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 R = a·b·c / ( 4·S ), S being the triangle's area , (1) is transformed to:

(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.

Copyright © 1996-2008 Alexander Bogomolny

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