<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] If you don't mind I give the details of @Vasea Igor's answer.
Define $f(a,b):=\frac{a^2}{b-1}+\frac{b^2}{a-1}$, where $a,b>1$. Then
$$
\frac{\partial}{\partial a}f(a,b)=\frac{2a(a-1)^2-(b-1)b^2}{(b-1)(a-1)^2}=\frac{2a^3-4a^2+2a-b^3+b^2}{(b-1)(a-1)^2},
$$
$$
\frac{\partial}{\partial b}f(a,b)=-\frac{a^2}{(b-1)^2}+\frac{2b}{a-1}=-\frac{a^3-a^2-2b^3+4b^2-2b}{(b-1)^2(a-1)}.
$$
The real solutions of the system
$$
2a^3-4a^2+2a-b^3+b^2=0,
$$
$$
a^3-a^2-2b^3+4b^2-2b=0
$$
are
$$
(a,b)={(2,2),(0,0),(1,0),(0,1),(1,1)}.
$$
Since $a,b>1$ the only candidate is $(a,b)=(2,2)$.
Now we compute $H(a,b)$ the Hesse Matrix of $f(a,b)$ at $a=2, b=2$.
$$
H(a,b)=\left[ \begin {array}{cc} \frac{2}{b-1}+2\,{\frac {{b}
^{2}}{ \left( a-1 \right) ^{3}}}&-2\,{\frac {a}{ \left( b-1 \right) ^{
2}}}-2\,{\frac {b}{ \left( a-1 \right) ^{2}}}\\ -2\,
{\frac {a}{ \left( b-1 \right) ^{2}}}-2\,{\frac {b}{ \left( a-1
 \right) ^{2}}}&2\,{\frac {{a}^{2}}{ \left( b-1 \right) ^{3}}}+\frac{2}{a-1}\end {array} \right] .
$$
Now we get 
$$
H(2,2)=\left[ \begin {array}{cc} 10&-8\\ -8&10\end {array}
 \right] .
$$
The eigenvalues of $H(2,2)$ are $18,2>0$. It means $H(2,2)$ is positive definite, which implies the function $f$ has a local minimum (and at the same time, global minimum) at $a=2, b=2$ which is $f(2,2)=8$.
</s>