Let A=2b1bb2+1b1b2 where b>0. Then the minimum value of det⁡(A)b is

We have, $A = [\begin{array}{ccc} 2 & b & 1 \\ b & b^{2} + 1 & b \\ 1 & b & 2 \end{array}]$ , where b>0

Now, det $(A) = 2 (2 b^{2} + 2 - b^{2}) - b (2 b - b) + 1 (b^{2} - b^{2} - 1)$

$= 2 (b^{2} + 2) - b^{2} - 1 = b^{2} + 3$

Now, $\frac{\det (A)}{b} = \frac{b^{2} + 3}{b} = b + \frac{3}{b}$

Since, $A.M. \geq G.M.$

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