**Problem A2**

Suppose that a, b, c are the sides of a triangle. Prove that:

a^{2}(b + c - a) + b^{2}(c + a - b) + c^{2}(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:

c^{2} ≥ (b + c - a)(c + a - b).

Similarly for a^{2} and b^{2}. 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