x, y, z are positive reals with product 1. Show that x3/( (1 + y)(1 + z) ) + y3/( (1 + z)(1 + x) ) + z3/( (1 + x)(1 + y) ) ≥ 3/4.
Solution
Assume x ≤ y ≤ z. Then 1/( (1+y)(1+z) ) ≤ 1/( (1+z)(1+y) ) ≤ 1/( (1+x)(1+y) ). So by Chebyshev's inequality we have Show that x3/( (1 + y)(1 + z) ) + y3/( (1 + z)(1 + x) ) + z3/( (1 + x)(1 + y) ) ≥ 1/3 (x3 + y3 + z3)(1/( (1+y)(1+z) ) + 1/( (1+z)(1+x) ) + 1/( (1+x)(1+y) ) = 1/3 (x3 + y3 + z3)(3 + x + y + z)/( (1+x)(1+y)(1+z) ). Put A = (x + y + z)/3. We have (a + h)3 + (a - h)3 = 2a3 + 6ah2 ≥ 2a3, so for fixed x + y + z, x3 + y3 + z3 is smallest when x = y = z. In other words 1/3 (x3 + y3 + z3) ≥ A3. By the AM/GM we have (1+x)(1+y)(1+z) <= (1+A)3. So 1/3 (x3 + y3 + z3)(3 + x + y + z)/( (1+x)(1+y)(1+z) ) ≥ A3(3 + 3A)/(1+A)3 = 3A3/(1+A)2. Now A ≥ (xyz)1/3 = 1, so 3A3/(1+A)2 ≥ 3(1 - 1/(1+A) )2 ≥ 3/4.
Note: some versions of this problem allow x, y, z to be negative. But then the result is false. Take x = N, y = -2N, z = -1/2N2, where N is large. Then xyz = 1, but the expression given is approx N3/(-2N) + (-2N)3/N = -8 1/2 N2 which is large and negative.
© John Scholes
jscholes@kalva.demon.co.uk
30 Aug 2002