The
following problem came from IMO 1983. The solution was worked out by a German
contestants who was awarded a special prize for his elegant solution.
Problem
Let a, b, c denote the lengths of the sides of a
triangle.
Prove that a2b (a - b) + b2c
(b - c) + c2a (c - a)
> 0.
Determine when equality holds.
Solution
Without loss of generality, assume a is the longest side of the
triangle.
Then a2b (a -
b) + b2c (b - c) + c2a
(c - a)
= a (c
- b)2 (b + c - a) + b (a
- b) (a - c) (a + b - c)
> 0
because (b + c - a) and (a + b - c)
are positive by the triangular inequality.
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