J87. Prove that for any acute triangle ABC , the following inequality holds:
1
− a 2 + b 2 + c 2
+ a 2
1
− b 2 + c 2
+ a 2
1
+ b 2
− c 2
≥
1
2 Rr
.
Proposed by Mircea Becheanu, Bucharest, Romania
First solution by Brian Bradie, VA, USA
Using the Law of Cosines and the formula
R = abc
4 rs
, we can rewrite the original inequality as a cos α b
+ cos β c
+ cos γ
≥ 4 s = 2( a + b + c ) , (1) where α , β and γ are the acute angles in the triangle. Using the Law of Sines, we can write c = a sin γ sin α and b = a sin β sin α
.
Substituting into (1) yields tan α + tan β + tan γ ≥ 2(sin α + sin β + sin γ ) .
(2)
On (0 ,
π
2
), tan inequality that x is convex and sin x is concave; it therefore follows from Jensen’s tan α + tan β + tan γ ≥ 3 tan sin α + sin β + sin γ ≤ 3 sin
α + β + γ
3
α + β + γ
3
= 3 tan
π
3
= 3 sin
π
3
= 3
√
3 , and
=
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 2 Rr = 2
S p
· abc
4 S
= abc
2 p
= abc a + b + c
, the given inequality is equivalent to b 2
1
+ c 2
− a 2
+ a 2
1
+ c 2
− b 2
+ a 2
1
+ b 2
− c 2
≥ abc a + b + c
.
(1)
Mathematical Reflections 3 (2008) 5

Let us observe that since ABC is an acute triangle the following is true b
2
+ c
2
− a
2
> 0 ⇒ 2( b − c )
2
( b
2
+ c
2
− a
2
) ≥ 0
⇔ ( b − c )
2
(2 b
2
+ 2 c
2
− 2 a
2
) ≥ 0
⇔ ( b − c )
2
⇔ ( b
2
− c
2
[(
)
2 b + c )
2
+ ( b − c )
2
+ ( b − c )
4
− 2 a
2
− 2 a
2
]
( b − c )
2
≥
≥
0
0
⇔ ( b − c )
4
⇔ [ a
2
− ( b
−
−
2 c ) a
2
2
( b − c )
]
2
≥ ( a
2
2
+ a
+ b
2
4
≥ a
4
−
− c
2
)( a
2
( b
+
2 c
−
2 c
−
2
)
2 b
2
)
⇔ a
2
− ( b − c )
2
≥ p
( a 2 + b 2
− c 2 )( a 2 + c 2
− b 2 )
⇔ ( a + b − c )( a + c − b ) ≥ p
( a 2 + b 2
− c 2 )( a 2 + c 2
− b 2 ) .
(2)
Thus, using the AM-GM inequality and using the result in (2) we have that:
1
2 b 2
1
+ c 2
− a 2
+ a 2
1
+ c 2
− b 2
≥
≥
1 p ( b 2 + c 2
− a 2 )( a 2 + c 2
− b 2 )
1
( b + c − a )( a + c − b )
.
(3)
Summing up inequality (3) and the two obtained by a circular permutation of the letters we obtain b 2
1
+ c 2
− a 2
+
+
≥ a
1
2
2 + c b 2
1
2
− b 2
1
1
+
+ c 2
− a 2 a 2
+
+ b a 2
1
( b + c − a )( a + c − b )
1
2
−
+
1
( c b
2
=
+ b 2
− c
+ c
2
1
2
−
+ a b 2
1
)(
1
2
+ c a +
1 a
2 b
2
−
+
−
+
( a + c − b )( a + b − c ) a + b + c
=
⇒
≥
( b + c − a )( a + c − b )( a + b − c ) b 2 + c
1
2
1
− a 2
+ a 2 + a + b + c c 2
− b 2
+ a 2
1
.
(4)
( b + c − a )( a + c − b )( a + b − c )
+ b 2
− c 2 a c c
2
1
)
2
+
− a b 2
2 + c 2
+
1 a 2
− b 2
1
+ b 2
− c 2
It is known that p
( b + c − a )( a + c − b ) ≤
( b + c − a ) + ( a + c − b )
2
= c.
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 = 4 R cos
A
2 cos
B
2 cos
C
2 and r = 4 R sin
A
2 sin
B
2 sin
C
,
2 where s denotes the semiperimeter of triangle ABC . It is simply to check, by using the preceding formulas that 4 sRr = abc .
Let f : (0 , that
π
2
) → R be the function defined by f ( x ) = f ′′ ( x ) = x + x sin 2 x + sin(2 x ) cos 3 x x cos x
> 0 ,
. A calculation shows and hence, f is a convex function. Using the Law of Cosines combined with
Jensen’s inequality for convex functions we get that
− a 2
1
+ b 2 + c 2
+ a 2
1
− b 2 + c 2
+
1 a 2 + b 2
− c 2
X
1
= cyclic
2 bc cos A
≥
1
2 abc
· 3 · cos a + b + c
3
A + B + C
3
=
2 s abc
=
1
2 Rr
,
=
1
X a
2 abc cyclic cos A and the problem is solved.
Fourth solution by Tarik Adnan Moon, Kushtia, Bangladesh
X cyc
− a 2
1
+ b 2 + c 2
≥
1
2 Rr
We know that, − a
2
+ b
2
+ c
2
= 2 bc · cos A . So,we need to prove that,
X cyc
1
2 bc · cos A
≥
1
2 Rr
Lemma 1: We know that, [ ABC ] = sr = abc
4 R
= ⇒ 4 sr =
After multiplying by 2 abc we get, abc
R
X a cos A
≥ cyc abc
=
Rr
4 sr
= 4 s r
By Cauchy-Schwarz inequaltiy we get,
X a · cos A cyc
!
X cyc a cos A
!
≥
X a
!
2 cyc
= 4 s
2
...
(1)
Mathematical Reflections 3 (2008) 7

Lemma 2: We know that,
X a · cos A cyc
!
=
2 sr
R
So, it is left to prove that,
X a · cos A cyc
!
=
2 sr
R
≤ s ⇔ R ≥ 2 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, a cos A = 2 R sin A · cos A = R · sin 2 A
Then,we use the identity, P sin 2 A = 4 Q sin A and using the extended law of sines we obtain, 4 R
2
Q sin A = bc sin A = 2[ ABC ]
From these three we obtain, P a cos A =
2[ ABC ]
R
=
2 sr
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