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:
$$\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$$

WLOG assume $a \leq b$, therefore, $a^2 \leq b^2$ and $\frac{1}{b - 1} \leq \frac{1}{a - 1}$. Therefore, by Rearrangement Inequality,
$$\frac{a^2}{b-1} + \frac{b^2}{a - 1} \geq \frac{a^2}{a-1} + \frac{b^2}{b - 1}$$
From here, one can simply follow Valentin's steps,
$$\frac{a^2}{a - 1} + \frac{b^2}{b - 1} = 4 + (a - 1) + \frac{1}{a - 1} + (b - 1) + \frac{1}{b - 1} \geq 4 + 2 + 2 = 8$$
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As Chris points out, $(a - 2)^2 \geq 0$ implies $a^2 - 4a + 4 \geq 0$ or $a^2 \geq 4(a - 1)$ and hence $\frac{a^2}{a - 1} \geq 4$. Thereby, cutting short few more steps.
