Find the minimum real number $\lambda$ so that the following relation holds ($x>y$): $\lambda\frac{x^3-y^3}{(x-y)^3}\geq \frac{x^2-y^2}{(x-y)^2}$ 
ّFind the minimum real number $\lambda$ so that the following relation holds for arbitrary real numbers $x,y$($x>y$): $$\lambda\frac{x^3-y^3}{(x-y)^3}\geq \frac{x^2-y^2}{(x-y)^2}$$

I simplified the relation into the following: $(\lambda-1)x^2+\lambda xy+(\lambda +1)y^2\geq0$ , but can't go further... 

For $\lambda$, any value above $0$ is possible.
$\lambda\dfrac{x^3-y^3}{(x-y)^3}\geq \dfrac{x^2-y^2}{(x-y)^2}$
$\lambda\dfrac{x^3-y^3}{(x-y)}\geq (x^2-y^2)$
$\lambda (x^2+xy+y^2) \geq x^2-y^2$

$\lambda  \geq \dfrac{x^2-y^2}{x^2+xy+y^2} \cdots(1)$

$(x-y)^2 \gt 0 \implies x^2+y^2 \gt 4xy$
Also $(x^2-y^2) \gt 0$
Since numerator and denominator of equation $(1)$ is postive, $\lambda$ is positive.
As $x$ becomes close to $y$, $\lambda$ tends to $0$
