\(M=\frac{a^2}{b-1}+\frac{b^2}{a-1}=\frac{a^2}{b-1}+4\left(b-1\right)+\frac{b^2}{a^2-1}+4\left(a-1\right)-4a-4b+8\)
\(\ge2\sqrt{\frac{a^2}{b-1}\cdot4\left(b-1\right)}+2\sqrt{\frac{b^2}{a-1}\cdot4\left(a-1\right)}-4a-4b+8=4a+4b-4a-4b+8=8\) (AM-GM)
Dấu "=" xảy ra <=> a=b=2