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Algebra Difficulty 5.0 AIME Prove it United States

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

Solution

Solution:
The condition is equivalent to z2+(x+y1)z=0z^{2} + (x + y - 1) z = 0. Since zz is positive, z=1xyz = 1 - x - y, so x+y+z=1x + y + z = 1. By the AM-GM inequality,
xyz(x+y+z3)3=127 x 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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