Find all real numbers $x,y > 1$ such that $\frac{x^2}{y-1}+\frac{y^2}{x-1} = 8$ 
Find all real numbers $x,y > 1$ such that $$\frac{x^2}{y-1}+\frac{y^2}{x-1} = 8$$

Attempt
We can rewrite this as $x^2(x-1)+y^2(y-1) = 8(x-1)(y-1)$. Then I get a multivariate cubic, which I find hard to find all solutions to.

user236182 was on the right track by using Cauchy-Schwarz. We have from Cauchy-Schwarz that $$((y-1)+(x-1)) \left (\dfrac{x^2}{y-1}+\dfrac{y^2}{x-1} \right) \geq \left (\sqrt{y-1}\sqrt{\dfrac{x^2}{y-1}}+\sqrt{x-1}\sqrt{\dfrac{y^2}{x-1}} \right)^2$$ and letting $a = x+y$ we get $8 \geq \dfrac{a^2}{a-2} \implies a < 2$ or $a = 4$. But $a \not < 2$ and thus $a = 4 = x+y$. Plugging this into $x^2(x-1)+y^2(y-1) = 8(x-1)(y-1)$ for $y$ gives a quadratic in $x$ giving the solution $x =2 = y$. 
