Question 66, Given $x, y>0, x^{2}+y \geq x^{3}+y^{2}$, find the maximum value of $x^{2}+y^{2$.


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Question 66, Given $x, y>0, x^{2}+y \geq x^{3}+y^{2}$, find the maximum value of $x^{2}+y^{2}$.
Question 66, Answer: Since $x(x-1)^{2} \geq 0 \Rightarrow x-x^{2} \geq x^{2}-x^{3}$, and according to the condition, $x^{2}+y \geq x^{3}+y^{2} \Rightarrow x^{2}-x^{3} \geq y^{2}-y$, so $x-x^{2} \geq y^{2}-y \Rightarrow x^{2}+y^{2} \leq x+y \Rightarrow\left(x^{2}+y^{2}\right)^{2} \leq(x+y)^{2} \leq 2\left(x^{2}+y^{2}\right) \Rightarrow x^{2}+y^{2} \leq 2$. Therefore, the maximum value of $x^{2}+y^{2}$ is 2, which is achieved when $x=y=1$.