Question

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

 

Answer 1

\(VT=\frac{\left(x^3-x^2y\right)+\left(xy^2-y^3\right)}{x-y}=\frac{x^2\left(x-y\right)+y^2\left(x-y\right)}{x-y}=\frac{\left(x^2+y^2\right)\left(x-y\right)}{x-y}=x^2+y^2\ge y^2\)

Answer 2

Xét VT=\(=\frac{x^2\left(x-y\right)+y^2\left(x-y\right)}{x-y}\\ =\frac{\left(x^2+y^2\right)\left(x-y\right)}{x-y}\\ =x^2+y^2\ge y^2\)

Dấu "=" xảy ra khi x=0