On a \problem of the Month"

In the problem of the month 1999 : 106], one was to prove that p a + b ; c + p b + c ; a + p c + a ; b p a + p b + p c , where a, b, c are sides of a triangle. It is to be noted that this inequality will follow immediately from the Majorization Inequality 1]. Here, if A and B are vectors (a 1 a 2 : : : a n), we say that A majorizes B and write it as A B. Then, if F is a convex function, If F is concave, the inequality is reversed. For the triangle inequality, we can assume without loss of generality that a b c. Then a + b ; c a, (a + b ; c) + (a + c ; b) a + b, and (a + b ; c) + (a + c ; b) + (b + c ; a) = a + b + c. Therefore, if F is concave, As to the substitution a = y + z, b = z + x, c = x + y which was used in the referred to solution and was called the Ravi Substitution, this transformation was known and used before he was born. Geometrically, x, y, z are the lengths which the sides are divided into by the points of tangency of the incircle. Thus, we have the following implications for any triangle inequality or identity: F (aa bb c) 0 (=) F (y + z z z + xx x + y) 0 , F (xx y y z) 0 (=) F ((s ; a) (s ; b) (s ; c)) 0 (here s is the semiperimeter). This transformation eliminates the troublesome triangle constraints and lets one use all the machinery for a set of three non-negative numbers.