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Algebra Difficulty 4.6 AIME Find the answer

x,yx, y are positive real numbers such that x+y2=xyx+y^{2}=x y. What is the smallest possible value of xx?

A number or a short expression. Spacing and $ signs are ignored.

Solution

4 Notice that x=y2/(y1)=2+(y1)+1/(y1)2+2=4x=y^{2} /(y-1)=2+(y-1)+1 /(y-1) \geq 2+2=4. Conversely, x=4x=4 is achievable, by taking y=2y=2.

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