AM-GM Equality Confusion with a Fixed Product
In an algebra worksheet, positive real numbers x,y,z satisfy x+y+z=6 and xyz=4. A candidate writes that AM-GM gives xyz ≤ ((x+y+z)/3)^3 = 8, with equality only at x=y=z=2. Because the actual product 4 is less than 8, the candidate argues that the equality case is irrelevant and the equations place no upper bound on x; by making y and z very small, x could approach 6. They conclude that no maximum exists. A second student notices that (x,y,z)=(4,1,1) satisfies both equations, so x=4 is attainable, but does not prove that x cannot exceed 4. The teacher asks the class to resolve whether x has a maximum and to identify the exact error in the AM-GM reasoning.
1. Find the maximum possible value of x, with proof.
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From y+z=6−x and yz=4/x. For positive y,z to exist with these sum and product, the quadratic t^2−(6−x)t+4/x=0 must have nonnegative discriminant: Δ=(6−x)^2−16/x≥0. Since x>0 and y+z>0 gives 0<x<6, multiply by x: x(6−x)^2≥16. Now x(6−x)^2−16=(x−4)(x^2−8x+4). On 0<x<6, the quadratic x^2−8x+4 is positive only for x<4−2√3≈0.536 and negative afterwards. If x≤4−2√3, the product is negative, so no solution. Hence x>0.536, the quadratic is negative, and the inequality forces x−4≤0, i.e. x≤4. At x=4, y+z=2 and yz=1, so y=z=1. Therefore the maximum is x=4.
2. Diagnose the AM-GM error and explain why x=4 is not ruled out by equality at 2.
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AM-GM gives only a global necessary condition xyz≤8; when xyz=4<8, it says the triple is not the equality case, not that every x is possible. The sum and product together still restrict x through the existence of y,z. The equality case x=y=z=2 is simply one feasible point for product 8, not a boundary for product 4. The second student's example (4,1,1) is valid: sum 6, product 4. The discriminant condition shows the feasible x values form an interval whose right endpoint is 4. The candidate's claim that x can approach 6 is false because y+z would be near 0 and yz would be near 0, not 4; the product requires y,z not both small.