jmc

Question

For positive real numbers x, y, and z, compute the maximum value of xyz(x + y + z)(x + y)^2 (y + z)^2.

Accepted Answer

By AM-GM, xz + (xy + y^2 + yz) 2 xz(xy + y^2 + yz) = 2 xyz(x + y + z).But xz + (xy + y^2 + yz) = (x + y)(y + z), so (x + y)(y + z) 2 xyz(x + y + z).Then (x + y)^2 (y + z)^2 4xyz(x + y + z), so xyz(x + y + z)(x + y)^2 (y + 2)^2 14.Equality occurs whenever xz = xy + y^2 + yz. For example, we can take x = 2, y = 1, and z = 3. Thus, the maximum value is 14.