\(A=\frac{2x^3+2y^3+2}{2xy+2}=\frac{x^3+x^3+1+y^3+y^3+1}{2xy+2}\ge\frac{3\sqrt[3]{x^6}+3\sqrt[3]{y^6}}{x^2+y^2+2}=\frac{3.2}{2+2}=\frac{3}{2}\)
\(A_{min}=2\) khi \(x=y=1\)
Lại có \(\left\{{}\begin{matrix}x;y\ge0\\x^2+y^2=2\end{matrix}\right.\) \(\Rightarrow0\le x;y\le\sqrt{2}\)
\(\Rightarrow x^2\left(x-\sqrt{2}\right)\le0\Rightarrow x^3\le x^2\sqrt{2}\)
Tương tự: \(y^3\le y^2\sqrt{2}\)
Mặt khác \(x;y\ge0\Rightarrow xy+1\ge1\)
\(\Rightarrow A\le\frac{a^2\sqrt{2}+b^2\sqrt{2}+1}{1}=1+2\sqrt{2}\)
\(A_{max}=1+2\sqrt{2}\) khi \(\left(a;b\right)=\left(0;\sqrt{2}\right)\) và hoán vị
