Prove that (1 + x/y)(1 + y/z)(1 + z/x) ≥ 2 + 2(x + y + z)/w for all positive reals x, y, z, where w is the cube root of xyz.
Solution
(1 + x/y)(1 + y/z)(1 + z/x) = 1 + x/y + y/x + y/z + z/y + z/x + x/z = (x + y + z)(1/x + 1/y + 1/z) - 1 ≥ 3(x + y + z)/w - 1, by the arithmetic geometric mean inequality,
= 2(x + y + z)/w + (x + y + z)/w - 1 ≥ 2(x + y + z) + 3 - 1, by the arithmetic geometric mean inequality.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002