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 $\frac{\det (A)}{b}$ is:

Calculate $\det (A)$:$\det (A)=2((b^2+1)2-b^2)-b(2b-b)+1(b^2-(b^2+1))$$\det (A)=2(2b^2+2-b^2)-b^2-1$$\det (A)=2(b^2+2)-b^2-1=2b^2+4-b^2-1=b^2+3$.We need the minimum value of $\frac{\det (A)}{b}=\frac{b^2+3}{b}=b+\frac{3}{b}$.Using AM-GM inequality for $b>0$:$\frac{b+\frac{3}{b}}{2}\ge \sqrt{b\cdot \frac{3}{b}}$$b+\frac{3}{b}\ge 2\sqrt{3}$.The minimum value is $2\sqrt{3}$.