Let a, b, c be positive real numbers with abc = 1. Prove that:
1/(a3(b + c)) + 1/(b3(c + a)) + 1/(c3(a + b)) ≥ 3/2.
Solution
Put a = 1/x, b = 1/y, c = 1/z. Then 1/(a3(b+c)) = x3yz/(y+z) = x2/(y+z). Let the expression given be E. Then by Cauchy's inequality we have (y+z + z+x + x+y)E ≥ (x + y + z)2, so E ≥ (x + y + z)/2. But applying the arithmetic/geometric mean result to x, y, z gives (x + y + z) ≥ 3. Hence result.
Thanks to Gerhard Woeginger for pointing out that the original solution was wrong.
(C) John Scholes
jscholes@kalva.demon.co.uk
2 Nov 1998
Last corrected/updated 25 Nov 03