DANIEL CAMPOS, ROBERTO BOSCH
as
A
A
B
A
B
A
B
2 B
2 C
2 tan
+ tan
+ tan
− 2 tan tan + tan tan + tan tan
2
2
2
2
2
2
2
2
2
2
X
A
B
tan − tan
=
.
2
2
cyc
2
Finally, note that
tan A + tan B − 2 cot
C
2
2 cos C2 sin2
=
cyc
cot A2 cot B2 cot C2
− cos A cos B
cos A cos B sin C2
=
cos C2 (1 − cos(A − B))
=
2 cos C2 sin2
=
so that
P
tan A + tan B − 2 cot C2
C
2
cos A cos B sin C2
A−B
2
cos A cos B sin C2
2 cos2 A2 cos2 B2 cos C2
cos A cos B sin C2
A
B
tan − tan
2
2
X 2 cos2 A cos2 B cos C
1
2
2
2
=
cot A2 cot B2 cot C2 cyc cos A cos B sin C2
X tan A tan B A
B 2
=
tan − tan
.
2
2
2
cyc
2
,
B
A
tan − tan
2
2
2
Thus, we have to prove that
X
B 2
A
≥ 0.
(tan A tan B − 2) tan − tan
2
2
cyc
This follows from the more general result for arbitrary real numbers x, y, z,
X
(2)
(tan A tan B − 2)(x − y)2 ≥ 0.
cyc
Indeed, rewrite this as
X tan A tan B tan C − 2 tan C
tan C
cyc
(x − y)2 =
X tan A + tan B − tan C
tan C
cyc
=
X tan A
cyc
≥
X
tan C
(x − y)2 +
(x − y)2
tan C
(y − z)2 − (x − y)2
tan A
2|(x − y)(y − z)| − (x − y)2
cyc
This expression is symmetric in x, y, z, so we can assume x ≥ y ≥ z, i.e x − y = p, y − z = q, with
p, q ≥ 0 and x − z = p + q. Hence we obtain
X
2|(x − y)(y − z)| − (x − y)2 = 2(p2 + 3pq + q 2 ) − 2(p2 + pq + q 2 ) = 4pq ≥ 0.
cyc
and the conclusion follows.
A REMARKABLE INEQUALITY
3
Refinements of original inequality.
Let s, R, r and K be the semiperimeter, circumradius , inradius and area of triangle ABC respectively.
Lemma 1 :
X
tan2
cyc
A
(4R + r)2 − 2s2
,
=
2
s2
Proof :
We have that sin A2 =
that
q
(s−b)(s−c)
bc
and cos A2 =
P
X
cyc
A
tan
2
2
=
=
=
=
q
s(s−a)
bc ,
hence tan2
A
2
=
(s−b)(s−c)
s(s−a) ,
and it follows
(s − b)2 (s − c)2
cyc
s(s − a)(s − b)(s − c)
!2
P
(s − b)(s − c) −
cyc
(−s2
+ ab + bc +
s2 r2
2
(4R + r) − 2s2
s2
ca)2
2K 2
s (s
− a + s − b + s − c)
s2 r2
− 2s2 r2
because ab + bc + ca = s2 + r2 + 4Rr [5].
Lemma 2 :
cos A cos B cos C =
s2 − (2R + r)2
,
4R2
Proof :
We have that
cos A cos B cos C =
=
=
=
1
(sin2 A + sin2 B + sin2 C − 2)
2 1 X 2
2
a − 8R
8R2 cyc
1
((a + b + c)2 − 2(ab + bc + ca) − 8R2 )
8R2
1
s2 − (2R + r)2
2
2
2
2
(4s
−
2(s
+
r
+
4Rr)
−
8R
)
=
,
8R2
4R2
as claimed.
Lemma 3 :
(1 − cos A)(1 − cos B)(1 − cos C) =
r2
,
2R2
4
DANIEL CAMPOS, ROBERTO BOSCH
Proof :
We have that
A
B
C
sin2 sin2
2
2
2
8(s − a)2 (s − b)2 (s − c)2
a2 b2 c2
2
4
8s r
r2
=
.
16s2 r2 R2
2R2
(1 − cos A)(1 − cos B)(1 − cos C) = 8 sin2
=
=
Hence the original inequality (1) can be rewritten as
2r2
(4R + r)2 − 2s2
≥
,
s2 − (2R + r)2
s2
and after some algebraic manipulations this is
2s4 − (24R2 + 16Rr + r2 )s2 + (64R4 + 96R3 r + 52R2 r2 + 12Rr3 + r4 ) ≥ 0.
This is equivalent to
1
1p
s2 ≤ 6R2 + 4Rr + r2 −
64R4 − 112R2 r2 − 64Rr3 − 7r4 = H(R, r)
4
4
because the other case is
1
1p
s2 ≥ 6R2 + 4Rr + r2 +
64R4 − 112R2 r2 − 64Rr3 − 7r4 .
4
4
If this inequality is true, then, by Blundon’s inequality [1][2], we have that
1
1p
4R2 + 4Rr + 3r2 ≥ s2 ≥ 6R2 + 4Rr + r2 +
64R4 − 112R2 r2 − 64Rr3 − 7r4
4
4
≥ 6R2 + 4Rr.
From Euler’s inequality R ≥ 2r, it is clear that this can’t hold.
In addition, it is simple to verify that H(R, r) ≤ 4R2 + 4Rr + 3r2 = Q(R, r), where Q(R, r) is the
best upper bound in quadratic forms [1][2]. Then a natural question arises, is it possible to use
p
(3)
s2 ≤ 2R2 + 10Rr − r2 + 2(R − 2r) R2 − 2Rr
(the best inequality for homogeneous functions), to find a refinement to (1)? We don’t know, but
certainly (3) can be rewritten in terms of trigonometric identities. Therefore, it should be possible
to find an equivalent inequality; the idea is to relate this result with (1).
Another possible refinement of the inequality follows directly from the previous proof, namely
!
Y 1 − cos A
B
C
A
B 2
C 2
A 2
≥ 1 + max
tan − tan
, tan − tan
, tan − tan
cos A
2
2
2
2
2
2
cyc
Assuming that tan A2 ≥ tan B2 ≥ tan C2 , we obtain from the proof of (1) that
Y 1 − cos A X 1 − cos A
A
B
B
C
−
≥ pq = tan − tan
tan − tan
cos A
1 + cos A
2
2
2
2
cyc
cyc
A REMARKABLE INEQUALITY
Hence
Y 1 − cos A
cyc
5
B
A
B
C
tan − tan
≥
+ tan − tan
1 + cos A
2
2
2
2
cyc
2
A
C
= 1 + tan − tan
2
2
!
A
B 2
B
C 2
C
A 2
= 1 + max
tan − tan
, tan − tan
, tan − tan
2
2
2
2
2
2
X 1 − cos A
cos A
A sharper inequality than (1) was found by Constantin Mateescu in [6]:
Q
Q
(1 − cos A) − cos A
Y 1 − cos A
X 1 − cos A cyc
cyc
Q
≥
+
.
cos
A
1
+
cos
A
(1
+
cos
A)
cyc
cyc
cyc
We present an elementary proof, distinct from those in [6]:
Denoting by x =
as
P
cos A, y =
P
cyc
cyc
cos A cos B and z =
Q
cos A we can rewrite Mateescu’s inequality
cyc
1−x+y−z
3 + x − y − 3z 1 − x + y − 2z
4 − 5z
≥
+
=
z
1+x+y+z
1+x+y+z
1+x+y+z
or equivalently
y 2 ≥ 2z(1 + x − 2z) = 2z(x2 + x − 2y),
where we have used the identity
1 = cos2 A + cos2 B + cos2 C + 2 cos A cos B cos C = x2 − 2y + 2z.
This again can be rewritten as
y 2 − 2xz ≥ 2z(x2 − 2y),
or
X
cos2 B cos2 C ≥ 2 cos A cos B cos C
cyc
X
cos2 A.
cyc
This can be slightly simplified (in the case of the acute-angled triangle, which is the only interesting
case) as
X cos A cos B
X
≥2
cos2 A.
cos
C
cyc
cyc
0
)
Assuming that the triangle is acute, we can write A = (π−A
,B=
2
inequality becomes
0
X sin A0 sin B 0
X
2 A
2
2
≥
2
sin
.
0
2
sin C2
cyc
cyc
(π−B 0 )
,
2
C=
(π−C 0 )
.
2
Then our
Let a, b, c be the lengths of the triangle A0 B 0 C 0 and s its semiperimeter. Write x = s − a, y = s − b,
z = s − c. Then
r
r
A0
(s − b)(s − c)
yz
sin
=
=
,
2
bc
(x + y)(x + z)
so that
0
0
sin A2 sin B2
z
=
,
C0
x
+
y
sin 2
6
DANIEL CAMPOS, ROBERTO BOSCH
and finally it’s not difficult to verify that the inequality
X z
X
yz
≥2
,
x+y
(x + y)(x + z)
cyc
cyc
is exactly third degree Schur’s inequality.
References
[1]
[2]
[3]
[4]
[5]
[6]
W.J. Blundon, Inequalities associated with the triangle, Canad. Math. Bull. 8, 1965, 615-626.
Bottemi et al, Geometric Inequalities. Wolters-Noordhoff, 1969, p. 51.
T. Andreescu, Mathematical Reflections: the next two years (2008 - 2009). XYZ Press, 2012, pp. 145-151.
A.W.Marshall, I.Olkin, Inequalities: Theory of Majorization and Its Applications, Academic Press, 1979.
T. Andreescu, D. Andrica, Complex numbers from A to ... Z, Birkha̋user, 2006.
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=404742&hilit=campos.
Daniel Campos
University of Chicago, USA.
[email protected]
Roberto Bosch
Archimedean Academy, FL, USA.
[email protected]