Áp dụng BĐT Cauchy-Schwarz ta có:

\(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\Rightarrow x+y\le\sqrt{2}\)

\(A=x^3+y^3=\frac{x^4}{x}+\frac{y^4}{y}\ge\frac{\left(x^2+y^2\right)^2}{x+y}\ge\frac{1}{\sqrt{2}}\)

Dấu '=' xảy ra khi \(\hept{\begin{cases}x=y\\x^2+y^2=1\end{cases}\Leftrightarrow x=y=\frac{1}{\sqrt{2}}}\)

Vậy \(A_{min}=\frac{1}{\sqrt{2}}\)khi \(x=y=\frac{1}{\sqrt{2}}\)

Từ giả thuyết suy ra:\(0\le x^2,y^2\le1\Rightarrow\hept{\begin{cases}x^3\le x\\y^3\le y\end{cases}}\)

\(A=x^3+y^3\le x+y\le\sqrt{2}\)

Dấu '=' xảy ra khi \(x=y=\frac{1}{\sqrt[3]{\sqrt{2}}}\)

Vậ5y \(A_{max}=\sqrt{2}\)khi \(x=y=\frac{1}{\sqrt[3]{\sqrt{2}}}\)

\(A=\dfrac{x^3+y^3+4}{xy+1}\ge\dfrac{x^3+y^3+4}{\dfrac{x^2+y^2}{2}+1}=\dfrac{x^3+y^3+4}{2}=\dfrac{\dfrac{1}{2}\left(x^3+x^3+1\right)+\dfrac{1}{2}\left(y^3+y^3+1\right)+3}{2}\)

\(\ge\dfrac{\dfrac{3}{2}\left(x^2+y^2\right)+3}{2}=3\)

\(A_{min}=3\) khi \(x=y=1\)

Do \(x^2+y^2=2\Rightarrow\left\{{}\begin{matrix}x\le\sqrt{2}\\y\le\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^3\le\sqrt{2}x^2\\y^3\le\sqrt{2}y^2\end{matrix}\right.\)

\(\Rightarrow A\le\dfrac{\sqrt{2}\left(x^2+y^2\right)+4}{xy+1}=\dfrac{4+2\sqrt{2}}{xy+1}\le\dfrac{4+2\sqrt{2}}{1}=4+2\sqrt{2}\)

\(A_{max}=4+2\sqrt{2}\) khi \(\left(x;y\right)=\left(0;\sqrt{2}\right);\left(\sqrt{2};0\right)\)

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A=x³+y³=(x+y)(x²-xy+y²)

=(x+y)²\(\ge\)0

Dấu "=" xảy ra khi x=-y