J87. Prove that for any acute triangle ABC, the following inequality

J87. Prove that for any acute triangle ABC, the following inequality holds:
1
1
1
1
+
+
≥
.
−a2 + b2 + c2 a2 − b2 + c2 a2 + b2 − c2
2Rr
Proposed by Mircea Becheanu, Bucharest, Romania
First solution by Brian Bradie, VA, USA
Using the Law of Cosines and the formula
R=
abc
,
4rs
we can rewrite the original inequality as
b
c
a
+
+
≥ 4s = 2(a + b + c),
cos α cos β cos γ
(1)
where α, β and γ are the acute angles in the triangle. Using the Law of Sines,
we can write
sin γ
sin β
c=a
and b = a
.
sin α
sin α
Substituting into (1) yields
tan α + tan β + tan γ ≥ 2(sin α + sin β + sin γ).
(2)
On (0, π2 ), tan x is convex and sin x is concave; it therefore follows from Jensen’s
inequality that
√
π
α+β+γ
= 3 tan = 3 3, and
tan α + tan β + tan γ ≥ 3 tan
3
3
√
π
3 3
α+β+γ
= 3 sin =
.
sin α + sin β + sin γ ≤ 3 sin
3
3
2
Hence, (2) holds with equality if and only if α = β = γ. Thus, the original
inequality holds with equality if and only if the triangle is an equilateral triangle.
Second solution by Mihai Miculita, Oradea, Romania
Because 2Rr = 2 Sp ·
b2
abc
4S
=
abc
2p
=
abc
a+b+c ,
the given inequality is equivalent to
1
1
abc
1
+ 2
+ 2
≥
. (1)
2
2
2
2
2
2
+c −a
a +c −b
a +b −c
a+b+c
Mathematical Reflections 3 (2008)
5
Let us observe that since ABC is an acute triangle the following is true
b2 + c2 − a2 > 0 ⇒ 2(b − c)2 (b2 + c2 − a2 ) ≥ 0
⇔ (b − c)2 (2b2 + 2c2 − 2a2 ) ≥ 0
⇔ (b − c)2 [(b + c)2 + (b − c)2 − 2a2 ] ≥ 0
⇔ (b2 − c2 )2 + (b − c)4 − 2a2 (b − c)2 ≥ 0
⇔ (b − c)4 − 2a2 (b − c)2 + a4 ≥ a4 − (b2 − c2 )2
⇔ [a2 − (b − c)2 ]2 ≥ (a2 + b2 − c2 )(a2 + c2 − b2 )
p
⇔ a2 − (b − c)2 ≥ (a2 + b2 − c2 )(a2 + c2 − b2 )
p
⇔ (a + b − c)(a + c − b) ≥ (a2 + b2 − c2 )(a2 + c2 − b2 ). (2)
Thus, using the AM-GM inequality and using the result in (2) we have that:
1
1
1
1
≥p
+
2
2
2
2 b2 + c2 − a2 a2 + c2 − b2
(b + c − a )(a2 + c2 − b2 )
1
≥
. (3)
(b + c − a)(a + c − b)
Summing up inequality (3) and the two obtained by a circular permutation of
the letters we obtain
1
1
1
1
1
1
+
+
=
+
b2 + c2 − a2 a2 + c2 − b2 a2 + b2 − c2
2 b2 + c2 − a2 a2 + c2 − b2
1
1
1
1
1
1
+
+
+
+
2 b2 + c2 − a2 a2 + b2 − c2
2 a2 + c2 − b2 a2 + b2 − c2
1
1
+
≥
(b + c − a)(a + c − b) (b + c − a)(a + b − c)
1
+
(a + c − b)(a + b − c)
a+b+c
=
(b + c − a)(a + c − b)(a + b − c)
1
1
1
⇒ 2
+ 2
+ 2
2
2
2
2
b +c −a
a +c −b
a + b2 − c2
a+b+c
≥
. (4)
(b + c − a)(a + c − b)(a + b − c)
It is known that
p
(b + c − a) + (a + c − b)
(b + c − a)(a + c − b) ≤
= c.
2
Multiplying the above inequality with its respective ones obtained by circular
permutation of letters we obtain
(b + c − a)(a + c − b)(a + b − c) ≤ abc. (5)
Mathematical Reflections 3 (2008)
6
Using (4) and (5) we readily obtain the desired inequality (1).
Third solution by Ovidiu Furdui, Cluj, Romania
We will use the following standard trigonometric formulae
s = 4R cos
B
C
A
cos cos
2
2
2
and r = 4R sin
A
B
C
sin sin ,
2
2
2
where s denotes the semiperimeter of triangle ABC. It is simply to check, by
using the preceding formulas that 4sRr = abc.
Let f : (0, π2 ) → R be the function defined by f (x) = cosx x . A calculation shows
that
x + x sin2 x + sin(2x)
f ′′ (x) =
> 0,
cos3 x
and hence, f is a convex function. Using the Law of Cosines combined with
Jensen’s inequality for convex functions we get that
X
1
1
1
1 X a
1
+
+
=
=
−a2 + b2 + c2 a2 − b2 + c2 a2 + b2 − c2
2bc cos A
2abc
cos A
cyclic
≥
1
·3·
2abc
a+b+c
3
cos A+B+C
3
=
cyclic
2s
1
=
,
abc
2Rr
and the problem is solved.
Fourth solution by Tarik Adnan Moon, Kushtia, Bangladesh
X
1
1
≥
2
2
2
−a + b + c
2Rr
cyc
We know that, −a2 + b2 + c2 = 2bc · cos A. So,we need to prove that,
X
1
1
≥
2bc · cos A
2Rr
cyc
Lemma 1: We know that, [ABC] = sr =
abc
4R
=⇒ 4sr =
abc
R
After multiplying by 2abc we get,
X
cyc
a
abc
4sr
≥
=
= 4s
cos A
Rr
r
By Cauchy-Schwarz inequaltiy we get,
!
!
X a
X
≥
a · cos A
cos A
cyc
cyc
Mathematical Reflections 3 (2008)
X
cyc
!2
a
= 4s2 ...(1)
7
Lemma 2: We know that,
X
cyc
a · cos A
!
=
2sr
R
So, it is left to prove that,
X
cyc
a · cos A
!
=
2sr
≤ s ⇔ R ≥ 2r
R
And we are done.
Some words about the lemmas:
Lemma 1: Straightforward, just need to use extended law of sines.
Lemma 2:We know that,P
a cos A = 2RQ
sin A · cos A = R · sin 2A
Then,we use the identity,
sin 2A = 4 sin A
Q
and using the extended law of sines we obtain, 4R2 sin A = bc sin A = 2[ABC]
P
From these three we obtain,
a cos A = 2[ABC]
= 2sr
R
R
Also solved by Andrea Munaro, Italy; Arkady Alt, San Jose, California, USA;
Daniel Campos Salas, Costa Rica; Daniel Lasaosa, Universidad Publica de
Navarra, Spain; G.R.A.20 Math Problems Group, Roma, Italy; Ivanov Andrey, Chisinau, Moldova; Athanasios Magkos, Kozani, Greece; Michel Bataille,
France; Ricardo Barroso Campos, Spain; Roberto Bosch Cabrera, Cuba; Samin
Riasat, Notre Dame College, Dhaka, Bangladesh; Vicente Vicario Garcia, Huelva,
Spain.
Mathematical Reflections 3 (2008)
8