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... 
 A: Recall that for a quadratic to be $\geq0$ you must have that discriminant $\Delta\leq 0$ and the leading coefficient $> 0$ in this case $\lambda>1$ and $\lambda^2y^2-4(\lambda^2-1)y^2\leq 0\iff \lambda^2-4\lambda^2+4\leq 0 $
A: Your work is interesting and I have a possible solution using that method,
First I'd like to check out this approach.
I would always factor first. I don't know if you did that so I'm working it out.
We are given that $x > y$ so $(x-y) >0 $ and all divisions by $(x-y)$ are defined. Multiplication by $(x-y)$ doesn't change the inequality.
$$ \lambda \frac{x^3 - y^3}{(x-y)^3} \geq \frac{x^2 - y^2}{(x-y)^2}$$
$$ \lambda \frac{(x - y)(x^2 + xy + y^2}{(x-y)^3} \geq \frac{(x-y)(x+y)}{(x-y)^2}$$
$$ \lambda (x^2 + xy + y^2) \geq (x-y)(x+y)$$
$$ (\lambda - 1) x^2 +  \lambda xy + (\lambda + 1)y^2 \geq 0$$
Right, we agree here. :-)
So here's a technique used in finding limits of functions of two variables:
I got down to a solution with this, below; needs completed calculation and check as this is first draft rough work.
Let $y = kx$. Since $x > y$ given in hypothesis, if $x$ is positive k < 1  (negative is fine) and if $x$ is negative, $k < -1$.
This covers all cases except $x = 0$. If $x = 0$, we calculate separately.
If $x = 0 $ your equation simplifies to 
$$(\lambda + 1)y^2 \geq 0$$ 
$y < 0 $ given $y < x$ but $$y^2 > 0$$. Therefore $(\lambda + 1) \geq 0$ and $\lambda \geq -1$
For y = kx, 
$$ (\lambda - 1) x^2 +  \lambda xy + (\lambda + 1)y^2 \geq 0$$
$$ (\lambda - 1) x^2 +  \lambda x(kx) + (\lambda + 1)(kx)^2 \geq 0$$
$$ \lambda x^2 - x^2 + k \lambda x^2 + \lambda k^2 x^2 + k^2 x^2 \geq 0$$
Separating out the terms with $\lambda$
$$ \lambda x^2  + k \lambda x^2 + \lambda k^2 x^2  \geq x^2 - k^2 x^2$$
$$\lambda x^2(1 + k + k^2) \geq x^2(1 - k^2)$$
We dealt with $x = 0$ above. For $x \neq 0$ we can divide
$$\lambda  \geq \frac{(1 - k^2)}{(1 + k + k^2)}$$
Minimize with respect to k subject to the restrictions above [if x is positive k < 1 (negative is fine) and if x is negative, $k<−1$]
$$\frac{(-2k)(1 + k + k^2) -(1-k^2)(1 + 2k)}{(1 + k + k^2)^2} = 0$$
Denominator is never zero
$$(-2k)(1 + k + k^2) -(1-k^2)(1 + 2k) = 0$$
$$ -2k -2k^2 - 2k^3 -1 - 2k + k^2 + 2k^3 = 0$$
$$  -k^2 -4k -1 = 0$$
$$ k = \frac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}$$   
$$\lambda  \geq \frac{(1 - k^2)}{(1 + k + k^2)}$$
Work it out but remember the restrictions on k above.
Again rough work, not proofread, correct as needed. thanks.

Note that if $x = 0$ and $y < 0$ (allowable by definition above) then using either the simplified form or the original form we get $\lambda = -1$.
This disproves some of the answers given.
My calculation above for $k$ when $y = kx , y < x$, may allow even lower value of  $\lambda$. I leave it to you to take further for now.  
A: 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 $0$
