Theo định lý Pitago: \(x^2+y^2=1\)
\(x^3+x^3+\dfrac{1}{2\sqrt{2}}\ge3\sqrt[3]{\dfrac{x^6}{2\sqrt{2}}}=\dfrac{3x^2}{\sqrt{2}}\)
Tương tự: \(y^3+y^3+\dfrac{1}{2\sqrt{2}}\ge\dfrac{3y^2}{\sqrt{2}}\)
\(\Rightarrow2\left(x^3+y^3\right)+\dfrac{1}{\sqrt{2}}\ge\dfrac{3}{\sqrt{2}}\left(x^2+y^2\right)=\dfrac{3}{\sqrt{2}}\)
\(\Rightarrow x^3+y^3\ge\dfrac{1}{\sqrt{2}}\)
Mặt khác: \(x^2+y^2=1\Rightarrow\left\{{}\begin{matrix}x^2< 1\\y^2< 1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}0< x< 1\\0< y< 1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^3< x^2\\y^3< y^2\end{matrix}\right.\) \(\Rightarrow x^3+y^3< x^2+y^2=1\)
