a, b, c are positive reals whose product does not exceed their sum. Show that a2 + b2 + c2 ≥ (√3) abc.
Solution
(a + b + c)2 - 3(a2 + b2 + c2) = -(a - b)2 - (b - c)2 - (c - a)2 ≤ 0, and we are given that a + b + c ≥ abc. Hence (abc)2/3 ≤ a2 + b2 + c2. So if abc √3 > a2 + b2 + c2, then abc < 3√3. But, by the arithmetic geometric mean inequality, we have a2 + b2 + c2 ≥ 3(abc)2/3. Hence we would also have 3(abc)2/3 ≤ a2 + b2 + c2 < abc √3 and hence abc > 3√3. Contradiction. So we must have abc √3 ≤ a2 + b2 + c2.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002