Date: Fri, 22 Oct 2000 13:14:10 -0300 (GMT)
From: martin lukarevski
Hi Alexander,
Four years ago a young woman posted following problem:
Can this trigonometric inequality be proved
| (1) | Sin(A)·Sin(B) / Sin2(C/2) + Sin(B)·Sin(C) / Sin2(A/2) + Sin(C)·Sin(A) / Sin2(B/2) ≠ 9, |
where A,B,C are angles in a triangle. I wish to say that this one was the hardest but in the same time the most interesting trigonometric inequality I ever come across. Yes, it can be proved but I believe not directly and not by means.
Let’s start with the proof. At the beginning we will transform (1) in an equivalent form using formulas:
Sin(A) = a / (2·R) and Sin2(A/2) = (a2 - (b – c )2) / ( 4·b·c )
And analogously for the other terms in (1) where a, b, c are the lenghts of the sides of the triangle and R
is its circumradius. With this and using
| (2) |
1 / ( a2·( a2 – (b – c)2 ) )
+ 1 /
( b2·(b2 – (c –
a)2 ) +
1/ ( c2·( c2 –
(a – b)2 ) )
 9
/ (16·S2)
|
which is still too difficult to be proved. So, another transformation follows. Now we may put:
x = (b + c – a )/2, y = ( c + a – b )/2, z = ( a + b – c )/2
Hence a = y + z, b = z + x, c = x + y, and x = p – a, y = p – b , z = p – c, where p is the semiperimeter of the triangle.
We must bear in mind that a,b,c are sidelengths and therefore x,y,z are all strictly positive. With this and using Heron’s formula
S2 = p·( p – a )·( p – b )·( p – c ),
(2) becomes:
| (3) |
1 / ( 4·x·y·( x + y )2 ) + 1 / (4·y·z·( y + z )2 ) + 1 / ( 4·z·x·( z + x )2 ) 9 / ( 16·x·y·z·( x + y + z ) )
|
Eqivalent to (3) is:
| (4) |
K = ( x + y
+ z )·( x /
( y + z )2
+ y /
( z + x )2
+ z /
( x + y )2
)
9 / 4
|
If we prove (4) for all x,y,z > 0 we are done. For that purpose we will use the following inequality which is standard exercise and can easily be verified:
| (*) |
For all x,y,z > 0, M = x / ( y + z ) + y / ( z + x ) + z / ( x + y ) 3 / 2 |
And now the last blow. If we use Cauchy-Schwarz-Bunyakovski inequality for the left side of (4) we obtain:
K M2 and M2 
9
/ 4, so K 9/4.
Q.E.D.
This comletes the proof.
Best regards,
Martin Lukarevski
from Skopje,Macedonia
P.S. I would like to know Alex, what is the source of this problem.
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