\(a+\frac{4}{b\left(a-b\right)^2}=a-b+b+\frac{4}{b\left(a-b\right)^2}\ge a-b+2\sqrt{\frac{4b}{b\left(a-b\right)^2}}=a-b+\frac{4}{a-b}\ge4\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=3\\b=1\end{matrix}\right.\)
b/ \(a-b+\frac{4}{\left(a-b\right)\left(b+1\right)^2}+b\ge2\sqrt{\frac{4\left(a-b\right)}{\left(a-b\right)\left(b+1\right)^2}}+b=\frac{4}{b+1}+b+1-1\ge4-1\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)