Non-negative reals x, y, z satisfy x^{2} + y^{2} + z^{2} + xyz = 4. Show that xyz ≤ xy + yz + zx ≤ xyz + 2.

**Solution**

Assume x ≥ y ≥ z. If z > 1, then x^{2} + y^{2} + z^{2} + xyz > 1 + 1 + 1 + 1 = 4. Contradiction. So z ≤ 1. Hence xy + yz + zx ≥ xy ≥ xyz.

Put x = u + v, y = u - v, so that u, v ≥ 0. Then the equation given becomes u^{2}(2 + z) + (2 - z)v^{2} + z^{2} = 4. So we we keep z fixed and reduce v to nil, then we must increase u. But xy + yz + zx - xyz = (u^{2} - v^{2})(1 - z) + 2zu, so decreasing v and increasing u has the effect of increasing xy + yz + zx - xyz. Hence xy + yz + zx - xyz takes its maximum value when x = y. But if x = y, then the equation gives x = y = √(2 - z). So to establish that xy + yz + zx - xyz ≤ 2 it is sufficient to show that 2 - z + 2z√(2 - z) ≤ 2 + z(2 - z). Evidently we have equality if z = 0. If z is non-zero, then the relation is equivalent to 2√(2 - z) ≤ 3 - z or (z - 1)^{2} ≥ 0. Hence the relation is true and we have equality only for z = 0 or 1.

© John Scholes

jscholes@kalva.demon.co.uk

11 May 2002