Question

Let x, y, z be real numbers such that x² + y² + z² - 2xyz = 1. Prove that (1 + x)(1 + y)(1 + z) ≤ 4 + 4xyz.

Answer 1

Write 1 + 2xyz = x² + y² + z² ⇔ 3 + 3xyz = 3/2(x² + y² + z²) + 3/2 

=>3 + 3xyz = (x² + y² + z²) + 1/2[(x² + 1) + (y² + 1) + (z² + 1)] ≥ x² + y² + z² + x + y + z 

(Use x² + y² + z² ≥ xy + yz + zx and AM-GM: x² + 1 ≥ 2x etc.) 

=> 3 + 3xyz ≥ xy + yz + zx + x + y + z. 

By adding 1 + xyz in both sides, 

we get 4 + 4xyz ≥ 1 + x + y + z + xy + yz + zx + xyz = (1 + x)(1 + y)(1 + z). 

Equality holds when x = y = z = 1.