$ x^{2} + y^{3} + y^{2} \geq x^{3} + y^{4} + y^{2} \geq x^{3} + 2y^{3} \Rightarrow x^{2} + y^{2} \geq x^{3} + y^{3}$Lại có
$ (x^{2} + y^{2} )^{2} = ( \sqrt{x} . \sqrt{x}^{3} + \sqrt{y}\sqrt{y}^{3})^{2} \leq ( x + y)( x^{3} + y^{3})$
$ \Rightarrow (x^{2} + y^{2})^{2} \leq (x+y)(x^{2}+y^{2})\Rightarrow x^{2} + y^{2} \leq x+y$
$ \Rightarrow ( x^2 + y^2)^2 \leq (x+y)^2 \leq 2(x^2+y^2)$
$ \Rightarrow x^2 + y^2 \leq 2 \Rightarrow x^3 + y^3\leq 2$
Dấu "=" xảy ra khi x=y=1