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.
 A: Note $a=x-1$ and $b=y-1$ and apply $r^2+s^2\geqslant2rs$. The solution is $x=y=2$.
A: As suggested by matt, let $a=x-1$, $b=y-1$ and apply the inequality $r+s\ge 2\sqrt{rs}$ for $r,s>0$ with equality if and only if $r=s$. Then $a,b>0$ and
$$\frac{(a+1)^2}{b}+\frac{(b+1)^2}{a}=\left(\frac{a^2}{b}+\frac{1}{b}\right)+\left(\frac{b^2}{a}+\frac{1}{a}\right)+2\left(\frac{a}{b}+\frac{b}{a}\right)$$
$$\ge \left(2\sqrt{\frac{a^2}{b}\cdot \frac{1}{b}}\right)+\left(2\sqrt{\frac{b^2}{a}\cdot \frac{1}{a}}\right)+2\left(2\sqrt{\frac{a}{b}\cdot \frac{b}{a}}\right)$$
$$=2\left(\frac{a}{b}+\frac{b}{a}\right)+4\ge 2\left(2\sqrt{\frac{a}{b}\cdot \frac{b}{a}}\right)+4=8$$
For equality to hold, we must have $\frac{a^2}{b}=\frac{1}{b}$ and $\frac{b^2}{a}=\frac{1}{a}$, i.e. $a=b=1$, i.e. $x=y=2$ (and indeed equality holds).
A: 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 $x =2 = y$. 
