<s>[INST] 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:
$$\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$$
 [/INST] Let $a = 1+x$ and $b = 1+y$. Then we need to prove that $$\dfrac{(x+1)^2}{y} + \dfrac{(y+1)^2}{x} \geq 8$$ i.e. $$\dfrac{x^2}{y} + 2 \dfrac{x}{y} + \dfrac1y + \dfrac{y^2}{x} + 2 \dfrac{y}{x} + \dfrac1x \geq 8$$
for $x,y \geq 0$.
Now apply AM-GM as shown below.
$$\dfrac{\dfrac{x^2}{y} + \dfrac{x}{y} + \dfrac{x}{y} + \dfrac1y + \dfrac{y^2}{x} + \dfrac{y}{x} + \dfrac{y}{x} + \dfrac1x}{8} \geq \sqrt[8]{\dfrac{x^2}{y} \times \dfrac{x}{y} \times \dfrac{x}{y} \times \dfrac1y \times \dfrac{y^2}{x} \times \dfrac{y}{x} \times \dfrac{y}{x} \times \dfrac1x} = 1$$
Hence, we get that $$\dfrac{x^2}{y} + 2 \dfrac{x}{y} + \dfrac1y + \dfrac{y^2}{x} + 2 \dfrac{y}{x} + \dfrac1x \geq 8$$ which is what we wanted to show.
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