Problem: Let x, y, and z be positive real numbers such that (x⋅y)+z=(x+z)⋅(y+z). What is the maximum possible value of xyz?
Solution
Solution: The condition is equivalent to z2+(x+y−1)z=0. Since z is positive, z=1−x−y, so x+y+z=1. By the AM-GM inequality, xyz≤(3x+y+z)3=271 with equality when x=y=z=31.
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Source: MathNet,
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