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) / Sin^{2}(C/2) + Sin(B)*Sin(C) / Sin^{2}(A/2) + Sin(C)*Sin(A) / Sin^{2}(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 Sin^{2}(A/2) = (a^{2} - (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 / ( a^{2}*( a^{2} - (b - c)^{2} ) ) + 1 / ( b^{2}*(b^{2} - (c - a)^{2} ) + 1/ ( c^{2}*( c^{2} - (a - b)^{2} ) ) 9 / (16*S^{2}) |
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
S^{2} = 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 M^{2} and M^{2} 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.