Let $A=\left[\begin{array}{ccc}2& b& 1\\ b& b^2+1& b\\ 1& b& 2\end{array}\right]$ where $b>0$ Then, the minimum value of $\frac{\det \left(A\right)}{b}$ is _______________

$A=\left[\begin{array}{ccc}2& b& 1\\ b& b^2+1& b\\ 1& b& 2\end{array}\right](b>0)$

$\left|A\right|=2(2b^2+2-b^2-B^2)-b(2b-b)+1(b^2-b^2-1)$

$\left|A\right|=2(b^2+2)-b^2-1$

$\left|A\right|=b^2+3$

$\frac{\left|A\right|}{b}=b+\frac{3}{b}\Rightarrow \frac{b+\frac{3}{b}}{2}\ge \sqrt{3}$

$b+\frac{3}{b}\ge 2\sqrt{3}$

So, the minimum value is $2\sqrt{3}$