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Algebra Difficulty 5.0 AIME Find the answer

Let x,yx, y, and zz be positive real numbers such that (x⋅y)+z=(x+z)⋅(y+z)(x \cdot y)+z=(x+z) \cdot(y+z). What is the maximum possible value of xyzx y z?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

The condition is equivalent to z2+(x+y−1)z=0z^{2}+(x+y-1) z=0. Since zz is positive, z=1−x−yz=1-x-y, so x+y+z=1x+y+z=1. By the AM-GM inequality, xyz≤(x+y+z3)3=127x y z \leq\left(\frac{x+y+z}{3}\right)^{3}=\frac{1}{27} with equality when x=y=z=13x=y=z=\frac{1}{3}.

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