Let A = $$\left[ {\matrix{ 2 & b & 1 \cr b & {{b^2} + 1} & b \cr 1 & b & 2 \cr } } \right]$$ where b > 0. Then the minimum value of ${{\det \left( A \right)} \over b}$ is -

Question

Let A = $$\left[ {\matrix{ 2 & b & 1 \cr b & {{b^2} + 1} & b \cr 1 & b & 2 \cr } } \right]$$ where b > 0.

Then the minimum value of ${{\det \left( A \right)} \over b}$ is -

$\sqrt 3 $