Question

[Duong Thi Nhuong]

Cho \(x\ne y\) . Chứng minh \(\dfrac{x^3+xy^2-x^2y-y^3}{x-y}\ge y^2\)

Answer 1

[Trần Thiên Kim]

Xét \(\dfrac{x^3+xy^2-x^2y-y^3}{x-y}-y^2\ge0\Leftrightarrow\dfrac{x\left(x^2+y^2\right)-y\left(x^2+y^2\right)}{x-y}-y^2\ge0\Leftrightarrow\dfrac{\left(x-y\right)\left(x^2+y^2\right)}{x-y}-y^2\ge0\Leftrightarrow x^2+y^2-y^2\ge0\Leftrightarrow x^2\ge0\left(đúng\right)\)

=> đpcm