Question

Cho x,y > 0 và \(x^2+y^3\ge x^3+y^4\).Chứng minh rằng: \(x^3+y^3\le2\).

Answer 1

\(x^2+y^3+y^2\ge x^3+y^4+y^2\ge x^3+2y^3\Rightarrow x^2+y^2\ge x^3+y^3\)

Mà \(\left(x^2+y^2\right)^2=\left(\sqrt{x}.x\sqrt{x}+\sqrt{y}.y\sqrt{y}\right)^2\le\left(x+y\right)\left(x^3+y^3\right)\)

\(\Rightarrow\left(x^2+y^2\right)^2\le\left(x+y\right)\left(x^2+y^2\right)\Rightarrow x^2+y^2\le x+y\)

\(\Rightarrow\left(x^2+y^2\right)^2\le\left(x+y\right)^2\le2\left(x^2+y^2\right)\)

\(\Rightarrow x^2+y^2\le2\Rightarrow x^3+y^3\le2\)

Dấu "=" khi \(x=y=1\)