18th Eötvös 1911

Real numbers a, b, c, A, B, C satisfy b² < ac and aC - 2bB + cA = 0. Show that B² ≥ AC.

Solution

2bB = aC + cA, so 4b²B² = (aC + cA)². But (aC - cA)² ≥ 0, so 4b²B² ≥ 4acAC. But ac > b² ≥ 0, so dividing by 4ac gives, b²/ac B² ≥ AC. But 0 ≤ b²/ac < 1, so B² ≥ AC. Note that we can have equality if A = B = C = 0.

16th Eötvös 1911

© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002