Prove that $\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$ I need to prove that for any real number $a>1$ and $b>1$ the following inequality is true:
$$\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$$

$AM-GM$:
$\frac{a^2}{b-1}+4(b-1)\geq 4a$
$\frac{b^2}{a-1}+4(a-1)\geq 4b$
$\Rightarrow \frac{a^2}{b-1}+4(b-1)+\frac{b^2}{a-1}+4(a-1)\geq 4a+4b$
$\Rightarrow \frac{a^2}{b-1}+\frac{b^2}{a-1}\geq 8$
