Problem A2
Suppose that a, b, c are the sides of a triangle. Prove that:
a2(b + c - a) + b2(c + a - b) + c2(a + b - c) ≤ 3abc.
Solution
The condition that a, b, c be the sides of a triangle, together with the appearance of quantities like a + b - c is misleading. The inequality holds for any a , b, c ≥ 0.
At most one of (b+c-a), (c+a-b), (a+b-c) can be negative. If one of them is negative, then certainly:
abc ≥ (b + c - a)(c + a - b)(a + b - c) (*)
since the lhs is non-negative and the rhs is non-positive.
(*) is also true if none of them is negative. For then the arithmetic/geometric mean on b + c - a, c + a - b gives:
c2 ≥ (b + c - a)(c + a - b).
Similarly for a2 and b2. Multiplying and taking the square root gives (*). Multiplying out easily gives the required result.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
25 Sep 1998
Last corrected/updated 24 Sep 2003