MHB Find Minimum Value of $(x-y)^2+\left( \sqrt{2-x^2}-\dfrac{9}{y} \right)^2$

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The objective is to minimize the expression $(x-y)^2+\left( \sqrt{2-x^2}-\dfrac{9}{y} \right)^2$ under the constraints $0<x<\sqrt{2}$ and $y>0$. The discussion includes various approaches to derive the minimum value, focusing on the relationship between $x$ and $y$. Techniques such as substitution and calculus are suggested to explore the function's behavior. The minimum value is sought through critical points and boundary analysis. Ultimately, the solution emphasizes the need for careful consideration of the defined constraints.
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Find the minimum value of $(x-y)^2+\left( \sqrt{2-x^2}-\dfrac{9}{y} \right)^2$ for $0<x<\sqrt{2}$ and $y>0$.
 
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anemone said:
Find the minimum value of $(x-y)^2+\left( \sqrt{2-x^2}-\dfrac{9}{y} \right)^2$ for $0<x<\sqrt{2}$ and $y>0$.

Solution suggested by other:
The given function is the square of the distance between a point of the quarter of the circle $x^2+y^2=2$ in the open first quadrant and a point of the half hyperbola $xy=9$ in that quadrant. Then tangents to the curves at $(1,\,1)$ and $(3,\,3)$ separate the curves, and both are perpendicular to $x=y$, so those points are at the minimum distance, and the answer is $(3-1)^2+(1-3)^2=8$.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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