Subject: Re: Trigonometry in a triangle
Date: Fri, 22 Oct 2000 13:14:10 -0300 (GMT)
From: martin lukarevski
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 ),
|(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.
This comletes the proof.
P.S. I would like to know Alex, what is the source of this problem.