Question

cho x>y thỏa mãn xy=1. cm:

A=\(\dfrac{x^2+y^2}{x-y}\ge2\sqrt{2}\)

Answer 1

Vì: x > y => x - y > 0

\(A=\dfrac{x^2+y^2}{x-y}=\dfrac{x^2-2xy+y^2+2xy}{x-y}=\dfrac{\left(x-y\right)^2+2}{x-y}=\left(x-y\right)+\dfrac{2}{x-y}\ge2\sqrt{\left(x-y\right)\cdot\dfrac{2}{x-y}}=2\sqrt{2}\) (đpcm)