a, b, c are real, and the sum of any two is greater than the third. Show that 2(a + b + c)(a2 + b2 + c2)/3 > a3 + b3 + c3 + abc.
wlog we may take a ≥ b ≥ c.
c (and hence also a and b) must be positive. For if c ≤ 0, then a ≥ b + c. Contradiction.
Multiplying out the required relation and cancelling, we must prove that 2(a2b + a2c + b2a + b2c + c2a + c2b) > a3 + b3 + c3 + 3abc. But we have a2(b + c) > a3, b2(c + a) > b3, c2(a + b) > c3. So it is sufficient to prove that a2b + a2c + b2a + b2c + c2a + c2b > 3abc. But a2b ≥ abc and ab2 ≥ abc. Also b2c + bc2 = bc(b + c) > abc. Similarly, ac(a + c) > abc. So we have in fact proved the slightly stronger result with abc replaced by 4abc/3.
13th Putnam 1953
© John Scholes
5 Mar 2002